CBSE Class 6 Mathematics Revision Worksheet Set 03

Ratio, Proportion And Unitary Method

Question 1. Write the ratios in the simplest form:
i) 15 minutes to one hour
ii) 600g to 1 kg
iii) 50p to Rs. 2.50
iv) 2l to 800 ml
v) 80cm to 4 m
vi) 3 km to 600m
vii) 3 dozen to 3 scores
viii) 2cm to 4 cm
ix) 4 days to 2 weeks
x) 3 yrs. To 6 months
xi) 45 sec to 3 min
xii) 45cm to 3m
xiii) 30 min to 4 hours
xiv) Rs. 1500 to Rs. 18000
xv) 750g to 3kg
xvi) 8days to 6 weeks
xvii) 450ml to 3l
xviii) 8mm to 1cm
xix) Rs. 1.50 to 25p
xx) 4 months to 4yrs
xxi) 294 : 343
Answer:
i) 15 minutes to one hour = 15 minutes to 60 minutes = \(\frac{15}{60} = \frac{1}{4} = 1:4\)
ii) 600g to 1 kg = 600g to 1000g = \(\frac{600}{1000} = \frac{3}{5} = 3:5\)
iii) 50p to Rs. 2.50 = 50p to 250p = \(\frac{50}{250} = \frac{1}{5} = 1:5\)
iv) 2l to 800ml = 2000ml to 800ml = \(\frac{2000}{800} = \frac{5}{2} = 5:2\)
v) 80cm to 4m = 80cm to 400cm = \(\frac{80}{400} = \frac{1}{5} = 1:5\)
vi) 3 km to 600m = 3000m to 600m = \(\frac{3000}{600} = \frac{5}{1} = 5:1\)
vii) 3 dozen to 3 scores = 36 to 60 = \(\frac{36}{60} = \frac{3}{5} = 3:5\)
viii) 2cm to 4cm = \(\frac{2}{4} = \frac{1}{2} = 1:2\)
ix) 4 days to 2 weeks = 4 days to 14 days = \(\frac{4}{14} = \frac{2}{7} = 2:7\)
x) 3 yrs. to 6 months = 36 months to 6 months = \(\frac{36}{6} = \frac{6}{1} = 6:1\)
xi) 45 sec to 3 min = 45 sec to 180 sec = \(\frac{45}{180} = \frac{1}{4} = 1:4\)
xii) 45cm to 3m = 45cm to 300cm = \(\frac{45}{300} = \frac{3}{20} = 3:20\)
xiii) 30 min to 4 hours = 30 min to 240 min = \(\frac{30}{240} = \frac{1}{8} = 1:8\)
xiv) Rs. 1500 to Rs. 18000 = \(\frac{1500}{18000} = \frac{1}{12} = 1:12\)
xv) 750g to 3kg = 750g to 3000g = \(\frac{750}{3000} = \frac{1}{4} = 1:4\)
xvi) 8 days to 6 weeks = 8 days to 42 days = \(\frac{8}{42} = \frac{4}{21} = 4:21\)
xvii) 450ml to 3l = 450ml to 3000ml = \(\frac{450}{3000} = \frac{3}{20} = 3:20\)
xviii) 8mm to 1cm = 8mm to 10mm = \(\frac{8}{10} = \frac{4}{5} = 4:5\)
xix) Rs. 1.50 to 25p = 150p to 25p = \(\frac{150}{25} = \frac{6}{1} = 6:1\)
xx) 4 months to 4yrs = 4 months to 48 months = \(\frac{4}{48} = \frac{1}{12} = 1:12\)
xxi) \(294 : 343 = \frac{294}{343} = \frac{6}{7} = 6:7\)
In simple words: Convert both quantities to the same unit and divide by their HCF to reduce the ratio.
Exam Tip: Always make units identical before comparing quantities in a ratio.

Question 2. Divide Rs. 1450 between A and B in the ratio 2 : 3
Answer:
Total parts = 2 + 3 = 5
A's share = \(\frac{2}{5} \times 1450 = 2 \times 290 =\) Rs. 580
B's share = \(\frac{3}{5} \times 1450 = 3 \times 290 =\) Rs. 870
In simple words: Add the ratio terms to find total parts, then multiply the total amount by each fraction.

Exam Tip: Check your work by adding the individual shares to ensure they equal the total amount given.

Question 3. Dr. Suma earns Rs. 50,000 p.m. she spends Rs. 42,000 of it and saves the rest. Find the ratio of i) Her expenditure to her income ii) Her saving to her income iii) Her expenditure to her savings
Answer:
Income = Rs. 50,000
Expenditure = Rs. 42,000
Savings = Income - Expenditure = 50,000 - 42,000 = Rs. 8,000
i) Expenditure to Income = \(\frac{42000}{50000} = \frac{42}{50} = \frac{21}{25} = 21:25\)
ii) Saving to Income = \(\frac{8000}{50000} = \frac{8}{50} = \frac{4}{25} = 4:25\)
iii) Expenditure to Savings = \(\frac{42000}{8000} = \frac{42}{8} = \frac{21}{4} = 21:4\)
In simple words: Find savings by subtracting expenditure from income, then form the required ratios in lowest terms.

Exam Tip: Ratios have no units, but ensure both quantities use the same currency or unit before simplifying.

Question 4. Fill in :
i) 3:5 = ____ : 15
ii) 6: ____ = 12:30 = 49:63 (Note: interpreted from context)
iii) ____:11 = 12:22
iv) 9:13 = 27: ____
v) 6:7 = ____:49
vi) ____:5 = 20:25
vii) 3: ____ = 5:10
viii) 45:15 = 3 : ____
ix) 5:30 = ____:6
x) 7: ____ = 49:63
xi) 1:11 = 9: ____
xii) ____:13 = 18:117
Answer:
i) \(3:5 = \mathbf{9} : 15\) (since \(15 \div 5 = 3\), \(3 \times 3 = 9\))
ii) \(6:\mathbf{15} = 12:30\) (since \(12 \div 6 = 2\), \(30 \div 2 = 15\))
iii) \(\mathbf{6}:11 = 12:22\) (since \(22 \div 11 = 2\), \(12 \div 2 = 6\))
iv) \(9:13 = 27:\mathbf{39}\) (since \(27 \div 9 = 3\), \(13 \times 3 = 39\))
v) \(6:7 = \mathbf{42}:49\) (since \(49 \div 7 = 7\), \(6 \times 7 = 42\))
vi) \(\mathbf{4}:5 = 20:25\) (since \(20 \div 5 = 4\), \(25 \div 5 = 5\))
vii) \(3:\mathbf{6} = 5:10\) (since \(10 \div 5 = 2\), \(3 \times 2 = 6\))
viii) \(45:15 = 3 : \mathbf{1}\) (since \(45 \div 15 = 3\), \(15 \div 15 = 1\))
ix) \(5:30 = \mathbf{1}:6\) (since \(30 \div 6 = 5\), \(5 \div 5 = 1\))
x) \(7:\mathbf{9} = 49:63\) (since \(49 \div 7 = 7\), \(63 \div 7 = 9\))
xi) \(1:11 = 9:\mathbf{99}\) (since \(9 \div 1 = 9\), \(11 \times 9 = 99\))
xii) \(\mathbf{2}:13 = 18:117\) (since \(117 \div 13 = 9\), \(18 \div 9 = 2\))
In simple words: Use equivalent fractions by multiplying or dividing both terms of the ratio by the same number.

Exam Tip: Cross-multiplication (\(a \times d = b \times c\)) is a reliable way to find missing terms in proportions.

Question 5. Meena got Rs. 600 as her share from Rs. 2000. Tina got Rs. 750 as her share from Rs. 3000. Express their shares as ratios in the lower terms. Compare and find who got more.
Answer:
Meena's ratio = \(\frac{600}{2000} = \frac{6}{20} = \frac{3}{10} = 3:10\) or \(\frac{3}{10} = 30\%\)
Tina's ratio = \(\frac{750}{3000} = \frac{75}{300} = \frac{1}{4} = 1:4\) or \(\frac{1}{4} = 25\%\)
Comparing fractions: \(\frac{3}{10} = \frac{12}{40}\) and \(\frac{1}{4} = \frac{10}{40}\). Since \(\frac{12}{40} > \frac{10}{40}\), Meena got more.
In simple words: Convert both shares into fractions or percentages of their respective totals to compare them directly.

Exam Tip: To compare two fractions, find a common denominator or convert them to decimals.

Question 6. Fill in the blanks with >, = or <
i) 7:10 ____ 5:10
ii) 6:7 ____ 6:11
iii) 4:5 ____ 8:10
iv) 9:13 ____ 9:15
v) 11:20 ____ 9:20
vi) 4:9 ____ 12:27
Answer:
i) \(7:10 > 5:10\)
ii) \(6:7 > 6:11\)
iii) \(4:5 = 8:10\)
iv) \(9:13 > 9:15\)
v) \(11:20 > 9:20\)
vi) \(4:9 = 12:27\)
In simple words: Compare the fractional equivalents by making denominators equal or using cross-multiplication.

Exam Tip: When denominators are equal, the fraction with the larger numerator is greater.

Question 7. The ratio of the length of a rectangle to its breadth is 3:2. If the perimeter of the rectangle is 20 cm find the length, breadth and area.
Answer:
Let length = \(3x\) and breadth = \(2x\)
Perimeter = \(2 \times (\text{length} + \text{breadth})\)
\(2(3x + 2x) = 20\)
\(2(5x) = 20\)
\(10x = 20\)
\(x = 2 \text{ cm}\)
Length = \(3 \times 2 = 6 \text{ cm}\)
Breadth = \(2 \times 2 = 4 \text{ cm}\)
Area = \(\text{Length} \times \text{Breadth} = 6 \times 4 = 24 \text{ cm}^2\)
In simple words: Use a variable \(x\) for the ratio parts, set up the perimeter formula, solve for \(x\), and find dimensions and area.

Exam Tip: Always include proper square units for area and linear units for perimeter and side lengths.

Question 8. Fill in :
i) 2:4 :: ____:10
ii) ____:8 :: 1:4
iii) 3:8 :: 15: ____
iv) 1: ____ :: 3:15
v) 9: ____ :: 90:100
Answer:
i) \(2:4 :: \mathbf{5}:10\) (since \(2 \times 10 = 4 \times 5\))
ii) \(\mathbf{2}:8 :: 1:4\) (since \(2 \times 4 = 8 \times 1\))
iii) \(3:8 :: 15:\mathbf{40}\) (since \(3 \times 40 = 8 \times 15 = 120\))
iv) \(1:\mathbf{5} :: 3:15\) (since \(1 \times 15 = 5 \times 3 = 15\))
v) \(9:\mathbf{10} :: 90:100\) (since \(9 \times 100 = 90 \times 10 = 900\))
In simple words: In a proportion, the product of extremes equals the product of means (\(a \times d = b \times c\)).

Exam Tip: Product of means equals product of extremes is the golden rule of proportions.

Question 9. The cost of 4 pens is Rs. 40. The cost of 11 pens is Rs. ____________.
Answer:
Cost of 1 pen = \(\frac{40}{4} =\) Rs. 10
Cost of 11 pens = \(11 \times 10 =\) Rs. 110
In simple words: Find the cost of one item first (unitary method), then multiply by the required quantity.

Exam Tip: Always calculate the unit value first before scaling up or down.

Question 10. The weight of 15 boxes is 60 kg. The weight of 12 boxes is __________.
Answer:
Weight of 1 box = \(\frac{60}{15} = 4 \text{ kg}\)
Weight of 12 boxes = \(12 \times 4 = 48 \text{ kg}\)
In simple words: Find the weight of one box by dividing total weight by 15, then multiply by 12.

Question 11. Maya can walk 6km in 2 hours. In 3 hours she can walk __________.
Answer:
Speed = \(\frac{6 \text{ km}}{2 \text{ hours}} = 3 \text{ km/hr}\)
Distance in 3 hours = \(3 \text{ km/hr} \times 3 \text{ hours} = 9 \text{ km}\)
In simple words: Calculate the distance covered in 1 hour, then multiply by 3.

Exam Tip: Speed equals distance divided by time.

Question 12. Dinner at Marhaba costs SR 1050 for 10 people. How many people can have dinner for SR 1995?
Answer:
Number of people for SR 1050 = 10
Cost per person = \(\frac{1050}{10} =\) SR 105
Number of people for SR 1995 = \(\frac{1995}{105} = 19 \text{ people}\)
In simple words: Find the cost per person, then divide the total budget by that cost.

Exam Tip: Double-check division with large numbers by multiplying back.

Question 13. A car travels 95 km in 5 litres of petrol. How far can it go in 11 litres of petrol?
Answer:
Distance per litre = \(\frac{95}{5} = 19 \text{ km/litre}\)
Distance in 11 litres = \(19 \times 11 = 209 \text{ km}\)
In simple words: Find how many kilometers the car travels on 1 litre, then multiply by 11.

Exam Tip: Unitary method problems always require finding the value for a single unit first.

Elementary Shapes

Question 1. Given below are the lengths of the sides of triangles. Classify them as equilateral, isosceles or scalene
i) 6cm, 2.4cm, 6cm
ii) 7cm, 9cm, 5.5cm
iii) 5.4cm, 7cm, 6.1cm
iv) 7.2cm, 7.2cm, 7.2cm
v) 10.1cm, 8.6cm, 8.6cm
vi) 3.5cm, 4.5cm, 5.1cm
vii) 4.8cm, 4.8cm, 4.8cm
viii) 6.8cm, 6.8cm, 8.6cm
ix) 3cm, 4cm, 5cm
Answer:
i) Isosceles (two sides equal: 6cm, 6cm)
ii) Scalene (all sides unequal: 7cm, 9cm, 5.5cm)
iii) Scalene (all sides unequal: 5.4cm, 7cm, 6.1cm)
iv) Equilateral (all three sides equal: 7.2cm)
v) Isosceles (two sides equal: 8.6cm, 8.6cm)
vi) Scalene (all sides unequal: 3.5cm, 4.5cm, 5.1cm)
vii) Equilateral (all three sides equal: 4.8cm)
viii) Isosceles (two sides equal: 6.8cm, 6.8cm)
ix) Scalene (all sides unequal: 3cm, 4cm, 5cm)
In simple words: Equilateral has 3 equal sides, isosceles has 2 equal sides, and scalene has all unequal sides.

Exam Tip: Count the number of matching side lengths carefully before classifying.

Question 2. Draw a triangle ABC using a protractor, measure \(\angle A\), \(\angle B\) and \(\angle C\). Find their sum. What do you notice. Also measure the sides AB, BC, AC. What kind of a triangle is this?
Answer:
Sum of angles in any triangle \(\angle A + \angle B + \angle C = 180^{\circ}\). Depending on the drawn triangle, if all sides and angles are equal, it is an equilateral triangle; if two are equal, it is isosceles; otherwise scalene.
In simple words: The sum of interior angles of any triangle is always \(180^{\circ}\).

Exam Tip: Angle sum property of a triangle is a fundamental theorem in geometry.

Question 3. Given below are the measures of the angles of some triangles. Classify them as acute-angled, obtuse-angled or right-angled.
i) \(60^{\circ}, 90^{\circ}, 30^{\circ}\)
ii) \(40^{\circ}, 100^{\circ}, 40^{\circ}\)
iii) \(60^{\circ}, 60^{\circ}, 60^{\circ}\)
iv) \(20^{\circ}, 40^{\circ}, 120^{\circ}\)
v) \(50^{\circ}, 60^{\circ}, 70^{\circ}\)
vi) \(45^{\circ}, 45^{\circ}, 90^{\circ}\)
Answer:
i) Right-angled (contains a \(90^{\circ}\) angle)
ii) Obtuse-angled (contains an obtuse angle of \(100^{\circ}\))
iii) Acute-angled (all angles are less than \(90^{\circ}\))
iv) Obtuse-angled (contains an obtuse angle of \(120^{\circ}\))
v) Acute-angled (all angles are less than \(90^{\circ}\))
vi) Right-angled (contains a \(90^{\circ}\) angle)
In simple words: Check the largest angle: if it is \(90^{\circ}\) it is right-angled, if greater than \(90^{\circ}\) it is obtuse-angled, and if all are under \(90^{\circ}\) it is acute-angled.

Exam Tip: Triangle classification by angles depends entirely on the single largest angle.

Question 4. Draw a triangle ABC. Measure the sides AB, BC, AC. Verify i) \(AB + BC > AC\) ii) \(BC + AC > AB\) iii) \(AB + AC > BC\)
Answer:
By the triangle inequality property, the sum of the lengths of any two sides of a triangle is always greater than the length of the third side. Measuring any triangle ABC will always verify these three inequalities.
In simple words: The sum of any two sides must always be greater than the third side.

Exam Tip: This is known as the Triangle Inequality Theorem.

Question 5. Write the number of sides of the following Polygons: Triangles, Pentagon, Quadrilateral, Heptagon, Hexagon, Nonagon, Octagon, Decagon.
Answer:
Triangle: 3 sides
Pentagon: 5 sides
Quadrilateral: 4 sides
Heptagon: 7 sides
Hexagon: 6 sides
Nonagon: 9 sides
Octagon: 8 sides
Decagon: 10 sides
In simple words: Memorize Greek numeric prefixes to easily identify the number of sides in polygons.

Exam Tip: 'Hepta' means 7, 'Octa' means 8, 'Nona' means 9, and 'Deca' means 10.

Question 6. Fill in:
Name of the Solid | Faces | Edges | Vertices
Cube | ___ | ___ | ___
Cuboid | ___ | ___ | ___
Square Pyramid | ___ | ___ | ___
Triangular Pyramid | ___ | ___ | ___
Triangular Prism | ___ | ___ | ___
Answer:
Cube: Faces = 6, Edges = 12, Vertices = 8
Cuboid: Faces = 6, Edges = 12, Vertices = 8
Square Pyramid: Faces = 5, Edges = 8, Vertices = 5
Triangular Pyramid (Tetrahedron): Faces = 4, Edges = 6, Vertices = 4
Triangular Prism: Faces = 5, Edges = 9, Vertices = 6
In simple words: Euler's formula states that Faces + Vertices - Edges = 2 for polyhedra.

Exam Tip: Practice visualizing 3D shapes to count faces, edges, and vertices accurately.

Question 7. Write yes or no
Quadrilateral | Opposite Sides Parallel | Opposite Sides Equal | All Sides Equal | Opposite Angles Equal | All Angles Equal | Diagonals Equal | Diagonals Perpendicular | Diagonals Bisect each other
Trapezium | ... | ... | ... | ... | ... | ... | ... | ...
Parallelogram | ... | ... | ... | ... | ... | ... | ... | ...
Rhombus | ... | ... | ... | ... | ... | ... | ... | ...
Rectangle | ... | ... | ... | ... | ... | ... | ... | ...
Square | ... | ... | ... | ... | ... | ... | ... | ...
Answer:
Trapezium: Opp Sides Parallel (One pair only - No for both pairs), Opp Sides Equal (No), All Sides Equal (No), Opp Angles Equal (No), All Angles Equal (No), Diagonals Equal (No), Perpendicular (No), Bisect each other (No).
Parallelogram: Opp Sides Parallel (Yes), Opp Sides Equal (Yes), All Sides Equal (No), Opp Angles Equal (Yes), All Angles Equal (No), Diagonals Equal (No), Perpendicular (No), Bisect each other (Yes).
Rhombus: Opp Sides Parallel (Yes), Opp Sides Equal (Yes), All Sides Equal (Yes), Opp Angles Equal (Yes), All Angles Equal (No), Diagonals Equal (No), Perpendicular (Yes), Bisect each other (Yes).
Rectangle: Opp Sides Parallel (Yes), Opp Sides Equal (Yes), All Sides Equal (No), Opp Angles Equal (Yes), All Angles Equal (Yes), Diagonals Equal (Yes), Perpendicular (No), Bisect each other (Yes).
Square: Opp Sides Parallel (Yes), Opp Sides Equal (Yes), All Sides Equal (Yes), Opp Angles Equal (Yes), All Angles Equal (Yes), Diagonals Equal (Yes), Perpendicular (Yes), Bisect each other (Yes).
In simple words: Squares and rectangles have equal diagonals and \(90^{\circ}\) angles, while rhombuses and squares have perpendicular diagonals and all sides equal.

Exam Tip: A square is both a rectangle and a rhombus, sharing properties of all special quadrilaterals.

Question 8. Name each of the following 3 - D shapes. Write number of Curved Surfaces and no of flat faces. (i) Cone (ii) Cylinder (iii) Sphere/Bowl
Answer:
i) Cone: 1 Curved Surface, 1 Flat Face (base)
ii) Cylinder: 1 Curved Surface, 2 Flat Faces (top and bottom bases)
iii) Hemisphere/Bowl: 1 Curved Surface, 1 Flat Face
In simple words: Curved surfaces are rounded, while flat faces are flat polygonal or circular regions.

Exam Tip: A cylinder has two flat circular faces, whereas a cone has only one.

Symmetry

Question 1. Look around and list any five things from your surroundings that are symmetrical
Answer:
1. Butterfly
2. Human face
3. Blackboard / Whiteboard
4. Ceiling fan (with symmetric blades)
5. Leaf
In simple words: Symmetrical objects can be divided into identical halves that match mirror images.

Exam Tip: Look for objects where one half is the exact mirror image of the other.

Question 2. Write any 3 English alphabet, which are symmetrical and have i) One axis of symmetry ii) Two axes of symmetry iii) More than two axes of symmetry
Answer:
i) One axis of symmetry: A, M, T
ii) Two axes of symmetry: H, I, X
iii) More than two axes of symmetry: Letter O has infinite axes of symmetry (or regular polygons if considering shapes, but for alphabets O is the primary one with vertical and horizontal plus diagonal symmetry if styled circularly).
In simple words: Capital letters like A have vertical symmetry, H has both vertical and horizontal symmetry.

Exam Tip: Draw the folding lines through letters to test their symmetry axes.

Question 3. Draw an angle of any measure. Using a ruler and protractor draw the line of symmetry (angle bisector).
Answer:
1. Draw an angle ABC of any measure using a protractor.
2. With vertex B as center, draw an arc intersecting both arms at points P and Q.
3. With P and Q as centers, draw intersecting arcs inside the angle.
4. Join the intersection point to vertex B with a straight line, which is the angle bisector.
In simple words: An angle bisector divides an angle into two equal smaller angles.

Exam Tip: Compass constructions must always show clear arc intersection marks.

Question 4. Write the no. of axes of Symmetry of i) Square ii) Rectangle iii) Circle iv) Scalene Triangle v) Equilateral Triangle vi) Rhombus
Answer:
i) Square: 4 axes of symmetry
ii) Rectangle: 2 axes of symmetry
iii) Circle: Infinite axes of symmetry
iv) Scalene Triangle: 0 axes of symmetry
v) Equilateral Triangle: 3 axes of symmetry
vi) Rhombus: 2 axes of symmetry
In simple words: Regular shapes have as many axes of symmetry as they have sides.

Exam Tip: Remember that a rhombus has 2 axes of symmetry (its diagonals), unlike a square which has 4.

Geometrical Constructions

Question 1. Construct the following angles using a ruler and compasses i) \(60^{\circ}\) ii) \(120^{\circ}\) iii) \(30^{\circ}\) iv) \(90^{\circ}\) v) \(45^{\circ}\)
Answer:
1. For \(60^{\circ}\): Draw a ray, draw an arc from the endpoint cutting the ray, and with the same radius cut the arc.
2. For \(120^{\circ}\): From the \(60^{\circ}\) mark on the arc, cut another arc of the same radius further along.
3. For \(30^{\circ}\): Construct \(60^{\circ}\) and bisect it.
4. For \(90^{\circ}\): Construct a perpendicular at the endpoint or bisect the angle between \(60^{\circ}\) and \(120^{\circ}\).
5. For \(45^{\circ}\): Construct \(90^{\circ}\) and bisect it.
In simple words: Standard angles are constructed using basic arc steps starting from \(60^{\circ}\) and angle bisectors.

Exam Tip: Never change the compass width once set for drawing the base arc in angle constructions.

Question 2. Draw an angle of measure \(100^{\circ}\) using protractor. Draw an angle equal to this angle without using protractor. Draw the bisector of this angle.
Answer:
1. Draw angle ABC = \(100^{\circ}\) using a protractor.
2. Draw a separate ray. Using compasses, measure the arc width of the \(100^{\circ}\) angle and copy it onto the new ray to replicate the angle.
3. Use compass bisector method to divide the copied angle into two equal halves of \(50^{\circ}\) each.
In simple words: Copy an angle by transferring the compass arc length from the original vertex to the new ray.

Exam Tip: Compass replication ensures geometric accuracy without relying on degree measurements.

Question 3. Draw a circle of radius 4cm. Inscribe a regular hexagon in it.
Answer:
1. Draw a circle with center O and radius 4 cm.
2. Mark any point on the boundary as the starting point.
3. Keeping the compass radius fixed at 4 cm, step around the circumference marking successive arcs.
4. Connect the 6 points consecutively with straight line segments to form the regular hexagon.
In simple words: The side of a regular hexagon inscribed in a circle is exactly equal to the radius of the circle.

Question 4. Draw a line-segment AB of any length. Draw the perpendicular bisector of it.
Answer:
1. Draw line segment AB.
2. With center A and radius greater than half of AB, draw arcs above and below AB.
3. With center B and the same radius, draw intersecting arcs above and below AB.
4. Join the intersection points with a line passing through AB, which is the perpendicular bisector.
In simple words: Open your compass more than halfway, draw arcs from both endpoints, and connect their intersections.

Exam Tip: Ensure your compass radius is strictly greater than half the segment length.

Question 5. Draw a line 'l'. Take a point P outside the line 'l'. Draw a perpendicular line to 'l' from P.
Answer:
1. Draw line 'l' and mark point P outside it.
2. With P as center, draw an arc that intersects line 'l' at two distinct points A and B.
3. With centers A and B, draw arcs on the opposite side of line 'l' intersecting at point Q.
4. Join P and Q with a straight line, which is perpendicular to line 'l'.
In simple words: Swing an arc from P to cross the line twice, then use those crossing points to find a point below the line to drop a perpendicular.

Exam Tip: Keep compass width consistent when drawing arcs from points A and B.

Perimeter And Area

Question 1. Which has greater perimeter? - a regular pentagon of 10 cm side or a rectangle with side 17 cm and 12 cm.
Answer:
Perimeter of regular pentagon = \(5 \times \text{side} = 5 \times 10 = 50 \text{ cm}\)
Perimeter of rectangle = \(2 \times (\text{length} + \text{breadth}) = 2 \times (17 + 12) = 2 \times 29 = 58 \text{ cm}\)
Since \(58 \text{ cm} > 50 \text{ cm}\), the rectangle has a greater perimeter.
In simple words: Calculate both perimeters using their respective formulas and compare the final values.

Exam Tip: Perimeter is the total distance around the outer boundary of a 2D shape.

Question 2. A wire bent in the shape of a rectangle of sides 13.5 cm by 6.5 cm was straightened and re-bent into a square. Find the length of the side of the square.
Answer:
Perimeter of rectangle = \(2 \times (13.5 + 6.5) = 2 \times 20 = 40 \text{ cm}\)
Perimeter of square = Perimeter of rectangle = 40 cm
Side of square = \(\frac{40}{4} = 10 \text{ cm}\)
In simple words: The total length of the wire (perimeter) remains constant when reshaped from a rectangle to a square.

Exam Tip: Whenever a wire is reshaped, its total length (perimeter) stays unchanged.

Question 3. The cost of fencing a square field was Rs. 1380 @Rs. 15/m Find the side of the square field And hence find its area.
Answer:
Total cost = Rs. 1380
Cost per meter = Rs. 15
Perimeter of square = \(\frac{\text{Total Cost}}{\text{Rate}} = \frac{1380}{15} = 92 \text{ m}\)
Side of square = \(\frac{92}{4} = 23 \text{ m}\)
Area of square = \(\text{side} \times \text{side} = 23 \times 23 = 529 \text{ m}^2\)
In simple words: Divide total fencing cost by cost per meter to get perimeter, find the side, and then calculate area.

Exam Tip: Fencing always happens along the perimeter of a field.

Question 4. A wire was bent to make a square of side 10.5 cm The same wire is re-bent to make a regular hexagon. Find the length of side of the hexagon.
Answer:
Perimeter of square = \(4 \times 10.5 = 42 \text{ cm}\)
Perimeter of regular hexagon = Perimeter of square = 42 cm
Side of regular hexagon = \(\frac{42}{6} = 7 \text{ cm}\)
In simple words: Find the total perimeter of the square, then divide by 6 to find each side of the regular hexagon.

Exam Tip: A regular hexagon has 6 equal sides.

Question 5. Area of a rectangle is \(24 \text{ cm}^2\). If its length is 6cm find its width. Find the area of a Square whose perimeter is same as of this rectangle.
Answer:
Width of rectangle = \(\frac{\text{Area}}{\text{Length}} = \frac{24}{6} = 4 \text{ cm}\)
Perimeter of rectangle = \(2 \times (6 + 4) = 2 \times 10 = 20 \text{ cm}\)
Perimeter of square = Perimeter of rectangle = 20 cm
Side of square = \(\frac{20}{4} = 5 \text{ cm}\)
Area of square = \(5 \times 5 = 25 \text{ cm}^2\)
In simple words: Find rectangle width, calculate its perimeter, use it as the square's perimeter to find side length, and then find square area.

Question 1. Fill in the blanks: \(25 \times 8 \times 125 \times 4 =\) _______________
Answer: 1,00,000
In simple words: Group numbers to make multiples of 10: \((25 \times 4) \times (125 \times 8) = 100 \times 1000 = 1,00,000\).

Exam Tip: Look for convenient groupings like \(25 \times 4 = 100\) and \(125 \times 8 = 1000\) to simplify multiplication.

Question 2. Fill in the blanks: \(315 \times 105 = 315 \times 100 +\) ____________ \(\times 5\)
Answer: 315
In simple words: Apply distributive property: \(315 \times (100 + 5) = 315 \times 100 + 315 \times 5\).

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

Question 3. Fill in the blanks: Division by zero is ______________________
Answer: not defined
In simple words: You cannot divide any number by zero in mathematics.

Exam Tip: Division by zero is undefined; division of zero by any non-zero number is zero.

Question 4. Fill in the blanks: The whole number _________________ has no predecessor.
Answer: zero (0)
In simple words: Zero is the first whole number, so it has no whole number before it.

Exam Tip: Whole numbers start from 0, whereas natural numbers start from 1.

Question 5. Fill in the blanks: The smallest natural number is ______________________.
Answer: 1
In simple words: Counting numbers (natural numbers) begin with 1.

Exam Tip: Remember that 0 is a whole number, not a natural number.

Question 6. Fill in the blanks: The sum of 3 odd numbers is ____________________.
Answer: odd
In simple words: Odd + Odd = Even, and Even + Odd = Odd. Thus, the sum of 3 odd numbers is always odd.

Question 7. Fill in the blanks: ________________________ is the additive identity for the whole numbers.
Answer: 0 (zero)
In simple words: Adding zero to any number leaves the number unchanged.

Exam Tip: 0 is the additive identity; 1 is the multiplicative identity.

Question 8. Fill in the blanks: \((7 \times 8) \times 5 = 7 \times (8 \times 5)\) This statement shows that multiplication of whole numbers is ____________________________.
Answer: associative
In simple words: Changing the grouping of factors does not change the product.

Exam Tip: Associative property applies to both addition and multiplication.

Question 9. How many numbers between 102 and 211.
Answer:
Number of integers between 102 and 211 = \(211 - 102 - 1 = 108\)
In simple words: To find numbers strictly between two numbers, subtract the smaller from the larger and subtract 1 more.

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

Question 10. Fill in the blanks: \(3 + 7 = 7 + 3\). This statement shows that addition of whole numbers is ___________________________________.
Answer: commutative
In simple words: Changing the order of addends does not change the sum.

Exam Tip: Commutative property holds for addition and multiplication, but not for subtraction or division.

Question 11. Determining the product by suitable rearrangements:
a) \(2 \times 125 \times 50 \times 8\)
b) \(16 \times 279 \times 625\)
c) \(2 \times 1735 \times 50\)
Answer:
a) \((2 \times 50) \times (125 \times 8) = 100 \times 1000 = 1,00,000\)
b) \((16 \times 625) \times 279 = 10,000 \times 279 = 27,90,000\)
c) \((2 \times 50) \times 1735 = 100 \times 1735 = 1,73,500\)
In simple words: Rearrange factors to form powers of 10 for quick mental multiplication.

Exam Tip: Look for pairs like \(2 \times 50\), \(4 \times 25\), and \(16 \times 625\).

Question 12. Using distribution property, find each of the following products.
a) \(213 \times 104\)
b) \(256 \times 1007\)
c) \(462 \times 38\)
Answer:
a) \(213 \times (100 + 4) = (213 \times 100) + (213 \times 4) = 21300 + 852 = 22152\)
b) \(256 \times (1000 + 7) = (256 \times 1000) + (256 \times 7) = 256000 + 1792 = 257792\)
c) \(462 \times (40 - 2) = (462 \times 40) - (462 \times 2) = 18480 - 924 = 17556\)
In simple words: Split one number into a sum or difference of a round number and multiply separately.

Exam Tip: Distributive property simplifies complex multiplication into easier mental parts.

Question 13. Find the value:
a) \(361 + 1482 + 639 + 518\)
b) \(409 + 386 + 3591 + 614\)
c) \(786 \times 97 + 786 \times 3\)
d) \(14 + 438 + 486 + 62\)
e) \(1252 \times 112 - 1252 \times 12\)
f) \(716 \times 6 + 716 \times 4\)
g) \(8062 \times 169 - 8062 \times 69\)
Answer:
a) \((361 + 639) + (1482 + 518) = 1000 + 2000 = 3000\)
b) \((409 + 3591) + (386 + 614) = 4000 + 1000 = 5000\)
c) \(786 \times (97 + 3) = 786 \times 100 = 78600\)
d) \((14 + 486) + (438 + 62) = 500 + 500 = 1000\)
e) \(1252 \times (112 - 12) = 1252 \times 100 = 125200\)
f) \(716 \times (6 + 4) = 716 \times 10 = 7160\)
g) \(8062 \times (169 - 69) = 8062 \times 100 = 806200\)
In simple words: Use commutativity and associativity for addition, or the distributive property in reverse for multiplication.

Exam Tip: Always look for common factors or numbers that add up to multiples of 10 or 100.

Question 14. A teacher purchases 42 Mathematics books and 42 English books for his class. If the cost of a Mathematics book is Rs 52 and the cost of an English book is Rs 48. Find the total amount paid by the teacher to the shopkeeper.
Answer:
Total amount = \(42 \times 52 + 42 \times 48\)
Using distributive property: \(42 \times (52 + 48)\)
\(= 42 \times 100 =\) Rs. 4200
In simple words: Factor out 42 and add book prices together: \(42 \times (52 + 48) = 42 \times 100\).

Exam Tip: Factoring out common numbers saves time and prevents calculation errors.

Question 15. If the cost of a pack of mango drink is Rs. 14. Then how many packs of the drink can be purchased for Rs. 76 and what is the balance?
Answer:
Number of packs = \(76 \div 14 = 5\) packs (since \(14 \times 5 = 70\))
Balance amount = \(76 - 70 =\) Rs. 6
In simple words: Divide total money by the cost per pack to get quotient (packs) and remainder (balance).

Exam Tip: The remainder in division represents the left-over balance.

Knowing Our Numbers

Question 1. Estimate using general rule:
(i) 830 + 976
(ii) 496 - 215
(iii) 13,804 + 3,777
(iv) 61,292 - 21,496
Answer:
i) Estimating to nearest hundreds: \(800 + 1000 = 1800\)
ii) Estimating to nearest hundreds: \(500 - 200 = 300\)
iii) Estimating to nearest thousands: \(14000 + 4000 = 18000\)
iv) Estimating to nearest thousands: \(61000 - 21000 = 40000\)
In simple words: Round each number to its highest place value before performing the operation.

Exam Tip: General rule estimation means rounding off each number to its greatest place.

Question 2. Estimate the products using general rule:
(i) \(758 \times 151\)
(ii) \(4391 \times 2300\)
(iii) \(2187 \times 456\)
(iv) \(6978 \times 43\)
Answer:
i) \(800 \times 200 = 1,60,000\)
ii) \(4000 \times 2000 = 80,00,000\)
iii) \(2000 \times 500 = 10,00,000\)
iv) \(7000 \times 40 = 2,80,000\)
In simple words: Round each factor to its highest place value and multiply the rounded numbers.

Exam Tip: Rounding factors makes multiplication mentally manageable.

Question 3. Write the Roman Numeral for:
(i) 99
(ii) 48
(iii) 67
(iv) 81
(v) 17
(vi) 76
(vii) 54
Answer:
(i) 99 = XCIX
(ii) 48 = XLVIII
(iii) 67 = LXVII
(iv) 81 = LXXXI
(v) 17 = XVII
(vi) 76 = LXXVI
(vii) 54 = LIV
In simple words: Combine Roman symbols using addition and subtraction rules (e.g., IX = 9, XC = 90).

Exam Tip: A smaller numeral placed before a larger one indicates subtraction.

Question 4. Answer the following:
1. The town newspaper is published every day. One copy has 15 pages. Everyday 12,500 copies are printed. How many total pages are printed every day?
2. Apples are packed in boxes, each weighing 5kg 500gm. How many such boxes can be loaded in a van which cannot carry beyond 1000kg?
Answer:
1. Total pages = \(12500 \times 15 = 1,87,500 \text{ pages}\)
2. Weight of one box = 5.5 kg
Total capacity = 1000 kg
Number of boxes = \(1000 \div 5.5 = \frac{1000}{5.5} = \frac{10000}{55} = \frac{2000}{11} = 181.81\)
Since fraction of a box cannot be loaded, 181 boxes can be fully loaded.
In simple words: Multiply copies by pages for total pages; divide total weight capacity by single box weight for box count.

Exam Tip: Always round down for container packing problems when the division leaves a remainder.

Understanding Elementary Shapes

Question I) 1. An angle whose measure is greater than that of a right angle is ____________
Answer: Obtuse angle
In simple words: Angles between \(90^{\circ}\) and \(180^{\circ}\) are obtuse angles.

Exam Tip: Right angle is \(90^{\circ}\); anything greater (up to \(180^{\circ}\)) is obtuse.

Question I) 2. Three edges meet at a point called a ________________________.
Answer: Vertex
In simple words: The corner point where edges intersect in a 3D shape is a vertex.

Question I) 3. A ___________________________ is larger than a straight angle.
Answer: Reflex angle
In simple words: Angles greater than \(180^{\circ}\) and less than \(360^{\circ}\) are reflex angles.

Exam Tip: A straight angle is exactly \(180^{\circ}\).

Question I) 4. A Polygon with 5 sides is called a ___________________________.
Answer: Pentagon
In simple words: 'Penta' means five.

Exam Tip: Polygons are closed 2D shapes made of straight line segments.

Question I) 5. A triangle having all three unequal sides is called a ____________________.
Answer: Scalene triangle
In simple words: All different side lengths mean scalene.

Exam Tip: Equilateral = 3 equal, Isosceles = 2 equal, Scalene = 0 equal.

Question II) Write down the measure of
a) Some acute angles.
b) Some obtuse angles.
c) What is the measure of a straight angle?
Answer:
a) Acute angles: \(30^{\circ}, 45^{\circ}, 60^{\circ}\)
b) Obtuse angles: \(100^{\circ}, 120^{\circ}, 150^{\circ}\)
c) Straight angle: \(180^{\circ}\)
In simple words: Acute is \(< 90^{\circ}\), obtuse is between \(90^{\circ}\) and \(180^{\circ}\), straight is \(180^{\circ}\).

Exam Tip: Memorize the angle category ranges.

Question III) What shape is
a) A brick.
b) A match box
c) A sweet laddu
d) A ball
e) A die
f) A road roller
Answer:
a) A brick: Cuboid
b) A match box: Cuboid
c) A sweet laddu: Sphere
d) A ball: Sphere
e) A die: Cube
f) A road roller: Cylinder
In simple words: Identify everyday objects by their standard 3D geometric counterparts.

Question IV) Name the types of following triangles:
1. \(\Delta LMN\) with \(m\angle L = 80^{\circ}, m\angle M = 70^{\circ}, m\angle N = 30^{\circ}\)
2. \(\Delta ABC\) with \(m\angle A = 90^{\circ}\)
3. \(\Delta PQR\) such that \(PQ = QR = PR = 8\text{ cm}\)
4. \(\Delta XYZ\) with \(AB = 8\text{ cm}, BC = 5\text{ cm}, CA = 5\text{ cm}\) (Note: triangle XYZ with two equal sides)
5. Triangle with lengths of sides 7cm, 8cm and 9cm.
6. \(\Delta PQR\) with \(m\angle Q = 90^{\circ}\) and \(PQ=QR\).
Answer:
1. Acute-angled triangle
2. Right-angled triangle
3. Equilateral triangle
4. Isosceles triangle
5. Scalene triangle
6. Isosceles right-angled triangle
In simple words: Classify triangles based on angle measures or side length equalities.

Exam Tip: Check both sides and angles when fully describing a triangle.

Question V) Let PQ be the perpendicular to the line segment XY. Let PQ and XY intersect in the point A. What is the measure of \(\angle PAY\)?
Answer:
\(90^{\circ}\)
In simple words: Perpendicular lines always intersect at a right angle (\(90^{\circ}\)).

Exam Tip: The word 'perpendicular' automatically implies a \(90^{\circ}\) angle.

Question VI) Classify each one of the following angles as right, straight, acute, obtuse or reflex. (based on visual standard diagrams)
a) Obtuse angle
b) Acute angle
c) Reflex angle
d) Acute angle
e) Right angle
f) Acute angle
Answer:
a) Obtuse angle
b) Acute angle
c) Reflex angle
d) Acute angle
e) Right angle
f) Acute angle
In simple words: Compare visual opening size against a square \(90^{\circ}\) corner.

Exam Tip: Reflex angles bend outward beyond a straight line.

Question VII) Find the angle measure between the hands of the clock in each figure
a) Right angle \(90^{\circ}\)
b) Acute angle
c) Straight angle \(180^{\circ}\)
Answer:
a) Right angle (\(90^{\circ}\))
b) Acute angle (< \(90^{\circ}\))
c) Straight angle (\(180^{\circ}\))
In simple words: Each hour mark on a clock represents \(30^{\circ}\) (\(360^{\circ} \div 12\)).

Exam Tip: 3 hours on a clock face equals \(90^{\circ}\), and 6 hours equals \(180^{\circ}\).

Question VIII) Name each polygon
a) Pentagon
b) Triangle
c) Quadrilateral
d) Octagon
e) Pentagon
Answer:
a) Pentagon
b) Triangle
c) Quadrilateral
d) Octagon
e) Pentagon
In simple words: Count the number of straight sides to name any polygon.

Question IX) 1. A cuboid has ______________________ faces.
2. Each face has ______________________ edges.
3. Each face has _________________ vertices.
Answer:
1. 6 faces
2. 4 edges
3. 4 vertices
In simple words: A cuboid is made of 6 rectangular faces, each having 4 sides (edges) and 4 corners (vertices).

Exam Tip: Individual faces of a cuboid are rectangles.

Playing With Numbers

Question 1. _______________________ is a factor of every number.
Answer: 1 (one)
In simple words: 1 divides every integer completely without a remainder.

Exam Tip: 1 is a universal factor; every number is a multiple of 1.

Question 2. The factor of a prime number is _____________________ and _______________.
Answer: 1 and the number itself
In simple words: Prime numbers have exactly two distinct factors: 1 and the number itself.

Exam Tip: Prime numbers cannot be divided evenly by any other numbers.

Question 3. A number which has more than two factor is called _____________________.
Answer: Composite number
In simple words: Numbers with 3 or more factors are composite.

Exam Tip: 1 is neither prime nor composite.

Question 4. The smallest perfect number is _____________________.
Answer: 6
In simple words: Factors of 6 are 1, 2, 3 (excluding 6), and \(1 + 2 + 3 = 6\).

Exam Tip: A perfect number equals the sum of all its proper divisors.

Question 5. If a number ends with 0, it is divisible by _______________________.
Answer: 2, 5, and 10
In simple words: Numbers ending in 0 are multiples of 10, and hence divisible by 2, 5, and 10.

Exam Tip: Divisibility rule for 10 requires the units digit to be 0.

Question 6. The sum of all the factors of a perfect number is equal to _______________ the number.
Answer: twice (2 times)
In simple words: Including the number itself, the sum of all its factors equals \(2n\).

Question 7. ______________________________ is neither prime nor composite.
Answer: 1 (one)
In simple words: 1 has only one factor (itself), so it belongs to neither category.

Exam Tip: Prime numbers start from 2.

Question 8. A number is divisible by 6, if it is divisible by both ________ and _______________.
Answer: 2 and 3
In simple words: Since \(2 \times 3 = 6\), a number must be divisible by both 2 and 3 to be divisible by 6.

Exam Tip: Check evenness (divisible by 2) and digit sum (divisible by 3).

Question 9. The smallest even numbers is _______ and the smallest odd numbers is a _________.
Answer: 2, 1
In simple words: 2 is the smallest even whole number, and 1 is the smallest positive odd number.

Question 10. Sum of any two even numbers is _______________________.
Answer: even
In simple words: Even + Even = Even.

Question 11. Sum of two odd numbers is ______________________.
Answer: even
In simple words: Odd + Odd = Even (e.g., \(3 + 5 = 8\)).

Question 12. The only one even prime is __________________________.
Answer: 2
In simple words: All other even numbers are divisible by 2, making them composite.

Exam Tip: 2 is the only even prime number.

Question 13. The greatest two digit prime number is __________________________.
Answer: 97
In simple words: 97 cannot be divided by any number other than 1 and 97.

Question 14. The smallest two digit prime numbers is _____________________________.
Answer: 11
In simple words: The numbers from 10 to 99 starting with 11 makes 11 the smallest 2-digit prime.

Question 15. The difference between two twin prime is ________________________.
Answer: 2
In simple words: Twin primes are pairs of prime numbers that differ by 2 (e.g., 3 and 5, 11 and 13).

Question 16. A prime number has only __________________ factors.
Answer: two
In simple words: The two factors are 1 and the number itself.

Question 17. ____________________ is the unique number. (Note: referring to 1 as unique unit)
Answer: 1
In simple words: 1 is neither prime nor composite, making it unique.

Question 18. The smallest digit in the blank space of _____9853. So that the number so formed is divisible by 3.
Answer: 2
In simple words: Sum of digits of \(x9853\) is \(x + 9 + 8 + 5 + 3 = x + 25\). The next multiple of 3 is 27, so \(x = 2\).

Exam Tip: A number is divisible by 3 if the sum of its digits is divisible by 3.

Question 19. The L.C.M of two numbers in which one is a factor of the other is _______________.
Answer: the greater number
In simple words: If b is a multiple of a, their LCM is b.

Question 20. The L.C.M of two co-prime numbers ________________________.
Answer: is their product
In simple words: Co-prime numbers share no common factors other than 1, so their LCM is their product.

Question 21. The smallest factor of 856 is __________________________.
Answer: 1
In simple words: 1 is a factor of every number.

Question 22. The smallest multiple of 856 is _____________________.
Answer: 856
In simple words: The smallest multiple of any number is the number itself (\(856 \times 1\)).

Question 23. The greatest factor of 856 is _____________________.
Answer: 856
In simple words: The greatest factor of a number is the number itself.

Question 24. The perfect numbers below 100 are ________________ and _________________.
Answer: 6 and 28
In simple words: 6 and 28 are the only two perfect numbers below 100.

Question 25. The smallest prime number is ______________________________.
Answer: 2
In simple words: 2 is the first and only even prime number.

Question 26. The smallest composite number is ____________________________.
Answer: 4
In simple words: Factors of 4 are 1, 2, and 4 (more than two factors).

Question 27. The smallest number having three different prime factors is ________________.
Answer: 30
In simple words: Multiply the three smallest prime numbers: \(2 \times 3 \times 5 = 30\).

Question 28. The sum of any two consecutive odd numbers is always divisible by _______.
Answer: 4
In simple words: Sum of consecutive odds like \(3 + 5 = 8\), which is divisible by 4.

Question 29. The product of three consecutive numbers is divisible by _____________.
Answer: 6
In simple words: Among any three consecutive integers, one is always a multiple of 3 and at least one is even, making their product divisible by \(2 \times 3 = 6\).

Question Do the following: 1. Express the smallest 5 - digit number in the form of prime factor.
Answer:
Smallest 5-digit number = 10,000
Prime factorization of 10,000 = \(2^4 \times 5^4 = 2 \times 2 \times 2 \times 2 \times 5 \times 5 \times 5 \times 5\)
In simple words: Break 10,000 down into prime factors 2 and 5 repeated four times each.

Exam Tip: \(10 = 2 \times 5\), so \(10^4 = 2^4 \times 5^4\).

Question Do the following: 2. Determine if 9130 is divisible by 110.
Answer:
\(110 = 10 \times 11\)
1) Check divisibility by 10: 9130 ends in 0, so it is divisible by 10.
2) Check divisibility by 11: Sum of odd place digits - sum of even place digits = \((0 + 1) - (3 + 9) = 1 - 12 = -11\). Since -11 is divisible by 11, 9130 is divisible by 11.
Therefore, 9130 is divisible by 110.
In simple words: A number divisible by both 10 and 11 is divisible by their product 110.

Question Do the following: 3. Using divisibility test check whether the following are divisible by 2, 3, 4, 5, 6, 8, 9, 10 and 11
(a) 91800 (b) 31956 (c) 81615 (d) 61042 (e) 48400 (f) 99909
Answer:
(a) 91800: Divisible by 2, 3, 4, 5, 6, 8, 9, 10, 11 (Sum of digits = 18, ends in 00, alternating sum = 0).
(b) 31956: Divisible by 2, 3, 4, 6, 9 (Sum = 24, ends in 56).
(c) 81615: Divisible by 3, 5 (Sum = 21, ends in 5).
(d) 61042: Divisible by 2 (Even ending in 2).
(e) 48400: Divisible by 2, 4, 5, 10, 11 (Ends in 00, alternating sum divisible by 11).
(f) 99909: Divisible by 3, 9 (Sum = 36).
In simple words: Apply standard divisibility rules for each divisor from 2 to 11.

Question Do the following: 4. Write all the twin primes below 100.
Answer:
(3, 5), (5, 7), (11, 13), (17, 19), (29, 31), (41, 43), (59, 61), (71, 73)
In simple words: Pairs of prime numbers with a difference of 2.

Question Do the following: 5. Write all the prime numbers below 70.
Answer:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67
In simple words: Numbers having only two factors (1 and themselves) up to 67.

Question Do the following: 6. Find the smallest number when divided by 28, 40 and 44 leave a remainder 8 in each case.
Answer:
Find LCM of 28, 40, and 44:
\(28 = 2^2 \times 7\)
\(40 = 2^3 \times 5\)
\(44 = 2^2 \times 11\)
\(\text{LCM} = 2^3 \times 5 \times 7 \times 11 = 8 \times 385 = 3080\)
Required number = \(\text{LCM} + 8 = 3080 + 8 = 3088\)
In simple words: Find the LCM of the divisors and add the common remainder.

Exam Tip: Whenever a common remainder is required, find the LCM and add the remainder.

Question Do the following: 7. Write two prime numbers whose sum is 100.
Answer:
\(3 + 97 = 100\) (or \(11 + 89 = 100\), \(17 + 83 = 100\), etc.)
In simple words: Find pairs of primes that add up to 100.

Question Do the following: 8. Write three pairs of prime numbers whose sum is an odd number.
Answer:
(2, 7) sum = 9; (2, 11) sum = 13; (2, 13) sum = 15
In simple words: Since 2 is the only even prime, pairing 2 with any odd prime always results in an odd sum.

Question Do the following: 9. Find the smallest four digit number which is exactly divisible by 12, 16, 24 and 36.
Answer:
Find LCM of 12, 16, 24, 36:
\(12 = 2^2 \times 3\)
\(16 = 2^4\)
\(24 = 2^3 \times 3\)
\(36 = 2^2 \times 3^2\)
\(\text{LCM} = 2^4 \times 3^2 = 16 \times 9 = 144\)
Smallest 4-digit number = 1000
\(1000 \div 144 = 6\) with remainder 136
Next multiple = \(144 \times 7 = 1008\)
In simple words: Find the LCM of the numbers and take its smallest 4-digit multiple.

Question Do the following: 10. Write all the composite numbers between 30 and 50.
Answer:
32, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 49
In simple words: Non-prime numbers between 30 and 50.

Question Do the following: 11. The length, breadth and height of a room are 8m25cm, 6m75cm and 4m50cm respectively. Determine the longest tape which can measure the three dimension of the room exactly.
Answer:
Convert dimensions to cm:
Length = 825 cm
Breadth = 675 cm
Height = 450 cm
Find HCF of 825, 675, and 450:
\(825 = 25 \times 33 = 3 \times 5^2 \times 11\)
\(675 = 25 \times 27 = 3^3 \times 5^2\)
\(450 = 25 \times 18 = 2 \times 3^2 \times 5^2\)
\(\text{HCF} = 3^1 \times 5^2 = 3 \times 25 = 75 \text{ cm}\)
In simple words: The longest tape is the HCF of the three dimensions.

Exam Tip: HCF is used when finding maximum size or length to measure multiple quantities exactly.

Question Do the following: 12. Telegraph pole occurs at equal distances of 220m along a road and heaps of stones are put at equal distances of 300m along the same road. The first heap is at the foot of the first pole. How far from it along the road is the next heap which lies at the foot of a pole.
Answer:
Find LCM of 220 and 300:
\(220 = 2^2 \times 5 \times 11\)
\(300 = 2^2 \times 3 \times 5^2\)
\(\text{LCM} = 2^2 \times 3 \times 5^2 \times 11 = 4 \times 3 \times 25 \times 11 = 3300 \text{ m}\)
In simple words: The next meeting point occurs at the LCM of their interval distances, which is 3300 m.

Basic Geometrical Ideas

Question I) (a) Name all the different angles shown in the figures : (b) Count the number of angles. (c) List the acute angles (d) List the obtuse angles (e) Identify the right angles and straight angles.
Answer:
Based on the provided geometric figure sets (PQR triangle, quadrilateral ABCD, intersecting lines at O, and circle sectors):
a) Angles formed by intersecting line segments and vertices.
b) Count varies by figure (e.g., triangle has 3 interior angles, quadrilateral has 4).
c) Acute angles measure less than \(90^{\circ}\).
d) Obtuse angles measure between \(90^{\circ}\) and \(180^{\circ}\).
e) Right angles are \(90^{\circ}\); straight angles are \(180^{\circ}\).
In simple words: Categorize each marked angle by measuring its opening relative to a right angle.

Question 2. List all the points which are in the exterior and interior of the given angle < 1
Answer:
Interior points lie inside the two arms of the angle; exterior points lie outside the region enclosed by the arms.
In simple words: Interior is inside the V-shape of the angle, exterior is outside.

Question 3. Draw a rough sketch of a quadrilateral PQRS state a) two pairs of opposite sides. b) two pairs of opposite angles. c) four pairs of adjacent angles d) four pairs of adjacent sides. e) Draw the diagonals and name them.
Answer:
a) Opposite sides: PQ & RS, PS & QR
b) Opposite angles: \(\angle P\) & \(\angle R\), \(\angle Q\) & \(\angle S\)
c) Adjacent angles: (\(\angle P, \angle Q\)), (\(\angle Q, \angle R\)), (\(\angle R, \angle S\)), (\(\angle S, \angle P\))
d) Adjacent sides: (PQ, QR), (QR, RS), (RS, SP), (SP, PQ)
e) Diagonals: PR and QS
In simple words: Opposite elements face each other; adjacent elements share a common vertex or side.

Question 4. Name the interior and exterior points of the angle marked \(x^{\circ}\) in the following:
Answer:
Interior points lie within the opening of angle \(x^{\circ}\); exterior points lie outside this angular region.
In simple words: Points inside the angular sweep are interior, others are exterior.

Question 5. a) Identify the triangles in the figure : b) Write the names of angles c) Write the names of line segments.
Answer:
a) Identify individual triangles formed by intersecting lines in the given diagram.
b) List all 3-letter angle names (e.g., \(\angle EFO\)).
c) List all line segments composing the polygon boundaries.
In simple words: Break complex geometric diagrams into individual triangles and line segments.

Question 6. Draw any circle and mark (a) its centre (b) three radii (c) a diameter (d) shade a minor sector (e) colour a minor arc (f) a chord (g) two points in its interior (h) two points in its exterior (j) three points on the circle
Answer:
(a) Center O
(b) Radii OA, OB, OC
(c) Diameter AB passing through center
(d) Region enclosed by two radii and an arc (minor sector shaded)
(e) Portion of circumference (minor arc colored)
(f) Line segment joining two points on the circle (chord)
(g) Two interior points inside the circle
(h) Two exterior points outside the circle
(j) Three points lying directly on the circumference
In simple words: Standard circle parts defined relative to center, radius, and circumference.

Integers

Question I) 1. -5 + (-11) = ______________
Answer: -16
In simple words: Adding two negative numbers results in a larger negative number.

Question I) 2. 8 + (-6) = ____________
Answer: 2
In simple words: Subtract the smaller absolute value from the larger and keep the sign of the larger number.

Question I) 3. (-26) + (-37) = ___________
Answer: -63
In simple words: Sum the magnitudes and apply the negative sign.

Question I) 4. Write the greatest negative integer _____________
Answer: -1
In simple words: -1 is closest to zero on the negative side, making it the largest negative integer.

Question I) 5. Write all integers between - 30 and -20 ___________
Answer: -29, -28, -27, -26, -25, -24, -23, -22, -21
In simple words: List all whole integer steps strictly between -30 and -20.

Question I) 6. Find the sum of - 45 and 30
Answer:
\(-45 + 30 = -15\)
In simple words: Subtract 30 from 45 and keep the negative sign of the larger number.

Question I) 7. Which is greater : -65 or -56 ?
Answer: -56
In simple words: Negative numbers closer to zero are greater.

Question I) 8. Which integer is neither positive nor negative ?
Answer: 0 (zero)
In simple words: Zero is neutral and has no sign.

Question II) 1. Which number will we reach if we move 4 numbers to the right of -2 ?
Answer: 2
In simple words: Moving right on the number line means adding: \(-2 + 4 = 2\).

Question II) 2. If we are at -6 on the number line, in which direction should we move to reach -1 ?
Answer: Right direction
In simple words: Moving from -6 to -1 increases the value, which is to the right.

Question II) 3. Using the number line write the integer a) 4 less than -1 b) 6 more than -6
Answer:
a) \(-1 - 4 = -5\)
b) \(-6 + 6 = 0\)
In simple words: 'Less than' means move left, 'more than' means move right.

Question II) 4. Use number line and add the following integers : a) \((-1) + (-8)\) b) \((-1) + (-2) + (-4)\) c) \(-8 - (-10)\)
Answer:
a) \(-1 - 8 = -9\)
b) \(-1 - 2 - 4 = -7\)
c) \(-8 + 10 = 2\) (subtracting a negative is adding its positive counterpart)
In simple words: Combine negatives by adding their values; subtracting a negative is equivalent to addition.

Question II) 5. Fill in the blanks with >, < or =
a) 54 - (-11) _________ 57 + (-4)
b) (-35) - (-52) _________ (-52) - (-35)
Answer:
a) Left side = \(54 + 11 = 65\); Right side = \(57 - 4 = 53\). Thus, \(65 > 53\) (\(>\)).
b) Left side = \(-35 + 52 = 17\); Right side = \(-52 + 35 = -17\). Thus, \(17 > -17\) (\(>\)).
In simple words: Simplify both sides of the expression first, then compare their values.

Question II) 6. Find : i) 60 - (-20) - (+10) ii) (-15) + 12 - 9 + 1
Answer:
i) \(60 + 20 - 10 = 80 - 10 = 70\)
ii) \(-15 + 12 - 9 + 1 = -3 - 9 + 1 = -12 + 1 = -11\)
In simple words: Remove parentheses by applying sign rules, then compute step-by-step from left to right.

Fractions

Question 1. What fraction of an hour is 30 minutes ?
Answer:
\(\frac{30 \text{ minutes}}{60 \text{ minutes}} = \frac{1}{2}\)
In simple words: 30 minutes is half of a 60-minute hour.

Question 2. Write the natural numbers from 5 to 15. What fraction of them are prime numbers?
Answer:
Natural numbers from 5 to 15: 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15 (Total = 11 numbers)
Prime numbers among them: 5, 7, 11, 13 (Total = 4 primes)
Fraction = \(\frac{4}{11}\)
In simple words: Count total numbers and prime numbers, then form the ratio fraction.

Question 3. Express as mixed fractions. a) \(\frac{30}{7}\) b) \(\frac{51}{9}\)
Answer:
a) \(\frac{30}{7} = 4\frac{2}{7}\) (since \(7 \times 4 = 28, 30 - 28 = 2\))
b) \(\frac{51}{9} = 5\frac{6}{9} = 5\frac{2}{3}\) (since \(9 \times 5 = 45, 51 - 45 = 6\))
In simple words: Divide numerator by denominator to get the whole number quotient and remainder numerator.

Question 4. Express as improper fractions. a) \(7\frac{3}{9}\) b) \(11\frac{1}{13}\)
Answer:
a) \(7\frac{3}{9} = \frac{7 \times 9 + 3}{9} = \frac{63 + 3}{9} = \frac{66}{9} = \frac{22}{3}\)
b) \(11\frac{1}{13} = \frac{11 \times 13 + 1}{13} = \frac{143 + 1}{13} = \frac{144}{13}\)
In simple words: Multiply whole number by denominator and add numerator.

Question 5. Find as equivalent fraction of \(\frac{56}{72}\) with denominator 18.
Answer:
\(\frac{56}{72} = \frac{56 \div 4}{72 \div 4} = \frac{14}{18}\)
In simple words: Divide both numerator and denominator by 4 so the denominator becomes 18.

Question 6. Simplify:
i) \(\frac{21}{35}\) ii) \(\frac{34}{85}\) iii) \(\frac{75}{120}\) iv) \(\frac{66}{75}\) v) \(\frac{112}{128}\) vi) \(\frac{48}{57}\)
Answer:
i) \(\frac{21}{35} = \frac{3}{5}\) (dividing by 7)
ii) \(\frac{34}{85} = \frac{2}{5}\) (dividing by 17)
iii) \(\frac{75}{120} = \frac{5}{8}\) (dividing by 15)
iv) \(\frac{66}{75} = \frac{22}{25}\) (dividing by 3)
v) \(\frac{112}{128} = \frac{7}{8}\) (dividing by 16)
vi) \(\frac{48}{57} = \frac{16}{19}\) (dividing by 3)
In simple words: Divide both numerator and denominator by their HCF to reduce fractions to lowest terms.

Question 7. Add:
a) \(\frac{5}{12}\) and \(\frac{19}{24}\)
b) \(\frac{3}{8}, \frac{1}{2},\) and \(\frac{5}{6}\)
c) \(2\frac{5}{6}\) and \(8\frac{1}{4}\)
Answer:
a) LCM of 12 and 24 is 24. \(\frac{5 \times 2}{24} + \frac{19}{24} = \frac{10 + 19}{24} = \frac{29}{24} = 1\frac{5}{24}\)
b) LCM of 8, 2, 6 is 24. \(\frac{9}{24} + \frac{12}{24} + \frac{20}{24} = \frac{9 + 12 + 20}{24} = \frac{41}{24} = 1\frac{17}{24}\)
c) Convert to improper: \(\frac{17}{6} + \frac{33}{4}\). LCM of 6 and 4 is 12. \(\frac{34}{12} + \frac{99}{12} = \frac{133}{12} = 11\frac{1}{12}\)
In simple words: Find a common denominator (LCM) before adding fraction numerators.

Question 8. The weight of two boxes together is \(5\frac{3}{4} \text{ kg}\). If one box weighs \(2\frac{5}{6} \text{ kg}\), find the weight of the other.
Answer:
Total weight = \(\frac{23}{4} \text{ kg}\)
First box weight = \(\frac{17}{6} \text{ kg}\)
Weight of other box = \(\frac{23}{4} - \frac{17}{6}\)
LCM of 4 and 6 is 12. \(\frac{69}{12} - \frac{34}{12} = \frac{35}{12} = 2\frac{11}{12} \text{ kg}\)
In simple words: Subtract the weight of the known box from the total combined weight.

Mensuration

Question 1. The length and breadth of a Rectangle are 11 cm and 9 cm. Find its area and Perimeter.
Answer:
Perimeter = \(2 \times (11 + 9) = 2 \times 20 = 40 \text{ cm}\)
Area = \(11 \times 9 = 99 \text{ cm}^2\)
In simple words: Perimeter is \(2(l + b)\) and Area is \(l \times b\).

Question 2. The area of a Rectangle is 144 sq cm and its length is 16 cm. Find the breadth of the Rectangle?
Answer:
Breadth = \(\frac{\text{Area}}{\text{Length}} = \frac{144}{16} = 9 \text{ cm}\)
In simple words: Divide total area by length to find breadth.

Question 3. Find the perimeter of a regular hexagon of side x cm.
Answer:
Perimeter = \(6 \times x = 6x \text{ cm}\)
In simple words: Multiply the side length by 6 since a regular hexagon has 6 equal sides.

Question 4. If the perimeter of a regular pentagon is 65 cm. Find its side?
Answer:
Side = \(\frac{\text{Perimeter}}{5} = \frac{65}{5} = 13 \text{ cm}\)
In simple words: Divide total perimeter by 5 for a regular pentagon.

Question 5. The perimeter of a triangle is 42 cm. If two of its sides are 16 cm and 12 cm. Find its third side?
Answer:
Third side = Perimeter - (Side 1 + Side 2) = \(42 - (16 + 12) = 42 - 28 = 14 \text{ cm}\)
In simple words: Subtract the sum of the two given sides from the total perimeter.

Question 6. A piece of string is 45 cm long. It is bent to form an equilateral triangle. Find the side of triangle.
Answer:
Side = \(\frac{45}{3} = 15 \text{ cm}\)
In simple words: Divide the total string length by 3.

Question 7. An athlete takes 5 rounds of a rectangular park 120 m long and 80m wide. Find the total distance covered by him.
Answer:
Perimeter of park = \(2 \times (120 + 80) = 2 \times 200 = 400 \text{ m}\)
Total distance for 5 rounds = \(5 \times 400 = 2000 \text{ m} = 2 \text{ km}\)
In simple words: Multiply the perimeter by the number of rounds.

Question 8. Find the area of a square whose perimeter is 260 cm.
Answer:
Side of square = \(\frac{260}{4} = 65 \text{ cm}\)
Area = \(65 \times 65 = 4225 \text{ cm}^2\)
In simple words: Find side length by dividing perimeter by 4, then square it to find area.

Question 9. The total cost of fencing a square park at Rs. 20 per metre is Rs. 2880. Find the side of the square park.
Answer:
Perimeter = \(\frac{2880}{20} = 144 \text{ m}\)
Side = \(\frac{144}{4} = 36 \text{ m}\)
In simple words: Divide total cost by rate to get perimeter, then divide by 4 to get side length.

Question 10. The floor of a room with dimensions 5 m and 3m is to be covered with square tiles. If each square tile is of side 25 cm. Find the number of tiles required.
Answer:
Area of room = \(500 \text{ cm} \times 300 \text{ cm} = 1,50,000 \text{ cm}^2\)
Area of one tile = \(25 \text{ cm} \times 25 \text{ cm} = 625 \text{ cm}^2\)
Number of tiles = \(\frac{150000}{625} = 240 \text{ tiles}\)
In simple words: Divide the total room area by the area of a single tile after converting units to cm.

Practical Geometry

Question 1. Draw a circle of radius 5.6cm
Answer:
1. Set compass width to 5.6 cm using a ruler.
2. Fix the needle at center point O and draw a complete rotation.
In simple words: Open compass to 5.6 cm and draw the circle.

Question 2. With the same centre O, draw two circles of radii 5cm and 2.5cm
Answer:
1. Fix center point O.
2. Draw a circle of radius 5 cm.
3. From the same center O, draw a concentric circle of radius 2.5 cm.
In simple words: Concentric circles share the exact same center point.

Question 3. Draw any circle and mark points P, Q and R such that a) P is on the circle b) Q is in the interior of the circle. c) R is in the exterior of the circle.
Answer:
1. Draw a circle with center O.
2. Place point P directly on the boundary line.
3. Place point Q anywhere inside the circle.
4. Place point R anywhere outside the circle boundary.
In simple words: Point P on boundary, Q inside, R outside.

Question 4. Draw any line segment PQ. Mark any point B on it. Through B, draw a perpendicular to PQ.
Answer:
1. Draw line segment PQ and mark point B on it.
2. With B as center, draw an arc cutting PQ at two points X and Y.
3. With X and Y as centers, draw intersecting arcs above point B.
4. Join the intersection point to B to form the perpendicular line.
In simple words: Use compass setup at point B to erect a \(90^{\circ}\) perpendicular line.

Question 5. Draw XY of length 8.3cm and find its axis of symmetry.
Answer:
1. Draw line segment XY = 8.3 cm.
2. Construct its perpendicular bisector using compasses; this line is its axis of symmetry.
In simple words: The perpendicular bisector of a line segment acts as its line of symmetry.

Question 6. Draw a line segment of length 10.5cm and construct its perpendicular bisector.
Answer:
1. Draw segment AB = 10.5 cm.
2. Open compass > half of 10.5 cm, draw arcs from A and B above and below the segment, and connect their intersections.
In simple words: Standard perpendicular bisector construction.

Question 7. With AB of length 6.2cm as diameter, draw a circle.
Answer:
1. Draw diameter AB = 6.2 cm.
2. Find midpoint O of AB.
3. With center O and radius 3.1 cm, draw the circle.
In simple words: Radius is half of diameter (\(6.2 \div 2 = 3.1 \text{ cm}\)).

Question 8. Draw a circle of radius 4.5cm. Draw any two of its chords. Construct the perpendicular bisectors of these chords. Where do they meet.
Answer:
1. Draw circle of radius 4.5 cm.
2. Draw two non-parallel chords.
3. Construct perpendicular bisectors for both chords.
4. They always meet at the center O of the circle.
In simple words: Perpendicular bisectors of any two chords of a circle always intersect at the circle's center.

Question 9. Draw a line segment of length 10.8cm. Using compasses, divide it into four equal parts. Verify by actual measurement.
Answer:
1. Draw segment of 10.8 cm.
2. Construct its perpendicular bisector to divide it into 2 equal parts (5.4 cm each).
3. Construct perpendicular bisectors of each half to get 4 equal parts (2.7 cm each).
In simple words: Repeatedly bisect the segments to divide length into 4 equal parts.

Question 10. Draw the perpendicular bisector of AB whose length is 8.3cm a) Take any point P on the bisector drawn. Examine whether PA = PB b) If M is the mid point of AB, what can you say about the length of MA and MB?
Answer:
a) Yes, PA = PB because any point on the perpendicular bisector is equidistant from the endpoints.
b) MA = MB = \(4.15 \text{ cm}\) (half of 8.3 cm).
In simple words: Points on a perpendicular bisector are always equidistant from both segment endpoints.

Question 11. Draw an angle of measure \(137^{\circ}\) and construct its bisector.
Answer:
1. Draw \(137^{\circ}\) angle using a protractor.
2. Use compass arc and intersection steps to construct its angle bisector dividing it into two \(68.5^{\circ}\) angles.
In simple words: Measure with protractor, then bisect using compasses.

Question 12. Draw a right angle and construct its bisector.
Answer:
1. Construct \(90^{\circ}\).
2. Bisect it to obtain two \(45^{\circ}\) angles.
In simple words: Bisecting a right angle gives \(45^{\circ}\).

Question 13. Draw an angle of measure \(152^{\circ}\) and divide into four equal parts
Answer:
1. Draw \(152^{\circ}\) angle.
2. Bisect it to get two \(76^{\circ}\) angles.
3. Bisect each \(76^{\circ}\) angle to get four \(38^{\circ}\) parts.
In simple words: Bisect the angle twice to divide it into four equal parts.

Question 14. Draw an angle of measure \(60^{\circ}\) and bisect it.
Answer:
1. Construct \(60^{\circ}\).
2. Bisect it to obtain two \(30^{\circ}\) angles.
In simple words: Compass bisector of \(60^{\circ}\) yields \(30^{\circ}\).

Question 15. Draw an angle of measure \(150^{\circ}\) and bisect it.
Answer:
1. Draw \(150^{\circ}\) angle.
2. Construct its angle bisector to divide it into two \(75^{\circ}\) parts.
In simple words: Bisecting \(150^{\circ}\) gives \(75^{\circ}\).

Question 16. Construct with ruler and compasses, angles of following measures. a) \(60^{\circ}\) b) \(120^{\circ}\) c) \(90^{\circ}\) d) \(45^{\circ}\) e) \(15^{\circ}\) f) \(30^{\circ}\)
Answer:
a) \(60^{\circ}\): Single arc cut from baseline.
b) \(120^{\circ}\): Double arc cut along baseline.
c) \(90^{\circ}\): Perpendicular bisector or intersection of \(60^{\circ}\) and \(120^{\circ}\).
d) \(45^{\circ}\): Bisect \(90^{\circ}\).
e) \(15^{\circ}\): Bisect \(30^{\circ}\).
f) \(30^{\circ}\): Bisect \(60^{\circ}\).
In simple words: Standard compass angle constructions using bisection.

Question 17. Draw a circle and any two of its diameter. What is the figure obtained? What figure is obtained if the diameter are perpendicular to each other?
Answer:
1. Joining the endpoints of two diameters forms a rectangle (or parallelogram).
2. If the two diameters are perpendicular to each other, joining their endpoints forms a square.
In simple words: Perpendicular diameters create a square when their endpoints are connected.

Question 1) Can you draw a triangle which has a) exactly one line of symmetry b) exactly two lines of symmetry? c) exactly three lines of symmetry? d) no lines of symmetry?
Answer:
a) Yes, an isosceles triangle has exactly one line of symmetry.
b) No, a triangle cannot have exactly two lines of symmetry.
c) Yes, an equilateral triangle has exactly three lines of symmetry.
d) Yes, a scalene triangle has no lines of symmetry.
In simple words: Isosceles has 1, equilateral has 3, scalene has 0, and triangles cannot have 2 lines of symmetry.

Question 2) Find the number of lines of symmetry in each of the following shapes? (referencing visual shapes on worksheet page 51)
Answer:
Symmetry lines vary by each listed symmetrical pattern (ranging from 1 to 4 or infinite axes depending on rotational and mirror design).
In simple words: Count how many folding lines divide each shape into identical halves.

Question 3) Consider the letters of English alphabets, A to Z. List among them the letters which have a) Vertical lines of symmetry b) Horizontal lines of symmetry c) No lines of symmetry
Answer:
a) Vertical lines of symmetry: A, H, I, M, O, T, U, V, W, X, Y
b) Horizontal lines of symmetry: B, C, D, E, H, I, K, O, X
c) No lines of symmetry: F, G, J, L, N, P, Q, R, S, Z
In simple words: Check capital letters for vertical mirror splits, horizontal mirror splits, or none.

Question 4) Complete the following table:
Sl# | Shape | Rough Figure | Number of lines of symmetry
1. | Equilateral triangle | ... | 3
2. | Square | ... | 4
3. | Rectangle | ... | 2
4. | Isosceles Triangle | ... | 1
5. | Rhombus | ... | 2
6. | Circles | ... | Infinite
7. | Parallelogram | ... | 0
8. | Scalene Triangle | ... | 0
Answer:
1. Equilateral triangle: 3 lines of symmetry
2. Square: 4 lines of symmetry
3. Rectangle: 2 lines of symmetry
4. Isosceles Triangle: 1 line of symmetry
5. Rhombus: 2 lines of symmetry
6. Circles: Infinite lines of symmetry
7. Parallelogram: 0 lines of symmetry
8. Scalene Triangle: 0 lines of symmetry
In simple words: Standard table mapping geometric shapes to their symmetry line counts.

Question 5) Write some application of symmetry in everyday life.
Answer:
1. Architecture and bridge design for structural balance.
2. Art, rangoli, and clothing patterns for aesthetic appeal.
3. Automobile design for aerodynamics and balance.
In simple words: Symmetry provides balance, beauty, and structural stability in design.

Q.1 write down the following in the exponential form:--
a) 3×a×a×b×b
b) 5×y×y×z×z×z

Q.2. Add the following :--
a) 15x and 6x
b) 4abc,-6abc,7 abc & 2 abc
c) 5xy,12xy &-3xy

Q.3 Evaluate the following if x=2,y=3&z=1
a) 15xy+3zx
b) 9xyz+11xy--3yz
c) 16xyz

Q.4 write the following by using literals and numbers:--
I) Difference of x cubed and y square.
Ii) Sum of four times of x and three times of y
III) X is decreased from 11.

IV) 6 times of y is subtracted from 3 times of x

Please refer to attached file for CBSE Class 6 Maths Practice Worksheets (2)