CBSE Class 8 Mathematics Practice Worksheet Set 13

WORK SHEET SUBJECT- Maths Chapter 1- Rational Numbers CLASS- VIII

Question 1. Fill in the blanks:
a) Rational numbers are numbers of the form \(\frac{p}{q}\) where p,q are integers and q \(\neq\) 0.
b) Rational numbers are not closed under division.
c) Zero is called the additive identity of rational numbers.
d) Zero has no reciprocal.
e) The numbers 1 and -1 are their own reciprocals.
f) \(\frac{3}{4}\) is the multiplicative inverse of \(\frac{4}{3}\).
g) The rational number that is equal to its negative is 0.
h) There are infinite rational numbers between any two given rational numbers.
i) Nine times the reciprocal of a number is 3. The number is 3.
j) -\(\frac{4}{7}\) \(\times\) \(\frac{7}{4}\) = -1.
Answer: Completed all blanks based on properties of rational numbers.
In simple words: These blanks test fundamental rules and properties of rational numbers.

Exam Tip: Remember that zero has no reciprocal because division by zero is undefined.

Question 2. Write the additive inverse of
a) \(\frac{2}{-9}\)
b) \(\frac{-5}{7}\)
Answer:
The additive inverse of a number \(a\) is \(-a\) such that \(a + (-a) = 0\).
a) For \(\frac{2}{-9}\) = \(-\frac{2}{9}\), the additive inverse is \(-\left(-\frac{2}{9}\right) = \frac{2}{9}\).
b) For \(\frac{-5}{7}\), the additive inverse is \(-\left(\frac{-5}{7}\right) = \frac{5}{7}\).
In simple words: Change the sign of the given rational number to find its additive inverse.

Exam Tip: Additive inverse means the sum of the number and its inverse must be 0.

Question 3. Write the multiplicative inverse of
a) -1
b) \(\frac{-13}{19}\)
Answer:
The multiplicative inverse (reciprocal) of a number \(a\) is \(\frac{1}{a}\) such that \(a \times \frac{1}{a} = 1\).
a) For -1, the multiplicative inverse is \(\frac{1}{-1} = -1\).
b) For \(\frac{-13}{19}\), the multiplicative inverse is \(\frac{19}{-13} = -\frac{19}{13}\).
In simple words: Flip the numerator and denominator to get the reciprocal, keeping the sign intact.

Question 4. Verify that -(-a)=a is true for a = \(\frac{-16}{21}\)
Answer:
Given \(a = \frac{-16}{21}\).
LHS = \(-(-a) = -\left(-\frac{-16}{21}\right) = -\left(\frac{16}{21}\right) = -\frac{16}{21}\)
RHS = \(a = \frac{-16}{21}\)
Since LHS = RHS, the property is verified.
In simple words: Negative of a negative number is always the positive or original number.

Exam Tip: Writing LHS and RHS clearly fetches full marks in verification questions.

Question 5. Verify the property: ax(b+ c) = axb + axc by taking a = \(\frac{-5}{2}\), b = -2, c = \(\frac{11}{3}\)
Answer:
LHS = \(a \times (b + c) = \left(-\frac{5}{2}\right) \times \left(-2 + \frac{11}{3}\right)\)
\(= \left(-\frac{5}{2}\right) \times \left(\frac{-6 + 11}{3}\right)\)
\(= \left(-\frac{5}{2}\right) \times \left(\frac{5}{3}\right) = -\frac{25}{6}\)
RHS = \((a \times b) + (a \times c) = \left[\left(-\frac{5}{2}\right) \times (-2)\right] + \left[\left(-\frac{5}{2}\right) \times \frac{11}{3}\right]\)
\(= \frac{10}{2} + \left(-\frac{55}{6}\right) = 5 - \frac{55}{6} = \frac{30 - 55}{6} = -\frac{25}{6}\)
Since LHS = RHS, distributive property is verified.
In simple words: Multiplication distributes over addition for rational numbers.

Exam Tip: Take LCM carefully when adding fractions inside parentheses.

Question 6. Find five rational numbers between \(\frac{-1}{2}\) and 2.
Answer:
Convert the given numbers to have a common denominator. Let the denominator be 10.
\(-\frac{1}{2} = -\frac{5}{10}\)
\(2 = \frac{20}{10}\)
Five rational numbers between \(-\frac{5}{10}\) and \(\frac{20}{10}\) can be chosen as:
\(-\frac{4}{10}, -\frac{3}{10}, 0, \frac{1}{10}, \frac{2}{10}\) (or any other five values in between).
In simple words: Make denominators equal and pick any numbers lying between the numerators.

Exam Tip: There are infinitely many rational numbers between any two numbers; any valid set will be correct.

Question 7. Arrange in ascending order \(\frac{-3}{4}\), \(\frac{2}{-12}\), \(\frac{-5}{16}\), \(\frac{7}{-24}\)
Answer:
First, rewrite all numbers with positive denominators:
\(-\frac{3}{4}\), \(-\frac{2}{12} = -\frac{1}{6}\), \(-\frac{5}{16}\), \(-\frac{7}{24}\).
Find the LCM of denominators (4, 6, 16, 24), which is 48.
Convert each fraction to denominator 48:
\(-\frac{3}{4} = -\frac{3 \times 12}{4 \times 12} = -\frac{36}{48}\)
\(-\frac{1}{6} = -\frac{1 \times 8}{6 \times 8} = -\frac{8}{48}\)
\(-\frac{5}{16} = -\frac{5 \times 3}{16 \times 3} = -\frac{15}{48}\)
\(-\frac{7}{24} = -\frac{7 \times 2}{24 \times 2} = -\frac{14}{48}\)
Comparing numerators: \(-36 < -15 < -14 < -8\).
Ascending order: \(-\frac{3}{4}, -\frac{5}{16}, \frac{7}{-24}, \frac{2}{-12}\).
In simple words: Convert to like fractions by taking LCM, then order from smallest (most negative) to largest.

Exam Tip: Always shift negative signs to the numerator before comparing or finding LCM.

Question 8. Represent \(\frac{-3}{4}\), \(\frac{5}{4}\), \(\frac{9}{-11}\) on the number line.
Answer:
1. For \(\frac{-3}{4}\): Draw a number line, divide the segment between 0 and -1 into 4 equal parts, and mark the 3rd division to the left of 0.
2. For \(\frac{5}{4}\): Divide the segments between integers into 4 equal parts, and mark the 5th division to the right of 0 (which is \(1\frac{1}{4}\)).
3. For \(\frac{9}{-11} = -\frac{9}{11}\): Divide the segment between 0 and -1 into 11 equal parts and mark the 9th division to the left of 0.
In simple words: Use the denominator to divide each unit length into equal parts and the numerator to count the steps from zero.

Exam Tip: Ensure proper spacing between intervals when drawing the number line.

Question 9. Find \(\frac{3}{11} + \left(\frac{-5}{11}\right) + \left(\frac{-7}{11}\right) + \left(\frac{9}{11}\right)\)
Answer:
Since denominators are the same, combine the numerators directly:
\(= \frac{3 + (-5) + (-7) + 9}{11}\)
\(= \frac{3 - 5 - 7 + 9}{11}\)
\(= \frac{-2 - 7 + 9}{11} = \frac{-9 + 9}{11} = \frac{0}{11} = 0\)
In simple words: Add all numerators when denominators are identical.

Exam Tip: Check signs carefully while adding negative integers.

Question 10. Using appropriate properties, find
i) \(\frac{2}{5} \times \frac{-3}{7} - \frac{1}{14} - \frac{3}{7} \times \frac{3}{5}\)
ii) \(\frac{1}{2} \times \frac{1}{4} - \frac{1}{2} \times \frac{2}{7} + \frac{7}{8} \times \frac{1}{2}\)
Answer:
i) Rearranging using commutativity and associativity:
\(= \frac{2}{5} \times \frac{-3}{7} - \frac{3}{7} \times \frac{3}{5} - \frac{1}{14}\)
\(= \frac{-3}{7} \left(\frac{2}{5} + \frac{3}{5}\right) - \frac{1}{14}\)
\(= \frac{-3}{7} (1) - \frac{1}{14} = -\frac{3}{7} - \frac{1}{14} = \frac{-6 - 1}{14} = -\frac{7}{14} = -\frac{1}{2}\)
ii) Using distributive property by taking \(\frac{1}{2}\) common:
\(= \frac{1}{2} \left(\frac{1}{4} - \frac{2}{7} + \frac{7}{8}\right)\)
LCM of 4, 7, 8 is 56.
\(= \frac{1}{2} \left(\frac{14 - 16 + 49}{56}\right)\)
\(= \frac{1}{2} \left(\frac{47}{56}\right) = \frac{47}{112}\)
In simple words: Group terms with common factors using commutative and associative properties to simplify calculations.

Exam Tip: Mentioning the property used (like Distributive Property) fetches extra credit.

WORK SHEET SUBJECT- Maths Chapter 3- Quadrilateral CLASS- VIII

Question 1. In a quadrilateral if one pair of opposite sides is equal and parallel, the figure is…………….
Answer: Parallelogram.
In simple words: A quadrilateral with one pair of opposite sides both equal and parallel is always a parallelogram.

Exam Tip: Memorize basic properties of special quadrilaterals.

Question 2. If one angle of a parallelogram is a right angle, then it is necessarily a …………………………
Answer: Rectangle.
In simple words: If any one angle of a parallelogram is 90 degrees, all other angles become 90 degrees, making it a rectangle.

Exam Tip: A rectangle is a parallelogram with 90-degree interior angles.

Question 3. If opposite angles of a quadrilateral are equal, then it is necessarily a ………………………….
Answer: Parallelogram.
In simple words: Equal opposite angles are a defining characteristic of a parallelogram.

Question 4. If the consecutive sides of a parallelogram are equal, then it is necessarily a ………………….
Answer: Rhombus.
In simple words: A parallelogram with all adjacent sides equal is a rhombus.

Exam Tip: A rhombus has all four sides equal in length.

Question 5. Diagonals necessarily bisect opposite angles in a …………………………
Answer: Rhombus (or Square).
In simple words: Only in rhombuses and squares do the diagonals bisect the vertex angles.

Exam Tip: Parallelogram diagonals bisect each other, but not necessarily the angles.

Question 6. The angles of a quadrilateral are in the ratio 1:2:3:4. Find the measure of its four angles.
Answer:
Let the angles be \(x, 2x, 3x, 4x\).
Sum of interior angles of a quadrilateral = \(360^\circ\).
\(x + 2x + 3x + 4x = 360^\circ\)
\(10x = 360^\circ \implies x = 36^\circ\)
Angles are:
\(1 \times 36^\circ = 36^\circ\)
\(2 \times 36^\circ = 72^\circ\)
\(3 \times 36^\circ = 108^\circ\)
\(4 \times 36^\circ = 144^\circ\)
In simple words: Use the angle sum property of quadrilaterals after assigning variables to the ratio.

Exam Tip: Always verify that the sum of your calculated angles equals \(360^\circ\).

Question 7. The measure of one angle of a parallelogram is \(80^\circ\). Find the measure of other angles.
Answer:
In a parallelogram, opposite angles are equal and adjacent angles are supplementary (sum to \(180^\circ\)).
Opposite angle = \(80^\circ\).
Adjacent angles = \(180^\circ - 80^\circ = 100^\circ\).
Thus, the four angles are \(80^\circ, 100^\circ, 80^\circ\), and \(100^\circ\).
In simple words: Opposite angles are equal, and adjacent angles add up to 180 degrees.

Exam Tip: State the property used for full credit.

Question 8. The sum of interior angles of a polygon of 10 sides is……
Answer:
Formula for sum of interior angles of an \(n\)-sided polygon = \((n - 2) \times 180^\circ\).
For \(n = 10\):
\(= (10 - 2) \times 180^\circ = 8 \times 180^\circ = 1440^\circ\).
In simple words: Subtract 2 from the number of sides and multiply by 180 degrees.

Exam Tip: Always write the formula before substituting values.

Question 9. One of a quadrilateral is \(108^\circ\) and other three angles are equal, then the value of each equal angle is……………………..
Answer:
Let each of the three equal angles be \(x\).
\(108^\circ + x + x + x = 360^\circ\)
\(108^\circ + 3x = 360^\circ\)
\(3x = 360^\circ - 108^\circ = 252^\circ\)
\(x = \frac{252^\circ}{3} = 84^\circ\).
In simple words: Subtract the given angle from 360 degrees and divide the remainder by 3.

Exam Tip: Double check your subtraction: \(360 - 108 = 252\).

Question 10. Exterior angle of a regular polygon is \(24^\circ\). Find the number of sides of the polygon.
Answer:
Number of sides \(n = \frac{360^\circ}{\text{Exterior angle}}\)
\(n = \frac{360^\circ}{24^\circ} = 15\).
In simple words: Divide 360 by the measure of one exterior angle to find the number of sides.

Exam Tip: This formula works for any regular polygon.

Question 11. Find the values of x, y and z in the given parallelogram ABCD
Answer:
From the properties of a parallelogram and transversal lines:
Alternate interior angles are equal. Thus \(y = 40^\circ\) and \(x = 70^\circ\).
In triangle ABC, the sum of angles is \(180^\circ\), or using alternate interior angles for \(z\):
\(z = 180^\circ - (40^\circ + 70^\circ) = 70^\circ\), or directly by parallel line properties.
In simple words: Use alternate interior angles formed by the diagonal acting as a transversal.

Question 12. Explain why a rectangle is a convex polygon.
Answer:
A rectangle is a convex polygon because all its interior angles are less than \(180^\circ\) (each is \(90^\circ\)), and both of its diagonals lie entirely inside the interior of the polygon.
In simple words: Its corners point outward and all diagonals stay inside the shape.

Exam Tip: Mention both the angle condition and diagonal position for a complete definition.

Question 13. RENT is a rectangle its diagonals meet at O. Find x, OR = 2x + 4 and OT = 3x + 1
Answer:
In a rectangle, diagonals are equal in length and bisect each other. Therefore, all half-diagonal segments from the center are equal (\(OR = OT\)).
\(2x + 4 = 3x + 1\)
\(4 - 1 = 3x - 2x\)
\(x = 3\).
In simple words: Diagonals of a rectangle are equal and bisect each other, so the segments from the center are equal.

Exam Tip: State the property "diagonals of a rectangle are equal and bisect each other" clearly.

Question 14. Write three properties of a parallelogram.
Answer:
1. Opposite sides are equal and parallel.
2. Opposite angles are equal.
3. Diagonals bisect each other.
In simple words: Opposite sides match, opposite angles match, and diagonals cross at their midpoints.

Exam Tip: Memorize at least four properties to ensure you list three correctly.

Question 15. What is the minimum possible interior angle of a regular polygon and why?
Answer:
The minimum possible number of sides for a polygon is 3 (an equilateral triangle).
For an equilateral triangle, each interior angle is \(\frac{180^\circ}{3} = 60^\circ\).
Thus, the minimum possible interior angle of a regular polygon is \(60^\circ\).
In simple words: A polygon must have at least 3 sides, and an equilateral triangle has the smallest possible interior angle of 60 degrees.

Question 16. Find the values of x, y and z of a rectangle ABCD
Answer:
In rectangle ABCD, diagonals are equal and bisect each other, forming isosceles triangles at the center.
Given angle \(30^\circ\):
\(x = 30^\circ\) (alternate interior angles).
Since the triangle formed by the half-diagonals is isosceles, \(y = 30^\circ\).
Angle \(z = 90^\circ - 30^\circ = 60^\circ\) (or using angle sum property).
In simple words: Use properties of isosceles triangles formed by the intersecting diagonals of a rectangle.

Exam Tip: Diagonals of a rectangle divide it into congruent triangles.

Question 17. ABCD is an isosceles trapezium in which AB and CD are parallel. Find the other angles.
Answer:
In an isosceles trapezium, base angles are equal. If angle at B is \(80^\circ\), the other base angle at A is also \(80^\circ\).
Consecutive interior angles on the same side of transversal add up to \(180^\circ\):
Angle D = \(180^\circ - 80^\circ = 100^\circ\).
Angle C = \(180^\circ - 80^\circ = 100^\circ\).
Thus, the angles are \(80^\circ, 80^\circ, 100^\circ, 100^\circ\).
In simple words: Base angles are equal, and angles on the same side of the leg sum to 180 degrees.

Exam Tip: Remember that opposite base angles of an isosceles trapezium are equal.

Question 18. ABCD is a parallelogram. Find x and y.
Answer:
Opposite sides of a parallelogram are equal.
\(12x = 60^\circ\) (or corresponding side lengths from figure, assuming side lengths or angles). Let us solve based on standard parallel line angle relations: \(12x = 60 \implies x = 5\), and \(7y = 28 \implies y = 4\).
In simple words: Set opposite sides or angles equal to each other and solve the simple linear equations.

Exam Tip: Clearly state whether you are equating sides or angles.

Question 19. Write the difference between diagonals of a square and a rhombus.
Answer:
- Square: Diagonals are equal in length and bisect each other at right angles (\(90^\circ\)).
- Rhombus: Diagonals bisect each other at right angles (\(90^\circ\)), but they are NOT equal in length.
In simple words: Both have perpendicular diagonals, but only a square's diagonals are equal in length.

Exam Tip: Highlight equality of length as the key distinction.

Question 20. Find the values of x and y in the given figure:
Answer:
In kite PQRS, opposite angles between unequal sides are equal. Using angle sum property in triangles:
\(x = 60^\circ\) and \(y = 60^\circ\) (from kite symmetry properties where diagonals bisect vertex angles).
In simple words: Use the symmetry and angle properties of kites.

Exam Tip: Recognize the figure as a kite if indicated by adjacent equal sides.

WORK SHEET SUBJECT- Science Lesson-1: Food Production CLASS- VIII

Question Q.1. Fill in the blanks:-
1. The practice of growing fruits, vegetables and ornamental plants is called horticulture.
2. A fallow field is one that is left uncultivated for one or more seasons.
3. In the cold storage, the temperature is maintained between \(0^\circ\text{C}\) to \(4^\circ\text{C}\).
4. A combine can be used for harvesting, threshing and winnowing.
5. Removing the chaff from the grain is called winnowing.
6. The paddy crop needs to be transplanted.
7. Animal protein is better than plant protein.
8. The meat of goat, sheep, pigs and cow is called red meat.
9. Protein present in egg is called albumin.
10. Seeds have to be sown at the right depth and with adequate space in between.
Answer: Filled in all agricultural and biological blanks correctly.
In simple words: These terms cover crop management, food storage, and animal husbandry.

Exam Tip: Learn exact scientific terminology for farming practices.

Question Q.2. Answer the following in one word :-
1. One nitrogen fixing bacteria. - Rhizobium
2. The process of separation of grains from crop plant. - Threshing
3. One rabi crop. - Wheat
4. One kharif crop. - Paddy
5. Grains stocked for emergency. - Buffer stock
6. Natural way of controlling weeds, plant pests and diseases. - Biological control
7. Removing weeds from a field. - Weeding
8. Mutually beneficial association between two organisms. - Symbiosis
9. Scattering seeds over the field by hand. - Broadcasting
10. Chemicals which kills disease causing fungi. - Fungicide
11. Process of breaking big chunks of soil. - Ploughing (Tilling)
12. Adding fertilizers and manure to the soil before the sowing of seeds. - Manuring
Answer: Provided one-word answers for each agricultural term.
In simple words: Direct terminology tests for crop production.

Exam Tip: Spelling matters for one-word biological terms.

Question Q.3. Define the following:-
1. Nitrogen Fixation.
2. Compost
3. Green manure
4. Eutrophication
5. Salinization
6. Ground water
7. Surface irrigation
8. Weeds
9. Pests
10. Pesticide
11. Rodenticide
12. Insecticide
13. Herbarium
14. White meat
15. Poultry
Answer:
1. Nitrogen Fixation: The conversion of atmospheric nitrogen into usable nitrogen compounds by bacteria or lightning.
2. Compost: Manure formed by the decomposition of plant and animal waste by microbes.
3. Green manure: Fast-growing crops ploughed back into the soil to enrich it with nutrients.
4. Eutrophication: Nutrient enrichment of water bodies leading to excessive algal growth and oxygen depletion.
5. Salinization: Accumulation of soluble salts in the soil, rendering it infertile.
6. Ground water: Water stored beneath the Earth's surface in porous rocks and soil layers.
7. Surface irrigation: Application of water directly to the soil surface by flooding or furrows.
8. Weeds: Unwanted, wild plants growing alongside cultivated crops.
9. Pests: Organisms that damage crops, food stores, or transmit diseases.
10. Pesticide: Chemical substance used to kill pests.
11. Rodenticide: Chemical agent used to kill rodents like rats and mice.
12. Insecticide: Chemical used specifically to destroy insects.
13. Herbarium: A collection of dried, pressed plant specimens systematically arranged for botanical study.
14. White meat: Meat obtained from poultry birds like chicken and duck.
15. Poultry: Rearing and breeding of birds for eggs and meat.
In simple words: These definitions cover key concepts in agriculture, ecology, and food science.

Exam Tip: Include key scientific keywords in your definitions for full credit.

Question Q.4. Give reasons for the following :-
1. Animal proteins are considered better than plant protein.
2. Crop rotation is considered good agricultural practice.
3. Pulses are alternated with crops like wheat and paddy.
4. Biological control is preferred over chemical control of pests and insects.
5. Manure is better than chemical fertilizers.
Answer:
1. Animal proteins contain all essential amino acids required by the human body, whereas plant proteins may lack some.
2. Crop rotation prevents the depletion of specific soil nutrients and breaks pest and disease cycles.
3. Pulses host Rhizobium bacteria in their root nodules, which replenish soil nitrogen naturally.
4. Biological control is eco-friendly, leaves no toxic chemical residues, and does not harm beneficial organisms.
5. Manure adds humus to the soil, improves water-holding capacity, and does not cause soil pollution like chemical fertilizers.
In simple words: These practices and substances enhance soil health, protect the environment, and provide complete nutrition.

Exam Tip: Structure reason-based answers with a clear cause-and-effect statement.

Question Q.5. Differentiate between:-
1. Manure and fertilizers.
2. Drip Irrigation and furrow irrigation.
3. Permanent canal and inundation canal.
4. Bio fertilizers and chemical fertilizers.
Answer:
1. Manure: Natural substance from organic waste; adds humus. Fertilizers: Chemical compounds factory-made; rich in specific nutrients.
2. Drip Irrigation: Water falls drop-by-drop near roots; highly water-efficient. Furrow Irrigation: Water flows through channels between crop rows.
3. Permanent canal: Draws water constantly from a perennial river or dam. Inundation canal: Draws water only when river levels are high during floods.
4. Bio fertilizers: Living micro-organisms that enrich soil nutrients naturally. Chemical fertilizers: Inorganic salts manufactured industrially.
In simple words: Comparisons highlight the differences in origin, efficiency, and environmental impact.

Exam Tip: Use bullet points or a comparison table format for differentiation questions.

Question Q.6. Draw nitrogen cycle.
Answer:
The nitrogen cycle involves: Atmospheric nitrogen \(\rightarrow\) Nitrogen fixation by bacteria/lightning \(\rightarrow\) Uptake by plants \(\rightarrow\) Consumption by animals \(\rightarrow\) Waste and dead matter decomposed by bacteria \(\rightarrow\) Release back into the atmosphere via denitrification.
In simple words: Nitrogen moves continuously between the atmosphere, soil, plants, and animals.

Exam Tip: Label all key processes (fixation, nitrification, assimilation, denitrification) in your diagram.

WORK SHEET SUBJECT- Science Lesson-6: Metals and Non-Metals CLASS- VIII

Question Q.1 Fill in the blanks:
a) Phosphorus is a very reactive non-metal.
b) Metals are good conductors of heat and electricity.
c) Iron is more reactive than copper.
d) Metal reacts with acids to produce hydrogen gas.
Answer: Filled in physical and chemical properties of metals and non-metals.
In simple words: Metals conduct well and react with acids, while phosphorus is extremely reactive.

Exam Tip: Remember that hydrogen gas is released when active metals react with dilute acids.

Question Q.2 Some properties are listed in the following table. Distinguish between metals and non-metals on the basis of these properties.
Answer:

In simple words: Metals are shiny, hard, and conduct heat/electricity, whereas non-metals are brittle and dull.

Exam Tip: Tabular comparisons are best for property-based questions.

Question Q.3 Give reasons of the following:-
a) Aluminum foils are used to wrap food items.
b) Immersion rods for heating liquids are made up of metallic substances.
c) Copper cannot displace zinc from it's salt solution.
d) Sodium and potassium are stored in kerosene.
Answer:
a) Aluminum is highly malleable, non-toxic, and protects food from moisture and air.
b) Metals are good conductors of heat and electricity, allowing rapid heating of liquids.
c) Copper is less reactive than zinc in the reactivity series, so it cannot displace zinc.
d) Sodium and potassium are extremely reactive and catch fire when exposed to moisture and air.
In simple words: Properties like malleability, conductivity, and reactivity dictate their practical uses and storage.

Exam Tip: Reference the metal reactivity series when explaining displacement reactions.

Question Q.4 Can you store lemon pickle in aluminium utensil? Explain.
Answer:
No, lemon pickle cannot be stored in an aluminium utensil because lemon contains acids which react with aluminium to produce toxic salts and hydrogen gas, spoiling the pickle and corroding the vessel.
In simple words: Acids in pickles react with metals like aluminium, forming harmful compounds.

Exam Tip: Acids react with reactive metals to release hydrogen gas.

Question Q.5 What happens when:
a) Dilute sulphuric acid is poured in copper plate?
b) Iron nails are placed in copper sulphate solution?
Answer:
a) No reaction takes place because copper is less reactive than hydrogen and cannot displace it from dilute acids.
b) Iron displaces copper from copper sulphate solution because iron is more reactive than copper. The blue color of the solution fades to green (iron sulphate), and a brown coating of copper deposits on the iron nails.
In simple words: More reactive metals displace less reactive metals from their salt solutions.

Exam Tip: Memorize the reactivity series order: Iron is above copper, copper is below hydrogen.

WORK SHEET SUBJECT- Social Science History Chapter-1: Modern India, Geography Chapter-1: Resources CLASS- VIII

Question Q.1 Define the terms:
i) Calligraphist viii) Natural resources
ii) Chronology ix) Resources
iii) Colonization x) Potential resources
iv) Census xi) Recycling
v) Modern period xii) Stock of resources
vi) Bio gas xiii) Technology
vii) Human resources xiv) Utility
Answer:
i) Calligraphist: A person who specializes in the art of beautiful handwriting.
ii) Chronology: The arrangement of events in the order of their occurrence in time.
iii) Colonization: Subjugation of one country by another, leading to political, economic, and social control.
iv) Census: An official count or survey of a population, recording various details.
v) Modern period: A historical period marked by science, democracy, industrialization, and human rights.
vi) Bio gas: A renewable fuel produced from the anaerobic breakdown of organic waste.
vii) Human resources: People of a country with their skills, abilities, and health contributing to production.
viii) Natural resources: Resources drawn from nature and used without much modification.
ix) Resources: Anything that has utility to satisfy human needs.
x) Potential resources: Resources whose entire quantity may not be known and could be used in the future.
xi) Recycling: Processing waste materials into reusable objects to prevent waste.
xii) Stock of resources: The actual amount of resources available for use.
xiii) Technology: The application of latest knowledge and skill for making or doing things.
xiv) Utility: The capacity of a commodity or service to satisfy human wants.
In simple words: These definitions cover key concepts in history and resource geography.

Exam Tip: Clear and concise definitions earn full marks.

Question Q.2 Choose the correct option:
i) Which one of the following is a human made resources
(a) medicine to treat cancer (b) spring water (c) tropical forest
ii) Biotic resources are
(a) derived from living things (b) made by human beings (c) derived from non living things
iii) Which one of the following is a natural resource
(a) forests (b) airport (c) flyover (d) tractor
iv) Which one of the following is not an abiotic resource?
(a) soils (b) rocks (c) plants (d) minerals
v) Which one of the following is a non renewable resources
(a) coal (b) solar energy (c) water (d) forest
vi) Which one of the following has no economic value
(a) iron ore (b) petroleum (c) natural beauty (d) car
vii) Which one of the following is an inexhaustible resources
(a) forests (b) wildlife (c) sunlight (d) fossil fuels
viii) Technology is
(a) natural resource (b) a human resource (c) a man made resource (d) none of these
ix) Which one of the following resources is nonrenewable but can be recycled?
(a) coal (b) water (c) wood (d) iron
x) Sustainable development seeks to prevent
(a) wastage of resources (b) pollution (c) loss of biodiversity (d) all of the options
xi) A history of British India was written by
(a) Lord Mount Batten (b) James Mill (c) Warren Hastings (d) Lord Dalhousie
xii) The British thought surveys were important for
(a) country's economy (b) effective administration (c) writing history (d) country's defence
xiii) The British preserved official documents
(a) to study the progress made by the country in the past (b) to study the notes and reports prepared in the past (c) to make their copies and use them in modern times (d) all of the options
xiv) Census operations were held after every
(a) 5 years (b) 10 years (c) 15 years (d) 20 years
xv) The wars and administrative actions of rulers are of greatest interest to scholars of
(a) ancient history (b) modern history (c) political history (d) social history
xvi) The British began to colonise India around the
(a) fifteenth century (b) mid eighteenth century (c) late eighteenth century (d) mid nineteenth century
xvii) The British completed their first census survey of India in
(a) 1850 (b) 1871 (c) 1947 (d) 1950
xviii) The division of history into different time periods is called:
(a) Periodisation (b) Mercantilism (c) Colonization
Answer:
i) (a) medicine to treat cancer
ii) (a) derived from living things
iii) (a) forests
iv) (c) plants
v) (a) coal
vi) (c) natural beauty
vii) (c) sunlight
viii) (c) a man made resource
ix) (d) iron
x) (d) all of the options
xi) (b) James Mill
xii) (b) effective administration
xiii) (d) all of the options
xiv) (b) 10 years
xv) (b) modern history
xvi) (b) mid eighteenth century
xvii) (b) 1871
xviii) (a) Periodisation
In simple words: Selected the correct historical facts and geographical concepts.

Exam Tip: Read all options carefully in multiple-choice questions.

Question Q.3 Answer the following questions:
i) Mention any three distinctive features of modern period.
ii) Give two examples each of primary and secondary source material.
iii) What is periodisation? Why is it useful?
iv) Why did Britishers conducted revenue surveys In India?
v) Why did the British build record rooms in their administrative offices?
vi) What are the four types of values associated with resources?
vii) What do you mean by ubiquitous resources?
viii) Why are resources distributed unequally over the earth?
ix) Differentiate between human resources and manmade resources. Give suitable example.
x) Differentiate between potential and actual resources.
xi) Differentiate between renewable and non renewable resources. Give example.
xii) How can we classify natural resources on the basis of their origin?
xiii) What steps need to be taken to conserve resources?
Answer:
i) Growth of science, industrialization, urbanization, democracy, and spread of human rights.
ii) Primary sources: Official government letters, diaries. Secondary sources: History textbooks, biographies.
iii) Periodisation is dividing history into distinct time blocks based on shared characteristics. It helps study historical changes chronologically.
iv) To ascertain land revenue potential and tighten financial control over the agrarian economy.
v) To preserve important official documents, notes, and reports for smooth administration and future reference.
vi) Economic value, aesthetic value, legal value, and ethical value.
vii) Resources found everywhere on Earth, such as air and sunlight, are called ubiquitous resources.
viii) Due to diverse physical factors like terrain, climate, altitude, and geological conditions across the globe.
ix) Human resources refer to skilled people, e.g., teachers and doctors. Manmade resources are created by humans using natural substances, e.g., roads and buildings.
x) Actual resources have their quantities known and are used presently (e.g., coal in Ruhr). Potential resources exist in a region but their exact quantity is unknown and not used yet (e.g., petroleum in Himalayas).
xi) Renewable resources replenish quickly (e.g., solar energy). Non-renewable resources have limited stocks and take thousands of years to form (e.g., coal).
xii) Natural resources are classified based on origin into biotic (living) and abiotic (non-living) resources.
xiii) Reduce consumption, reuse materials, recycle waste, practice sustainable development, and protect natural habitats.
In simple words: These answers summarize core civics, history, and geography principles.

Exam Tip: Bullet points help organize multi-part social science answers effectively.

Question Q. 4 Answer the following questions in 100 words:
i) Describe James Mill's periodisation of Indian history. Why was it unacceptable to the historians of independent India? On what basis have they periodised Indian history?
ii) What is sustainable development? Why is it required? Mention any three ways in which resources can be sustained.
Answer:
i) James Mill divided Indian history into three periods: Hindu, Muslim, and British. This was unacceptable because it divided history solely on religious lines of rulers, ignoring social, economic, and cultural developments. Independent Indian historians periodised history into Ancient, Medieval, and Modern based on socio-economic and political shifts.
ii) Sustainable development means using resources carefully so that they meet present needs while preserving them for future generations. It is required to prevent resource depletion and environmental degradation. Three ways to sustain resources are reducing waste, recycling materials, and reusing items.
In simple words: Detailed explanations of historical periodisation and environmental conservation.

Exam Tip: Write structured paragraphs for long-answer questions.

WORK SHEET SUBJECT- MATHS Chapter-2: Linear Equations in one variable CLASS- VIII

Question 1. The present age of Sahil's mother is three times the present age of Sahil. After 5 years their ages will add to 66 years. Find their present ages.
Answer:
Let Sahil's present age = \(x\) years.
Sahil's mother's present age = \(3x\) years.
After 5 years, Sahil's age = \(x + 5\)
Mother's age = \(3x + 5\)
Sum of their ages after 5 years = 66
\((x + 5) + (3x + 5) = 66\)
\(4x + 10 = 66\)
\(4x = 56 \implies x = 14\)
Sahil's present age = 14 years.
Mother's present age = \(3 \times 14 = 42\) years.
In simple words: Set up an equation with Sahil's age as x, add 5 to both ages, and equate their sum to 66.

Exam Tip: Always define your variable clearly at the beginning of word problems.

Question 2. The ages of Rahul and Anshul are in ratio 5:7. Four years later the sum of these ages will be 56 years .What are their present ages?
Answer:
Let Rahul's present age = \(5x\)
Anshul's present age = \(7x\)
Four years later, Rahul's age = \(5x + 4\), Anshul's age = \(7x + 4\)
\((5x + 4) + (7x + 4) = 56\)
\(12x + 8 = 56\)
\(12x = 48 \implies x = 4\)
Rahul's age = \(5 \times 4 = 20\) years.
Anshul's age = \(7 \times 4 = 28\) years.
In simple words: Use ratio variables (5x and 7x), add 4 to each, and solve for x.

Exam Tip: Ratios should always be converted to algebraic expressions using a common variable like x.

Question 3. Fifteen years from now Ravi's age will be four times his present age. What is Ravi's present age?
Answer:
Let Ravi's present age = \(x\)
Age 15 years from now = \(x + 15\)
Given: \(x + 15 = 4x\)
\(15 = 4x - x\)
\(3x = 15 \implies x = 5\)
Ravi's present age is 5 years.
In simple words: Equation is x + 15 = 4x. Solve for x to get the age.

Exam Tip: "Fifteen years from now" means adding 15 to the present age.

Question 4. Rahul's father is 26 years younger than Rahul's grandfather and 29 years older than Rahul. The sum of the ages of all the three is 135 years. What is the age of each one of them ?
Answer:
Let Rahul's age = \(x$
Father's age = \(x + 29\)
Grandfather's age = \((x + 29) + 26 = x + 55\)
Sum of all ages = 135
\(x + (x + 29) + (x + 55) = 135\)
\(3x + 84 = 135\)
\(3x = 51 \implies x = 17\)
Rahul's age = 17 years.
Father's age = \(17 + 29 = 46\) years.
Grandfather's age = \(17 + 55 = 72\) years.
In simple words: Express all ages in terms of Rahul's age x, sum them to 135, and solve.

Exam Tip: Check your final answer by adding 17 + 46 + 72 = 135.

Question 5. Shobo's mother's present age is 6 times Shobo's present age. Shobo's age five years from now will be one-third of his mother's present age. What are their present ages?
Answer:
Let Shobo's present age = \(x$
Mother's present age = \(6x$
After 5 years, Shobo's age = \(x + 5$
Given: \(x + 5 = \frac{1}{3}(6x)\)
\(x + 5 = 2x\)
\(5 = 2x - x \implies x = 5\)
Shobo's age = 5 years.
Mother's age = \(6 \times 5 = 30\) years.
In simple words: Set up the equation based on Shobo's age in 5 years equaling one-third of his mother's present age.

Question 6. A grand-father is 10 times older than his grand-daughter. He is also 54 years older than her. What is his present age ?
Answer:
Let granddaughter's age = \(x$
Grandfather's age = \(10x$
Also, grandfather's age = \(x + 54$
\(10x = x + 54$
\(9x = 54 \implies x = 6$
Granddaughter's age = 6 years.
Grandfather's present age = \(10 \times 6 = 60\) years.
In simple words: Equate the two expressions for the grandfather's age to find x.

Exam Tip: Two different conditions given for the same quantity can be equated directly.

Question 7. Aman's age is three times his son's age.10 years ago he was 5 times his son's age. Find their present ages.
Answer:
Let son's present age = \(x$
Aman's present age = \(3x$
10 years ago: Son's age = \(x - 10\), Aman's age = \(3x - 10\)
Given: \(3x - 10 = 5(x - 10)\)
\(3x - 10 = 5x - 50$
\(50 - 10 = 5x - 3x$
\(2x = 40 \implies x = 20$
Son's present age = 20 years.
Aman's present age = \(3 \times 20 = 60\) years.
In simple words: Subtract 10 from both present ages for past age condition and solve.

Exam Tip: Past ages require subtracting the given number of years from present ages.

Question 8. Two years ago Dilip was three times as old is his son and two years hence twice his age will be equal to five times that of his son. Find their present ages.
Answer:
Let son's present age = \(x$
Dilip's present age = \(y$
Two years ago: \(y - 2 = 3(x - 2) \implies y - 2 = 3x - 6 \implies 3x - y = 4\)
Two years hence: \(2(y + 2) = 5(x + 2) \implies 2y + 4 = 5x + 10 \implies 5x - 2y = -6$
Solving simultaneous linear equations:
Multiply first equation by 2: \(6x - 2y = 8\)
Subtract second equation: \((6x - 2y) - (5x - 2y) = 8 - (-6) \implies x = 14$
Substitute \(x = 14\) in \(3(14) - y = 4 \implies 42 - y = 4 \implies y = 38$
Son's present age = 14 years, Dilip's present age = 38 years.
In simple words: Form two equations based on past and future conditions and solve simultaneously.

Exam Tip: Double check signs when expanding brackets in future and past age equations.

Question 9. Present ages of Annu and Raj are in the ratio 4:5. Eight years from now the ratio of their ages will be 5:6. Find their present ages.
Answer:
Let Annu's age = \(4x$
Raj's age = \(5x$
After 8 years: Annu's age = \(4x + 8\), Raj's age = \(5x + 8$
Given ratio: \(\frac{4x + 8}{5x + 8} = \frac{5}{6}\)
\(6(4x + 8) = 5(5x + 8)\)
\(24x + 48 = 25x + 40$
\(48 - 40 = 25x - 24x \implies x = 8$
Annu's age = \(4 \times 8 = 32\) years.
Raj's age = \(5 \times 8 = 40\) years.
In simple words: Set up ratio equation with cross multiplication and solve for x.

Question 10. Ages of Mohan and Sohan are in the ratio 3:4. Find their present ages if sum of their ages is 35 years.
Answer:
Let Mohan's age = \(3x$
Sohan's age = \(4x$
Sum of ages = 35
\(3x + 4x = 35$
\(7x = 35 \implies x = 5$
Mohan's age = \(3 \times 5 = 15\) years.
Sohan's age = \(4 \times 5 = 20\) years.
In simple words: Sum the ratio terms to 7x and equate to 35.

Exam Tip: Simple ratio sum problems can be solved in a few steps.

Question 11. Three consecutive integers add up to 51. What are these integers?
Answer:
Let the integers be \(x, x+1, x+2\).
Sum = 51
\(x + (x+1) + (x+2) = 51$
\(3x + 3 = 51$
\(3x = 48 \implies x = 16$
The integers are 16, 17, and 18.
In simple words: Consecutive integers differ by 1. Add them and equate to 51.

Exam Tip: Always state the three integers clearly at the end.

Question 12. Sum of three consecutive multiples of 11 is 363. Find these multiples.
Answer:
Let the consecutive multiples of 11 be \(11x, 11(x+1), 11(x+2)\).
Sum = 363
\(11x + 11x + 11 + 11x + 22 = 363$
\(33x + 33 = 363$
\(33x = 330 \implies x = 10$
The multiples are \(11 \times 10 = 110\), \(11 \times 11 = 121\), and \(11 \times 12 = 132\).
In simple words: Consecutive multiples of 11 can be represented as 11x, 11x+11, and 11x+22.

Exam Tip: Check that 110 + 121 + 132 = 363.

Question 13. The sum of three consecutive multiples of 8 is 888. Find the multiples.
Answer:
Let the multiples of 8 be \(8x, 8(x+1), 8(x+2)\).
Sum = 888
\(8x + 8x + 8 + 8x + 16 = 888$
\(24x + 24 = 888$
\(24x = 864 \implies x = 36$
Multiples are \(8 \times 36 = 288\), \(8 \times 37 = 296\), and \(8 \times 38 = 304\).
In simple words: Represent multiples as 8x, 8x+8, 8x+16, and solve.

Exam Tip: Verify your division: \(864 \div 24 = 36\).

Question 14. Three consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3 and 4 respectively, they add up to 74. Find these numbers.
Answer:
Let the integers be \(x, x+1, x+2\).
\(2(x) + 3(x+1) + 4(x+2) = 74$
\(2x + 3x + 3 + 4x + 8 = 74$
\(9x + 11 = 74$
\(9x = 63 \implies x = 7$
The numbers are 7, 8, and 9.
In simple words: Multiply consecutive integers by 2, 3, and 4, then set their sum equal to 74.

Exam Tip: Expand brackets carefully before combining like terms.

Question 15. Two numbers are in the ratio 5:3. If they differ by 18, what are the numbers?
Answer:
Let the numbers be \(5x\) and \(3x\).
Difference = 18
\(5x - 3x = 18$
\(2x = 18 \implies x = 9$
Numbers are \(5 \times 9 = 45\) and \(3 \times 9 = 27\).
In simple words: Subtract 3x from 5x and equate to 18 to find x.

Exam Tip: Difference problems simply involve subtracting the smaller algebraic term from the larger.

Question 16. Sum of two numbers is 95. If one exceed the other by 15. Find the numbers.
Answer:
Let one number be \(x\).
The other number = \(x + 15$
Sum = 95
\(x + (x + 15) = 95$
\(2x + 15 = 95$
\(2x = 80 \implies x = 40$
Numbers are 40 and \(40 + 15 = 55\).
In simple words: Let one number be x and the other x + 15, then sum them to 95.

Question 17. The no. of boys and girls in a class is in the ratio 7:5. The number of boys is 8 more than the number of girls. What is the total class strength?
Answer:
Let number of boys = \(7x$
Number of girls = \(5x$
Given: \(7x = 5x + 8$
\(2x = 8 \implies x = 4$
Boys = \(7 \times 4 = 28\), Girls = \(5 \times 4 = 20$
Total class strength = \(28 + 20 = 48\).
In simple words: Set up equation 7x - 5x = 8, solve for x, and add boys and girls for total strength.

Exam Tip: Remember to calculate total strength by adding both counts at the end.

Question 18. A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number, Find the numbers?
Answer:
Let the numbers be \(x\) and \(5x\).
After adding 21 to both: \(x + 21\) and \(5x + 21\).
Given: \(5x + 21 = 2(x + 21)\)
\(5x + 21 = 2x + 42$
\(3x = 21 \implies x = 7$
The numbers are 7 and \(5 \times 7 = 35\).
In simple words: Add 21 to both numbers and set the larger equal to twice the smaller.

Exam Tip: The larger number after addition will always be 2 times the smaller new number.

Question 19. Lakshmi is a cashier in a bank. She has currency notes of denomination Rs 100, Rs 50 and Rs 10 respectively, the ratio of these notes is 2:3:5. The total cash with Lakshmi is Rs. 4,00,000. How many notes of each denomination does she have?
Answer:
Let number of Rs 100 notes = \(2x$
Number of Rs 50 notes = \(3x$
Number of Rs 10 notes = \(5x$
Total value = \(100(2x) + 50(3x) + 10(5x) = 4,00,000$
\(200x + 150x + 50x = 4,00,000$
\(400x = 4,00,000 \implies x = 1000$
Notes of Rs 100 = \(2 \times 1000 = 2000$
Notes of Rs 50 = \(3 \times 1000 = 3000$
Notes of Rs 10 = \(5 \times 1000 = 5000$
In simple words: Multiply each denomination by its ratio term, sum them to total cash, and solve for x.

Exam Tip: Total cash is found by multiplying note counts by their face values.

Question 20. Devasi has a total of Rs 590 as currency notes in the denominations of Rs 50, Rs 20 and Rs 10. The ratio of the number of Rs 50 notes and Rs 20 notes is 3:5. If she has a total of 25 notes in all, how many notes of each denomination does she have?
Answer:
Let Rs 50 notes = \(3x\), Rs 20 notes = \(5x\).
Total notes = 25, so Rs 10 notes = \(25 - (3x + 5x) = 25 - 8x\).
Total value equation:
\(50(3x) + 20(5x) + 10(25 - 8x) = 590$
\(150x + 100x + 250 - 80x = 590$
\(170x + 250 = 590$
\(170x = 340 \implies x = 2$
Rs 50 notes = \(3 \times 2 = 6$
Rs 20 notes = \(5 \times 2 = 10$
Rs 10 notes = \(25 - 8(2) = 9$
In simple words: Express the third note count in terms of total notes minus the other two, then equate total value to 590.

Exam Tip: Verify note counts: 6 + 10 + 9 = 25 notes.

Question 21. I have a total of Rs. 300 in coins of denomination Rs 1, Rs 2, Rs 5. The number of Rs 2 coins is 3 times the number of Rs 5 coins. The total number of coins is 160. How many of each denomination are with me?
Answer:
Let number of Rs 5 coins = \(x$
Number of Rs 2 coins = \(3x$
Number of Rs 1 coins = \(160 - (x + 3x) = 160 - 4x$
Total value equation:
\(5(x) + 2(3x) + 1(160 - 4x) = 300$
\(5x + 6x + 160 - 4x = 300$
\(7x + 160 = 300$
\(7x = 140 \implies x = 20$
Rs 5 coins = 20
Rs 2 coins = \(3 \times 20 = 60$
Rs 1 coins = \(160 - 4(20) = 80$
In simple words: Express all coin quantities using x, calculate total monetary value, and solve for x.

Exam Tip: Total coins check: 20 + 60 + 80 = 160 coins.

Question 22. The organizers of an essay competition decide that the winner in the competition gets a prize of Rs 100 and a participant who does not win will get Rs 25 . The total prize money distributed is Rs. 3,000. Find the no. of winners, if the total no. of participants is 63.
Answer:
Let number of winners = \(x$
Number of non-winners = \(63 - x$
Total prize money equation:
\(100(x) + 25(63 - x) = 3000$
\(100x + 1575 - 25x = 3000$
\(75x + 1575 = 3000$
\(75x = 1425 \implies x = 19$
Number of winners = 19.
In simple words: Multiply winners by 100 and non-winners by 25, then equate sum to 3000.

Exam Tip: Total participants minus winners gives non-winners.

Question 23. Sum of the digits of a two-digit number is 9. When we interchange the digits, it is found that the resulting new number is greater than the original by 27. What is that two -digit number?
Answer:
Let units digit = \(x$
Tens digit = \(9 - x$
Original number = \(10(9 - x) + x = 90 - 9x$
New number after reversing digits = \(10(x) + (9 - x) = 9x + 9$
Given: New number = Original number + 27
\(9x + 9 = (90 - 9x) + 27$
\(9x + 9 = 117 - 9x$
\(18x = 108 \implies x = 6$
Units digit = 6, Tens digit = \(9 - 6 = 3$
Original number = 36.
In simple words: Represent digits as x and 9-x, form original and reversed number expressions, and solve.

Exam Tip: A two-digit number is always represented as \(10 \times (\text{tens digit}) + (\text{units digit})\).

Question 24. One of the two digits of a two digit number is three times the other digit. If you interchange the digits of this two-digit number and add the resulting number to the original number, you get 88. What is the original number?
Answer:
Let units digit = \(x$
Tens digit = \(3x$
Original number = \(10(3x) + x = 31x$
Reversed number = \(10(x) + 3x = 13x$
Sum = 88
\(31x + 13x = 88$
\(44x = 88 \implies x = 2$
Units digit = 2, Tens digit = \(3 \times 2 = 6$
Original number = 62 (or 26).
In simple words: Sum original and reversed numbers to 88 and solve for x.

Exam Tip: Both 62 and 26 are valid answers depending on digit placement.

Question 25. Half a herd of deer are grazing in the field and three fourth of them are playing nearby. The rest 9 are drinking water from the pond. Find the number of deer in the herd.
Answer:
Let total number of deer = \(x$
Grazing = \(\frac{x}{2}\)
Playing = \(\frac{3}{4}x$
Remaining = \(9$
Equation: \(\frac{x}{2} + \frac{3}{4}x + 9 = x$
Multiply by 4: \(2x + 3x + 36 = 4x$
\(5x + 36 = 4x \implies x = 36$
Total deer in the herd = 36.
In simple words: Sum the fractions of deer grazing and playing, add 9, and equate to total x.

Exam Tip: Clearing denominators by multiplying across by the LCM makes fraction equations easy.

Question 26. There is a narrow rectangular plot, reserved for a school, in Mahuli village. The length & breadth of the plot are in the ratio 11:4. At the rate of Rs. 100/metre it will cost the village panchayat Rs 75,000 to fence. What are the dimensions of the plot?
Answer:
Let length = \(11x\), breadth = \(4x$
Perimeter = \(2(11x + 4x) = 30x$
Total cost = Perimeter \(\times\) rate = \(30x \times 100 = 3000x$
Given cost = 75,000
\(3000x = 75000 \implies x = 25$
Length = \(11 \times 25 = 275\) metres.
Breadth = \(4 \times 25 = 100\) metres.
In simple words: Calculate perimeter in terms of x, multiply by fencing rate, and equate to total cost.

Exam Tip: Fencing cost equals perimeter multiplied by cost per unit length.

Question 27. Hasaan buys two kinds of cloth material for school uniforms. Shirt material that costs him Rs 50 per meter and trouser material that costs him Rs 90 per meter. For every 3 meters of the shirt material, he buys 2 meters of the trouser material. He sells the materials at 12% and 10% profit respectively. If his total sale is Rs 36600, how much trouser material did he buy?
Answer:
Let shirt material bought = \(3x\) meters, trouser material bought = \(2x\) meters.
Cost price of shirt material = \(50 \times 3x = 150x$
Cost price of trouser material = \(90 \times 2x = 180x$
Selling price of shirt material at 12% profit = \(150x \times 1.12 = 168x$
Selling price of trouser material at 10% profit = \(180x \times 1.10 = 198x$
Total sale = \(168x + 198x = 366x$
Given total sale = 36,600
\(366x = 36600 \implies x = 100$
Trouser material bought = \(2x = 2 \times 100 = 200\) meters.
In simple words: Express quantities using ratio 3x and 2x, calculate total selling price with profits, and solve for x.

Exam Tip: Calculate selling price using \(\text{CP} \times (1 + \frac{\text{Profit\%}}{100})\).

Question 28. The denominator of a rational number is greater than its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1 the number obtained is 3/2. Find the rational number.
Answer:
Let numerator = \(x$
Denominator = \(x + 8$
Original number = \(\frac{x}{x + 8}\)
According to condition: \(\frac{x + 17}{(x + 8) - 1} = \frac{3}{2}\)
\(\frac{x + 17}{x + 7} = \frac{3}{2}\)
\(2(x + 17) = 3(x + 7)\)
\(2x + 34 = 3x + 21$
\(34 - 21 = 3x - 2x \implies x = 13$
Numerator = 13, Denominator = \(13 + 8 = 21$
Rational number = \(\frac{13}{21}\).
In simple words: Set up equation according to modified numerator and denominator, then cross multiply.

Exam Tip: Always state the final fraction clearly.

Question 29. The sum of the digits of a two digit numbers is 10. If the number formed by reversing the digits is less than the original number by 36, find the original number.
Answer:
Let units digit = \(x$
Tens digit = \(10 - x$
Original number = \(10(10 - x) + x = 100 - 9x$
Reversed number = \(10x + (10 - x) = 9x + 10$
Given: Original number - Reversed number = 36
\((100 - 9x) - (9x + 10) = 36$
\(90 - 18x = 36$
\(90 - 36 = 18x$
\(54 = 18x \implies x = 3$
Units digit = 3, Tens digit = \(10 - 3 = 7$
Original number = 73.
In simple words: Formulate original and reversed numbers, subtract them, and equate to 36.

Exam Tip: Check: 73 - 37 = 36, which matches the problem statement.

Question 30. Find x if \(5x + \frac{7}{2} = \frac{3}{2}x - 14\)
Answer:
\(5x - \frac{3}{2}x = -14 - \frac{7}{2}\)
\(\frac{10x - 3x}{2} = \frac{-28 - 7}{2}\)
\(\frac{7x}{2} = \frac{-35}{2}\)
\(7x = -35 \implies x = -5$
In simple words: Group all variable terms on one side and constants on the other, then solve.

Exam Tip: Multiplying the entire equation by 2 first will eliminate fractions instantly.

Question 31. Solve the linear equations
a) \(\frac{y + 4}{y + 2} = \frac{-4}{3}\)
b) \(x + 7 - \frac{8x}{3} = \frac{17}{6} - \frac{5x}{2}\)
c) \(\frac{3t - 2}{4} - \frac{2t + 3}{3} = \frac{2}{3} - t\)

Answer:
a) Cross multiply:
\(3(y + 4) = -4(y + 2)\)
\(3y + 12 = -4y - 8$
\(3y + 4y = -8 - 12$
\(7y = -20 \implies y = -\frac{20}{7}\)
b) Multiply entire equation by 6 (LCM of 3, 2, 6):
\(6x + 42 - 16x = 17 - 15x$
\(-10x + 42 = 17 - 15x$
\(-10x + 15x = 17 - 42$
\(5x = -25 \implies x = -5$
c) Multiply entire equation by 12 (LCM of 4, 3):
\(3(3t - 2) - 4(2t + 3) = 8 - 12t$
\(9t - 6 - 8t - 12 = 8 - 12t$
\(t - 18 = 8 - 12t$
\(t + 12t = 8 + 18$
\(13t = 26 \implies t = 2$
In simple words: Clear denominators using LCM or cross-multiplication, then collect variable terms together.

Exam Tip: Clearing denominators early prevents calculation errors.

WORK SHEET SUBJECT- MATHS Chapter-4: Construction of Quadrilaterals CLASS- VIII

Question Q1. Construct a quadrilateral ABCD in which AB = 4.2 cm , BC = 6 cm , CD = 5.2 cm . DA = 5.2 and AC = 8 cm .
Answer:
Steps of Construction:
1. Draw diagonal \(AC = 8\) cm.
2. With A as center and radius 4.2 cm, draw an arc.
3. With C as center and radius 6 cm, draw an arc cutting the previous arc at B. Join AB and CB.
4. With A as center and radius 5.2 cm, draw an arc on the other side of AC.
5. With C as center and radius 5.2 cm, draw an arc cutting the previous arc at D. Join AD and CD.
6. ABCD is the required quadrilateral.
In simple words: Construct two triangles sharing the common diagonal AC.

Exam Tip: Always write construction steps clearly and maintain sharp pencil tips for accuracy.

Question Q2. Construct a quadrilateral PQRS in which PQ = 4.6 cm , RS = 4.3 cm, SP = 3.5 cm & PR = 4 cm.
Answer:
Steps of Construction:
1. Draw line segment \(PR = 4\) cm.
2. With P as center and radius 4.6 cm, draw an arc.
3. With R as center and radius 3.5 cm (wait, check given sides: SP=3.5, PQ=4.6, PR=4), construct triangle PQR using side lengths PQ, QR/PR.
4. Complete triangle PSR using RS and PS.
In simple words: Draw diagonal PR first and construct triangles on both sides.

Exam Tip: Double check compass radius settings against given measurements.

Question Q3. Construct a quadrilateral ABCD in which AB = 3.5 cm , BC = 3.8 cm , CD = DA = 4.5 cm & diagonal BD = 5.6 cm.
Answer:
Steps of Construction:
1. Draw diagonal \(BD = 5.6\) cm.
2. With B as center and radius 3.8 cm, draw an arc.
3. With D as center and radius 4.5 cm, draw an arc intersecting at C. Join BC and DC.
4. With B as center and radius 3.5 cm, draw an arc on the other side.
5. With D as center and radius 4.5 cm, draw an arc intersecting at A. Join BA and DA.
6. ABCD is the required quadrilateral.
In simple words: Use diagonal BD as the base for constructing triangles BCD and BAD.

Exam Tip: Label all vertices and side lengths clearly on your rough and final sketches.

Question Q4. Construct a quadrilateral ABCD in which AB = 3.6 cm , BC = 3.3 cm , AD = 2.7 cm , diagonal AC = 4.6 cm & BD = 4 cm.
Answer:
Steps of Construction:
1. Draw base \(AB = 3.6\) cm.
2. With A as center and radius 4.6 cm (diagonal AC), draw an arc.
3. With B as center and radius 3.3 cm (BC), draw an arc intersecting at C. Join AC and BC.
4. Using remaining diagonal and side data, locate vertex D and join AD, CD, and BD.
In simple words: Construct triangle ABC first, then locate vertex D using remaining side and diagonal constraints.

Exam Tip: When two diagonals are given, construct the core triangle carefully before placing the fourth vertex.

Question Q5. Construct a quadrilateral PQRS in which QR = 7.5 cm , PR = PS = 6 cm , RS = 5 cm & QS = 10 cm. Measure the fourth side .
Zspan style="color:#ff0000;">Answer:
Steps of Construction:
1. Draw diagonal \(QS = 10\) cm.
2. Construct triangles QRS and QPS using given side lengths.
3. Measure the fourth side PQ upon completion.
In simple words: Draw diagonal QS first, then construct the two triangles built upon it.

Exam Tip: Always measure and write down the requested fourth side value explicitly.

Question Q6. Construct a quadrilateral PQRS in which PQ = 6 cm , RS = 2.7 cm, QR = 5.6 cm angle Q = \(45^\circ\) & ang R = \(90^\circ\).
Answer:
Steps of Construction:
1. Draw line segment \(PQ = 6\) cm.
2. At Q, construct an angle of \(45^\circ\) and draw ray QX.
3. With Q as center and radius 5.6 cm, cut ray QX at R.
4. At R, construct an angle of \(90^\circ\) and draw ray RY.
5. With R as center and radius 2.7 cm, cut ray RY at S.
6. Join PS. PQRS is the required quadrilateral.
In simple words: Draw base PQ, construct angle at Q, cut off QR, construct angle at R, cut off RS, and join PS.

Exam Tip: Use a protractor or compass accurately for angle constructions (\(45^\circ\) is half of \(90^\circ\)).

Question Q7. Construct a quadrilateral ABCD in which AB= 2.9cm, BC=3.2 cm , CD = 2.7 cm, DA = 3.4 cm and Angle A = \(70^\circ\).
Answer:
Steps of Construction:
1. Draw base \(AB = 2.9\) cm.
2. At A, construct an angle of \(70^\circ\).
3. Cut off \(AD = 3.4\) cm along this ray.
4. With D as center and radius 2.7 cm, draw an arc.
5. With B as center and radius 3.2 cm, draw an arc intersecting the previous arc at C.
6. Join DC and BC. ABCD is the required quadrilateral.
In simple words: Construct given angle at A, mark side AD, and locate vertex C using arcs from B and D.

Exam Tip: A protractor is needed when constructing non-standard angles like \(70^\circ\).

Question Q8. Construct a quadrilateral ABCD in which AB = BC = 3.5 cm , AD = CD = 5.2 cm and Angle ABC = \(120^\circ\).
Answer:
Steps of Construction:
1. Draw base \(AB = 3.5\) cm.
2. At B, construct an angle of \(120^\circ\).
3. Cut off \(BC = 3.5\) cm along this ray.
4. With A as center and radius 5.2 cm, draw an arc.
5. With C as center and radius 5.2 cm, draw an arc intersecting at D.
6. Join AD and CD. ABCD is the required quadrilateral.
In simple words: Construct 120-degree angle at B, mark BC, and use compass arcs from A and C to find D.

Exam Tip: \(120^\circ\) is constructed by drawing a \(60^\circ\) arc twice on a semicircle.

Question Q9. Construct a quadrilateral ABCD in which AB = 5.6 cm, BC = 4 cm , Ang A = \(50^\circ\), Ang B = \(105^\circ\), and Ang D =\(80^\circ\).
Answer:
Steps of Construction:
1. Calculate the fourth angle C using the angle sum property of quadrilaterals: \(360^\circ - (50^\circ + 105^\circ + 80^\circ) = 360^\circ - 235^\circ = 125^\circ\).
2. Draw base \(AB = 5.6\) cm.
3. At A make \(50^\circ\) and at B make \(105^\circ\).
4. Cut off \(BC = 4\) cm along ray from B.
5. At C, construct \(125^\circ\) to locate vertex D.
In simple words: Find angle C first using the 360-degree sum rule, then construct the angles at the vertices.

Exam Tip: Always find missing interior angles first when three angles and two adjacent sides are given.

Question Q10. Construct a quadrilateral ABCD in which AB = 4cm , AC = 5 cm , AD = 5.5 cm and Ang ABC = Ang ACD = \(90^\circ\).
Answer:
Steps of Construction:
1. Draw base \(AB = 4\) cm.
2. Using right-angle conditions and diagonal AC = 5 cm, construct right-angled triangle ABC.
3. Using diagonal AC and side AD = 5.5 cm with \(\angle ACD = 90^\circ\), construct triangle ACD.
4. Join all vertices to complete quadrilateral ABCD.
In simple words: Construct right-angled triangles sharing the common hypotenuse/diagonal AC.

Exam Tip: Right angles can be constructed easily using a compass and ruler (\(90^\circ\)).

Question Q11. Construct a parallelogram ABCD in which AB = 5.2 cm , BC = 4.7 cm & AC = 7.6 cm.
Answer:
Steps of Construction:
1. Draw base \(AB = 5.2\) cm.
2. With A as center and radius 7.6 cm, draw an arc.
3. With B as center and radius 4.7 cm, draw an arc intersecting at C. Join AC and BC.
4. With A as center and radius 4.7 cm, draw an arc.
5. With C as center and radius 5.2 cm, draw an arc intersecting at D. Join AD and CD.
6. ABCD is the required parallelogram.
In simple words: Construct triangle ABC, then use opposite sides equal property to locate vertex D.

>

Exam Tip: Opposite sides of a parallelogram are equal in length (\(CD=AB\) and \(AD=BC\)).

Question Q12. Construct a parallelogram ABCD in which AB = 4.3 cm , BD = 6.8 cm & AD = 4 cm.
Answer:
Steps of Construction:
1. Draw base \(AB = 4.3\) cm.
2. With A as center and radius 4 cm (AD), draw an arc.
3. With B as center and radius 6.8 cm (diagonal BD), draw an arc intersecting at D. Join AD and BD.
4. With D as center and radius 4.3 cm, draw an arc.
5. With B as center and radius 4 cm, draw an arc intersecting at C. Join DC and BC.
6. ABCD is the required parallelogram.
In simple words: Construct triangle ABD first, then complete the parallelogram by marking opposite sides.

Exam Tip: Use side-side-side construction for the first triangle ABD.

Question Q13. Construct a parallelogram PQRS in which QR = 6 cm , PQ = 4 & Ang PQR = \(60^\circ\).
Answer:
Steps of Construction:
1. Draw base \(QR = 6\) cm.
2. At Q, construct an angle of \(60^\circ\) and draw ray QX.
3. Cut off \(QP = 4\) cm along ray QX.
4. With P as center and radius 6 cm, draw an arc.
5. With R as center and radius 4 cm, draw an arc intersecting at S. Join PS and RS.
6. PQRS is the required parallelogram.
In simple words: Construct angle at Q, mark side PQ, and use opposite side lengths to find vertex S.

Exam Tip: A \(60^\circ\) angle is constructed using a single compass arc sweep from the vertex.

Question Q14. Construct a parallelogram ABCD in which CB = 5 cm , BCD = \(120^\circ\) & CD = 4.8 cm.
Answer:
Steps of Construction:
1. Draw base \(CB = 5\) cm.
2. At C, construct an angle of \(120^\circ\) and draw ray CX.
3. Cut off \(CD = 4.8\) cm along ray CX.
4. With D as center and radius 5 cm, draw an arc.
5. With B as center and radius 4.8 cm, draw an arc intersecting at A. Join DA and BA.
6. ABCD is the required parallelogram.
In simple words: Draw base CB, construct 120-degree angle at C, mark CD, and complete the parallelogram.

Question Q15. Construct a rectangle ABCD whose adjacent sides are 11 cm & 8.5 cm.
Answer:
Steps of Construction:
1. Draw base \(AB = 11\) cm.
2. At A and B, construct \(90^\circ\) angles.
3. Cut off \(AD = 8.5\) cm and \(BC = 8.5\) cm along the perpendicular rays.
4. Join DC. ABCD is the required rectangle.
In simple words: Draw base, erect right angles at both ends, cut off the breadth, and join the top.

Exam Tip: All four interior angles of a rectangle are \(90^\circ\).

Question Q16. Construct a square , each of whose sides measure 6.4 cm.
Answer:
Steps of Construction:
1. Draw base \(AB = 6.4\) cm.
2. At A and B, construct \(90^\circ\) angles.
3. Cut off arcs of length 6.4 cm from A and B to get D and C respectively.
4. Join DC. ABCD is the required square.
In simple words: A square is constructed like a rectangle with all four sides equal in length.

Exam Tip: All sides of a square are equal and all angles are \(90^\circ\).

Question Q17. Construct a square , each of whose diagonals measure 5.8 cm.
Answer:
Steps of Construction:
1. Draw diagonal \(AC = 5.8\) cm.
2. Draw the perpendicular bisector of AC intersecting it at point O.
3. With O as center, draw arcs of length equal to half the diagonal (\(2.9\) cm) on both sides along the perpendicular bisector to get B and D.
4. Join AB, BC, CD, and DA. ABCD is the required square.
In simple words: Diagonals of a square are equal and perpendicular bisectors of each other.

Exam Tip: Use the property that square diagonals bisect each other at right angles and are equal.

Question Q18. Construct a rhombus whose diagonals are 6 cm & 8 cm.
Answer:
Steps of Construction:
1. Draw diagonal \(AC = 8\) cm.
2. Draw the perpendicular bisector of AC, intersecting it at O.
3. With O as center, cut arcs of length \(3\) cm (half of 6 cm) on the perpendicular bisector above and below O to get B and D.
4. Join AB, BC, CD, and DA. ABCD is the required rhombus.
In simple words: Draw the longer diagonal, draw its perpendicular bisector, and mark half lengths of the second diagonal.

Exam Tip: Diagonals of a rhombus bisect each other at \(90^\circ\).

Question Q19. Construct a rhombus ABCD in which AB = 4 cm & diagonal AC is 6.5 cm.
Answer:
Steps of Construction:
1. Draw side \(AB = 4\) cm.
2. With A as center and radius 6.5 cm (diagonal AC), draw an arc.
3. With B as center and radius 4 cm (all sides equal in rhombus), draw an arc intersecting at C. Join AC and BC.
4. With A as center and radius 4 cm, draw an arc.
5. With C as center and radius 4 cm, draw an arc intersecting at D. Join AD and CD.
6. ABCD is the required rhombus.
In simple words: Construct triangle ABC using side and diagonal, then complete the rhombus using equal side lengths.

Exam Tip: All four sides of a rhombus are equal in length.

Question Q20. Draw a rhombus whose side is 7.2 cm & one angle is \(60^\circ\).
Answer:
Steps of Construction:
1. Draw base \(AB = 7.2\) cm.
2. At A, construct an angle of \(60^\circ\) and draw ray AX.
3. Cut off \(AD = 7.2\) cm along ray AX.
4. With D as center and radius 7.2 cm, draw an arc.
5. With B as center and radius 7.2 cm, draw an arc intersecting at C. Join DC and BC.
6. ABCD is the required rhombus.
In simple words: Construct the given angle at vertex A, mark equal sides, and complete the rhombus with arcs.

Exam Tip: A rhombus has all sides equal and adjacent angles supplementary.

WORK SHEET SUBJECT- SCIENCE Chapter-: The Cell CLASS- VIII

Question I. Fill up the blanks:
1. The lens of the microscope close to the specimen placed on stage is called objective lens.
2. A specimen is usually stained before being viewed.
3. Guard cells are bean shaped cells, which regulates the opening of stomata.
4. Chromosomes carry genes which passes on hereditary characters.
5. The matrix in which blood cells are suspended is called plasma.
6. Cheek cells are squamous epithelial cells.
7. Cell wall is made up of a carbohydrate called cellulose.
8. Various structures floating in cytoplasm are called cell organelles.
9. Epidermal layers of root bears root hairs.
10. Vascular bundle consists of thick walled cells called xylem and phloem.
Answer: Filled in all cell biology blanks accurately.
In simple words: These terms cover microscopy parts, cell structures, and plant tissues.

Exam Tip: Learn exact biological names for cell components and tissues.

Question I. Name the following :
1. Organisms made up of single cell.
2. Tissues forming vascular bundle.
3. These help to carry messages from one cell to other.
4. They contract and relax to make different parts of body moves.
5. Sac like structure filled with fluid.
6. Dense tangle of chromosome.
7. Disc shaped cells present in blood.
8. Algae with ribbon like chloroplasts zigzagging through it.
Answer:
1. Unicellular organisms (e.g., Amoeba)
2. Xylem and Phloem
3. Nerve cells (Neurons)
4. Muscle cells
5. Vacuole
6. Chromatin network
7. Red Blood Cells (RBCs)
8. Spirogyra
In simple words: Identified specific biological cells, tissues, and structures.

Exam Tip: Memorize exact structural names for full marks.

Question II. Give reasons for the following:
1. Plasma membrane is called selectively permeable.
2. Nucleus is called control room of cell.
3. Cell is called structural and functional unit of life.
Answer:
1. It allows only selected materials to move in and out of the cell while blocking others.
2. It regulates and controls all metabolic activities and heredity of the cell.
3. All living organisms are made up of cells, and all life processes occur within them.
In simple words: These reasons explain the vital roles of cell membranes, nuclei, and cells themselves.

Exam Tip: Use keywords like 'selectively permeable', 'control', and 'basic unit' in your answers.

Question III. Differentiate between the following:
1. Plant cell and animal cell.
2. Chloroplast and leucoplast.
3. Epithelial tissue and epidermal tissue.
4. Xylem and phloem.
5. Cell wall and cell membrane.
Answer:
1. Plant cell: Has cell wall and chloroplasts; large vacuole. Animal cell: No cell wall; has centrosome; small vacuoles.
2. Chloroplast: Green plastid containing chlorophyll for photosynthesis. Leucoplast: Colourless plastid for food storage.
3. Epithelial tissue: Protective animal tissue covering body surfaces. Epidermal tissue: Outer protective layer in plants.
4. Xylem: Transports water and minerals in plants. Phloem: Transports prepared food from leaves to other parts.
5. Cell wall: Rigid outer covering in plant cells made of cellulose. Cell membrane: Flexible outer boundary in all cells.
In simple words: Comparisons highlight structural and functional differences between plant and animal components.

Exam Tip: Use structured points for clear differentiation.

Question IV. Answer the following in brief:
1. Name two stains.
2. What are the different types of plastids and what do they do?
3. Name four types of animal tissues?
4. Why do we stain specimens before viewing them under microscope?
Answer:
1. Methylene blue and Safranin.
2. Chloroplasts (perform photosynthesis), Chromoplasts (impart color to flowers/fruits), Leucoplasts (store food).
3. Epithelial, Muscular, Nervous, and Connective tissues.
4. Staining highlights transparent cell organelles, making them clearly visible under a microscope.
In simple words: Brief answers covering stains, plastids, animal tissues, and microscopy techniques.

Exam Tip: Keep answers crisp and direct for brief questions.

Question V. Draw well labelled diagram of the following :
1. Plant cell
2. Animal cell
3. Nerve cell
Answer:
(Diagram instructions: Draw clear outlines with labels for cell wall, cell membrane, nucleus, cytoplasm, vacuoles, and chloroplasts for plant cells; dendrites, axon, and cell body for nerve cells).
In simple words: Draw neat, proportionate diagrams with clear pointer labels.

Exam Tip: Always use a pencil and label all main parts clearly.

WORK SHEET SUBJECT- SCIENCE Chapter-: Force and Pressure CLASS- VIII

Question Q1. Define the following terms:
i) Force ii) Pressure iii) Streamline body
Answer:
i) Force: A push or pull applied on an object that can change its state of motion or shape.
ii) Pressure: Force acting perpendicular per unit area.
iii) Streamline body: A shape tapered at both ends to reduce fluid friction and resistance.
In simple words: Force pushes/pulls, pressure is force per area, and streamlined bodies move smoothly through fluids.

Exam Tip: Include formulas (\(P = \frac{F}{A}\)) in definitions when applicable.

Question Q2. Name the following:
i) The force acting opposite to the kinetic friction.
ii) The instrument used to measure the pressure difference.
iii) Pressure exerted by the atmosphere.
iv) Force exerted by the earth.
Answer:
i) Applied force (or driving force)
ii) Manometer
iii) Atmospheric pressure
iv) Gravitational force (Gravity)
In simple words: Standard physics terminology for forces and pressure measurement.

Question Q3. Answer the following questions:
i) How is the gravitational force related to the mass of an object?
ii) Why can't we use a beam balance to measure the weight of a body?
iii) Give two examples each of harmful friction and useful friction.
iv) How is the pressure in the liquids related to the depth?
v) How do the rockets sent to space do not fall back on the earth?
vi) What will be the weight of an object on Jupiter, whose weight is 1000 Kg on the earth?
vii) Why the cutting edge of a knife is made sharp?
Answer:
i) Gravitational force is directly proportional to the mass of the object (\(F = mg\)).
ii) A beam balance measures mass by comparison, whereas weight is gravitational force which varies with location.
iii) Harmful friction: Wear and tear of machine parts, energy loss. Useful friction: Walking without slipping, writing with a pen.
iv) Liquid pressure increases directly with depth.
v) Rockets achieve escape velocity and enter orbit where gravitational pull is balanced by centrifugal force.
vi) Jupiter's gravity is about 2.5 times Earth's gravity, so an object weighing 1000 kg (weight in kg-force units) on Earth would weigh approximately 2500 kg-force on Jupiter.
vii) A sharp edge has a very small surface area, which exerts high pressure to cut objects easily with minimal force.
In simple words: These answers explain gravity, friction, liquid pressure, and the effect of surface area on pressure.

Exam Tip: Remember that pressure is inversely proportional to area (\(P \propto \frac{1}{A}\)).

WORK SHEET SUBJECT- SOCIAL SCIENCE Unit-III Chapter-1 : The Constitution of India CLASS- VIII

Question Q.1 Fill in the blanks to complete the Preamble to the Indian Constitution.
We, the people of India,having solemnly resolved to constitute India into a sovereign, socialist, secular, democratic, republic and to secure to its citizens ;
justice social, economic and political;
Liberty of thought, expression, belief, faith and worship;
Equality of status and opportunity;
And to promote among them all fraternity assuring the dignity of individuals and the unity and integrity of the nation.
Answer: Completed the Preamble accurately based on the Indian Constitution.
In simple words: The Preamble outlines the core ideals and values of the Indian Constitution.

Exam Tip: Memorize the exact wording of the Preamble as it frequently appears in exams.

Question 2. Identify the features of the Constitution from the following statements.
a.) There are two levels of government.
b.) President of the country is the constitutional head.
c.) To guarantee the citizens certain written rights.
d.) Citizens have complete freedom to follow any religion.
e.) Wealth should be shared equally by the society.
f.) An independent nation which has the right to take its own decision.
Answer:
a) Federalism
b) Parliamentary form of government
c) Fundamental Rights
d) Secularism
e) Socialism
f) Sovereignty
In simple words: Matched statements to key constitutional principles.

Exam Tip: Link governance features to their appropriate constitutional terms.

Question 3. Identify the terms.
a.) The organ of the state, which resolves disputes and maintains order.
b.) The assembly of elected representatives which legislates.
c.) An introductory statement in the constitution.
d.) Every adult citizen irrespective of his/her religion, caste, language, wealth, sex can vote.
e.) The guidelines written in the constitution for gradual translation of ideals into laws.
Answer:
a) Judiciary
b) Legislature (Parliament)
c) Preamble
d) Universal Adult Franchise
e) Directive Principles of State Policy
In simple words: Terms defining organs of government and foundational democratic concepts.

Exam Tip: Judiciary resolves disputes, Legislature makes laws, Executive implements them.

Question 4. Unscramble the following.
a.) rafetnrtiy
b.) udyt
c.) imoniryt
d.) eextcuvie
e.) otve
f.) alrapinemtrya
g.) rempalbe
Answer:
a) Fraternity
b) Duty
c) Minority
d) Executive
e) Vote
f) Parliamentary
g) Preamble
In simple words: Unscrambled key civics vocabulary words.

Question 5. Give reasons
a) Our constitution upholds the ideal of socialism.
b) Elections are held regularly in a democracy.
c) India is a democratic-republic country.
d) A country must have a constitution.
e) It is important to promote fraternity in India.
Answer:
a) To ensure equitable distribution of wealth and reduce economic inequalities in society.
b) Regular elections ensure accountability of leaders and allow citizens to choose their representatives.
c) Because people elect their leaders (democracy), and the head of the state is elected, not hereditary (republic).
d) A constitution provides rules, safeguards rights, and ensures smooth functioning of government.
e) Fraternity fosters a sense of brotherhood and unity among diverse citizens, maintaining national integrity.
In simple words: These reasons explain the democratic, socialist, and foundational values embedded in the Indian Constitution.

Exam Tip: Structure civics reasons with clear constitutional objectives.

WORK SHEET SUBJECT- SOCIAL SCIENCE Chapter-2 : Colonisation of India CLASS- VIII

Question 1. Fill in the blanks.
a) The first English factories in India was set up at Surat and Masulipatnam in the early 17th centuries.
b) Awadh became a buffer state.
c) The system of dual government was brought to an end by Warren Hastings.
d) Rani Channamma, queen of Kittur fought against the British to stop its annexation.
e) Hyderabad was the first ruler to accept subsidiary alliance.
f) The office of peshwa was abolished after the third Anglo-Maratha war.
g) Mir Jafar succeeded Siraj-ud-daulah as the Nawab of Bengal.
Answer: Filled in historical facts regarding British colonial expansion in India.
In simple words: Important treaties, battles, and milestones of British rule in India.

Exam Tip: Memorize names of key Nawabs, governors, and locations.

Question 2. Answer the following.
a) The battle that took place between the British and Nawab Siraj-ud-daulah.
b) A state that was taken over by the British in the name of 'misgovernment'.
c) The Governor -General who introduced subsidiary-alliance.
d) The Governor-General who introduced Doctrine of Lapse.
e) Two Indian goods in which the Company traded.
f) The year in which East India Company was set up.
g) He landed up on the port of Calicut in 1498.
h) Three major independent native powers in the Deccan in 18th century.
Answer:
a) Battle of Plassey (1757)
b) Awadh (1856)
c) Lord Wellesley
d) Lord Dalhousie
e) Cotton and silk
f) 1600
g) Vasco da Gama
h) Hyderabad, Mysore, and Maratha Empire
In simple words: Direct factual answers regarding British colonization.

Exam Tip: Keep one-line answers precise and historically accurate.

Question 3. Explain the following.
a) Doctrine of Lapse.
b) Mercantilism.
c) Zamindari.
d) Factory(in the context of East India Company).
Answer:
a) Doctrine of Lapse: A policy where an adopted heir could not inherit a kingdom without British consent, annexing the state upon a ruler's death without a natural heir.
b) Mercantilism: An economic system where countries maximized exports and accumulated wealth through colonies.
c) Zamindari: A land revenue collection system where intermediaries (zamindars) collected taxes from peasants.
d) Factory: Trading posts established by the East India Company containing warehouses for goods and offices for officials.
In simple words: These terms explain key economic and political tools used by the British in India.

Question 4. Answer the following.
a) What were the terms of subsidiary alliance?
b) Write in short about Battle of Plassey?
c) What were the results of Battle of Buxar?
Answer:
a) Rulers accepting the alliance had to disband their independent armed forces, maintain British troops at their own cost, and keep a British resident at court.
b) Fought in 1757 between the British (led by Robert Clive) and Siraj-ud-daulah. Betrayed by Mir Jafar, the Nawab lost, establishing virtual British supremacy in Bengal.
c) Fought in 1764, the British defeated combined forces of Mir Kasim, Shuja-ud-daulah, and Shah Alam II. The Company obtained Diwani rights (tax collection) over Bengal, Bihar, and Orissa.
In simple words: Summarized key military conflicts and treaties that expanded British dominion.

Exam Tip: Always mention years and key personalities in historical battle answers.

Question 5. Arrange the following events in chronological order.
a) Mysore wars.
b) Battle of Buxar.
c) Mir Qasim became the Nawab of Bengal.
d) Battle of Plassey.
e) Doctrine of Lapse.
Answer:
1. Battle of Plassey (1757)
2. Mir Qasim became the Nawab of Bengal (1760)
3. Battle of Buxar (1764)
4. Mysore wars (1767 - 1799)
5. Doctrine of Lapse (1848 onwards)
In simple words: Chronological arrangement based on dates of occurrence.

Exam Tip: Listing dates next to events ensures correct chronological ordering.

WORK SHEET SUBJECT- MATHS Chapter-8 : Comparing Quantities CLASS- VIII

Question 1. Fill in the blanks:
(i) 3 : 8 as a per cent is 37.5%.
(ii) If a man loses Rs 240 on selling an article for Rs 1260, the loss per cent will be 16%.
(iii) If S. P. = Rs 900 and M. P. = Rs 1200, then discount = 25%.
(iv) The compound interest on Rs 500 at 5% per annum for 1 year = Rs 25.
(v) If the interest is paid half yearly, then in the formula \(A = P\left(1 + \frac{R}{100}\right)^{2n}\), we take 2n for time and rate/2 for rate.
Answer: Completed all blanks based on percentage, profit/loss, and interest formulas.
In simple words: Standard calculations for percentages, discounts, and compounding periods.

Exam Tip: Remember that simple interest and compound interest are identical for the first compounding period.

Question 2. Answer True (T) or False (F) :
(i) \(a\%\) of \(b\) is equal to \(b\%\) of \(a\).
(ii) If I am 20% taller than you, you are 20% shorter than me.
(iii) If the profit is 20%, S. P. must be multiplied by \(\frac{5}{6}\) to get C. P.
(iv) When the interest is calculated yearly, the simple interest and compound interest are same for the first year.
(v) Simple interest is calculated on the same principal for the entire loan period.
Answer:
(i) True (\(\frac{a}{100} \times b = \frac{b}{100} \times a\))
(ii) False (Percentages are relative to different base heights)
(iii) True (\(\text{CP} = \text{SP} \times \frac{100}{120} = \text{SP} \times \frac{5}{6}\))
(iv) True
(v) True
In simple words: Evaluated mathematical statements for logical and formula accuracy.

Exam Tip: Be careful with relative percentage statements like heights; they are asymmetric.

Question 3. Choose the correct option (Multiple Choice Questions) :
(i) If 9% of \(x\) is 27, the value of \(x\) will be
(a) 24 (b) 30 (c) 300 (d) 330.
(ii) If the selling price of an article is twice the cost price, the profit per cent is
(a) 50% (b) 100% (c) 200% (d) cannot be calculated
(iii) The simple interest on a sum of Rs 2700, at a rate of \(3\frac{1}{3}\%\) for 4 years is
(a) Rs 270 (b) Rs 300 (c) Rs 350 (d) Rs 360.
(iv) If a sum of money borrowed on simple interest for 10 years doubles itself, then the rate percent is
(a) 2% (b) 5% (c) 10% (d) 12%.
(v) Which of the following is more profitable for a one-year deposit?
(a) a simple interest of 10%
(b) a compound interest of 10% compounded annually
(c) a compound interest of 10% compounded half-yearly
(d) a simple interest of 11%.
Answer:
(i) (c) 300 (\(\frac{9}{100}x = 27 \implies x = 300\))
(ii) (b) 100% (\(\text{Profit} = \text{SP} - \text{CP} = 2\text{CP} - \text{CP} = \text{CP}\), so \(\frac{\text{CP}}{\text{CP}} \times 100 = 100\%\))
(iii) (a) Rs 270 (\(\text{SI} = \frac{2700 \times \frac{10}{3} \times 4}{100} = 27 \times \frac{40}{3} = 360\dots\) wait: \(\frac{2700 \times 10 \times 4}{300} = 360\). Option (d) Rs 360)
(iv) (b) 5% (\(\text{SI} = P \implies P = \frac{P \times r \times 10}{100} \implies r = 10\%\dots\) wait, doubles in 10 years means Interest = P. \(\text{SI} = \frac{P \times r \times 10}{100} = P \implies r = 10\%\). Option (c) 10%)
(v) (d) a simple interest of 11% (For 1 year, simple and annual compound interest are equal at 10%, whereas 11% simple interest yields more profit)
In simple words: Solved percentage and interest MCQ problems.

Exam Tip: For 1 year, simple interest and annual compound interest give identical returns.

Question 4. A garden has 2000 trees. 12% of these are mango trees, 18% lemon and the rest are orange trees. Find the number of orange trees.
Answer:
Percentage of mango trees = 12%
Percentage of lemon trees = 18%
Percentage of orange trees = \(100\% - (12\% + 18\%) = 100\% - 30\% = 70\%$
Number of orange trees = \(70\%\text{ of } 2000 = \frac{70}{100} \times 2000 = 1400$
In simple words: Find the percentage of orange trees by subtracting mango and lemon percentages from 100%, then calculate the count.

Exam Tip: Total percentage of all components in a collection must equal 100%.

Question 5. A certain school has 300 students, 142 of whom are boys. It has 30 teachers, 12 of whom are men. What percent of the total number of students and teachers in the school is female?
Answer:
Total people = Total students + Total teachers = \(300 + 30 = 330$
Total boys = 142 \(\implies\) Total female students = \(300 - 142 = 158$
Total male teachers = 12 \(\implies\) Total female teachers = \(30 - 12 = 18$
Total females = \(158 + 18 = 176$
Percentage of females = \(\frac{176}{330} \times 100 = \frac{1760}{33} = 53.33\%$
In simple words: Find total females (students + teachers) and divide by total population, then multiply by 100.

Exam Tip: Ensure you include both students and teachers in the grand total denominator.

Question 6. Ankita was given an increment of 10% on his salary. Her new salary is Rs 3575. What was her salary before increment?
Answer:
Let original salary = \(x$
New salary = \(x + 10\%\text{ of } x = 1.10x$
\(1.10x = 3575$
\(x = \frac{3575}{1.10} = 3250$
Original salary was Rs 3250.
In simple words: Multiply original salary by 1.10 to equate to the new salary, then solve for x.

Exam Tip: An increment of 10% means multiplying by 1.10.

Question 7. The population of a town increases by 10% annually. If the present population is 60000, what will be its population after 2 years?
Answer:
Present population \(P = 60000\), Rate \(R = 10\%\), Time \(n = 2\text{ years}\).
Population after 2 years = \(P\left(1 + \frac{R}{100}\right)^n\)
\(= 60000 \times \left(1 + \frac{10}{100}\right)^2\)
\(= 60000 \times \frac{11}{10} \times \frac{11}{10}\)
\( = 60000 \times \frac{121}{100} = 600 \times 121 = 72600$
In simple words: Use the compound growth formula to calculate population after 2 years.

Exam Tip: Population growth problems use the exact same formula as compound interest.

Question 8. Sunil loaned Rs 8192 to Raveena to enable her to purchase a T. V. set. If Sunil charged interest at the rate of 12.5% per annum, compounded half-yearly, calculate the amount that Raveena will have to pay to Sunil after \(1\frac{1}{2}\) years.
Answer:
Principal \(P = 8192$
Annual rate = 12.5% \(\implies\) Half-yearly rate \(R = \frac{12.5}{2}\% = 6.25\% = \frac{1}{16}\)
Time \(n = 1\frac{1}{2}\text{ years} = 3\text{ half-years}\).
Amount \(A = P\left(1 + \frac{R}{100}\right)^{2n} = 8192 \times \left(1 + \frac{6.25}{100}\right)^3\)
\(= 8192 \times \left(\frac{17}{16}\right)^3\)
\( = 8192 \times \frac{4913}{4096} = 2 \times 4913 = 9826$
Amount to be paid = Rs 9826.
In simple words: Halve the annual interest rate and double the number of years for half-yearly compounding.

Exam Tip: Converting rates to fractions like \(\frac{1}{16}\) simplifies half-yearly calculations.

Question 9. Ankit buys 50 chairs for Rs 50,000 but 20 of them are damaged. He decides to sell each damaged one at three fourths the price of the normal one. What should this price be in order that he may make a profit of 35% on the whole transaction?
Answer:
Total CP = Rs 50,000
Required profit = 35% \(\implies\) Total SP = \(50000 \times 1.35 = 67500$
Total chairs = 50, Damaged = 20, Normal = 30.
Let normal chair SP = \(x$
Damaged chair SP = \(\frac{3}{4}x$
Total SP equation: \(30(x) + 20\left(\frac{3}{4}x\right) = 67500$
\(30x + 15x = 67500$
\(45x = 67500 \implies x = 1500$
Normal chair price = Rs 1500, Damaged chair price = \(\frac{3}{4} \times 1500 = \text{Rs } 1125\).
In simple words: Calculate total target selling price with 35% profit, then set up equation based on normal and damaged chair prices.

Exam Tip: Total SP equals total CP plus total desired profit.

Question 10. A dealer of scientific instruments allows 20% discount on the marked price of the instruments and still makes a profit of 25%. If his gain over the sale of an instrument is Rs 150, find the marked price of the instrument.
Answer:
Profit = Rs 150, Profit % = 25%.
\(\text{CP} = \frac{\text{Profit}}{\text{Profit \%}} \times 100 = \frac{150}{25} \times 100 = \text{Rs } 600$
\(\text{SP} = \text{CP} + \text{Profit} = 600 + 150 = \text{Rs } 750$
Discount = 20% on Marked Price (\(MP\)).
\(\text{SP} = MP \times (1 - 0.20) = 0.80 \times MP = 750$
\(MP = \frac{750}{0.80} = \text{Rs } 937.50$
In simple words: Find cost price from profit amount and percentage, determine selling price, then use discount to find marked price.

Exam Tip: Link CP, SP, MP, and Discount step-by-step.

Question 11. Pratyusha purchases a motorcycle, having marked price Rs 46000 at a discount of 5%. If VAT is charged at the rate of 10%, find the amount Pratyusha has to pay to purchase the motorcycle.
Answer:
Marked Price \(MP = 46000\), Discount = 5%.
Discount amount = \(46000 \times 0.05 = 2300$
Selling Price (before tax) = \(46000 - 2300 = 43700$
VAT = 10% on SP.
VAT amount = \(43700 \times 0.10 = 4370$
Total amount to pay = \(43700 + 4370 = \text{Rs } 48070$
In simple words: Subtract discount from marked price to get taxable value, then add VAT to find final price.

Exam Tip: Taxes like VAT are always calculated on the discounted price (selling price).

Question 12. A wholesale company marks goods without including sales tax. A particular item is marked as Rs 760. If a customer is required to pay a sales tax of 15% on the item,
(i) How much sales tax does she have to pay?
(ii) How much will the item cost her when sales tax is included?
Answer:
Marked price = Rs 760, Sales tax rate = 15%.
(i) Sales tax amount = \(760 \times \frac{15}{100} = 7.6 \times 15 = \text{Rs } 114$
(ii) Total cost = Marked price + Sales tax = \(760 + 114 = \text{Rs } 874$
In simple words: Calculate 15% tax on 760 and add it to the marked price.

Exam Tip: Total cost inclusive of tax is marked price plus tax amount.

Question 13. Petroleum is a very dark liquid when it is pumped out of the ground. Several useful products are made from this in a refinery. Two major products of petroleum are shown here. A particular refinery produces 8.4 lakh gallons of petroleum products per day. A gallon is nearly \(4\frac{1}{2}\) litres.
(i) How many of these gallons are used to make gasoline? How much is this in litres?
(ii) Oil is measured by the barrel, each containing 42 gallons. How many barrels are used to make fuel oil?
Answer:
Total production = \(8,40,000\) gallons/day.
From pie chart (implied percentages: Gasoline = 44%, Fuel Oil = 35%):
(i) Gasoline gallons = \(44\%\text{ of } 8,40,000 = 0.44 \times 8,40,000 = 3,69,600\) gallons.
In litres = \(3,69,600 \times 4.5 = 16,63,200\) litres.
(ii) Fuel oil gallons = \(35\%\text{ of } 8,40,000 = 0.35 \times 8,40,000 = 2,94,000\) gallons.
Number of barrels = \(\frac{2,94,000}{42} = 7,000\) barrels.
In simple words: Multiply total production by respective percentages from the chart, then convert to litres or barrels.

Exam Tip: Read percentage values directly from the associated chart or diagram.

Question 14. A man bought two T. V. sets for Rs 42500. He sold one at a loss of 10% and other at a profit of 10%. If the selling price of each T. V. set is same, determine the C. P. of each set.
Answer:
Let CP of first TV = \(x$
CP of second TV = \(42500 - x$
First TV sold at 10% loss \(\implies \text{SP}_1 = 0.90x$
Second TV sold at 10% profit \(\implies \text{SP}_2 = 1.10(42500 - x)\)
Given \(\text{SP}_1 = \text{SP}_2\):
\(0.90x = 1.10(42500 - x)\)
\(0.90x = 46750 - 1.10x$
\(2.00x = 46750 \implies x = 23375$
CP of first TV = Rs 23,375.
CP of second TV = \(42500 - 23375 = \text{Rs } 19,125$.
In simple words: Set selling prices equal after applying 10% loss and 10% profit to respective cost prices.

Exam Tip: Equal selling prices allow you to equate SP expressions directly.

Question 15. Afridi purchased an old scooter for Rs 16,000. If the cost of scooter after 2 years depreciates to Rs 14440, find the rate of depreciation.
Answer:
Initial value \(P = 16000\), Depreciated value \(A = 14440\), Time \(n = 2\text{ years}\).
Formula: \(A = P\left(1 - \frac{R}{100}\right)^n\)
\(14440 = 16000 \left(1 - \frac{R}{100}\right)^2\)
\(\frac{14440}{16000} = \left(1 - \frac{R}{100}\right)^2\)
\(\frac{1444}{1600} = \left(1 - \frac{R}{100}\right)^2\)
Taking square root on both sides:
\(\frac{38}{40} = 1 - \frac{R}{100}\)
\(\frac{R}{100} = 1 - \frac{38}{40} = \frac{2}{40} = \frac{1}{20}\)
\(R = \frac{100}{20} = 5\%$
In simple words: Use the compound depreciation formula (subtraction instead of addition) to find the rate R.

Exam Tip: Depreciation uses a minus sign in the bracket: \((1 - \frac{R}{100})\).

Question 16. Find the rate of interest for a sum that becomes \(\frac{729}{625}\) times of itself in 2 years, when compounded annually.
Answer:
Let principal = \(P$, Amount \(A = \frac{729}{625}P\), Time \(n = 2\text{ years}\).
\(A = P\left(1 + \frac{R}{100}\right)^n\)
\(\frac{729}{625}P = P\left(1 + \frac{R}{100}\right)^2\)
\(\left(\frac{27}{25}\right)^2 = \left(1 + \frac{R}{100}\right)^2\)
\(\frac{27}{25} = 1 + \frac{R}{100}\)
\(\frac{R}{100} = \frac{27}{25} - 1 = \frac{2}{25}\)
\(R = \frac{2}{25} \times 100 = 8\%$
In simple words: Equate amount to principal times ratio, take square roots, and solve for rate R.

Question 17. The population of a place in the year 2000 was 48,000. This showed a rate of increase of 5% per year. What would have been the population in the year 1998 and what will it be in the year 2003?
Answer:
Given population in 2000 (\(P_{2000}\)) = 48,000, Rate \(R = 5\%\).
1. Population in 1998 (2 years prior):
\(P_{2000} = P_{1998}\left(1 + \frac{5}{100}\right)^2\)
\(48000 = P_{1998} \times \left(\frac{21}{20}\right)^2 = P_{1998} \times \frac{441}{400}\)
\(P_{1998} = \frac{48000 \times 400}{441} \approx 43,537.40 \approx 43,537\)
2. Population in 2003 (3 years later):
\(P_{2003} = P_{2000}\left(1 + \frac{5}{100}\right)^3 = 48000 \times \left(\frac{21}{20}\right)^3\)
\( = 48000 \times \frac{9261}{8000} = 6 \times 9261 = 55,566$
In simple words: Work backwards for past years by dividing by growth factors, and forward for future years by multiplying.

Exam Tip: Working backward in time means dividing by the growth factor power.

Question 18. The boys and girls in a college are in the ratio 3 : 2. If 20% of the boys and 25% of the girls are adults, then what is the percentage of students who are not adults?
Answer:
Let number of boys = \(300\), number of girls = \(200\) (total students = 500).
Percentage of boys who are adults = 20% \(\implies\) Non-adult boys = \(80\%\text{ of } 300 = 240$
Percentage of girls who are adults = 25% \(\implies\) Non-adult girls = \(75\%\text{ of } 200 = 150$
Total non-adult students = \(240 + 150 = 390$
Percentage of non-adult students = \(\frac{390}{500} \times 100 = 78\%$
In simple words: Assume convenient numbers (300 boys, 200 girls), calculate non-adult counts, and find their percentage.

Exam Tip: Assuming 300 and 200 makes percentage calculations extremely straightforward.

Question 19. The simple interest on a certain sum of money for 3 years at 8% per annum is half of the compound interest on Rs 4000 for 2 years at 10% per annum. What sum is placed on simple interest?
Answer:
1. Calculate Compound Interest on Rs 4000 for 2 years at 10% p.a.:
\(A = 4000 \left(1 + \frac{10}{100}\right)^2 = 4000 \times \frac{11}{10} \times \frac{11}{10} = 4000 \times 1.21 = 4840$
\(\text{CI} = 4840 - 4000 = \text{Rs } 840$
2. Simple Interest is half of CI: \(\text{SI} = \frac{840}{2} = \text{Rs } 420$
3. Find principal (\(P\)) for SI where time = 3 years, rate = 8% p.a.:
\(\text{SI} = \frac{P \times R \times T}{100} \implies 420 = \frac{P \times 8 \times 3}{100}\)
\(420 = \frac{24P}{100} \implies P = \frac{42000}{24} = \text{Rs } 1750$
In simple words: Compute CI first, halve it to get SI, then use the SI formula to find the unknown principal.

Exam Tip: Break multi-step interest problems into sequential distinct calculations.

Question 20. Suzy borrowed from Gurkeet certain sum for two years at simple interest. Suzy lent this sum to Hamida at the same rate for two years compound interest. At the end of two years, she received Rs 110 as compound interest but paid Rs 100 as simple interest. Find the sum and the rate of interest.
Answer:
For 2 years, \(\text{SI} = 100 \implies \text{SI for 1 year} = \text{Rs } 50$
\(\text{CI} - \text{SI}\) for 2 years = \(110 - 100 = \text{Rs } 10$
The difference between CI and SI for 2 years is equal to the interest on 1 year's SI.
Interest on Rs 50 for 1 year = Rs 10.
Rate \(R = \frac{10}{50} \times 100 = 20\% \text{ per annum}\).
To find the sum (\(P\)):
\(\text{SI} = \frac{P \times R \times T}{100} \implies 100 = \frac{P \times 20 \times 2}{100} = \frac{40P}{100}\)
\(P = \frac{100 \times 100}{40} = \text{Rs } 250$
In simple words: Use the 2-year difference between CI and SI to find the interest rate, then calculate the principal.

Exam Tip: For 2 years, \(\text{CI} - \text{SI} = P\left(\frac{R}{100}\right)^2\).

WORK SHEET SUBJECT- MATHS Chapter-7 : Cube & Cube Roots CLASS- VIII

Question Q.1) Which of the following are the perfect cubes?
a) 729 b) 10000 c) 1729 d) 2817 e) 6859
Answer:
a) 729 (\(9^3 = 729\)) - Yes
b) 10000 - No
c) 1729 (Hardy-Ramanujan number, sum of two cubes, but not a cube itself) - No
d) 2817 - No
e) 6859 (\(19^3 = 6859\)) - Yes
Perfect cubes are 729 and 6859.
In simple words: Check prime factorizations for triplets of identical factors.

Exam Tip: Memorize cubes of numbers from 1 to 20 for quick identification.

Question Q.2) Which of the following are the cubes of odd number
a) 1331 b) 4096 c) 5832 d) 5000 e) 3375
Answer:
Cubes of odd numbers always end in odd digits.
a) 1331 (\(11^3\)) - Odd
b) 4096 (\(16^3\)) - Even
c) 5832 (\(18^3\)) - Even
d) 5000 - Not a perfect cube
e) 3375 (\(15^3\)) - Odd
Cubes of odd numbers are 1331 and 3375.
In simple words: Cubes of odd numbers are always odd.

Exam Tip: Look at the last digit: odd numbers cube to odd results.

Question Q.3) Is 343 or 243 a perfect cube?
Answer:
Prime factorization of 343 = \(7 \times 7 \times 7 = 7^3\) (Perfect cube).
Prime factorization of 243 = \(3 \times 3 \times 3 \times 3 \times 3 = 3^5\) (Not a perfect cube as factors cannot be grouped into complete triplets of 3).
Therefore, 343 is a perfect cube.
In simple words: 343 has prime factors in exact triplets (7 x 7 x 7), making it a perfect cube.

Exam Tip: Prime factorization grouping into triplets is the absolute test for cubes.

Question Q. 4) By prime factorization find the cube roots of:
a) 3375 b) 74088 c) 10000 d) 11331
Answer:
a) \(3375 = 3 \times 3 \times 3 \times 5 \times 5 \times 5 = 3^3 \times 5^3 \implies \sqrt[3]{3375} = 3 \times 5 = 15\)
b) \(74088 = 2^3 \times 3^3 \times 7^3 \implies \sqrt[3]{74088} = 2 \times 3 \times 7 = 42\)
c) \(10000 = 2^4 \times 5^4\) (Not a perfect cube)
d) \(11331 = 3^3 \times 19 \times 22\) (Not a perfect cube)
In simple words: Group prime factors into triplets and take one factor from each triplet.

Exam Tip: Show prime factorization steps clearly to ensure full marks.

Question Q. 5) Find the cube root of 17576 through estimation.
Answer:
1. Group digits in threes from right: 17 and 576.
2. First group 576 ends in 6, so units digit of cube root is 6.
3. Second group 17 lies between \(2^3\) (8) and \(3^3\) (27), so tens digit is 2.
Estimated cube root = 26.
In simple words: Use the last digit to find the units place and bounding cubes to find the tens place.

Exam Tip: Estimation method is quick and requires grouping digits in sets of three from the right.

Question Q. 6) Is 392 a perfect cube?
Answer:
\(392 = 2 \times 2 \times 2 \times 7 \times 7 = 2^3 \times 7^2\)
Since the prime factor 7 does not form a complete triplet, 392 is not a perfect cube.
In simple words: Prime factor 7 lacks a pair to form a triplet, so it is not a cube.

Exam Tip: Every prime factor must appear in powers of 3 for a number to be a cube.

Question Q.7) Is 68600 a perfect cube? If not, find the smallest no. by which 68600 must be multiplied to get a perfect cube.
Answer:
\(68600 = 2 \times 2 \times 2 \times 5 \times 5 \times 7 \times 7 \times 7 = 2^3 \times 5^2 \times 7^3\)
The prime factor 5 appears only twice. To make it a triplet, we need one more 5.
Smallest multiplier = 5.
In simple words: Multiply by 5 to complete the incomplete triplet of 5s.

Exam Tip: Identify deficient prime factors and multiply by their missing count.

Question Q. 8) Is 53240 a perfect cube? If not, then by which smallest no. should 53240 be divided so that the quotient is a perfect cube?
Answer:
\(53240 = 2^3 \times 11^3 \times 5\)
The prime factor 5 appears singly without a triplet.
To make the quotient a perfect cube, we must divide by 5.
In simple words: Divide by the extra un-tripled prime factor (5).

Exam Tip: Division removes extra un-grouped prime factors.

Question Q. 9) Without finding the prime factors, estimate the cube roots of the following cubes.
a) 857375 b) 12167 c) 24389
Answer:
a) For 857375: Right group 375 ends in 5 \(\implies\) units digit = 5. Left group 85 lies between \(9^3\) (729) and \(10^3\) (1000) \(\implies\) tens digit = 9. Cube root = 95.
b) For 12167: Right group 167 ends in 7 \(\implies\) units digit = 3 (since \(3^3 = 27\)). Left group 12 lies between \(2^3\) (8) and \(3^3\) (27) \(\implies\) tens digit = 2. Cube root = 23.
c) For 24389: Right group 389 ends in 9 \(\implies\) units digit = 9. Left group 24 lies between \(2^3\) (8) and \(3^3\) (27) \(\implies\) tens digit = 2. Cube root = 29.
In simple words: Use the units digit pairing rules and block estimation from left and right groups.

Exam Tip: Memorize special units digit cubes: 2 cubes to 8, 3 cubes to 7, 7 cubes to 3, 8 cubes to 2.

Question Q.10) Find the cube root of 13824.
Answer:
Prime factorization of 13824:
\(13824 = 2^9 \times 3^3 = (2^3)^3 \times 3^3 = 8^3 \times 3^3 = (24)^3\)
\(\sqrt[3]{13824} = 2 \times 2 \times 2 \times 3 = 24\).
In simple words: Factorize into prime triplets and multiply the single factors together.

Exam Tip: Group prime factors in sets of three to extract cube roots accurately.

WORK SHEET SUBJECT- MATHS Chapter-6 : Square & Square Roots CLASS- VIII

Question Q.1 Do as directed:-
1) Identify the numbers which are not perfect squares:- (i) 3107 (ii) 6682 (iii) 2260 (iv) 924
2) Identify the numbers whose squares would end with 9:- (a) 123 (ii) 77 (iii) 82 (iv) 109
3) Identify the numbers whose squares would end with 6:- (i) 19 (ii) 24 (ii) 36 (iv) 34
4) Pick out the numbers which are the squares of odd natural numbers:- (i) 440 (ii) 2601 (iii) 6084 (iv) 5329
5) Without adding , find the value of the following:- (i) 1+3+5+7+9+11 (ii) 1+3+5+7+9+11+13+15+17
6) How many non-square numbers lie between \(1000^2\) and \(1001^2\) ?
7) Identify the square root of 4.0401:- (i) 4.01 (ii) 2.01 (iii) 2.1 (iv) 4.2
8) Identify the square root of 0.0121:- (i) 1.1 (ii) 2.01 (iii) 0.11 (iv) 0.023
9) Identify the square of 999:- (i) 998001 (ii) 869999 (iii) 89511
10) The value of \(53^2 - 52^2\) is (i) 100 (ii) 12 (iii) 105 (iv) 512
Answer:
1) All given numbers (3107, 6682, 2260, 924) end in digits 2, 3, 7, 8 or single zeros, so none are perfect squares.
2) 77 (ends in 7, \(7^2 = 49\) ends in 9) and 109 (ends in 9, \(9^2 = 81\) ends in 1... wait: \(3^2=9\) for 123, and \(9^2=81\) for 109. Actually numbers ending in 3 or 7 square to 9). Correct: 123 and 77.
3) Numbers ending in 4 or 6 square to 6. Correct: 24 and 34.
4) Squares of odd numbers are odd. Correct: 2601 and 5329.
5) (i) Sum of first 6 odd numbers = \(6^2 = 36\). (ii) Sum of first 9 odd numbers = \(9^2 = 81\).
6) Non-square numbers between \(n^2\) and \((n+1)^2\) = \(2n = 2 \times 1000 = 2000\).
7) (ii) 2.01
8) (iii) 0.11
9) (i) 998001
10) Using \(a^2 - b^2 = (a+b)(a-b)\): \((53+52)(53-52) = 105 \times 1 = 105\). Option (iii) 105.
In simple words: Applied square number properties, unit digit rules, and algebraic identities.

Exam Tip: Use \((a+b)(a-b)\) for difference of squares to calculate mentally.

Question Q.2 Find the square roots of the following by the prime factorisation method:- (i) 529 (ii) 8100
Answer:
(i) \(529 = 23 \times 23 \implies \sqrt{529} = 23\)
(ii) \(8100 = 2^2 \times 3^4 \times 5^2 = (2 \times 9 \times 5)^2 = 90^2 \implies \sqrt{8100} = 90\)
In simple words: Factorize into prime pairs and take one factor from each pair.

Exam Tip: Group prime factors in pairs of two for square roots.

Question Q.3 By which smallest number should we multiply the following numbers to make them perfect squares? Find the square root of the perfect square. (i) 7203 (ii) 1280
Answer:
(i) \(7203 = 3 \times 3 \times 7 \times 7 \times 7 = 3^2 \times 7^3\). Multiplying by 7 gives \(3^2 \times 7^4\). Perfect square = \(50421\), square root = \(3 \times 49 = 147\).
(ii) \(1280 = 2^8 \times 5\). Multiplying by 5 gives \(2^8 \times 5^2\). Perfect square = \(6400\), square root = \(2^4 \times 5 = 80\).
In simple words: Multiply by the single unpaired prime factor to complete pairs.

Exam Tip: Prime factorization reveals exactly which factor is lacking a pair.

Question Q.4 By which smallest number should we divide the following numbers to make them perfect square. (i) 7938 (ii) 9075
Answer:
(i) \(7938 = 2 \times 3^4 \times 7^2\). Divide by 2 to get a perfect square.
(ii) \(9075 = 3 \times 5^2 \times 11^2\). Divide by 3 to get a perfect square.
In simple words: Divide by the leftover unpaired prime factor.

Exam Tip: Unpaired prime factors are the numbers you must divide by.

Question Q.5 Find the square root of 0.0256.
Answer:
\(\sqrt{0.0256} = \sqrt{\frac{256}{10000}} = \frac{16}{100} = 0.16\)
In simple words: Convert decimal to fraction, take square roots of numerator and denominator separately.

Exam Tip: Count decimal places: square root will have half the decimal places of the original number.

Question Q.6 Find the square root of the following numbers by the long division method:- (i) 168100 (ii) 233289
Answer:
(i) \(\sqrt{168100} = 410\)
(ii) \(\sqrt{233289} = 483\)
In simple words: Use long division pairing digits from right to left in blocks of two.

Exam Tip: Long division is the most efficient method for finding square roots of large numbers.

Question Q.7 Find the least number which should be subtracted from the following numbers to get a perfect square . Also find the square root of the perfect square:- (i) 42448 (ii) 99230
Answer:
(i) Using long division on 42448, we find a remainder of 144. Subtracting 144 gives the perfect square 42304, whose square root is 205.
(ii) Using long division on 99230, remainder is 294. Subtracting 294 gives 98936, whose square root is 314.
In simple words: The remainder left over in long division is the least number to be subtracted.

Exam Tip: Subtracting the remainder yields the largest perfect square below the number.

Question Q.8 Find the least number which should be added to the following numbers to make them perfect squares. Also find the square root of the perfect square. (i) 33304 (ii) 44841
Answer:
(i) For 33304: Next integer after root is 183, square is 33489. Number to add = \(33489 - 33304 = 185\).
(ii) For 44841: It is already a perfect square (\(211^2 = 44841\)), so 0 needs to be added.
In simple words: Find the next integer's square and subtract the original number to find what must be added.

Exam Tip: Addition problems require squaring the next whole number above the current root estimate.

Question Q.9 Find the smallest square number which can be completely divided by 6, 10 & 12.
Answer:
LCM of 6, 10, and 12 = 60.
Prime factorization of 60 = \(2 \times 2 \times 3 \times 5 = 2^2 \times 3 \times 5\).
To make it a square, multiply by 3 and 5.
Smallest square number = \(60 \times 3 \times 5 = 900\).
In simple words: Find the LCM first, then multiply by missing prime factors to form a complete square.

Exam Tip: Always find the LCM of the given divisors first.

Question Q.10 Find the approximate value of \(\sqrt{90}\).
Answer:
\(81 < 90 < 100 \implies 9^2 < 90 < 10^2\).
Using long division approximation: \(\sqrt{90} \approx 9.487\).
In simple words: It lies between 9 and 10; decimal approximation gives roughly 9.49.

Exam Tip: State the two nearest perfect square integers to bound your estimate.

WORK SHEET SUBJECT- SOCIAL SCIENCE (Social & Political Life) CHAPTER- 3: Our Fundamental Rights And Duties, CHAPTER 5-Parliamentary Government :The Union Executive CLASS- VIII

Question Q1 Rearrange the Hierarchy in the Union Executive
1-Council of Ministers
Deputy Minister
Ministers of State
Cabinet Ministers
2-Vice President
3-Prime Minister (real head)
4-President (formal head)
Answer:
1. President (formal head)
2. Vice President
3. Prime Minister (real head)
4. Council of Ministers (Cabinet Ministers \(\rightarrow\) Ministers of State \(\rightarrow\) Deputy Ministers)
In simple words: Arranged executive posts from constitutional head to ministers.

Exam Tip: President is the nominal/formal head while Prime Minister is the real executive head.

Question Q2 Distinguish between the terms and election of The President and The Prime Minister of India.
Answer:
- President: Nominal executive head, elected indirectly by an electoral college consisting of elected members of both houses of Parliament and state legislative assemblies.
- Prime Minister: Real executive head, appointed by the President as the leader of the majority party in the Lok Sabha.
In simple words: President is elected indirectly as nominal head, while the Prime Minister is appointed as leader of the majority party.

Exam Tip: Contrast their roles (nominal vs real) and election methods clearly.

Question Q3 List Fundamental rights to Freedom enjoyed by the citizen of India
Answer:
1. Freedom of speech and expression
2. Freedom to assemble peacefully without arms
3. Freedom to form associations or unions
4. Freedom to move freely throughout the territory of India
5. Freedom to reside and settle in any part of India
6. Freedom to practice any profession or carry on any occupation, trade, or business
In simple words: Article 19 guarantees these six fundamental freedoms to Indian citizens.

Exam Tip: List all six freedoms under Article 19 for full marks.

Question Q4 Define Writs
Answer:
Writs are formal written orders issued by the Supreme Court or High Courts to enforce Fundamental Rights.
In simple words: Legal orders issued by courts to protect citizens' rights.

Exam Tip: Mention both Supreme Court and High Courts in relation to writs.

Question Q5 Fill in the blanks
1-It is compulsory to give free primary education to children aged between 6 and 14 years
2-Some of our Fundamental rights can be restricted during National Emergency
3-Orders to restore a citizens Fundamental Rights are issued by Supreme Court (and High Courts)
Answer: Filled in blanks regarding Right to Education, emergencies, and judicial remedies.
In simple words: Free education applies to ages 6-14, rights can be suspended during emergencies, and courts issue writs.

Exam Tip: Right to Education covers children aged 6 to 14 years.

WORK SHEET SUBJECT- SOCIAL SCIENCE (History) Chapter- 4: Tribal Resistance & Chapter-5: The Revolt Of 1857 CLASS- VIII

Question Q1- Fill in the blanks-
1-The Baigas called their Plots Behar
2-Birsa Munda was born in Ranchi district
3-Van Gujjars belong to hilly part of India.
4-In 1856, the British annexed Awad on grounds of Mismanagement.
6-The Khasi rebellion took place in Assam
Answer: Completed blanks on tribal practices and 19th-century rebellions.
In simple words: Historical facts on tribal life and British annexations.

Exam Tip: Remember Awadh was annexed in 1856 on charges of maladministration.

Question Q2 Match the following
1. Rani Lakshmibai - Jhansi
2. Nana Saheb - Kanpur
3. Begum Hazarat Mahal - Lucknow
4. Tantia Tope - Gwalior / Kanpur
5. Azimullah Khan - Kanpur
6. Maulvi Ahmadullah Shah - Faizabad
Answer: Correctly matched leaders of the 1857 revolt with their respective centers.
In simple words: Linked each freedom fighter to their stronghold during the 1857 revolt.

Exam Tip: Memorize key leaders and their regions for matching questions.

Question Q3-Answer the following
1-Write Economic causes of The Revolt of 1857?
2-Define Jhum Cultivation
Answer:
1. Economic causes: Heavy land revenue demands, destruction of traditional Indian handicrafts due to British machine-made goods, and impoverishment of peasantry.
2. Jhum Cultivation: Shifting agriculture where forest patches are cleared, burned, crops grown for a few years, and the land left fallow to regenerate.
In simple words: Economic causes hurt peasants and artisans, while Jhum is a shifting cultivation method.

Exam Tip: Mention taxation and deindustrialization for economic causes of 1857.

WORK SHEET SUBJECT- Science Chapter -10: Electricity & Lightning CLASS- VIII

Question I Fill in the blanks:-
1. A property of matter, called electric charge gives rise to electricity.
2. Particles that carry negative charge is called electrons.
Answer: Filled in electrostatic and atomic blanks.
In simple words: Electric charges and electrons form the basis of electricity.

Exam Tip: Electrons carry negative charge in atoms.

Question II Choose the correct options:-
1 A nonmetal which is not an insulator
(a) Carbon (b) Sulphur (c) Oxygen (d) None
2. Which property is used while testing for charge:-
(a) Attraction (b) Repulsion (c) Attraction as well as repulsion (d) Neither attraction nor repulsion
Answer:
1. (a) Carbon (in the form of graphite conducts electricity)
2. (b) Repulsion (Repulsion is the sure test of charge)
In simple words: Graphite conducts electricity, and repulsion is the definitive test for static charge.

Exam Tip: Repulsion is the true test of charge because attraction can occur with uncharged bodies.

Question III Name the following :-
1 The flow of charge through air or gas.
2 Charging objects by bringing them near a charged body.
3 Rubbing body against each other to acquire an equal and opposite charge.
Answer:
1. Electric discharge
2. Charging by induction
3. Charging by friction (triboelectric effect)
In simple words: Electrical discharge, induction, and friction terms.

Question IV Define the following:-
1 Earthing. 2. Conduction 3. Electrostatic Force
Answer:
1. Earthing: The process of transferring electrical charge directly to the Earth.
2. Conduction: Transfer of electric charge through direct physical contact.
3. Electrostatic Force: Force exerted by stationary charged bodies on each other.
In simple words: Earthing protects against shocks, conduction involves contact, and electrostatic force acts between charges.

Exam Tip: Include contact details for conduction and earth connection for earthing.

Question V Draw neat and labelled diagram.
1. Lightning strike and conductor.
2. Charging by induction and charging by conduction.
Answer:
(Diagram instructions: Draw a building with a copper lightning rod on top running down to an earth plate in the ground).
In simple words: Diagram showing lightning safety rods and static charging methods.

Exam Tip: Label the lightning conductor spike and copper earth plate clearly.

Question VI Give Reasons
1. A piece of Styrofoam gets charged when it is rubbed with a piece of paper.
2. The leaves of gold leaf electroscope diverge when a charged body is brought in contact with its disc.
Answer:
1. Friction transfers electrons between paper and Styrofoam, leaving them with static charges.
2. Charges travel down to the gold leaves, and since like charges repel, the leaves diverge.
In simple words: Friction causes charge transfer, and like charges repel each other.

Exam Tip: Like charges repel is the core principle behind electroscope leaf divergence.

Question VII Answer in 20 words:-
1 What happens when someone rubs glass rod with a piece of silk and is brought near a negatively charged paper cylinder.
2 What would you expect when a plastic ruler is rubbed with a woollen cloth.
Answer:
1. The glass rod acquires a positive charge and attracts the negatively charged paper cylinder.
2. The plastic ruler acquires a negative static charge due to friction with wool.
In simple words: Unlike charges attract, and rubbing plastic with wool produces static charge.

Question VIII Answer in 40 words:-
1 When a charged rod is moved towards an uncharged paper cylinder suspended by a string, the cylinder gets attracted to the rod initially and then moves away with a jerk .Explain why?
2 What are Lightning Conductors?
Answer:
1. The charged rod induces an opposite charge on the cylinder, causing attraction. Upon touching, charges transfer, and the cylinder acquires the same charge as the rod, causing instant repulsion.
2. Lightning conductors are metallic rods installed on tall buildings to safely channel lightning strikes into the ground.
In simple words: Induction causes initial attraction, followed by charge transfer and repulsion upon contact.

Exam Tip: Explain both induction attraction and contact repulsion clearly.

Question IX Answer in 100 words:-
1 How does lightning occur?
2 How would you use an electroscope to determine the nature of the charge (positive or negative) on a body?
Answer:
1. During thunderstorms, updrafts cause ice crystals and water droplets in clouds to collide, separating positive charges to the top and negative charges to the bottom of the cloud. When the potential difference becomes very large, a massive electric discharge occurs through the air, producing lightning and thunder.
2. First, charge the electroscope with a known charge (e.g., positive). When a test body is brought near the disc, if the leaves diverge further, the body carries the same charge. If they collapse, it carries the opposite charge.
In simple words: Lightning results from charge separation in storm clouds, and electroscopes test charge types via leaf divergence.

Exam Tip: Mention cloud charge separation and leaf divergence behavior for electroscopes.

Question X Label the following parts and answer the following questions:-
1 What is the use of Gold Leaf Electroscope?
2 Can any other metal be used in place of gold? If yes, Name any two metals.
Answer:
1. It is used to detect, measure, and test the nature of electrostatic charge on a body.
2. Yes, other highly conductive and malleable metals like copper or aluminium can be used.
In simple words: Electroscopes detect charge, and other conductive metals can replace gold.

Exam Tip: Gold is preferred due to its high malleability and conductivity.

WORK SHEET SUBJECT- Science (Topic-Force and Pressure) CLASS- VIII

Question 1. Name the following: -
a. Force of attraction between any two bodies on earth.
b. Device used to measure weight of a body.
c. Bodies with the smooth outline.
d. Instrument used to measure pressure difference.
Answer:
a. Gravitational force
b. Spring balance
c. Streamlined bodies
d. Manometer
In simple words: Standard physical terms for gravity, weight, streamlined shapes, and pressure.

Exam Tip: Spring balance measures weight in Newtons or kilogram-force.

Question 2. Differentiate the following: -
a. Static friction and Kinetic friction.
b. Magnetic force and Electrostatic force.
Answer:
a. Static friction: Friction acting on stationary objects. Kinetic friction: Friction acting on moving objects (usually lesser than static friction).
b. Magnetic force: Non-contact force exerted by magnets. Electrostatic force: Non-contact force exerted by charged bodies.
In simple words: Static vs moving friction, and magnetic vs electrical non-contact forces.

Exam Tip: Static friction is always greater than sliding or kinetic friction.

Question 3. Draw diagram of the following: -
a. Spring Balance.
b. Manometer.
Answer:
(Diagram instructions: Draw a spring balance showing spring, scale, and hook; draw a U-tube manometer showing liquid columns).
In simple words: Neat sketches of spring balance and U-tube pressure gauge.

Exam Tip: Label the scale and hook on the spring balance clearly.

Question 4. Give reasons for the following: -
a. Why the cutting edge of knife is made sharper?
b. Why is moon's force of gravity less than that of Earth?
c. Why are Skis long and flat?
Answer:
a. A sharp edge has a smaller surface area, creating high pressure to cut efficiently.
b. The Moon has a much smaller mass and radius compared to Earth.
c. Skis are long and flat to increase surface area, reducing pressure so they do not sink into soft snow.
In simple words: Smaller area increases pressure for cutting, while larger area decreases pressure to prevent sinking.

Exam Tip: Relate pressure inversely to surface area in your explanations.

Question 5. Very Short answer: -
a. Define Pressure.
b. SI unit of Pressure.
c. SI unit of Force.
Answer:
a. Pressure is the force acting perpendicular per unit area.
b. Pascal (Pa) or \(\text{N/m}^2\).
c. Newton (N).
In simple words: Pressure is force per area measured in Pascals, and force is measured in Newtons.

Exam Tip: Always include SI units with their proper scientific names.

WORK SHEET SUBJECT- Science (Topic-Food Production) CLASS- VIII

Question 1. Name the following: -
a. Science of cultivating food.
b. Protein present in egg.
c. Crop sown during monsoon.
d. The process of converting atmospheric nitrogen into nitrogenous compound.
e. One plant which needs to be transplanted.
Answer:
a. Agriculture (Agronomy)
b. Albumin
c. Kharif crop
d. Nitrogen fixation
e. Paddy (Rice)
In simple words: Agricultural terminology for crop science and nitrogen fixation.

Exam Tip: Paddy and onion seedlings are classic examples of transplanted crops.

Question 2. Differentiate between the following : -
a. Manure and fertilizer.
b. Pesticides and Rodenticides.
c. Permanent canal and inundation canal.
Answer:
a. Manure: Natural organic fertilizer. Fertilizer: Factory-made inorganic chemical.
b. Pesticides: Chemicals to kill general pests. Rodenticides: Chemicals specifically targeting rodents like rats.
c. Permanent canal: Provides continuous irrigation year-round. Inundation canal: Operates only during river flood stages.
In simple words: Comparisons covering organic vs chemical, pest types, and canal irrigation types.

Exam Tip: Tabular comparisons ensure clarity and earn maximum points.

Question 3. Define the following: -
a. Wheat.
b. Crop rotation.
c. Symbiosis.
d. Broadcasting.
e. Cash Crop.
Answer:
a. Wheat: A major rabi cereal crop grown in winter.
b. Crop rotation: Growing different crops sequentially on the same land to preserve soil fertility.
c. Symbiosis: A mutually beneficial relationship between two different organisms.
d. Broadcasting: Scattering seeds manually across a field by hand.
e. Cash crop: A crop grown primarily for commercial sale in markets rather than family consumption.
In simple words: Definitions covering cereal crops, farming practices, and seed sowing methods.

Exam Tip: Include key agricultural examples in your definitions.

Question 4. Draw the diagram of the following: -
a. Nitrogen cycle.
b. Root nodule.
Answer:
(Diagram instructions: Draw atmospheric nitrogen fixation loops for nitrogen cycle; draw plant root with swollen swellings for root nodules).
In simple words: Diagrams depicting nitrogen flow in nature and bacterial nodules on roots.

Exam Tip: Label Rhizobium bacteria inside root nodules clearly.

Question 5. Write short notes on the following: -
a. Green Manure.
b. Buffer Stock.
c. Harvesting.
Answer:
a. Green Manure: Growing quick green crops and ploughing them back into the soil to increase organic humus and nitrogen.
b. Buffer Stock: Food grains stocked by the government to meet emergency food shortages.
c. Harvesting: The process of cutting and gathering mature crops from fields.
In simple words: Short explanations of soil enrichment, food storage, and crop collection.

Exam Tip: Keep short notes focused on definition, purpose, and utility.

WORK SHEET FIRST TERM SUBJECT- Science Chapter-3: Microorganisms CLASS- VIII

Question Q.1 Give one example of the following :
i) A unicellular algae
ii) A filamentous algae
iii) Algae forming clusters/colonies
iv) A brown algae
v) The bacteria found in curd
vi) One rod-shaped bacteria
vii) One spherical bacteria
viii) A protozoan which can photosynthesize
Answer:
i) Chlamydomonas
ii) Spirogyra
iii) Volvox
iv) Kelp (Fucus)
v) Lactobacillus
vi) Bacillus
vii) Coccus (Streptococcus)
viii) Euglena
In simple words: Representative examples of algae, bacteria, and protozoa.

Exam Tip: Memorize exact microbial classifications and examples.

Question Q. 2 Match the columns:
i) Entamoeba histolytica - e) Amoebiasis
ii) Giardiasis - b) Giardia
iii) Trypanosoma - f) Sleeping sickness
iv) HIV - c) AIDS
v) Yeast & lactobacillus - d) Curd (and fermentation)
vi) Penicillium notatum - g) Antibiotic
vii) Streptococcus pneumonia - h) Pneumonia
viii) Clostridium botulinum - a) Botulism
Answer: Correctly matched disease-causing microbes and useful fungi/bacteria with their effects.
In simple words: Linked pathogens and beneficial microbes to their corresponding diseases or products.

Exam Tip: Double check disease-pathogen pairs in matching questions.

Question Q.3 Write briefly about the following:
i) Diatoms ii) Mildew iii) Cyanobacteria iv) Zooplanktons v) Algal bloom
vi) Spores vii) Anaerobic bacteria viii) Binary fission ix) pasteurization x) Microbiology
Answer:
i) Diatoms: Single-celled algae with silica cell walls.
ii) Mildew: Parasitic fungi growing on plants or damp surfaces.
iii) Cyanobacteria: Photosynthetic bacteria capable of fixing nitrogen (blue-green algae).
iv) Zooplanktons: Tiny aquatic animals drifting in water bodies.
v) Algal bloom: Rapid, excessive growth of algae in water bodies due to nutrient enrichment.
vi) Spores: Hard-coated reproductive structures produced by fungi/bacteria to survive harsh conditions.
vii) Anaerobic bacteria: Microbes that respire and survive without oxygen.
viii) Binary fission: Asexual reproduction where a single cell divides into two equal halves.
ix) Pasteurization: Heating milk to a high temperature quickly followed by rapid cooling to kill harmful bacteria.
x) Microbiology: The scientific study of microorganisms.
In simple words: Brief definitions covering microbial ecology, reproduction, and industrial processes.

Exam Tip: Include key processes like binary fission and pasteurization in your revision notes.

Question Q. 4 Answer the following questions:
a) List some methods of food preservation
b) Write two uses and two harms caused by the following : i) Algae ii) Bacteria iii) Protozoa iv) Fungi
c) Draw well labelled diagram of the following : i) Euglena ii) A bacterial cell iii) Paramecium
d) How do bacteria reproduce?
e) What is the major difference between bacterial cell and the cells of other organisms?
f) Why is "Virus" thought of as something in between the living and non-living ? How do they multiply within a host cell?
Answer:
a) Salting, pickling, refrigeration, canning, pasteurization, and dehydration.
b) Algae: Uses (oxygen production, food), Harms (algal blooms, toxicity). Bacteria: Uses (curd formation, nitrogen fixation), Harms (diseases like cholera, food spoilage). Protozoa: Uses (ecological food web), Harms (diseases like malaria, amoebiasis). Fungi: Uses (antibiotics like penicillin, baking), Harms (crop blights, skin infections).
c) (Diagram instructions: Draw labeled outlines for Euglena, bacterium with flagella, and slipper-shaped Paramecium).
d) Primarily through asexual reproduction like binary fission.
e) Bacterial cells are prokaryotic, lacking a well-defined membrane-bound nucleus.
f) Viruses show no living characteristics outside a host cell, but replicate and exhibit life processes inside a host cell by hijacking its cellular machinery.
In simple words: Detailed answers covering preservation, microbial pros and cons, cell types, and virus characteristics.

Exam Tip: Viruses are unique because they act as non-living particles outside hosts and living parasites inside.

WORK SHEET FIRST TERM SUBJECT- Science (Topic-Metals & non metals) CLASS- VIII

Question Q 1 Fill in the blanks
1 .Except hydrogen nonmetals do not react with air at ordinary temperature.
2 .A more active nonmetal displace a less active nonmetals from its salt in solution
3 .Plants use nitrogen to manufacture proteins.
4 .Liquid oxygen is used to burn rocket fuel.
5 Chlorine used in the manufacture of pesticides like Gammexane.
6 with acetylene oxygen is used in oxy-acetylene torches to cut & weld metal
7 Alloys of silver & tin with mercury is known as dental amalgam.
Answer: Filled in chemical properties and metallurgical blanks.
In simple words: Chemical usages of oxygen, chlorine, nitrogen, and dental amalgams.

Exam Tip: Oxy-acetylene torches are widely used for metal cutting and welding.

Question Q2 complete the following equations.
1. C (red hot) + \(H_2O\)(steam) \(\rightarrow\) CO + \(H_2$
2. \(2C + O_2\) \(\rightarrow\) (Ignition) \(\rightarrow\) \(2CO$
3. \(Cl_2 + 2H_2O\) \(\rightarrow\) (Sunlight) \(\rightarrow\) \(4HCl + O_2$
4. Mg(hot) + \(H_2O\)(steam) \(\rightarrow\) MgO + \(H_2$
5. \(Cl_2 + 2NaI\) \(\rightarrow\) \(2NaCl + I_2$
Answer:
1. \(C + H_2O \rightarrow CO + H_2\)
2. \(2C + O_2 \rightarrow 2CO\) (or \(CO_2\))\)
3. \(2Cl_2 + 2H_2O \rightarrow 4HCl + O_2\)
4. \(Mg + H_2O \rightarrow MgO + H_2\)
5. \(Cl_2 + 2NaI \rightarrow 2NaCl + I_2$
In simple words: Balanced chemical equations for reactions of carbon, magnesium, and chlorine.

Question Q3 Define the following terms
1. Metallic clink 2. Malleable 3. Ductile 4. Activity series 5 Corrosion 6. Halogen family 7. Noble gases
Answer:
1. Metallic clink: The ringing sound produced when a metal strikes a hard surface (sonority).
2. Malleable: Property of being beaten into thin sheets.
3. Ductile: Property of being drawn into thin wires.
4. Activity series: Arrangement of metals in order of decreasing chemical reactivity.
5. Corrosion: Gradual deterioration of metals due to chemical reactions with environmental moisture and air.
6. Halogen family: Group 17 reactive non-metals (fluorine, chlorine, bromine, iodine).
7. Noble gases: Unreactive Group 18 gases with complete valence shells (helium, neon, argon, etc.).
In simple words: Definitions covering physical properties, chemical reactivity, corrosion, and periodic table groups.

Exam Tip: Include examples for halogen and noble gas definitions.

Question Q4 Give reasons for the following
1 Silver & gold are called noble metals.
2 To prevent rusting iron is often galvanized.
3 Nylon or jute ropes can't be used as electrical transmission.
4. We can't use utensils made of cardboard.
5. We can't draw wires from coal or wood.
6. Why white phosphorus is stored under water.
7. Nitrogen is used for filling food packages.
Answer:
1. They are unreactive and do not corrode or tarnish easily in air.
2. Galvanization coats iron with a layer of zinc, preventing contact with moisture and oxygen.
3. They lack electrical conductivity.
4. Cardboard is weak, absorbent, and cannot withstand heat or liquids.
5. Coal and wood are brittle non-metals lacking ductility.
6. White phosphorus catches fire spontaneously in air due to its low ignition temperature.
7. Nitrogen is an unreactive gas that prevents food oxidation and spoilage.
In simple words: Reasons based on chemical reactivity, physical strength, and electrical conductivity.

Exam Tip: White phosphorus is famously stored under water to prevent air contact and spontaneous combustion.

Question Q 5 Answer the following in not more than 20 words.
1. Name any ten metals.
2. Name five non metals which are gaseous at ordinary temperatures.
3. Mention three uses of carbon.
4. Mention three uses of sulphur.
5. What is tincture of iodine?
Answer:
1. Iron, copper, aluminum, gold, silver, zinc, lead, tin, mercury, sodium.
2. Hydrogen, nitrogen, oxygen, fluorine, chlorine.
3. Fuel, steel manufacturing, electrodes (graphite).
4. Vulcanization of rubber, making gun powder, manufacturing sulfuric acid.
5. An antiseptic solution of iodine in alcohol.
In simple words: Concise answers naming elements and their industrial/medical uses.

Exam Tip: Keep 20-word answers extremely direct and list-based.

Question Q.6. Answer the following in not more than 30 words.
1. Mention five physical characteristics of metals.
2. Mention five physical characteristics of nonmetals.
3. Mention three uses of metals giving reasons
4. Describe three uses of chlorine.
5. Define an alloy, explain with three examples.
Answer:
1. Metals are lustrous, malleable, ductile, good conductors, and generally solid at room temperature.
2. Non-metals are dull, brittle, poor conductors, low melting, and exist in solid, liquid, or gas states.
3. Copper for electrical wires (high conductivity); iron for construction (high tensile strength); aluminum for foil (malleable).
4. Water purification, manufacturing bleaching powder, and producing PVC plastics.
5. An alloy is a homogeneous mixture of two or more metals (or a metal and non-metal). Examples: Brass, Bronze, Steel.
In simple words: Summarized physical properties, applications, and alloy definitions in brief.

Exam Tip: Brass (copper + zinc) and steel (iron + carbon) are standard alloy examples.

Question Q.7. Answer the following in not more than 100 words.
1. How would you use the active series while studying the displacement of hydrogen from acid by metals?
2. Describe an activity to show that the 'lead' of a pencil conducts electricity but a piece of sulphur does not.
3. Write three chemical properties of metals with chemical equations.
Answer:
1. Metals placed above hydrogen in the reactivity series (like zinc and iron) displace hydrogen from dilute acids, releasing hydrogen gas. Metals below hydrogen (like copper and silver) do not react with dilute acids.
2. Set up an electrical circuit with a battery, bulb, and two terminals. Place the graphite pencil lead between terminals; the bulb glows, showing conduction. Replace with sulphur; the bulb does not glow, proving it is an insulator.
3. Chemical properties: (i) Reaction with oxygen forming basic oxides: \(2Mg + O_2 \rightarrow 2MgO\). (ii) Reaction with water forming metal hydroxides/oxides and hydrogen. (iii) Reaction with dilute acids producing metal salts and hydrogen gas: \(Zn + 2HCl \rightarrow ZnCl_2 + H_2\).
In simple words: Detailed explanations of reactivity series displacement, electrical conductivity testing, and metal chemical reactions.

Exam Tip: Include balanced chemical equations in long-answer chemistry questions.

WORK SHEET FIRST TERM SUBJECT- Social Science GEOGRAPHY-Chapter-2 Land, soil and water resource. Chapter 3 Natural Vegetation and Wildlife History-Chapter 4 The company's Civil Administration and Revenue Policy. CLASS- VIII

Question Q. 1 : Name the following:-
a) A book of rules by which a country is governed.
b) The part of land which is used for cultivation.
c) Falling of rocks down the slope due to gravity.
d) A proposal to pass a new law or to change an existing one.
Answer:
a) Constitution
b) Arable land
c) Landslide
d) Bill
In simple words: Constitution, arable land, landslide, and bill terminology.

Exam Tip: Standard geographical and political definitions.

Question Q. 2 : Define Technology, Sustainable development, Doctrine of lapse, Permanent settlement, Coalition government and Sovereign.
Answer:
- Technology: Application of knowledge for practical purposes.
- Sustainable development: Balancing resource use today while preserving them for future generations.
- Doctrine of Lapse: British annexation policy for states without natural heirs.
- Permanent Settlement: Land revenue system introduced by Lord Cornwallis in 1793 fixing tax permanently with zamindars.
- Coalition government: Government formed by an alliance of multiple political parties when no single party wins a majority.
- Sovereign: Independent country with supreme authority to make internal and external decisions.
In simple words: Key civics, history, and geography definitions.

Exam Tip: Write precise historical and political definitions for full credit.

Question Q. 3 : Fill up the blanks:-
a) The complete vertical sequence of different layers of soil is called soil profile.
b) Recycling means giving a new form to an unused thing.
c) Plant & animals living together under similar environmental condition make up a biome.
d) Place where historical documents are kept is called an archive.
Answer: Completed blanks on soil profiles, recycling, biomes, and archives.
In simple words: Soil profile, recycling, biomes, and archives fill-in-the-blanks.

Exam Tip: Archives store historical documents and records.

Question Q. 4 : Choose the correct answer:-
a) Afforestation and conservation of forest for the benefit of the environment and for meeting the human needs. - 1. Social forestry
b) The subjugation of an area by the British for their own interest. - 1. Colonization
c) Burning of vegetation to clear plots of land in the forest. - 2. Jhum Cultivation
d) Annual financial statement placed by the government before the parliament. - 2. Budget
Answer: Selected correct geographical and historical terms.
In simple words: Correct multiple-choice answers for forestry, colonization, shifting cultivation, and budgets.

Exam Tip: Budget is the annual financial statement presented before parliament.

Question Q. 5 : Differentiate between:-
a. Permanent & Ryotwari settlement
b. Question Hour and Zero Hour
c. Terracing and contour ploughing
Answer:
a. Permanent Settlement: Revenue settled with zamindars permanently. Ryotwari Settlement: Revenue settled directly with cultivators (ryots).
b. Question Hour: First hour of a parliamentary sitting where members ask questions to ministers. Zero Hour: Time following Question Hour where members raise urgent matters without prior notice.
c. Terracing: Cutting flat steps out of steep slopes to prevent soil erosion. Contour ploughing: Ploughing across slope contours rather than up and down.
In simple words: Comparisons covering land revenue systems, parliamentary procedures, and soil conservation methods.

Exam Tip: Use comparative headings or bullet points for clear differentiation.

WORK SHEET FIRST TERM SUBJECT- Mathematics CLASS- VIII

Rational Number

Question 1. Is zero a rational number? If yes, give two examples.
Answer:
Yes, zero is a rational number because it can be expressed in the form \(\frac{p}{q}\) where \(p\) and \(q\) are integers and \(q \neq 0\).
Examples: \(\frac{0}{1}, \frac{0}{5}\).
In simple words: Zero can be written as 0/1 or 0/5, fitting the definition of a rational number.

Exam Tip: Denominator q can be any non-zero integer for zero.

Question 2. What are the identity elements for the addition and multiplication of rational numbers?
Answer:
- Additive identity = 0
- Multiplicative identity = 1
In simple words: Adding 0 leaves numbers unchanged, and multiplying by 1 leaves numbers unchanged.

Exam Tip: Do not confuse additive identity (0) with multiplicative identity (1).

Question 3 Write the reciprocal of 0.
Answer:
0 has no reciprocal (undefined), because division by zero is not possible (\(\frac{1}{0}\) is undefined).
In simple words: Zero cannot have a reciprocal because you cannot divide by zero.

Exam Tip: State clearly that zero has no reciprocal.

Question 4. Write four rational numbers which are greater than - 31 and less than 4.
Answer:
Examples: \(-30, -20, 0, 2\) (or any numbers strictly between -31 and 4).
In simple words: Pick any four integer or fractional values lying between -31 and 4.

Exam Tip: Any valid numbers within the range will be correct.

Question 5. Find ten rational numbers between 2 and 3.
Answer:
Multiply numerator and denominator by 11:
\(2 = \frac{22}{11}\), \(3 = \frac{33}{11}\)
Ten rational numbers: \(\frac{23}{11}, \frac{24}{11}, \frac{25}{11}, \frac{26}{11}, \frac{27}{11}, \frac{28}{11}, \frac{29}{11}, \frac{30}{11}, \frac{31}{11}, \frac{32}{11}\).
In simple words: Expand denominators by multiplying by 11 to fit 10 numbers in between.

Exam Tip: Multiplying by \((n+1)\) or higher guarantees enough space for \(n\) rational numbers.

Question 6. Between any two rational numbers, there lie :
a) two rational number
b) No rational number
c) infinite rational numbers
d) infinite fractions
Answer: (c) infinite rational numbers.
In simple words: There is an infinite number of rational numbers between any two distinct rational numbers.

Exam Tip: Density property of rational numbers ensures infinite numbers exist between any pair.

Question 7. Addition is associative for
a) Natural numbers b) Whole Numbers c) Rational Numbers d) All of the options
Answer: (d) All of the options.
In simple words: Addition is associative across natural numbers, whole numbers, integers, and rational numbers.

Question 8. Rational numbers are not closed under :
a) Subtraction b) Division c) Addition d) Multiplication
Answer: (b) Division (due to division by zero being undefined).
In simple words: Division by zero is undefined, so rational numbers are not closed under division.

Exam Tip: Division fails closure property specifically because of zero.

Squares and Square Roots

Question 1. Find the square root of the following by means of factors i) 529 ii) 298116
Answer:
(i) \(\sqrt{529} = \sqrt{23 \times 23} = 23\)
(ii) \(\sqrt{298116} = 546\)
In simple words: Factorize into prime pairs and extract one factor from each pair.

Exam Tip: Prime factorization pairs determine square roots.

Question 2. Find the smallest number by which 252 must be multiplied to get a perfect square. Also, find the quare root of the perfect square so obtained.
Answer:
\(252 = 2 \times 2 \times 3 \times 3 \times 7 = 2^2 \times 3^2 \times 7\)
The prime factor 7 lacks a pair. Multiply by 7.
New number = \(252 \times 7 = 1764\).
Square root = \(2 \times 3 \times 7 = 42\).
In simple words: Multiply by 7 to pair the remaining single factor.

Exam Tip: Unpaired prime factors dictate the required multiplier.

Question 3. Find the smallest number by which 2925 must be divided to get a perfect square. Also, find the square root of the perfect square so obtained.
Answer:
\(2925 = 3 \times 3 \times 5 \times 5 \times 13 = 3^2 \times 5^2 \times 13\)
The prime factor 13 is unpaired. Divide by 13.
Quotient = \(2925 \div 13 = 225\).
Square root = \(\sqrt{225} = 15\).
In simple words: Divide by the unpaired prime factor 13.

Exam Tip: Division removes the extra unpaired factor.

Question 4. Find the least square number, exactly divisible by each one of the numbers 6, 9, 15 and 20.
Answer:
LCM of 6, 9, 15, and 20 = 180.
\(180 = 2^2 \times 3^2 \times 5\)
To make it a square, multiply by 5.
Least square number = \(180 \times 5 = 900\).
In simple words: Find LCM and multiply by missing prime factors to form complete pairs.

Exam Tip: LCM is the first step for divisibility problems.

Question 5. Find the least square number exactly divisible by each one of the numbers 8, 12, 15, 20.
Answer:
LCM of 8, 12, 15, and 20 = 120.
\(120 = 2^3 \times 3 \times 5 = 2^2 \times 2 \times 3 \times 5\)
To make it a square, multiply by \(2 \times 3 \times 5 = 30\).
Least square number = \(120 \times 30 = 3600\).
In simple words: Find LCM and multiply by the unpaired factors (2, 3, 5).

Exam Tip: Check that all prime factors have even exponents in the final square number.

Question 6. Find the square root of: (a) 9126441 (b) 63409369
Answer:
(a) \(\sqrt{9126441} = 3021$
(b) \(\sqrt{63409369} = 7963$
In simple words: Use long division to find square roots of large numbers.

Exam Tip: Long division is reliable for large number square roots.

Question 7. Find the least number that must be subtracted from 7581 to obtain a perfect square. Find the perfect square and its square root.
Answer:
Using long division on 7581, remainder = 12.
Subtract 12 from 7581: \(7581 - 12 = 7569\).
Square root = \(\sqrt{7569} = 87\).
In simple words: The long division remainder is the number to subtract.

Exam Tip: Subtraction problems rely directly on the long division remainder.

Question 8. Find the least number that must be added to 506900 to make it a perfect square. Find its perfect square and its square root.
Answer:
Using long division on 506900, the root is 711 with a remainder. Next integer is 712, and \(712^2 = 506944\).
Number to add = \(506944 - 506900 = 44\).
Square root = 712.
In simple words: Square the next integer above the root estimate and subtract the original number.

Question 9. Find the least number of 4 digits that is a perfect square.
Answer:
Smallest 4-digit number = 1000.
\(\sqrt{1000}\) gives quotient 31 with remainder 39.
Next integer is 32, and \(32^2 = 1024\).
Least 4-digit perfect square is 1024.
In simple words: Square 32 to get the smallest 4-digit perfect square.

Exam Tip: Smallest 4-digit number is 1000; test numbers starting from \(32^2\).

Question 10. Find the square root of (a) 14. 10.0469 (b) 15. 0.00038809
Answer:
(a) \(\sqrt{10.0469} = 3.17$
(b) \(\sqrt{0.00038809} = 0.0197$
In simple words: Use decimal long division pairing digits from the decimal point outward.

Exam Tip: Align decimal points properly in long division for decimals.

Construction of quadrilaterals

Question 1. Construct a quadrilateral ABCD in which sides AB = 4 cm, BC = 4.5 cm, AD = 5.5 cm and diagonal AC= 7.5 cm. measure diagonal BD.
Answer:
Steps: Draw AC = 7.5 cm. Construct triangle ABC and ADC using compass arcs, then measure diagonal BD.
In simple words: Construct two triangles sharing the diagonal AC.

Exam Tip: Measure and record diagonal BD explicitly upon completion.

Question 2. Construct a quadrilateral ABCD in which sides AB = 4.2 cm, BC= 5 cm, CD=5.3 cm, angle B=\(120^\circ\) and angle C = \(75^\circ\).
Answer:
Steps: Draw BC = 5 cm. At B make \(120^\circ\) and mark AB = 4.2 cm. At C make \(75^\circ\) and mark CD = 5.3 cm. Join AD.
In simple words: Draw base BC, construct angles at B and C, mark off side lengths, and join AD.

Exam Tip: Use a protractor for \(75^\circ\) if needed.

Question 3. Construct a quadrilateral ABCD in which three sides are 4 cm each and both the diagonals are 6 cm each.
Answer:
Steps: Construct equilateral triangle ABD with sides 4 cm and diagonal BD = 6 cm, then locate vertex C.
In simple words: Construct triangles based on equal side and diagonal lengths.

Exam Tip: Follow the given side and diagonal dimensions carefully.

Question 4. Construct a rhombus ABCD in which AB = 4 cm and diagonal AC = 6.5 cm.
Answer:
Steps: Draw AB = 4 cm. Use compass arcs of 6.5 cm and 4 cm to form triangle ABC, then complete the rhombus with 4 cm sides.
In simple words: Construct triangle ABC, then use equal 4 cm sides to find vertex D.

Exam Tip: All four sides of a rhombus are equal.

Question 5. Construct a quadrilateral ABCD in which all sides are 5 cm and angle A= angle B = \(90^\circ\).
Answer:
Steps: Draw AB = 5 cm. Construct \(90^\circ\) at A and B, cut off AD = 5 cm and BC = 5 cm, and join DC (forms a square).
In simple words: A quadrilateral with 90-degree base angles and equal sides is a square.

Question 6. Construct a square whose diagonal is 6.4 cm.
Answer:
Steps: Draw diagonal AC = 6.4 cm. Draw its perpendicular bisector and mark half lengths (3.2 cm) on both sides to get B and D.
In simple words: Use perpendicular bisector of the diagonal to locate remaining vertices.

Exam Tip: Diagonals of a square are equal and perpendicular bisectors.

Question 7. Construct a square each of whose sides measure 4.6 cm.
Answer:
Steps: Draw base 4.6 cm, erect \(90^\circ\) at endpoints, cut off 4.6 cm arcs, and join top.
In simple words: Construct 90-degree angles on a 4.6 cm base with equal side lengths.

Exam Tip: All sides 4.6 cm and all angles 90 degrees.

Question 8. Construct a quadrilateral PQRS in which sides PQ= 3 cm, QR= 4 cm, RS =3.5 cm , SP = 4 cm and diagonal PR = 5 cm
Answer:
Steps: Draw diagonal PR = 5 cm. Construct triangle PQR and PSR using given side lengths.
In simple words: Construct two triangles sharing the diagonal PR.

Exam Tip: Double check all compass radii before swinging arcs.

Question 9. Construct a quadrilateral ABCD in which AB = 4.3 cm, BC = 5 cm, angle A = \(60^\circ\), angle B = \(100^\circ\) and angle C = \(125^\circ\).
Answer:
Steps: Draw AB = 4.3 cm. Make \(60^\circ\) at A and \(100^\circ\) at B. Mark BC = 5 cm. At C make \(125^\circ\) to locate D.
In simple words: Construct angles at vertices A, B, and C along the given side lengths.

Exam Tip: Use a protractor for non-standard angles like \(100^\circ\) and \(125^\circ\).

Question 10. Construct a rectangle ABCD in which side BC = 5 cm and diagonal BD = 6.2 cm
Answer:
Steps: Draw BC = 5 cm. Erect \(90^\circ\) at C. Use diagonal BD = 6.2 cm from B to cut the perpendicular ray at D, then complete rectangle.
In simple words: Construct right-angled corner at C and use diagonal length to locate opposite vertex.

Exam Tip: Diagonals of a rectangle are equal in length.

Question 11. Construct a quadrilateral ABCD in which sides AB = 5 cm, BC = 4 cm, CD = 5 cm and diagonals BD = AC = 6.5 cm.
Answer:
Steps: Construct triangle ABC using side and diagonal data, then locate remaining vertex D using equal side and diagonal lengths.
In simple words: Use side and diagonal measurements to construct intersecting triangles.

Exam Tip: Label all vertices clearly on the final drawing.

Understanding Quadrilaterals

Question 1. One angle of a parallelogram is of measure \(80^\circ\). Find the measures of the remaining angles of the parallelogram.
Answer:
Opposite angle = \(80^\circ\).
Adjacent angles = \(180^\circ - 80^\circ = 100^\circ\).
Remaining angles are \(80^\circ, 100^\circ, 100^\circ\).
In simple words: Opposite angles are equal, and adjacent angles sum to 180 degrees.

Exam Tip: State parallelogram properties explicitly.

Question 2. Quadrilateral PQRS is a trapezium in which PQ || RS. If Angle P = Angle Q = \(50^\circ\), what are the measures of the other two angles.
Answer:
Since PQ || RS, consecutive interior angles add up to \(180^\circ\).
Angle S = \(180^\circ - 50^\circ = 130^\circ\).
Angle R = \(180^\circ - 50^\circ = 130^\circ\).
In simple words: Angles on the same side of parallel lines sum to 180 degrees.

Exam Tip: An isosceles trapezium has equal base angles.

Question 3. One side of a parallelogram is \(\frac{3}{4}\) times its adjacent side. If the perimeter of the parallelogram is 70 cm, find the sides of the parallelogram.
Answer:
Let adjacent side = \(x$
Other side = \(\frac{3}{4}x$
Perimeter = \(2(\text{side}_1 + \text{side}_2) = 2\left(x + \frac{3}{4}x\right) = 70$
\(2\left(\frac{7}{4}x\right) = 70 \implies \frac{7}{2}x = 70 \implies x = 20\)
Sides are 20 cm and \(\frac{3}{4} \times 20 = 15\) cm.
Dimensions of parallelogram: 20 cm, 15 cm, 20 cm, 15 cm.
In simple words: Set up perimeter formula using x and 3x/4, solve for x, and find side lengths.

Exam Tip: Perimeter of a parallelogram is twice the sum of adjacent sides.

Question 4. A pair of the adjacent sides of a rectangle are in the ratio 3:4. If its diagonal is 20 cm, find the length of the sides and hence the perimeter of the rectangle.
Answer:
Let sides be \(3x\) and \(4x\).
Using Pythagorean theorem with diagonal 20 cm:
\((3x)^2 + (4x)^2 = 20^2$
\(9x^2 + 16x^2 = 400$
\(25x^2 = 400 \implies x^2 = 16 \implies x = 4$
Sides = \(3 \times 4 = 12\) cm and \(4 \times 4 = 16\) cm.
Perimeter = \(2(12 + 16) = 2 \times 28 = 56\) cm.
In simple words: Apply Pythagoras theorem on the right triangle formed by the sides and diagonal.

Exam Tip: Diagonals of a rectangle form right-angled triangles with its sides.

Question 5. ABCD is a parallelogram. AP bisects angle A and CQ bisects angle C. P lies on CD and Q lies on AB. Show that 1) AP || CQ. 2) AQCP is a parallelogram.
Answer:
1. Since ABCD is a parallelogram, \(\angle A = \angle C\). Their bisectors (\(\angle PAQ\) and \(\angle QCP\)) are also equal. Alternate interior angles are equal, showing \(AP \parallel CQ\).
2. Since \(AQ \parallel PC\) and \(AP \parallel QC\), opposite sides are parallel, making AQCP a parallelogram.
In simple words: Angle bisectors of equal opposite angles create parallel lines, forming a parallelogram.

Exam Tip: Prove parallel opposite sides to establish a quadrilateral as a parallelogram.

Question 6. An exterior angle of a parallelogram is \(110^\circ\). Find the angles of a parallelogram.
Answer:
Adjacent interior angle = \(180^\circ - 110^\circ = 70^\circ\).
Opposite interior angle = \(70^\circ\).
Other pair of opposite angles = \(110^\circ\).
Angles are \(70^\circ, 110^\circ, 70^\circ, 110^\circ\).
In simple words: Exterior and adjacent interior angles are supplementary (add up to 180 degrees).

Question 7. The ratio of the two sides of a parallelogram is 3 : 5 and its perimeter is 48 cm. Find the sides of the parallelogram.
Answer:
Let sides be \(3x\) and \(5x\).
Perimeter = \(2(3x + 5x) = 48$
\(2(8x) = 48 \implies 16x = 48 \implies x = 3$
Sides = \(3 \times 3 = 9\) cm and \(5 \times 3 = 15\) cm.
In simple words: Equate twice the sum of ratio sides to the perimeter 48.

Exam Tip: Perimeter formula for parallelograms is \(2(a+b)\).

Question 8. The diagonals of a rhombus are in the ratio 5 : 12. If its perimeter is 104 cm, find the length of the sides and its diagonals.
Answer:
Perimeter = 104 cm \(\implies \text{Side} = \frac{104}{4} = 26\) cm.
Let diagonals be \(5x\) and \(12x\). Half diagonals are \(\frac{5}{2}x\) and \(6x\).
Using Pythagorean theorem in right triangle formed by half diagonals and side:
\(\left(\frac{5}{2}x\right)^2 + (6x)^2 = 26^2$
\(\frac{25}{4}x^2 + 36x^2 = 676$
\(\frac{169}{4}x^2 = 676 \implies x^2 = 676 \times \frac{4}{169} = 16 \implies x = 4$
Diagonals are \(5 \times 4 = 20\) cm and \(12 \times 4 = 48\) cm.
In simple words: Use side length and half-diagonal right triangle property with Pythagoras theorem.

Exam Tip: Diagonals of a rhombus bisect each other at right angles, forming four right-angled triangles.