CBSE Class 6 Mathematics Revision Worksheet Set 18

ASSIGNMENT- 1

Knowing Our Numbers

Q.1. How are the numbers 1000 and 500 represented in Roman numerals?
Answer: 1000 is represented by M and 500 is represented by D.
In simple words: The standard Roman numeral symbols for 1000 and 500 are M and D respectively.

Exam Tip: Memorize basic Roman numerals: I = 1, V = 5, X = 10, L = 50, C = 100, D = 500, M = 1000.

Q.2. Arrange 7724, 7247, 2474, 7742 in ascending order.
Answer: Comparing the given numbers from smallest to largest:
\(2474 < 7247 < 7724 < 7742\)
In simple words: Ascending order means arranging numbers from the smallest to the largest value.

Exam Tip: Compare digits starting from the leftmost place value.

Q.3. Write the expression for “eighteen is added to the product of fifteen and seven”.
Answer: \((15 \times 7) + 18\)
In simple words: First multiply 15 and 7, then add 18 to that product.

Exam Tip: Follow the order of operations; product is calculated before addition.

Q.4. Estimate to the nearest hundred \(151 \times 72\).
Answer: 151 rounded to the nearest hundred is 200.
72 rounded to the nearest hundred is 100.
Estimated product = \(200 \times 100 = 20000\)
In simple words: Round each number to the nearest hundred and multiply the rounded values.

Exam Tip: Look at the tens digit for rounding to the nearest hundred; if it is 5 or more, round up.

Q.5. Write 50,998 in expanded form.
Answer: \(50,000 + 900 + 90 + 8\)
In simple words: Write each digit multiplied by its respective place value.

Exam Tip: Do not include place values with a digit of 0 unless required.

Q.6. Write Hindu-Arabic number for LXXIX.
Answer: LXXIX = \(50 + 10 + 10 + (10 - 1) = 70 + 9 = 79\)
In simple words: L is 50, two X's make 20, and IX is 9. Adding them gives 79.

Exam Tip: Smaller numeral placed before a larger numeral denotes subtraction.

Q.7. Write the number name for 88,88,808.
Answer: Eighty-eight lakh eighty-eight thousand eight hundred eight.
In simple words: Read the number according to the Indian numbering system.

Exam Tip: Place commas correctly: 88,88,808.

Q.8. Using the given digits 9, 0, 4, 3, 6 without repetition and make the greatest and the smallest 5- digit numbers.
Answer: Greatest number = 96430
Smallest number = 30469
In simple words: For greatest, arrange digits in descending order. For smallest, arrange in ascending order, keeping 0 in the second place.

Exam Tip: A 5-digit number cannot start with 0.

Q.9. Arrange the numbers 93124, 94213, 98234, 94823, 93128 in descending order.
Answer: \(98234 > 94823 > 94213 > 93128 > 93124\)
In simple words: Descending order means arranging numbers from the largest to the smallest value.

Exam Tip: Compare digits from left to right carefully.

Q.10. You have the following digits 4, 5, 6, 0, 7 and 8. Using them, make the greatest and the smallest numbers each with 6- digit.
Answer: Greatest 6-digit number = 876540
Smallest 6-digit number = 405678
In simple words: Descending order gives the greatest number; ascending order (with 4 first) gives the smallest.

Exam Tip: Always place the smallest non-zero digit at the extreme left for the smallest number.

Q.11. A bus covers the distance of 160 km 575m every day. Find the total distance covered by it in 7 days.
Answer: Distance covered in 1 day = 160 km 575 m = 160.575 km
Distance covered in 7 days = \(160.575 \times 7\)
\(= 1124.025 \text{ km}\) or 1124 km 25 m
In simple words: Multiply the daily distance by 7 to get the total distance for a week.

Exam Tip: Convert units uniformly before multiplication or multiply km and m separately.

Q.12. Shekhar is a famous cricket player. He has so far scored 6,980 runs in test matches. He wishes to complete 10,000 runs. How many more runs does he need?
Answer: Total target runs = 10,000
Runs scored so far = 6,980
Runs needed = \(10000 - 6980 = 3,020\)
In simple words: Subtract the runs already scored from the target runs.

Exam Tip: Double check subtraction with regrouping across zeros.

Q.13. A vessel has 4l 500 ml of curd. In how many glasses, each of 25 ml capacity, can it be filled?
Answer: Total curd = 4 l 500 ml = \(4 \times 1000 + 500 = 4500 \text{ ml}\)
Capacity of one glass = 25 ml
Number of glasses = \(\frac{4500}{25} = 180\)
In simple words: Convert litres to millilitres and divide by the capacity of one glass.

Exam Tip: Ensure all quantities are in the same unit before dividing.

ASSIGNMENT- 2

Whole Numbers

Q.1. What is the successor and the predecessor of 599?
Answer: Successor of 599 = \(599 + 1 = 600\)
Predecessor of 599 = \(599 - 1 = 598\)
In simple words: Successor is 1 more than the number, and predecessor is 1 less.

Exam Tip: Successor means next number; predecessor means previous number.

Q.2. How many numbers are there between 92 and 103?
Answer: Numbers between 92 and 103 are 93, 94, 95, 96, 97, 98, 99, 100, 101, 102.
Total numbers = \(103 - 92 - 1 = 10\)
In simple words: Count all whole numbers lying strictly between the two given numbers.

Exam Tip: Formula for numbers between two integers \(a\) and \(b\) is \(b - a - 1\).

Q.3. Calculate the product of \(4 \times 233 \times 25\) using rearrangement.
Answer: \(4 \times 233 \times 25 = (4 \times 25) \times 233\)
\(= 100 \times 233 = 23,300\)
In simple words: Group numbers that multiply to give powers of 10 for easier calculation.

Q.4. Calculate \(761 \times 999 + 761\) using property.
Answer: \(761 \times 999 + 761 \times 1 = 761 \times (999 + 1)\)
\(= 761 \times 1000 = 761,000\)
In simple words: Use the distributive property of multiplication over addition by taking 761 common.

Exam Tip: Recognize that a standalone number has an implied multiplier of 1.

Q.5. Which is the greater number among 803838 and 830838?
Answer: 830838 is greater than 803838.
In simple words: Comparing the ten-thousands place, 3 is greater than 0.

Exam Tip: Compare corresponding digits from left to right.

Q.6. Solve by suitable rearrangement \(1762 + 553 + 1338 + 547\).
Answer: \(= (1762 + 1338) + (553 + 547)\)
\(= 3100 + 1100 = 4200\)
In simple words: Group numbers whose units digits add up to 10 to make addition simpler.

Exam Tip: Associative and commutative properties allow grouping numbers flexibly.

Q.7. Write the natural number which has no predecessor.
Answer: 1
In simple words: 1 is the first natural number, so it has no natural number before it.

Exam Tip: Whole numbers start at 0, but natural numbers start at 1.

Q.8. Solve: \(8 \times 62 \times 8 - 64 \times 60\)
Answer: \(8 \times 8 \times 62 - 64 \times 60\)
\(= 64 \times 62 - 64 \times 60\)
\(= 64 \times (62 - 60)\)
\(= 64 \times 2 = 128\)
In simple words: Simplify the first term and use the distributive property to factor out 64.

Exam Tip: Look for common factors to simplify expressions quickly.

Q.9. Find the value of \(54279 \times 92 + 8 \times 54279\) using property.
Answer: \(54279 \times (92 + 8)\)
\(= 54279 \times 100 = 5,427,900\)
In simple words: Factor out 54279 using the distributive property.

Exam Tip: Distributive property states \(a \times b + a \times c = a \times (b + c)\).

Q.10. Multiply \(25 \times 96\) using suitable property.
Answer: \(25 \times (100 - 4)\)
\(= (25 \times 100) - (25 \times 4)\)
\(= 2500 - 100 = 2400\)
In simple words: Write 96 as \(100 - 4\) and distribute 25.

Exam Tip: Splitting numbers near multiples of 10 makes mental math much easier.

Q.11. The total number of children in VI-A are 45 and in VI-B are 47. If each child pays Rs. 135 as donation amount, how much money is collected by both sections?
Answer: Total children = \(45 + 47 = 92\)
Donation per child = Rs. 135
Total money collected = \(92 \times 135 = \text{Rs. } 12,420\)
In simple words: Find total students first and multiply by the donation amount per student.

Exam Tip: Alternatively, calculate \(45 \times 135 + 47 \times 135\).

Q.12. Rita pays Rs 1500 to her Piano teacher and Rs 750 to the donation club every month. Find the total amount paid by her in 1 year.
Answer: Monthly payment = \(1500 + 750 = \text{Rs. } 2250\)
Amount paid in 1 year (12 months) = \(2250 \times 12 = \text{Rs. } 27,000\)
In simple words: Add monthly expenses and multiply by 12 months.

Q.1. Explain how 28 is a perfect number?
Answer: Proper divisors of 28 are 1, 2, 4, 7, and 14.
Sum of proper divisors = \(1 + 2 + 4 + 7 + 14 = 28\)
Since the sum of its proper divisors equals the number itself, 28 is a perfect number.
In simple words: A perfect number is equal to the sum of all its factors excluding itself.

Exam Tip: Other examples of perfect numbers are 6 and 496.

Q.2. Determine the prime factorization. (i) 468 (ii) 7325
Answer: (i) Prime factorization of 468:
\(468 = 2 \times 234 = 2 \times 2 \times 117 = 2 \times 2 \times 3 \times 39 = 2^2 \times 3^2 \times 13\)
(ii) Prime factorization of 7325:
\(7325 = 5 \times 1465 = 5 \times 5 \times 293 = 5^2 \times 293\)
In simple words: Continuously divide by prime numbers until only prime factors remain.

Exam Tip: Check divisibility by 2, 3, 5, 7, 11 systematically.

Q.3. Find the smallest 4-digit number which is divisible by 18, 24 and 32.
Answer: First find the LCM of 18, 24, and 32.
\(18 = 2 \times 3^2\)
\(24 = 2^3 \times 3\)
\(32 = 2^5\)
LCM = \(2^5 \times 3^2 = 32 \times 9 = 288\)
Smallest 4-digit number is 1000.
Divide 1000 by 288: \(1000 = 288 \times 3 + 136\)
Smallest 4-digit divisible number = \(1000 + (288 - 136) = 1000 + 152 = 1152\)
In simple words: Find the LCM, then find the smallest multiple of that LCM which has 4 digits.

Exam Tip: Add the difference between the divisor and the remainder to the base number.

Q.4. Find the greatest number of 6 digits exactly divisible by 24, 15 and 36.
Answer: Find LCM of 24, 15, and 36.
\(24 = 2^3 \times 3\)
\(15 = 3 \times 5\)
\(36 = 2^2 \times 3^2\)
LCM = \(2^3 \times 3^2 \times 5 = 8 \times 9 \times 5 = 360\)
Greatest 6-digit number is 999,999.
Divide 999,999 by 360, remainder is 279.
Greatest 6-digit divisible number = \(999,999 - 279 = 999,720\)
In simple words: Find the LCM and subtract the remainder obtained when dividing the largest 6-digit number by this LCM.

Exam Tip: Subtract the remainder directly from the dividend to get the exact multiple.

Q.5. Find first four common multiples of 3, 4 and 9.
Answer: LCM of 3, 4, and 9 is 36.
First four common multiples are the multiples of 36:
\(36 \times 1 = 36\)
\(36 \times 2 = 72\)
\(36 \times 3 = 108\)
\(36 \times 4 = 144\)
In simple words: Find the LCM first, then list its first four multiples.

Exam Tip: Common multiples of a set of numbers are always multiples of their LCM.

Q.6. The students in a class can be divided into groups of 2, 3, 5 and 6. What is the least number of children this can have.
Answer: Find the LCM of 2, 3, 5, and 6.
\(2 = 2\)
\(3 = 3\)
\(5 = 5\)
\(6 = 2 \times 3\)
LCM = \(2 \times 3 \times 5 = 30\)
In simple words: The least number of children is the LCM of the group sizes.

Q.7. Check the divisibility of 3340 by (i) 2 (ii) 3 (iii) 6
Answer: Number = 3340
(i) Divisibility by 2: Since the units digit is 0 (even), 3340 is divisible by 2.
(ii) Divisibility by 3: Sum of digits = \(3 + 3 + 4 + 0 = 10\). Since 10 is not divisible by 3, 3340 is not divisible by 3.
(iii) Divisibility by 6: Since 3340 is not divisible by 3, it is not divisible by 6 (even though it is divisible by 2).
In simple words: Check rules for 2, 3, and 6 individually.

Exam Tip: A number is divisible by 6 only if it is divisible by both 2 and 3.

Q.8. Check the divisibility of 88169308 by 11.
Answer: Sum of digits at odd places from right = \(8 + 3 + 6 + 8 = 25\)
Sum of digits at even places from right = \(0 + 9 + 1 + 8 = 18\)
Difference = \(25 - 18 = 7\)
Since 7 is not divisible by 11, the number 88169308 is not divisible by 11.
In simple words: Subtract the sum of even-place digits from the sum of odd-place digits. If the difference is 0 or a multiple of 11, it is divisible.

Exam Tip: Always count places from right to left starting at 1.

Q.9. The length, breadth and height of room are 625cm, 675cm and 500cm respectively, find the longest tape to measure the dimensions of the room exactly.
Answer: Find the HCF of 625, 675, and 500.
\(625 = 5^4\)
\(675 = 27 \times 25 = 3^3 \times 5^2\)
\(500 = 4 \times 125 = 2^2 \times 5^3\)
HCF = \(5^2 = 25 \text{ cm}\)
In simple words: The longest tape is the HCF of the three room dimensions.

Exam Tip: "Longest" or "maximum" measure usually indicates calculating the HCF.

Q.10. Find the greatest number such that if 245 and 1029 be divided by it, then remainder in each case is 5.
Answer: Subtract the remainder 5 from both numbers:
\(245 - 5 = 240\)
\(1029 - 5 = 1024\)
Now, find the HCF of 240 and 1024.
\(240 = 2^4 \times 3 \times 5\)
\(1024 = 2^{10}\)
HCF = \(2^4 = 16\)
In simple words: Subtract 5 from each number first, then find the HCF of the resulting numbers.

Exam Tip: Subtracting the remainder first turns the problem into an exact divisibility HCF problem.

ASSIGNMENT- 4

Basic Geometrical Ideas

Q.1. Construct a segment AB = 1.8 cm long. Construct another segment PQ whose length is thrice as that of AB. Measure its length.
Answer:
1. Draw a line segment \(AB = 1.8 \text{ cm}\).
2. Using a ruler and compass, mark off three consecutive segments equal to AB along a straight line to get PQ.
Length of PQ = \(3 \times 1.8 \text{ cm} = 5.4 \text{ cm}\).
In simple words: Draw a line segment three times the length of AB, which equals 5.4 cm.

Exam Tip: Always verify construction measurements with a ruler.

Q.2. From the given figure, find out: (i) all pairs of intersecting lines. (ii) all points of intersection.
Answer:
(i) Pairs of intersecting lines: Line AB and Line CD, Line AB and Line EF, Line CD and Line EF.
(ii) Points of intersection: Point O and Point M.
In simple words: Intersecting lines cross each other at a common point called the point of intersection.

Exam Tip: Name lines using any two points on them.

Q.3. In the concentric circles given below, find the radius and diameter of the bigger circle.
Answer: Radius of the bigger circle = 2 cm.
Diameter of the bigger circle = \(2 \times 2 \text{ cm} = 4 \text{ cm}\).
In simple words: Diameter is always twice the radius.

Exam Tip: Concentric circles share the same center but have different radii.

Q.4. If AB = 2.4 cm and CD = 2.5 cm, construct a segment whose length is equal to 3AB - 2CD.
Answer:
Length = \(3(2.4) - 2(2.5)\)
\(= 7.2 - 5.0 = 2.2 \text{ cm}\).
Construct a line segment of length 2.2 cm.
In simple words: Calculate the final length arithmetic first, then draw the segment of that length.

Q.5. How many diameters of a circle can be drawn? Where do they intersect? Is the point of intersection of diameter in the interior or exterior of the circle?
Answer:
1. Infinitely many diameters can be drawn.
2. They all intersect at the center of the circle.
3. The point of intersection (the center) lies in the interior of the circle.
In simple words: A circle has infinite diameters, and they all cross at the exact center inside the circle.

Exam Tip: The center is always in the interior region.

Q.6. Name from drawn figure. (i) name its diameter. (ii) name its radius. (iii) name its arc. (iv) is the figure half of a circle?
Answer:
(i) Diameter: AB
(ii) Radius: OA or OB
(iii) Arc: Arc AB (semi-circular arc)
(iv) Yes, the figure represents a semi-circle (half of a circle).
In simple words: A diameter divides a circle into two equal halves called semi-circles.

Q.7. (i) Name all the triangles formed in the figure. (ii) Which two points lie on sides BC and AB respectively? (iii) Name any two line segments inside the triangle ABC
Answer:
(i) Triangles formed: Triangle ABC, Triangle APC, Triangle APB, Triangle PQC, Triangle PQB (depending on internal partitions).
(ii) Point P lies on side BC (or internal), Point Q lies on side AB.
(iii) Line segments inside: AP, PQ, PC.
In simple words: Identify all distinct triangular shapes and internal line segments from the diagram.

Exam Tip: Use three vertex letters to name any triangle.

Q.8. From the drawn circle, name the following : (i) a chord. (ii) a point in the interior and a point in the exterior. (iii) A sector. (iv) A segment. (v) An arc.
Answer:
(i) Chord: DE (or BC)
(ii) Interior point: A; Exterior point: G
(iii) Sector: Region bounded by two radii and an arc (e.g., OAH)
(iv) Segment: Region bounded by chord DE and arc DE
(v) Arc: Arc BC or Arc HE
In simple words: Identify standard parts of a circle such as chords, sectors, and segments based on standard definitions.

Exam Tip: Remember a sector is like a slice of pie (radii + arc), while a segment is cut off by a chord.

ASSIGNMENT- 5

INTEGERS

Q.1. Write the integers which lie between -4 and 6.
Answer: -3, -2, -1, 0, 1, 2, 3, 4, 5
In simple words: Integers strictly between -4 and 6 do not include -4 and 6 themselves.

Exam Tip: Exclude the boundary numbers when asked for numbers "between" two values.

Q.2. Write four integers greater than -15.
Answer: -14, -10, 0, 5 (Any four integers greater than -15)
In simple words: Any number to the right of -15 on the number line is greater.

Exam Tip: For negative numbers, smaller absolute value means greater value.

Q.3. On the number line, find the number which is 6 places right to (-3).
Answer: \(-3 + 6 = 3\)
In simple words: Moving right on the number line means adding.

Exam Tip: Right direction = positive addition; left direction = negative subtraction.

Q.4. Add using a number line \((-5) + 3\).
Answer: Start at 0, move 5 units to the left to reach -5, then move 3 units to the right to land at -2.
Result = -2
In simple words: Start at -5 and move 3 steps to the right.

Exam Tip: Addition of a positive integer means moving right on the number line.

Q.5. Find the value of: \((-50) + (-200) + 200\).
Answer: \((-50) + [(-200) + 200]\)
\(= (-50) + 0 = -50\)
In simple words: -200 and +200 cancel each other out, leaving -50.

Exam Tip: Look for additive inverses to simplify expressions.

Q.6. Find the integer x such that \(-4 + x = 0\).
Answer: \(x = 4\)
In simple words: x must be the opposite (additive inverse) of -4.

Exam Tip: The sum of a number and its additive inverse is always zero.

Q.7. Is the sum of two negative integers is positive or negative?
Answer: Negative.
In simple words: Adding two negative debts results in a larger negative debt.

Exam Tip: Sum of two negative integers is always negative.

Q.8. Write the number which is: (i) 4 less than -3 (ii) 7 more than 3
Answer:
(i) \(-3 - 4 = -7\)
(ii) \(3 + 7 = 10\)
In simple words: "Less than" means subtract, and "more than" means add.

Exam Tip: Pay close attention to wording regarding direction of operations.

Q.9. Add (-52) and 60. Now, subtract (-8) from the result. What is the solution?
Answer:
Step 1: \((-52) + 60 = 8\)
Step 2: Subtract (-8) from 8: \(8 - (-8) = 8 + 8 = 16\)
In simple words: Subtracting a negative number is the same as adding its positive counterpart.

Exam Tip: Double negative becomes a positive sign: \(-(-) = +\).

Q.10. Give the successor and predecessor of 217.
Answer: Successor = \(217 + 1 = 218\)
Predecessor = \(217 - 1 = 216\)
In simple words: Add 1 for successor and subtract 1 for predecessor.

Exam Tip: Applicable to both positive and negative integers.

Q.11. What is the sign of the sum of two positive integers?
Answer: Positive.
In simple words: Adding two positive quantities always yields a positive result.

Exam Tip: Positive plus positive is always positive.

Q.12. Find the value of \((-1) + (-2) + 23 - (42)\)
Answer: \(= -1 - 2 + 23 - 42\)
\(= -3 + 23 - 42\)
\(= 20 - 42 = -22\)
In simple words: Combine negative terms and positive terms step-by-step.

Exam Tip: Group all negative numbers together and all positive numbers together before combining.

Q.13. Replace * by \(>, <\) or \(=\): \((-10) - (-10) * (-10) + (-10)\)
Answer:
Left side: \(-10 - (-10) = -10 + 10 = 0\)
Right side: \(-10 + (-10) = -20\)
Comparing 0 and -20: \(0 > -20\)
So, replace * with \(>\).
In simple words: Simplify both sides independently, then compare the resulting integers.

Exam Tip: Zero is always greater than any negative integer.

Q.14. Add the integers: \((-9) + 4 + (-6) + 3\)
Answer: \(= (-9 - 6) + (4 + 3)\)
\(= -15 + 7 = -8\)
In simple words: Add all negatives together and all positives together, then find the net result.

Exam Tip: Grouping like signs prevents calculation errors.

Q.15. Find the value of: (i) \(-(26) - 48 + (-54) - (-58)\) (ii) \(-(-25) + (-50) - (+65) - (+30)\)
Answer:
(i) \(-26 - 48 - 54 + 58\)
\(= -128 + 58 = -70\)
(ii) \(25 - 50 - 65 - 30\)
\(= 25 - 145 = -120\)
In simple words: Remove parentheses by applying sign rules, then combine terms.

Exam Tip: A negative sign outside a parenthesis reverses all signs inside.

ASSIGNMENT- 6

Understanding elementary shapes

Q.1. Name the angle whose measure is more than \(90^\circ\) and less than \(180^\circ\).
Answer: Obtuse angle.
In simple words: An angle between \(90^\circ\) and \(180^\circ\) is called an obtuse angle.

Exam Tip: Acute \(< 90^\circ\), Right \(= 90^\circ\), Obtuse between \(90^\circ\) and \(180^\circ\), Straight \(= 180^\circ\).

Q.2. Which direction will you face if you start facing west and make \(\frac{3}{4}\) of a revolution from left?
Answer: North.
In simple words: Turning \(\frac{3}{4}\) of a revolution counter-clockwise (from left) starting from West leads to North.

Exam Tip: Each quarter turn represents \(90^\circ\) or \(\frac{1}{4}\) of a revolution.

Q.3. Is the given angle acute or reflex?
Answer: Reflex angle.
In simple words: An angle greater than \(180^\circ\) is a reflex angle.

Q.4. What is the diameter of the circle, if the radius is 3.6 cm.
Answer: Diameter = \(2 \times \text{radius} = 2 \times 3.6 \text{ cm} = 7.2 \text{ cm}\).
In simple words: Diameter is twice the length of the radius.

Exam Tip: \(d = 2r\).

Q.5. Name the triangle in which: (i) One angle is obtuse angle. (ii) Two sides are equal.
Answer:
(i) Obtuse-angled triangle
(ii) Isosceles triangle
In simple words: Triangles are classified by angles (acute, right, obtuse) and sides (scalene, isosceles, equilateral).

Exam Tip: An isosceles triangle has two equal sides and two equal opposite angles.

Q.6. When two lines are drawn having distance 5cm equally between them, then what do we call them?
Answer: Parallel lines.
In simple words: Lines that remain at a constant perpendicular distance apart are parallel.

Exam Tip: Parallel lines never intersect.

Q.7. In the drawn rhombus, name the type of angles: (i) \(\angle A\) and \(\angle C\) (ii) \(\angle A\) and \(\angle D\)
Answer:
(i) \(\angle A\) and \(\angle C\) are opposite angles (acute or obtuse pairs, equal in measure).
(ii) \(\angle A\) and \(\angle D\) are adjacent angles (supplementary, summing to \(180^\circ\)).
In simple words: Opposite angles in a rhombus are equal, and adjacent angles add up to \(180^\circ\).

Exam Tip: Consecutive angles of any parallelogram or rhombus are supplementary.

Q.8. Where will the hour hand of a clock stop, if it starts from 8 and turns: (i) Two right angles (ii) Three right angles
Answer:
(i) Two right angles = \(180^\circ\) (or 6 hours elapsed). Starting from 8, moving 6 hours forward stops at 2.
(ii) Three right angles = \(270^\circ\) (or 9 hours elapsed). Starting from 8, moving 9 hours forward stops at 5.
In simple words: Each right angle corresponds to 3 hours on a clock face.

Exam Tip: 1 hour on a clock face equals \(30^\circ\).

Q.9. Find the faces, edges and corners of a: (i) Cylinder (ii) Prism
Answer:
(i) Cylinder: 3 faces (2 flat circular, 1 curved), 2 curved edges, 0 corners (vertices).
(ii) Triangular Prism: 5 faces (2 triangular, 3 rectangular), 9 edges, 6 corners.
In simple words: Count flat/curved surfaces, line segments where faces meet, and corner points.

Exam Tip: Use Euler's formula \(F + V - E = 2\) for polyhedrons.

Q.10. When two lines drawn from adjacent vertices cut each other in the interior part of a square at \(90^\circ\) then what do we call them?
Answer: Diagonals of the square.
In simple words: The lines joining opposite or adjacent structural points meeting at right angles inside are diagonals.

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

Q.11. Construct a segment RS = 2.8cm long, construct another segment PM whose length is twice as that of RS. Measure its length.
Answer:
1. Draw line segment \(RS = 2.8 \text{ cm}\).
2. Construct PM with length \(2 \times 2.8 \text{ cm} = 5.6 \text{ cm}\).
Measured length of PM = 5.6 cm.
In simple words: Draw a segment double the length of RS.

Exam Tip: Double 2.8 cm accurately by standard multiplication.

Q.12. Write the names of - (i) All angles formed with vertex O. (ii) All pairs of adjacent angles with vertex O.
Answer:
(i) Angles formed with vertex O: \(\angle AOB, \angle BOC, \angle COD, \angle AOC, \angle BOD, \angle AOD\).
(ii) Pairs of adjacent angles: \((\angle AOB, \angle BOC)\), \((\angle BOC, \angle COD)\), \((\angle AOB, \angle AOC)\), etc.
In simple words: List individual adjacent angle pairs sharing a common arm and vertex.

Exam Tip: Adjacent angles share a common vertex and a common arm.

Q.13. In the given adjoining figure, list the points which: (i) are in the interior of \(\angle RST\) (ii) are in the exterior of \(\angle RST\) (iii) lie on \(\angle RST\)
Answer:
(i) Interior points: C, B, F (depending on specific diagram layout).
(ii) Exterior points: A, E.
(iii) Points on the angle: R, S, T, G.
In simple words: Points inside the arms are interior, outside are exterior, and on the rays are on the angle.

Exam Tip: Rays forming the angle contain points lying directly on the angle.

ASSIGNMENT- 7

Fractions

Q.1. What fraction of numbers from 2 to 15 are prime numbers?
Answer: Numbers from 2 to 15: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15 (Total = 14 numbers).
Prime numbers in this range: 2, 3, 5, 7, 11, 13 (Total = 6 primes).
Required fraction = \(\frac{6}{14} = \frac{3}{7}\)
In simple words: Count prime numbers and divide by the total count of numbers in the given range.

Exam Tip: Always reduce fractions to their lowest terms.

Q.2. Form the fraction of 750g to a kilogram and reduce it to the lowest term.
Answer: 1 kilogram = 1000 g.
Fraction = \(\frac{750}{1000}\)
Reducing to lowest terms by dividing numerator and denominator by 250:
\(\frac{750 \div 250}{1000 \div 250} = \frac{3}{4}\)
In simple words: Convert both quantities to grams and simplify the fraction.

Exam Tip: Ensure both units are identical before forming a fraction.

Q.3. Find fraction of 25 paise to Rs 2.
Answer: Rs 2 = \(2 \times 100 = 200\) paise.
Fraction = \(\frac{25}{200}\)
Reducing to lowest terms: \(\frac{25 \div 25}{200 \div 25} = \frac{1}{8}\)
In simple words: Convert rupees to paise so both units match, then simplify.

Exam Tip: Rs 1 = 100 paise.

Q.4. Convert \(\frac{54}{5}\) into mixed fraction.
Answer: Divide 54 by 5: Quotient = 10, Remainder = 4.
Mixed fraction = \(10\frac{4}{5}\)
In simple words: Divide numerator by denominator to get the whole number and remainder.

Exam Tip: Format is Quotient \(\frac{\text{Remainder}}{\text{Divisor}}\).

Q.5. Find whether the given set of fractions is like or unlike? \(\frac{2}{12}, \frac{3}{14}, \frac{4}{12}, \frac{14}{12}\)
Answer: Unlike fractions.
In simple words: Fractions with different denominators are called unlike fractions.

Exam Tip: Like fractions have the exact same denominator.

Q.6. Put the following set of fractions in descending order: \(\frac{1}{2}, \frac{1}{21}, \frac{1}{5}, \frac{1}{32}, \frac{1}{50}, \frac{1}{9}, \frac{1}{15}\)
Answer: When numerators are equal, the fraction with the smaller denominator is greater.
Descending order: \(\frac{1}{2} > \frac{1}{5} > \frac{1}{9} > \frac{1}{15} > \frac{1}{21} > \frac{1}{32} > \frac{1}{50}\)
In simple words: For equal numerators, smaller denominator means a larger fraction value.

Exam Tip: Compare denominators directly when numerators are all 1.

Q.7. Solve the following: (i) \(\frac{1}{7} + \frac{2}{7}\) (ii) \(\frac{13}{9} - \frac{5}{9}\)
Answer:
(i) \(\frac{1 + 2}{7} = \frac{3}{7}\)
(ii) \(\frac{13 - 5}{9} = \frac{8}{9}\)
In simple words: Since denominators are the same, combine numerators directly.

Exam Tip: Keep the common denominator unchanged.

Q.8. Simplify: \(8\frac{1}{4} - 2\frac{5}{6}\)
Answer: Convert mixed fractions to improper fractions:
\(8\frac{1}{4} = \frac{33}{4}\), \(2\frac{5}{6} = \frac{17}{6}\)
LCM of 4 and 6 is 12.
\(\frac{33 \times 3}{12} - \frac{17 \times 2}{12} = \frac{99 - 34}{12} = \frac{65}{12} = 5\frac{5}{12}\)
In simple words: Convert to improper fractions, find the common denominator, subtract, and convert back.

Exam Tip: Always find the LCM of denominators for addition/subtraction.

Q.9. Find the value of: (i) \(\frac{5}{7} + \frac{1}{6}\) (ii) \(\frac{5}{8} - \frac{1}{2}\)
Answer:
(i) LCM of 7 and 6 is 42.
\(\frac{5 \times 6}{42} + \frac{1 \times 7}{42} = \frac{30 + 7}{42} = \frac{37}{42}\)
(ii) LCM of 8 and 2 is 8.
\(\frac{5}{8} - \frac{4}{8} = \frac{1}{8}\)
In simple words: Take the LCM of denominators to make them common before adding or subtracting.

Exam Tip: Check if one denominator is a multiple of the other to simplify LCM finding.

Q.10. Reema ate \(\frac{2}{3}\) of a bread. Her sister ate \(\frac{1}{3}\) the part. How much did they eat altogether?
Answer: Total eaten = \(\frac{2}{3} + \frac{1}{3} = \frac{2 + 1}{3} = \frac{3}{3} = 1\) (whole bread).
In simple words: Add both fractions since denominators are equal.

Exam Tip: A fraction equal to 1 represents a whole quantity.

Q.11. Represent \(\frac{1}{4}\) on a number line.
Answer:
1. Draw a number line from 0 to 1.
2. Divide the segment between 0 and 1 into 4 equal parts.
3. The first tick mark after 0 represents \(\frac{1}{4}\).
In simple words: Divide the unit interval into 4 equal segments and mark 1 part from 0.

Exam Tip: Denominator indicates the number of equal parts into which the unit is divided.

Q.12. Is \(\frac{4}{7}\) equal to \(\frac{16}{28}\)? Solve and show.
Answer: Simplify \(\frac{16}{28}\) by dividing numerator and denominator by 4:
\(\frac{16 \div 4}{28 \div 4} = \frac{4}{7}\)
Since both fractions simplify to the same value, they are equal.
In simple words: Reducing the second fraction gives the first fraction, proving they are equivalent.

Exam Tip: Cross-multiplication (\(4 \times 28 = 7 \times 16 = 112\)) also proves equivalence.

Q.13. Colour the fraction as indicated: \(\frac{1}{3}\)
Answer: In a figure divided into 3 equal parts, shade exactly 1 part.
In simple words: Color one out of three equal sections.

Exam Tip: The numerator indicates how many parts to color.

ASSIGNMENT- 8

Data Handling

Q.1. The total number of animals in five zoos is as follows: Zoo A : 50, Zoo B : 80, Zoo C : 120, Zoo D : 90, Zoo E : 50. Prepare a pictograph taking scale \(1 * = 10\) animals and answer questions: (i) Which zoo has maximum animals and which has minimum. (ii) Which zoo has more animals B or C and by how much? (iii) Which two zoos have same number of animals?
Answer:
Pictograph representation (using symbol * = 10 animals):
Zoo A: ***** (5 symbols)
Zoo B: ******** (8 symbols)
Zoo C: ************ (12 symbols)
Zoo D: ********* (9 symbols)
Zoo E: ***** (5 symbols)
(i) Maximum animals: Zoo C (120); Minimum animals: Zoo A and Zoo E (50 each).
(ii) Zoo C has more animals than Zoo B by \(120 - 80 = 40\) animals (4 symbols).
(iii) Zoo A and Zoo E have the same number of animals (50 each).
In simple words: Draw symbols according to the given scale and compare numerical values.

Exam Tip: Always state the key/scale clearly on a pictograph.

Q.2. The numbers of people using different modes of transport are shown below (Scale: 1 face = 10 people): Car: 6 faces, Scooter: 4 faces, Cycle: 3 faces, Bus: 8 faces, Auto: 3 faces. Answer: (i) Most popular mode of transport? (ii) How many people like to go by bus? (iii) How many more people like to go by bus than cycle? (iv) Which two modes of transport are equally liked by people?
Answer:
(i) Bus (80 people, 8 faces) is the most popular.
(ii) Bus: \(8 \times 10 = 80\) people.
(iii) Bus (80) - Cycle (\(3 \times 10 = 30\)) = \(80 - 30 = 50\) more people.
(iv) Cycle (30) and Auto (30) are equally liked.
In simple words: Multiply the number of icons by the scale value (10) to find total counts.

Exam Tip: Read symbol counts carefully from data tables.

Q.3. The data for various heights found in a class is given: 4, 5, 2, 2, 5, 4, 3, 4, 5, 5, 3, 2, 4, 2, 2, 2, 2, 3. Make a table and enter the data using tally marks.
Answer:
Counting frequencies of each height:
- Height 2: 7 occurrences (Tally: |||| ||)
- Height 3: 3 occurrences (Tally: |||)
- Height 4: 4 occurrences (Tally: ||||)
- Height 5: 4 occurrences (Tally: ||||)
Total frequency = \(7 + 3 + 4 + 4 = 18\).
In simple words: Tally marks group counts in sets of five for easy counting.

Exam Tip: Every fifth tally mark crosses the previous four (\(\cancel{||||}\)).

Q.4. The number of magazines sold by shopkeeper on seven days is shown below. Draw a bar graph choosing a suitable scale: Sun: 75, Mon: 50, Tue: 30, Wed: 50, Thu: 45, Fri: 20, Sat: 65.
Answer:
Scale: 1 unit length = 10 magazines on the vertical axis.
Bar heights:
- Sun: 7.5 units
- Mon: 5 units
- Tue: 3 units
- Wed: 5 units
- Thu: 4.5 units
- Fri: 2 units
- Sat: 6.5 units
In simple words: Draw equal width rectangular bars with heights proportional to magazine counts.

Exam Tip: Maintain equal spacing between bars on a bar graph.

Q.5. In the bar graph shown, likings of fruits by students: Apple: 11, Guava: 3, Orange: 7, Banana: 6. Answer: (i) Which fruit is liked by most students? (ii) How many students like orange? (iii) How many more students like banana than guava? (iv) How many more students like apple than orange?
Answer:
(i) Apple (11 students) is liked by the most students.
(ii) Orange: 7 students.
(iii) Banana (6) - Guava (3) = \(6 - 3 = 3\) more students.
(iv) Apple (11) - Orange (7) = \(11 - 7 = 4\) more students.
In simple words: Read bar heights directly from the vertical axis.

Exam Tip: Verify axis labels and increments before reading values.

ASSIGNMENT- 9

Decimals

Q.1. Represent 7.2 on a number line.
Answer:
1. Draw a number line and locate the integers 7 and 8.
2. Divide the interval between 7 and 8 into 10 equal parts.
3. Count 2 small subdivisions to the right of 7 to mark 7.2.
In simple words: 7.2 lies between 7 and 8, specifically two tenths past 7.

Exam Tip: Each small division between whole numbers represents 0.1.

Q.2. Write \(900 + 90 + \frac{9}{10}\) in decimal form.
Answer: \(990.9\)
In simple words: Combine hundreds, tens, and tenths place values into standard decimal notation.

Exam Tip: Missing place values (like ones) must be filled with a 0.

Q.3. Convert 5.25 in the fractional form. Also reduce it to its lowest form.
Answer: Fractional form = \(\frac{525}{100}\)
Reducing by dividing numerator and denominator by 25:
\(\frac{525 \div 25}{100 \div 25} = \frac{21}{4}\) (or \(5\frac{1}{4}\))
In simple words: Remove the decimal point by dividing by 100, then simplify.

Q.4. Write seventy seven point two three in decimal form.
Answer: 77.23
In simple words: Write whole number part before the decimal point and fractional digits after.

Exam Tip: Digits after the decimal point are read individually.

Q.5. Convert \(\frac{11}{1000}\) in the decimal form.
Answer: 0.011
In simple words: Three zeros in the denominator mean three decimal places.

Exam Tip: Pad with zeros if necessary before placing digits.

Q.6. Which is greater, 7.35 or 7.05?
Answer: 7.35 is greater than 7.05.
In simple words: Comparing the tenths place, 3 is greater than 0.

Exam Tip: Compare digits from left to right across decimal places.

Q.7. Write using decimals: (i) 85cm (ii) 575 paise
Answer:
(i) \(85 \text{ cm} = \frac{85}{100} \text{ m} = 0.85 \text{ m}\)
(ii) \(575 \text{ paise} = \frac{575}{100} \text{ Rs.} = \text{Rs. } 5.75\)
In simple words: Divide by 100 to convert cm to metres or paise to rupees.

Exam Tip: 1 m = 100 cm and Re 1 = 100 paise.

Q.8. Convert : (i) 14.35m into m and cm (ii) 56.904kg into kg and g
Answer:
(i) 14.35 m = 14 m and 35 cm
(ii) 56.904 kg = 56 kg and 904 g
In simple words: The integer part represents the larger unit, and the decimal part represents the smaller unit.

Q.9. Add: 92.5, 16.542 and 1.711.
Answer: Align decimal points:
\(92.500\)
\(+ 16.542\)
\(+ \phantom{0}1.711\)
\(= 110.753\)
In simple words: Pad with trailing zeros to match decimal places, then add vertically.

Exam Tip: Always align decimal points before adding or subtracting.

Q.10. Subtract: 19.05 from 25.56.
Answer: \(25.56 - 19.05 = 6.51\)
In simple words: Subtract column by column from right to left with aligned decimal points.

Exam Tip: Ensure the larger number is placed on top in subtraction.

Q.11. Reema had Rs. 7.25 with her. She spent Rs. 1.50. How much money is left with her?
Answer: Initial amount = Rs. 7.25
Amount spent = Rs. 1.50
Money left = \(7.25 - 1.50 = \text{Rs. } 5.75\)
In simple words: Subtract the spent amount from the initial amount.

Exam Tip: Borrow correctly across decimal places during subtraction.

Q.12. Romi has to cover a distance of 20km 50m. She walked 1 km and went by bus for 15km and hired an auto for rest of the distance. How much distance did she cover by auto?
Answer: Total distance = 20 km 50 m = 20.050 km
Distance covered by walking and bus = \(1 \text{ km} + 15 \text{ km} = 16 \text{ km}\) (or 16.000 km)
Distance by auto = \(20.050 - 16.000 = 4.050 \text{ km}\) (4 km 50 m)
In simple words: Subtract the distances covered by foot and bus from the total distance.

Exam Tip: 50 m expressed in km is 0.050 km.

Q.13. Natu bought 500g potatoes, 250g capsicums, 700g onions, 500g tomatoes, 100g gingers and 300g radish. Find the total weight of the vegetables in terms of decimal.
Answer: Total weight in grams = \(500 + 250 + 700 + 500 + 100 + 300 = 2350 \text{ g}\)
In kg (decimal form) = \(\frac{2350}{1000} = 2.350 \text{ kg}\)
In simple words: Sum all weights in grams and divide by 1000 to convert to kilograms.

Exam Tip: 1 kg = 1000 g.

Q.14. Bulbul spent Rs. 190.25 on her books, Rs. 275 on grocery and Rs. 85.50 on clothes. Find the total money spent by her.
Answer: Total money spent = \(190.25 + 275.00 + 85.50 = \text{Rs. } 550.75\)
In simple words: Add all individual expenditures together.

Exam Tip: Align decimal points carefully when adding currency values.

ASSIGNMENT- 10

PRACTICAL GEOMETRY

Q.1. Given AB of length 5cm. Construct PQ such that the length of PQ is twice that of AB.
Answer:
1. Draw a line segment \(AB = 5 \text{ cm}\).
2. Draw a line and use compasses to mark off two lengths equal to AB end-to-end to form PQ.
Length of PQ = \(2 \times 5 \text{ cm} = 10 \text{ cm}\).
In simple words: Construct a segment twice the length of AB.

Exam Tip: Use compasses for precise transfer of lengths.

Q.2. With the same centre, draw two circles of radius 6cm and 5cm.
Answer:
1. Mark a point O as center.
2. With O as center and radius 5 cm, draw a circle.
3. With the same center O and radius 6 cm, draw another circle.
In simple words: Draw two concentric circles sharing the exact same center point.

Q.3. Draw any line segment CD. Take any point M not on it. Through M, draw a perpendicular to CD. (Use rules and compasses)
Answer:
1. Draw line segment CD and mark point M outside it.
2. With M as center, draw an arc intersecting CD at two points, say P and Q.
3. With P and Q as centers and radius greater than half of PQ, draw two arcs intersecting each other on the other side of CD at point N.
4. Join M and N. MN is perpendicular to CD.
In simple words: Construct a perpendicular from an external point using standard compass arcs.

Exam Tip: Ensure compass width remains unchanged when drawing intersecting arcs from P and Q.

Q.4. Draw a line segment of length 10cm. Using compasses divide it into four equal parts.
Answer:
1. Draw segment AB = 10 cm.
2. Draw perpendicular bisector of AB to divide it into two 5 cm halves.
3. Draw perpendicular bisectors of each 5 cm half to divide them further, resulting in four equal parts of 2.5 cm each.
In simple words: Bisect the segment repeatedly to divide it into 4 equal segments.

Exam Tip: Successive bisection of 10 cm gives 5 cm, then 2.5 cm.

Q.5. Draw CD of length 8cm and find its axis of symmetry.
Answer:
1. Draw segment CD = 8 cm.
2. Construct the perpendicular bisector of CD using compasses.
3. This perpendicular bisector is the axis of symmetry.
In simple words: The axis of symmetry for a line segment is its perpendicular bisector.

Exam Tip: The axis divides the segment into two equal 4 cm symmetrical halves.

Q.6. Construct an angle of \(90^\circ\) and bisect it.
Answer:
1. Draw a ray and construct a \(90^\circ\) angle using a compass.
2. With the vertex as center, draw an arc cutting both arms of the \(90^\circ\) angle.
3. From those intersection points, draw intersecting arcs inside the angle.
4. Draw a ray from the vertex through this intersection to get a \(45^\circ\) angle.
In simple words: Construct a right angle and draw its angle bisector to get \(45^\circ\).

Exam Tip: Angle bisector divides the angle into two equal halves.

Q.7. Draw an angle of \(70^\circ\) and construct an angle whose measure is twice of the given angle.
Answer:
1. Draw an angle of \(70^\circ\) using a protractor.
2. Construct an angle equal to \(70^\circ\) adjacent to it using a compass, or calculate \(2 \times 70^\circ = 140^\circ\).
3. Draw an angle of \(140^\circ\) using a protractor.
In simple words: Draw an angle of \(140^\circ\) (which is \(2 \times 70^\circ\)).

Exam Tip: Double of \(70^\circ\) is \(140^\circ\), which is an obtuse angle.

Q.8. Draw a line segment of 5.6cm. From it cut 3.6cm. How much part of the line is remaining?
Answer: Remaining length = \(5.6 \text{ cm} - 3.6 \text{ cm} = 2.0 \text{ cm}\).
In simple words: Subtract the cut length from the total segment length.

Exam Tip: Measure accurately with a ruler.

ASSIGNMENT- 11

Mensuration

Q.1. The lid of a rectangular box of sides 40cm by 10cm is sealed all round with tape. What is the length of the tape required?
Answer: Length = 40 cm, Breadth = 10 cm.
Length of tape = Perimeter of rectangle = \(2 \times (\text{Length} + \text{Breadth})\)
\(= 2 \times (40 + 10) = 2 \times 50 = 100 \text{ cm}\).
In simple words: The tape length equals the perimeter of the rectangular lid.

Exam Tip: Perimeter of a rectangle = \(2(l + b)\).

Q.2. What is the length of the wooden strip required to frame a photograph of length and breadth 32cm and 21cm respectively?
Answer: Length = 32 cm, Breadth = 21 cm.
Perimeter = \(2 \times (32 + 21) = 2 \times 53 = 106 \text{ cm}\).
In simple words: The framing strip length is equal to the perimeter of the photograph.

Q.3. A piece of string is 30cm long. What will be the length of each side if the string is used to form: (i) Square (ii) An equilateral triangle
Answer:
(i) For a square (4 equal sides): Side length = \(\frac{30}{4} = 7.5 \text{ cm}\).
(ii) For an equilateral triangle (3 equal sides): Side length = \(\frac{30}{3} = 10 \text{ cm}\).
In simple words: Divide the total string length by the number of sides of the polygon.

Exam Tip: Regular polygons have sides of equal length.

Q.4. Find the distance covered by Rahul if he takes four rounds of a rectangular park whose length is 25m and breadth is 10m.
Answer: Perimeter of park = \(2 \times (25 + 10) = 2 \times 35 = 70 \text{ m}\).
Distance for 4 rounds = \(4 \times 70 = 280 \text{ m}\).
In simple words: Multiply the perimeter by the number of rounds.

Exam Tip: Total distance = Number of rounds \(\times\) Perimeter.

Q.5. A table top measures 12m 25cm by 9m 50cm. What is the perimeter of tabletop?
Answer: Length = 12.25 m, Breadth = 9.50 m.
Perimeter = \(2 \times (12.25 + 9.50) = 2 \times 21.75 = 43.50 \text{ m}\) (43 m 50 cm).
In simple words: Convert dimensions to decimals and apply the rectangle perimeter formula.

Exam Tip: Keep units consistent in decimals before calculating.

Q.6. Kanchan runs around a square park of side 90m and Vinita runs around a rectangular park whose length is 120m and breadth is 80m. Who covers the longer distance and by how much?
Answer:
Kanchan's distance (Perimeter of square) = \(4 \times 90 = 360 \text{ m}\).
Vinita's distance (Perimeter of rectangle) = \(2 \times (120 + 80) = 2 \times 200 = 400 \text{ m}\).
Vinita covers a longer distance by \(400 - 360 = 40 \text{ m}\).
In simple words: Calculate the perimeter for both parks and compare their values.

Exam Tip: Perimeter of square = \(4 \times \text{side}\).

Q.7. The length and breadth of three rectangles are given below. (i) 9m and 6m (ii) 17m and 3m (iii) 14m and 4m. Find which rectangle has greatest area and which has least.
Answer:
Area = Length \(\times\) Breadth
(i) Area = \(9 \times 6 = 54 \text{ m}^2\)
(ii) Area = \(17 \times 3 = 51 \text{ m}^2\)
(iii) Area = \(14 \times 4 = 56 \text{ m}^2\)
Greatest area: Rectangle (iii) [\(56 \text{ m}^2\)]; Least area: Rectangle (ii) [\(51 \text{ m}^2\)].
In simple words: Multiply length by breadth for each rectangle and compare the areas.

Exam Tip: Area of rectangle = \(l \times b\), measured in square units.

Q.8. A tile is a square of side 20cm, how many such tiles would be required to cover the floor of a square bathroom of side 3m?
Answer:
Side of bathroom = 3 m = 300 cm.
Area of bathroom floor = \(300 \times 300 = 90,000 \text{ cm}^2\).
Area of one tile = \(20 \times 20 = 400 \text{ cm}^2\).
Number of tiles = \(\frac{90,000}{400} = 225 \text{ tiles}\).
In simple words: Divide the total floor area by the area of a single tile.

Exam Tip: Convert all dimensions to centimeters before calculating areas.

Q.9. The area of a square of side 16cm is the same as that of a rectangle of length 32cm. Find its breadth.
Answer:
Area of square = \(16 \times 16 = 256 \text{ cm}^2\).
Area of rectangle = \(\text{Length} \times \text{Breadth} = 32 \times \text{Breadth}\).
\(32 \times \text{Breadth} = 256\)
Breadth = \(\frac{256}{32} = 8 \text{ cm}\).
In simple words: Equate the rectangle's area to the square's area and solve for breadth.

Q.10. A room has rectangular floor of size 300m by 200m. A square carpet of side 180m is laid in the room. Find the area of floor which is not carpeted.
Answer: Area of room floor = \(300 \times 200 = 60,000 \text{ m}^2\).
Area of carpet = \(180 \times 180 = 32,400 \text{ m}^2\).
Uncarpeted area = \(60,000 - 32,400 = 27,600 \text{ m}^2\).
In simple words: Subtract the area of the carpet from the total room floor area.

Exam Tip: Area not covered = Total area - Covered area.

Q.11. Find the area of the following figures by splitting into rectangles.
Answer: Split the compound L-shape or stair-step figure into non-overlapping rectangles, calculate individual areas, and add them together.
Total Area = Sum of areas of constituent rectangles.
In simple words: Divide complex shapes into simpler rectangles and sum their individual areas.

Exam Tip: Double check dimensions of each split section.

ASSIGNMENT- 12

Algebra

Q.1 Give expression for the following statements: (i) added to xm, (ii) x subtracted from 4, (iii) 2 subtracted from the sum of x and y, (iv) multiplied by -5, (v) Product of 2 and x divided by 3, (vi) n multiplied by 2 and 1 subtracted from the product, (vii) -5 divided by z, (viii) a increased by twice of b, (ix) Multiply x by y and then add 7 to it, (x) Subtract a from b and then multiply the difference by 7, (xi) 170 increased by the product of x and y, (xii) Three times the difference of 30 and a, (xiii) Multiplied by the sum of x and y by 3 and divide the product by z, (xiv) Add 3 and x and subtract from the product of y and z, (xv) Z multiplied by 7 and result subtracted from 85.
Answer:
(i) \(xm + a\) (assuming 'a' is added)
(ii) \(4 - x\)
(iii) \((x + y) - 2\)
(iv) \(-5x\)
(v) \(\frac{2x}{3}\)
(vi) \(2n - 1\)
(vii) \(\frac{-5}{z}\)
(viii) \(a + 2b\)
(ix) \(xy + 7\)
(x) \(7(b - a)\)
(xi) \(170 + xy\)
(xii) \(3(30 - a)\)
(xiii) \(\frac{3(x + y)}{z}\)
(xiv) \(yz - (x + 3)\)
(xv) \(85 - 7z\)
In simple words: Translate everyday arithmetic statements into algebraic expressions step-by-step.

Exam Tip: "Subtracted from" reverses the order of subtraction terms.

Q.2 Write the statement for the given expression: (i) 3X + 5 (ii) 6 - 4X (iii) 2Y + 2 (iv) \(\frac{x}{2} + 1\) (v) 5 - 3p (vi) -2m + 5
Answer:
(i) Five added to three times X.
(ii) Four times X subtracted from six.
(iii) Two added to twice of Y.
(iv) One added to half of x.
(v) Three times p subtracted from five.
(vi) Five added to negative two times m.
In simple words: Write algebraic expressions back into standard verbal statements.

Exam Tip: Use terms like "sum", "product", "increased by", or "decreased by".

Q.3. If the cost of x metres of cloth is Rs y. Find the cost of 4 metres of cloth.
Answer: Cost of x metres = Rs y.
Cost of 1 metre = \(\frac{y}{x}\).
Cost of 4 metres = \(\frac{4y}{x}\) Rs.
In simple words: Use unitary method: find the cost of 1 metre first, then multiply by 4.

Exam Tip: Unitary method involves finding the value of a single unit first.

Q.4. Aditi's marks in maths are 15 more than two-third of her Hindi marks. If she scores 'x' marks in Hindi, find her marks in maths.
Answer: Hindi marks = x.
Two-third of Hindi marks = \(\frac{2}{3}x\).
Maths marks = \(\frac{2}{3}x + 15\).
In simple words: Translate the statement directly into an algebraic expression.

Exam Tip: "More than" indicates addition.

Q.5. If the cost of one shirt is Rs a and the cost of 1 trouser is Rs. b, then find the cost of 6 shirts and 7 trousers.
Answer: Cost of 6 shirts = \(6a\)
Cost of 7 trousers = \(7b\)
Total cost = Rs. \((6a + 7b)\)
In simple words: Multiply quantity by unit cost for each item and add them together.

Exam Tip: Unlike algebraic terms cannot be combined into a single term.

Q.6. Complete the table and find the solution of \(2x + 10 = 30\)

Answer: Solution of the equation is \(x = 10\) (since \(2(10) + 10 = 30\)).
In simple words: Substitute each value of x into the expression until it equals 30.

Exam Tip: Trial and error method helps find unknown values in simple equations.

ASSIGNMENT- 13

Ratio And Proportion

Q.1. Find the ratio of the following: (i) 21 hours to 49 hours (ii) 75 cm to 3 metres (iii) A dozen to a score
Answer:
(i) \(\frac{21}{49} = \frac{3}{7}\) or \(3:7\)
(ii) 3 metres = 300 cm. Ratio = \(\frac{75}{300} = \frac{1}{4}\) or \(1:4\)
(iii) 1 dozen = 12, 1 score = 20. Ratio = \(\frac{12}{20} = \frac{3}{5}\) or \(3:5\)
In simple words: Convert quantities to the same units and simplify the fraction.

Exam Tip: Ratios have no units because units cancel out.

Q.2. Find the ratio of 250g to 5kg.
Answer: 5 kg = \(5 \times 1000 = 5000 \text{ g}\).
Ratio = \(\frac{250}{5000} = \frac{25}{500} = \frac{1}{20}\) or \(1:20\)
In simple words: Convert kilograms to grams before forming the ratio.

Exam Tip: 1 kg = 1000 g.

Q.3. Out of total 1450 students in a school, 1000 went for the picnic. Find the ratio of: (i) Students who went to the picnic to total students. (ii) Students who did not go to the picnic to those who went for the picnic.
Answer:
Total students = 1450
Went to picnic = 1000
Did not go = \(1450 - 1000 = 450\)
(i) Ratio (went to total) = \(\frac{1000}{1450} = \frac{20}{29}\) or \(20:29\)
(ii) Ratio (did not go to went) = \(\frac{450}{1000} = \frac{9}{20}\) or \(9:20\)
In simple words: Calculate each required count and simplify the resulting fraction.

Exam Tip: Order matters in ratios: numerator corresponds to the first mentioned quantity.

Q.4. Write the middle terms and the extreme terms, if the numbers form the proportion in 1 kg: 40 kg and 25g: 625 g.
Answer:
Proportion: \(1 : 40 :: 25 : 625\)
Extreme terms: 1 and 625
Middle terms: 40 and 25
In simple words: Outer numbers are extremes, and inner numbers are middle terms.

Q.5. A scooter needs 3 litres petrol to cover 90 km. How many litres of petrol is required to cover 120 km?
Answer: Petrol for 90 km = 3 litres.
Petrol for 1 km = \(\frac{3}{90} = \frac{1}{30}\) litres.
Petrol for 120 km = \(120 \times \frac{1}{30} = 4\) litres.
In simple words: Find petrol needed for 1 km using the unitary method, then multiply by 120.

Exam Tip: Unitary method is ideal for direct proportion problems.

Q.6. A car travels 90 km in \(2\frac{1}{2}\) hours. (i) How much time is required to cover 40 km with the same speed? (ii) Find the distance covered in 3 hours with the same speed.
Answer:
Speed = \(\frac{\text{Distance}}{\text{Time}} = \frac{90}{2.5} = 36 \text{ km/h}\).
(i) Time to cover 40 km = \(\frac{\text{Distance}}{\text{Speed}} = \frac{40}{36} = \frac{10}{9}\) hours (or \(1\frac{1}{9}\) hours / 1 hour 6 minutes 40 seconds).
(ii) Distance in 3 hours = \(\text{Speed} \times \text{Time} = 36 \times 3 = 108 \text{ km}\).
In simple words: Find the speed first, then use speed formulas to calculate time and distance.

Exam Tip: Speed = Distance \(\div\) Time.

Q.7. Check whether the following are in proportion or not: (i) 2:9 and 18:81 (ii) Rs. 10 to Rs. 15 and 4 to 6.
Answer:
(i) \(\frac{2}{9}\) and \(\frac{18}{81} = \frac{2}{9}\). Since ratios are equal, they are in proportion.
(ii) Ratio 1: \(\frac{10}{15} = \frac{2}{3}\). Ratio 2: \(\frac{4}{6} = \frac{2}{3}\). Since ratios are equal, they are in proportion.
In simple words: If both ratios simplify to the exact same fraction, they form a proportion.

Exam Tip: Alternatively, check if product of extremes equals product of means (\(a \times d = b \times c\)).

Q.8. A horse and cart together cost Rs 2880. Find the cost of each if their costs are in ratio 5 : 3.
Answer: Total ratio parts = \(5 + 3 = 8\).
Cost of horse = \(\frac{5}{8} \times 2880 = 5 \times 360 = \text{Rs. } 1,800\)
Cost of cart = \(\frac{3}{8} \times 2880 = 3 \times 360 = \text{Rs. } 1,080\)
In simple words: Divide total cost by the sum of ratio parts, then multiply by individual ratio values.

Exam Tip: Sum of parts represents the total whole quantity.

Q.9. Give two equivalent ratios of 6:4.
Answer:
Multiply or divide terms by the same non-zero number:
1. Multiply by 2: \(12:8\)
2. Multiply by 3: \(18:12\) (or divide by 2: \(3:2\))
In simple words: Equivalent ratios are obtained by multiplying or dividing numerator and denominator by the same number.

Exam Tip: Equivalent ratios represent the same fractional value.

Q.10. Lakshmi earns Rs 1,44,000 in 15 months. (i) How much does she earn in 7 months? (ii) In how many months will she earn Rs 2,40,000?
Answer:
Earnings in 1 month = \(\frac{144,000}{15} = \text{Rs. } 9,600\).
(i) Earnings in 7 months = \(7 \times 9600 = \text{Rs. } 67,200\).
(ii) Months to earn Rs 2,40,000 = \(\frac{240,000}{9600} = 25 \text{ months}\).
In simple words: Find monthly earnings first using the unitary method.

Exam Tip: Divide total earnings by monthly rate to find total months.

Q.11. Fill in the blanks: \(\frac{14}{21} = \frac{\Box}{3} = \frac{6}{\Box}\)
Answer:
Simplify \(\frac{14}{21} = \frac{2}{3}\).
First blank: \(\frac{\Box}{3} = \frac{2}{3} \implies \Box = 2\).
Second blank: \(\frac{6}{\Box} = \frac{2}{3} \implies 2\Box = 18 \implies \Box = 9\).
Completed expression: \(\frac{14}{21} = \frac{2}{3} = \frac{6}{9}\)
In simple words: Find equivalent fractions by maintaining the same numerator-to-denominator ratio (\(2:3\)).

Exam Tip: Cross-multiplication makes solving for missing boxes in proportions straightforward.