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Practical Geometry Class 6 Extra Questions and Answers CBSE

September 22, 2024

Extra Questions for Class 6 Maths revision becomes an easy task during the annual exam. Students who find it challenging to understand the basics of the chapter must practice all the essential questions to understand better. These crucial questions and answers on Knowing Our Numbers Class 6 Maths can aid the students’ preparation through the concepts of this chapter 1. https://meritbatch.com/knowing-our-numbers-class-6-extra-questions/

NCERT Class 6 Maths Chapter 14 Extra Questions and Answers

Extra Questions for Class 6 Maths Chapter 14 Practical Geometry

Practical Geometry Class 6 Extra Questions Very Short Answer Type

Practical Geometry Class 6 Extra Questions Question 1.
If AB = 3.6 and CD = 1.6 cm, construct a line segment equal to \(\overline { AB }\) + \(\overline { CD }\) and measure the total length.
Class 6 Maths Chapter 14 Extra Questions
Solution:
Step I: Draw a ray OX.
Step II : With centre 0 and radius equal to the length of AB (3.6 cm) mark a point P on the ray.

Step III: With centre P and radius equal to the length of CD (1.6 cm) mark another point Q on the ray.
Thus OQ is the required segment such that OQ = 3.6 cm + 1.6 cm = 5.2 cm.

Class 6 Practical Geometry Extra Questions Question 2.
Construct a perpendicular to a given line segment at point on it.
Solution:
Step IDraw a line \(\overleftrightarrow { PQ }\) and take any point A on it.
Class 6 Maths Ch 14 Extra Questions
Step II : With centre A draw an arc which meets PQ at C and D.
Step III : Join AB and produce.
Step IV: With centres C and D and radius equal to half of the length of the previous arc, draw two arcs which meets each other at B.
Thus AB is the required perpendicular to \(\overleftrightarrow { PQ }\).

Question 3.
Construct an angle of 60° and bisect it.
Solution:
Step I: Draw a line segment \(\overline { AB }\).
Class 6 Maths Practical Geometry Extra Questions
Step II: With centre B and proper radius, draw an arc which meets AB at C.
Step III : With C as centre and the same radius as in step II, draw an arc cutting the previous arc at D.
Step IV : Join B to D and produce.
Step V : Draw the bisector BE of ∠ABD.
Thus BE is the required bisector of ∠ABD.

Practical Geometry Class 6 Worksheets With Answers Question 4.
Draw an angle of 120° and hence construct an angle of 105°.
Solution:
Step I : Draw a line segment \(\overline { OA }\).
Questions On Practical Geometry For Class 6
Step II : With centre O and proper radius, draw an arc which meets OA at C.
Step III : With centre C and radius same, mark D and E on the previous arc.
Step IV : Join O to E and produce.
Step V : ∠EOA is the required angle of 120°.
Step VI : Construct an angle of 90° which meets the previous arc at F.
Step VII : With centre E and F and proper radius, draw two arcs which meet each other at G.
Step VIII : Join OG and produce.
Thus ∠GOA is the required angle of 105°.

Practical Geometry Class 6 Questions And Answers Question 5.
Using compasses and ruler, draw an angle of
75° and hence construct an angle of 37 \(\frac { { 1 }^{ \circ } }{ 2 }\).
Solution:
Step I: Draw a line segment OA.
Step II : Construct ∠BOA = 90° and ∠EOA = 60°
Step III : Draw OC as the bisector of ∠BOE , which equal to

Step IV : Draw the bisector OD of ∠COA.
Extra Questions For Class 6 Maths Practical Geometry
Thus ∠DOA is the required angle of 37 \(\frac { { 1 }^{ \circ } }{ 2 }\) .

Extra Questions On Practical Geometry For Class 6 Question 6.
Draw ∆ABC. Draw perpendiculars from A, B and C respectively on the sides BC, CA and AB. Are there perpendicular concurrent? (passing through the same points).
Solution:
Step I: Draw any ∆ABC.
Step II : Draw the perpendicular AD from A to BC.
Practical Geometry Class 6 Worksheet With Answers
Step III : Draw the perpendicular BE from B to AC.
Step IV : Draw the perpendicular CF from C to AB.
We observe that the perpendiculars AD, BE and CF intersect each other at P.
Thus, P is the point of intersection of the three perpendiculars.

Class 6 Maths Extra Questions

Decimals Class 6 Extra Questions and Answers CBSE

September 22, 2024

NCERT Class 6 Maths Chapter 8 Extra Questions and Answers

Extra Questions for Class 6 Maths Chapter 8 Decimals

Decimals Class 6 Extra Questions Very Short Answer Type

Decimal Questions For Class 6 Question 1.
Write the following in decimals.
5 hundreds 3 tens 8 ones 4 tenths
Solution:
5 Hundreds + 3 Tens + 8 Ones + 4 Tenths
= 5 x 100 + 3 x 10 + 8 x 1 + 4 x \(\frac { 1 }{ 10 }\)
= 500 + 30 + 8 + \(\frac { 4 }{ 10 }\)
= 538 + \(\frac { 4 }{ 10 }\) = 538.4

Decimals Class 6 Extra Questions Question 2.
Write 14.3 in place value table.
Solution:

Hundreds	Tens	Ones	Tenths
0	1	4	3

Class 6 Decimals Extra Questions Question 3.
Write the following in decimals.

Solution:

Questions On Decimals For Class 6 Question 4.
Write the following as decimals.
(a) Three tens and eight-tenths
(b) Fifteen point seven
Solution:
(a) Three tens and eight-tenths = 3 x 10 + 8 x = 30 + 0.8 = 30.8
(b) Fifteen point seven = 15.7

Decimals Class 6 Worksheet Question 5.
Write the following as decimals.

Solution:
(a) 200 + 50 + 6 + \(\frac { 2 }{ 10 }\) = 256 + 0.2 = 256.2
(b) 150 + 30 + \(\frac { 8 }{ 10 }\) = 180 + 0.8 = 180.8

Class 6 Maths Decimals Extra Questions Question 6.
Write in decimals:

Solution:

Decimals Questions For Class 6 Question 7.
Write as fractions in lowest form.
(a) 0.05
(b) 20.25
Solution:
Decimals Class 6 Extra Questions Maths Chapter 8

Question 8.
Write the following in decimals.
Three hundred six and seven hundredths.
Solution:
Three hundred six and seven hundredths
Decimals Class 6 Extra Questions Maths Chapter 8

Question 9.
Which is greater 0.4 or 0.5?
Solution:
0.4 = 4 and 0.5 = ±
Here \(\frac { 5 }{ 10 }\) > \(\frac { 4 }{ 10 }\)
∴ o.5 > 0.4

Question 10.
Express the following as rupees using decimals:
(a) 7 paise
(b) 625 paise
Solution:
Decimals Class 6 Extra Questions Maths Chapter 8

Decimals Class 6 Extra Questions Short Answer Type

Question 11.
Represent 1.3,3.8 and 4.1 on the number line.
Solution:

A represents 1.3
B represents 3.8
and C represents 4.1

Question 12.
Read the numbers from the place value table and write them in decimals;
Decimals Class 6 Extra Questions Maths Chapter 8
Solution:
(a) 4 Thousands + 0 Hundreds + 2 Tens + 3 Ones + 6 Tenths + 0 Hundredths
= 4 x 1000 + 0 x 100 + 2 x 10 + 3 x 1 + 6 x \(\frac { 1 }{ 10 }\) + 0 x \(\frac { 1 }{ 100 }\)
10 100
= 4000 + 0 + 20 + 3 + 0.6 + 0 = 4023.6

(b) 2 Thousands + 8 Hundreds + 8 Tens + 0 Ones + 3 Tenths + 4 Hundredths
= 2 x 1000 + 8 x 100 + 8 x 10 + 0 x 1 + 3 x \(\frac { 1 }{ 10 }\) + 4 x \(\frac { 1 }{ 100 }\)
= 2000 + 800 + 80 + 0.3 + 0,04
= 2880 + 0.34 = 2880.34

(c) 6 Thousands + 4 Hundreds + 2 Tens + 8 Ones + 4 Tenths + 3 Hundredths
= 6 x 1000 + 4 x 100 + 2 x 10 + 8 x 1 + \(\frac { 4 }{ 10 }\) + \(\frac { 3 }{ 100 }\)
= 6000 + 400 + 20 + 8 + 0.4 + 0.03
= 6428 + 0.43 = 6428.43

Question 13.
Write each of the following as decimals:
Decimals Class 6 Extra Questions Maths Chapter 8
Solution:

Question 14.
Convert the following unlike decimals into like decimals:

Solution:
LCM of 10, 50, 5 and 20 = 100
Decimals Class 6 Extra Questions Maths Chapter 8

Question 15.
Write the following decimals in their expanded form:
(a) 27.65
(b) 102.05
(c) 36.36
(d) 0.507
Solution:
(a) 27.65
Decimals Class 6 Extra Questions Maths Chapter 8
(b) 102.05

Question 16.
Write each of the decimals as a mixed fraction.
(a) 95.8
(b) 15.78
(c) 0.015
(d) 19.91
Solution:
Decimals Class 6 Extra Questions Maths Chapter 8

Question 17.
Express each of the following in terms of litres (L) using decimals:
(a) 625 mL
(b) 760 mL
(c) 11 L 125 mL
(d) 7 L 350 mL
Solution:
(a) We know that 1000 mL = 1 L
∴ 625mL = \(\frac { 625 }{ 1000 }\)L = 0.625 L
Thus, 625 mL = 0.625 L

(b) We know that 1000 mL = 1 L
∴ 760 mL = \(\frac { 760 }{ 1000 }\) L
= 0.760 L = 0.76 L
Thus, 760 mL = 0.76 L

(c) We know that 1000 mL = 1 L
∴ 11L 125L = 11 L + \(\frac { 125 }{ 1000 }\) L
= (11 +0.125) L = 11.125 L
Thus, 11 L 125 mL = 11.125 L

(d) We know that 1000 mL = 1 L
∴ 7 L 350 mL = 7 L + \(\frac { 350 }{ 1000 }\) L
= (7 + 0.350) L = 7.350 L = 7.35 L
Thus, 7L 350 mL = 7.35 L

Question 18.
Write 3.03, 2.75 and 2.5 in ascending order.
Solution:
The given decimals are unlike.
∴ Their corresponding like decimals are 3.03, 2.75 and 2.50.
Now neglecting the decimals, we have 303, 275 and 250.
Since, 303 > 275 > 250,
we have 3.03 > 2.75 > 2.50
∴ Ascending order is 2.50 < 2.75 < 3.03

Question 19.
Find the value of the following:
(а) 15 – 9.363
(B) 5.28 – 1.4 + 3.116
Solution:
Decimals Class 6 Extra Questions Maths Chapter 8 Q16
Hence, 5.28 – 1.4 + 3.116 = 6.996

Decimals Class 6 Extra Questions Higher Order Thinking Skills (HOTS)

Question 20.
Mr. Ranjan purchased 15.500 kg rice, 25.750 kg flour and 3.250 kg sugar. Find the total weight of his purchases. Is it 50 kg or less? If less, how much less?
Solution:
Weight of rice = 15.500 kg
Weight of flour = 25.750 kg
Weight of sugar = 3.250 kg
Total weight of this purchases = 15.500 kg + 25.750 kg + 3.250 kg
= 44.500 kg
We see that the total weight of his purchases is less than 50 kg.
Decimals Class 6 Extra Questions Maths Chapter 8
Thus, the total weight 44.500 kg is 5.500 kg less than 50 kg.

Question 21.
Ten years old Rahul can carry a maximum weight of 15 kg. If he bought 4 kg 900 g of apples, 2 kg 600 g of grapes and 5 kg 300 g of mangoes. Can he carry the total weight that he bought. If yes, then how much more weight he can carry with him?
Solution:
Weight of apples = 4 kg 900 g
Weight of grapes = 2 kg 600 g
Weight of mangoes = 5 kg 300 g
∴ Total weight of his purchases = 4 kg 900 g + 2 kg 600 g + 5 kg 300 g
= 4.900 kg + 2.600 kg + 5.300 kg
= 12.800 kg
But Rahul can carry a maximum weight of 15 kg.
Thus more weight that he can carry with him = 15 kg – 12.800 kg
Decimals Class 6 Extra Questions Maths Chapter 8
= 2.200 kg

Class 6 Maths Extra Questions

Understanding Elementary Shape Class 6 Extra Questions Maths Chapter 5

Understanding Elementary Shape Class 6 Extra Questions and Answers CBSE

September 22, 2024

NCERT Class 6 Maths Chapter 5 Extra Questions and Answers

Extra Questions for Class 6 Maths Chapter 5 Understanding Elementary Shape

Understanding Elementary Shape Class 6 Extra Questions Very Short Answer Type

Understanding Elementary Shapes Class 6 Extra Questions Question 1.
Which of the following line-segments is longer?

Solution:
By using divider, \(\overline { CD }\) seems to be longer than \(\overline { AB }\) .

Understanding Elementary Shapes Class 6 Worksheet Question 2.
How many line segments are used in making a triangle?

Solution:
Three line segments are used to make a triangle.

Understanding Elementary Shapes Class 6 Worksheet Pdf Question 3.
What is the measure of straight angle?
Solution:
The measure of straight angle is 180°.

Questions On Understanding Elementary Shapes For Class 6 Question 4.
What is complete angle?
Solution:
The angle for one revolution is called a complete angle.

Understanding Elementary Shapes Extra Questions Question 5.
Find the number of right angle turned through by the hour hand of a clock when it goes from 3 to 6.

Solution:
When the hour hand of a clock goes from 3 to 6, it turns through a right angle.

Chapter 5 Maths Class 6 Extra Questions Question 6.
Draw the rough sketch of the following:
(a) Acute angle
(b) Obtuse angle
(c) Reflex angle
Solution:

(a) Acute angle

(b) Obtuse angle

Class 6 Maths Chapter 5 Worksheet Question 7.
Identify the types of angle from the given figures:

Solution:
(a) Obtuse angle
(b) Acute angle
(c) Right angle
(d) Straight angle
(e) Reflex angle

Class 6 Maths Understanding Elementary Shapes Extra Questions Question 8.
What are the degree measures of the following angles?
(a) Right angle
(b) A complete angle
(c) Straight angle
Solution:
(a) Degree measure of a right angle is 90°.
(b) Degree measure of a complete angle is 360°.
(c) Degree measure of a straight angle is 180°.

Understanding Elementary Shapes Class 6 Questions Question 9.
What are the types of the given triangles on the basis of angles?

Solution:
(a) Acute angled triangle.
(b) Obtuse angled triangle.
(c) Right angled triangle.

Elementary Shapes Class 6 Worksheet Question 10.
What are the types of the following triangles on the basis of sides?

Solution:
(a) Scalene triangle.
(b) Equilateral triangle.
(c) Isosceles triangle.

Basic Geometrical Ideas Class 6 Extra Questions Short Answer Type

Question 11.
In the given figure, name the following angles as acute, obtuse, right, straight or reflex.
(a) ∠QOY
(b) ∠YOP
(c) ∠ROX
(d) ∠QOX
(e) ∠POQ

Solution:
(a) ∠QOY = acute angle.
(b) ∠YOP = obtuse angle.
(c) ∠ROX = right angle.
(d) ∠QOX = reflex angle.
(e) ∠POQ = straight angle.

Question 12.
In the given figure, find the measure of the angles marked with a, b, c, d, e and f.
Solution:
∠a = 180° -129° = 51°
∠b = 180° – (51° + 92°)
= 180° – 143° = 37°
∠c = 180° – 88° = 92°
∠d = 180° – 152° = 28°
∠e = 180° – 143° = 37°
∠f= 180° – (∠e + ∠d)
= 180° – (37° + 28°)
= 180°- 65°= 115°
∠g = 180° – ∠f = 180° – 115° = 65°

Question 13.
Classify the given triangles whose sides are indicated on them.

Solution:
(a) All sides are different. So, it is a scalene triangle.
(a) Lengths of two sides of the triangle are same. So, it is an isosceles triangle.
(b) All sides are unequal and one angle is right angle. So it is scalene right angled triangle.
(c) Two sides of this triangle are equal. So, it is an isosceles triangle.

Question 14.
Complete each of the following, so as to make a true statement:
(a) A ……….. is a rectangle with a pair of adjacent sides equal.
(b) A parallelogram with a pair of adjacent sides equal is called a ………. .
(c) A quadrilateral having exactly one pair of parallel sides is called a …….. .
(d) A quadrilateral having both pairs of opposite sides parallel, is called a ……… .
(e) A parallelogram whose each angle is a right angle is called a ………… .
Solution:
(a) Square
(b) Rhombus
(c) Trapezium
(d) Parallelogram
(e) Rectangle.

Basic Geometrical Ideas Class 6 Extra Questions Long Answer Type

Question 15.
Verify the ‘Euler’s formula’ V + F = E + 2 for the given figures.
(a) A triangular prism having 5 faces, 9 edges and 6 vertices.
(b) A rectangular prism with 6 faces, 12 edges and 8 vertices.
(c) A pentagonal prism with 7 faces, 15 edges and 10 vertices.
(d) A tetrahedron -with 4 faces, 6 edges and 4 vertices.
Solution:
(a) Here, F = 5, E = 9 and V = 6
∴ V + F = E + 2
⇒ 6 + 5 = 9 + 2
⇒ 11 = 11
Hence, verified.

(b) Here, F = 6, E = 12 and V = 8
∴ V + F = E + 2
⇒ 8 + 6 = 12 + 2
⇒ 14 = 14
Hence, verified.

(d) Here, F = 4, E = 6 and V = 4
∴ V + F = E + 2
⇒ 4 + 4 = 6 + 2
⇒ 8 = 8
Hence, verified.

Question 16.
Complete the given table for prisms:

Prism	Number Of faces	Number Of edges	Number Of vertices
Triangular	–	–	–
Quadrilateral	–	–	–
Pentagonal	–	–	–
Hexagonal	–	–	–
Octagonal	–	–	–
Decagonal	–	–	–

Solution:

Prism	Number Of faces	Number Of edges	Number Of vertices
Triangular	5	9	6
Quadrilateral	6	12	8
Pentagonal	7	15	10
Hexagonal	8	18	12
Octagonal	10	24	16
Decagonal	12	30	20

Basic Geometrical Ideas Class 6 Extra Questions Multiple Choice Type

Question 17.
Number of right angles turned by the hour hand of a clock when it goes from 3 to 6.
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
Correct option is (a).

Question 18.
A quadrilateral having a pair of unequal opposite sides is called
(a) Parallelogram
(b) Square
(c) Rectangle
(d) Trapezium
Solution:
The correct option is (d).

Basic Geometrical Ideas Class 6 Extra Questions Higher Order Thinking Skills (Hots)

Question 19.
In the given figure, find the values of x, y, z, s and m.

Solution:
Given that ∠A = 40°
(i) ∠DAB + ∠ABC = 180° (adjacent angles)
⇒ 40° + ∠ABC = 180°
⇒ ∠ABC = 180° – 40° = 140°
Hence, ∠x = 140°

(ii) ∠x + ∠y = 180° (adjacent angles)
⇒ 140° + ∠y = 180°
⇒ ∠y = 180° – 140° = 40°
Hence, ∠y = 40°

(iii) ∠y + ∠z – 180° (adjacent angles)
⇒ 40° + ∠z = 180°
⇒ ∠z = 180° – 40° = 140°
Hence, ∠z = 140°

(iv) ∠x + ∠s = 180° (straight angles)
⇒ 140° + ∠s = 180°
⇒ ∠s = 180° – 140° = 40°
Hence, ∠s = 40°

(v) ∠m + ∠z = 180° (straight angles)
⇒ ∠m + 140° = 180°
⇒ ∠m = 180° – 140° = 40°

Question 20.
Find the value of x from the given figure and hence find the measure of each angle of the triangle.

Solution:
(i) Sum of the three angles of a triangle = 180°
∴ 2x + 30° + 60° – x + 3x – 10° = 180°
⇒ (2x – x + 3x) + (30° + 60° – 10°) = 180°
⇒ 4x + 80° = 180°
⇒ 4x = 180° – 80°
⇒ 4x = 100°
∴ x = \(\frac { { 100 }^{ 0 } }{ 4 }\) = 25°
∴ Measure of the angles are:
(i) (2x + 30)° – 2 x 25° + 30° = 80°
(ii) (60 – x)° = 60° – 25° = 35°
(iii) (3x – 10)° = 3 x 25° – 10° = 75° – 10° = 65°
Hence, x = 25° and the angles of the triangles are: 80°, 35° and 65°.

Class 6 Maths Extra Questions

Basic Geometrical Ideas Class 6 Extra Questions and Answers CBSE

September 22, 2024

NCERT Class 6 Maths Chapter 4 Extra Questions and Answers

Extra Questions for Class 6 Maths Chapter 4 Basic Geometrical Ideas

Basic Geometrical Ideas Class 6 Extra Questions Very Short Answer Type

Class 6 Maths Chapter 4 Extra Questions Question 1.
Draw a rough sketch of:
(a) open curve
(b) closed curve
Solution:
Class 6 Basic Geometrical Ideas Extra Questions

Basic Geometrical Ideas Class 6 Worksheet With Answers Question 2.
Draw a rough sketch of closed curve made up of line segments.
Solution:
Required curve is ABCD closed with the line segments \(\overline { AB }\), \(\overline { BC }\), \(\overline { CD }\) and \(\overline { DA }\).

Basic Geometrical Ideas Class 6 Questions And Answers Question 3.
Draw two different angles having common point and a common arm.
Solution:
∠AOB and ∠COB are two different angles with common point O and common arm \(\overrightarrow { OB }\) .
IMG

Extra Questions On Basic Geometrical Ideas For Class 6 Question 4.
Identify the points which are:
(i) in the interior
(ii) in the exterior
(iii) on the closed curve in the given figure.
Chapter 4 Maths Class 6 Extra Questions
Solution:
(i) Points P, Q and R are in the interior of the closed curve.
(ii) points S and T are in the exterior of the closed curve.
(iii) U and V are on the closed curve.

Extra Questions For Class 6 Maths Chapter 4 Question 5.
Identify the following in the given figure:
(a) Sector
(b) Chord
(c) Diameter
(d) Segment.
Questions On Basic Geometrical Ideas For Class 6
Solution:
(a) OPR (shaded) is the sector of the circle.
(b) \(\overline { MN }\) is the chord.
(c) \(\overline { PQ }\) is the diameter.
(d) MXN (shaded) is the segment.

Ncert Class 6 Maths Chapter 4 Extra Questions Question 6.
In the given figure, name all the possible triangles.
Class 6 Geometry Questions
Solution:
Possible triangles are:
(i) ∆ABC
(ii) ∆ABD
(iii) ∆ABE
(iv) ∆ACD
(a) ∆ACE
(vi) ∆ADE

Basic Geometrical Ideas Class 6 Worksheet With Answers Pdf Question 7.
Name all the angles in the given figure.
Solution:
In the given figure, the names of all the angles are:

(i) ∠ABC
(ii) ∠BCD
(iii) ∠CDA
(iv) ∠DAB

Basic Geometrical Ideas Extra Questions Question 8.
In the given figure, name all the line segments:
Class 6 Chapter 4 Maths Extra Questions
Solution:
In the given figure, the name of the line segments are:
\(\overline { AB}\), \(\overline { BC }\), \(\overline { CD }\), \(\overline { DE }\), \(\overline { EA }\), \(\overline { DA }\), \(\overline { DB }\) and \(\overline { EC }\).

Basic Geometrical Ideas Class 6 Extra Questions Short Answer Type

Cbse Class 6 Maths Chapter 4 Extra Questions Question 9.
Using the given figure, name the following:
Class 6 Maths Chapter 4 Important Questions
(a) Line containing point M.
(b) Line passing through four points.
(c) Line passing through three points.
(d) Two pairs of intersecting lines.
Solution:
(a) \(\overleftrightarrow { MC }\) is the line containing the point M.
(b) \(\overleftrightarrow { AN }\) is the line passing through four points A, B, C and N.
(c) \(\overleftrightarrow { PQ }\) is the line passing through three points P, B and Q.
(d) Pairs for intersecting lines are
(i) \(\overleftrightarrow { AN }\) and \(\overleftrightarrow { PQ }\)
(ii) \(\overleftrightarrow { AN }\) and \(\overleftrightarrow { MC }\)

Basic Geometrical Concepts Class 6 Worksheet Question 10.
On the given line, some points are given, write down the names of all segments.

Solution:
Segments are:
\(\overline { PQ }\) , \(\overline { PR }\) , \(\overline { PS }\) , \(\overline { PT }\) , \(\overline { QR }\) , \(\overline { QS }\) , \(\overline { QT }\) , \(\overline { RS }\) , \(\overline { RT }\) , \(\overline { ST }\) .

Class 6 Maths Chapter 4 Worksheet Question 11.
How many lines can pass through
(i) one given point?
(ii) two given points?
(iii) three non-collinear points
Solution:
(i) Infinite number of lines can be passed through one given point.
(ii) Only one line can pass through two given points.
(iii) Three lines can pass through three non- collinear points.
Class 6 Maths Ch 4 Extra Questions

Basic Geometry Questions For Class 6 Question 12.
Look at the given figure and answer the following:
Basic Geometrical Ideas Class 6 Extra Questions Very Short Answer Type
(a) Name the sides of the polygon ABCDEF.
(b) Name any two pairs of adjacent sides. .
(c) Name all the segments which intersect each other at one point.
(d) Name all the diagonals of the given polygon.
Solution:
(a) The sides of the polygon are:
\(\overline { AB }\), \(\overline { BC }\), \(\overline { CD }\), \(\overline { DE }\), \(\overline { EF }\) and \(\overline { FA }\).
(b) \(\overline { AB }\) and \(\overline { BC }\), \(\overline { BC }\) and \(\overline { CD }\)are the pairs of adjacent sides.
(c) \(\overline { AD }\), \(\overline { BE }\) and \(\overline { CF }\) intersect each other at O.
(d) Name of the diagonals are:
\(\overline { AD }\), \(\overline { BE }\) and \(\overline { CF }\).

Question 13.
Fill in the blanks.
(a) ……… is the largest chord of a circle.
(b) ……… divides the circle into two equal semi circles.
(c) Rectangle is a ……… curve.
(d) Triangle has ……… angles and three ……… .
(e) Only one line can be drawn through ……… points.
(f) ……… number of lines can pass through a given point.
(g) A closed figure made up of entirely of line segments is called a ……… .
(h) A curve which does not cross itself is called a ……… curve.
(i) The length of boundary of a circle is called its ……… .
Solution:
(a) Diameter
(b) Diameter
(c) Closed
(d) three, sides
(e) two
(f) Infinite
(g) polygon
(h) simple
(i) circumference

Basic Geometrical Ideas Class 6 Extra Questions Long Answer Type

Question 14.
Draw the medians of a AABC and answer the following:
(а) Name the three medians.
(b) Do the medians intersect each other at the same point?
(c) What is that point called?
(d) Can this point be outside the triangle?
Basic Geometrical Ideas Class 6 Extra Questions Very Short Answer Type
Solution:
(a) Names of the medians are \(\overline { AD }\), \(\overline { BE }\) and \(\overline { CF }\) .
(b) Yes, the medians intersect each other at the same point G.
(c) The point of intersection of the medians of a triangle is called ‘Centroid’.
(d) No, this point cannot be out of the triangle.

Basic Geometrical Ideas Class 6 Extra Questions Multiple Choice Type

Question 15.
Least number of line segments required to make a polygon is
(a) 1
(b) 2
(c) 3
(c) 4
Solution:
Correct option is (c).

Question 16.
How many lines can be drawn through given two points?
(a) Only one
(b) 2
(c) 4
(d) Countless
Solution:
Correct option is (a).

Question 17.
How many vertices are there in a triangle?
(a) 1
(b) 2
(c) 3
(c) 4
Solution:
Correct option is (c).

Question 18.
Say ‘true’ or ‘False’.
(а) Two diameters of a circle will necessarily intersect.
(b) The centre of a circle always lies in the interior.
(c) The diameter is half of the radius of a circle.
(d) Longer chord is nearer to the centre of the circle.
Solution:
(a) True
(b) True
(c) False
(d) True

Basic Geometrical Ideas Class 6 Extra Questions Higher Order Thinking Skills (HOTS)

Question 19.
Draw an equilateral ∆ABC of any size. Draw AD as its median and an altitude AM.
(i) Does AD coincide with AM?
(ii) Name the point on the median which divides it in the ratio 1:2.
(iii) What is the measure of ∠ADC and ∠ADB?
(iv) Are D and M the same points?
Solution:
(i) Yes, AD coincides with AM.
Basic Geometrical Ideas Class 6 Extra Questions Very Short Answer Type
(ii) The point on the median which divides it in the ratio 1 : 2 is called centroid of the triangle.
(iii) ∠ADC = ∠ADB = 90°
(iv) Yes, D and M are the same points.

Question 20.
In the given figure, l, m and n are three parallel lines, x andy intersect these lines.
(i) Name the points lying on the line x.
(ii) Name the points lying on the liney.
(iii) Name the points inside the quadrilateral ABED.
(iv) Name the points outside the quadrilaterals ABED and BCFE.
(v) Name the lines passing through three points.
Solution:
(i) A, B and C lie on the line x.
(ii) D, E and F lie on the liney.
(iii) Q is the point inside ▢ABED
Basic Geometrical Ideas Class 6 Extra Questions Very Short Answer Type
(iv) Points R and S are outside the quadrilaterals ABED and BCFE.
(v) Lines x andy pass through the three points A, B, C and D, E, F respectively.

Class 6 Maths Extra Questions

Playing With Numbers Class 6 Extra Questions and Answers CBSE

September 22, 2024

NCERT Class 6 Maths Chapter 3 Extra Questions and Answers

Extra Questions for Class 6 Maths Chapter 3 Playing With Numbers

Playing With Numbers Class 6 Extra Questions Very Short Answer Type

Playing With Numbers Class 6 Questions And Answers Question 1.
What is the sum of any two
(a) even numbers
(b) odd numbers?
Solution:
(a) The sum of any two even numbers is even.
Example: 4 (even) + 6 (even) = 10 (even)
(b) The sum of any two odd numbers is even.
Example: 5 (odd) + 7 (odd) = 12 (even)

Playing With Numbers Class 6 Extra Questions Question 2.
Which of the following numbers is divisible by 3?
(a) 1212
(b) 625
Solution:
(a) Given number = 1212
Sum of the digits = 1 + 2 + 1 + 2 = 6, which is divisible by 3.
Hence, 1212 is also divisible by 3.
(b) Given number = 625
Sum of the digits = 6 + 2 + 5 = 13, which is not divisible by 3.
Hence, 625 is not divisible by 3.

Class 6 Maths Chapter 3 Extra Questions Question 3.
If the LCM and HCF of any two numbers are 15 and 4 respectively, find the product of the numbers.
Solution:
We know that the product of the number = LCM x HCF = 15 x 4 = 60
Hence, the product of the given numbers = 60.

Playing With Numbers Class 6 Worksheet Question 4.
Find the HCF of 5 and 7.
Solution:
Given numbers are 5 and 7. We observe that 5 and 7 are co-prime numbers.
Hence, the HCF is 1.

Extra Questions For Class 6 Maths Chapter 3 Question 5.
Write first 3 multiples of 25.
Solution:
We have 25 x 1 = 25; 25 x 2 = 50; 25 x 3 = 75
Hence, the required multiples are 25, 50 and 75.

Questions On Playing With Numbers For Class 6 Question 6.
What are the possible factors of (a) 12 (b) 18?
Solution:
(a) Possible factors of 12 are:
12 = 1 x 12; 12 = 2 x 6; 12 = 3 x 4
Hence, the factors of 12 are 1, 2, 3, 4, 6 and 12.

(b) Possible factors of 18 are:
18 = 1 x 18; 18 = 2 x 9; 18 = 3 x 6
Hence, the factors of 18 are 1, 2, 3, 6, 9 and 18.

Playing With Numbers Class 6 Worksheet With Answers Pdf Question 7.
Write first three multiples of 11.
Solution:
First three multiples of 11 are:
11 x 1 = 11; 11 x 2 = 22; 11 x 3 = 33
Hence, the required multiples are: 11,22 and 33.

Class 6 Playing With Numbers Extra Questions Question 8.
Write pairs of twin prime numbers less than 20.
Solution:
Pairs of twin prime numbers are: (3, 5), (5, 7), (11, 13), (17, 19).

Class 6 Maths Chapter 3 Worksheet Question 9.
Write the number which is even as well as prime.
Solution:
2 is the only even number which is prime number also.

Chapter 3 Maths Class 6 Extra Questions Question 10.
What is the fundamental theorem of arithmetic?
Solution:
Every number greater than 1 has exactly one prime factorisation.

Playing With Numbers Class 6 Extra Questions Short Answer Type

Extra Questions On Playing With Numbers Class 6 Question 11.
Simplify: 32 + 96 ÷ (7 + 9)
Solution:
Given that: 32 + 96 ÷ (7 + 9)
= 32 + 96 ÷ 16 (Using BODMAS)
= 32 + 6 = 38

Ncert Class 6 Maths Chapter 3 Extra Questions Question 12.
Simplify: 18 + {1 + (5 – 3) x 5}
Solution:
Given that: 18 + {1 + (5 – 3) x 5} (Using BODMAS)
= 18 + {1 + 2 x 5} = 18 + {1 + 10}
= 18 + 11 = 29.

Playing With Numbers Class 6 Worksheet With Answers Question 13.
Without actual division, show that 11 is a factor of 1,10,011.
Solution:
Here 1,10,011 = 1,10,000 + 11
= 11 x 10,000 + 11 x 1
= 11 x (10,000 + 1)
= 11 x 10,001
It is clear that 11 is a factor of 11 x 10,001.
Hence, 11 is a factor of 1,10,011.

Playing With Numbers Class 6 Questions Question 14.
The sum of two numbers is 25 and their product is 144. Find the numbers.
Solution:
The product of two numbers is 144.
∴ The possible factors are 1 x 144, 2 x 72, 3 x 48, 4 x 36, 6 x 24, 8 x 18, 9 x 16, 12 x 12
Here, we observe that out of these factors, we take 9 and 16.
Product = 9 x 16 = 144 and sum = 9 + 16 = 25
Hence, the required numbers are 9 and 16.

Class 6 Maths Ch 3 Extra Questions Question 15.
Is 80136 divisible by 11?
Solution:
Sum of the digits at odd places = 6 + 1 + 8 = 15
Sum of the digits at even places = 3 + 0 = 3
Difference of the two sums = 15 – 3 = 12,
which is neither 0 nor the multiple of 11.
Hence, 80136 is not divisible by 11.

Class 6 Chapter 3 Maths Extra Questions Question 16.
The HCF and LCM of two numbers are 6 and 120 respectively. If one of the numbers is 24, find the other number.
Solution:
Given that: HCF = 6
LCM = 120
Let the two numbers be a and b, where a = 24, b = ?
We know that: a x b = HCF x LCM
⇒ 24 x b = 6 x 120
⇒ b = \(\frac { 6 x 120 }{ 24 }\)
⇒ b = 30
Hence, the other number is 30.

Cbse Class 6 Maths Chapter 3 Extra Questions Question 17.
Find the LCM of 12 and 30.
Solution:
Given numbers are 12 and 30
12 = 2 x 2 x 3;
30 = 2 x 3 x 5
∴ LCM = 2 x 2 x 3 x 5 = 60
Hence, the LCM of 12 and 30 = 60.

Question 18.
Find the smallest 4-digit number which is divisible by 18, 24 and 32.
Solution:
Given numbers are 18, 24 and 32, we have
Playing With Numbers Class 6 Extra Questions Maths Chapter 3
Thus, LCM = 2 x 2 x 2 x 2 x 2 x 3 x 3 = 288
The smallest 4-digit number = 1000
Now, we write multiples of 288, till we get a 4-digit number.
288 x 1 = 288, 288 x 2 = 576,
288 x 3 = 864, 288 x 4 = 1152
Hence, 1152 is the required number.

Question 19.
Find the greatest number which divides 82 and 132 leaving 1 and 6, respectively as remainders.
Solution:
Given numbers are 82 and 132 and the remainders are 1 and 6 respectively.
We have, 82 – 1 = 81 and 132 – 6 = 126
So, we need to find the HCF of 81 and 126
Playing With Numbers Class 6 Extra Questions Maths Chapter 3
Common factor is 3 (occurring twice).
∴ HCF = 3 x 3 = 9
Hence, the required number is 9.

Question 20.
Find the greatest number that will divide 455, 582 and 710 leaving remainders 14, 15 and 17 respectively.
Solution:
Given numbers are 455, 582 and 710 and the respective remainders are 14, 15 and 17.
We have 455 – 14 = 441, 582 – 15 = 567 and 710 – 17 = 693.
Now let us find their HCF.
Playing With Numbers Class 6 Extra Questions Maths Chapter 3
Common factors are 3 and 7.
∴ HCF = 3 x 7 = 21
Hence, the required number is 21.

Playing With Numbers Class 6 Extra Questions Long Answer Type

Question 21.
Simplify the following:
40 + [20 – {28 ÷ 7 – 3 + (30 – 5 of 4)}]
Solution:
Using BODMAS Rule, we have
40 + [20 – {28 ÷ 7 – 3 + (30 – 5 of 4)}]
= 40 + [20 – {28 ÷ 7 – 3 + (30 – 20)}]
= 40 + [20 – {28 ÷ 7 – 3 + 10}]
= 40 + [20 – [4 – 3 + 10}]
= 40 + [20-11] = 40 + 9 = 49.

Question 22.
Three sets of English, Hindi, and Urdu books are to be stacked in such a way that the books are stored subjectwise and the height of each stack is the same. The numbers of English, Hindi and Urdu books are 336, 192 and 144 respectively. Assuring that the books have that same thickness, determine the number of stacks of English, Hindi and Urdu books.
Solution:
To arrange the books in the required way,
we have to find the greatest number that divides 336, 192 and 144 exactly.
So, HCF of 336, 192 and 144 is
Playing With Numbers Class 6 Extra Questions Maths Chapter 3
Common factors are 2 x 2 x 2 x 2 x 3 = 48
∴ HCF = 48, i.e. each stack contains 48 books.
∴ Number of stacks of English books = 336 + 48 = 7
Number of stacks of Hindi books = 192 4- 48 = 4
Number of stacks of Urdu books = 144 4- 48 = 3

Question 23.
Which of the following statements are true?
(а) 1371 is divisible by 3
(b) 1155 is not divisible by 9
(c) 1478 is not divisible by 4
(d) 2470 is divisible by 5
(e) If a number is divisible by 9, it is also divisible by 3.
(f) If a number is divisible by 3, it is also divisible by 9.
(g) The sum of any two odd numbers is even.
(h) If a number is divisible by 8, it must be divisible by 6.
(i) If a number is divisible by 3 and 6, it is divisible by 18.
(j) 1758 is not divisible by 8.
Solution:
(a) Yes, 1371 is divisible by 3. So it is true statement.
(b) Yes, 1155 is not divisible by 9. So it is true statement.
(c) Yes, 1478 is not divisible by 4. So it is true statement.
(d) Yes, 2470 is divisible by 5. So it is true statement.
(e) Yes, it is true statement. if No, it is not true statement.
(g) Yes, it is true statement.
(h) No, it is not true statement.
(i) No, it is not true statement.
(j) Yes, it is true statement.

Playing With Numbers Class 6 Extra Questions Multiple Choice Type

Question 24.
Which of the following numbers is divisible by 11?
(a)112111
(b) 928389
(c) 12011
(d) 11111

Question 25.
Match column I with column II.
Column I Column II
(a) A number divisible by 11 (i) 2
(b) HCF of two consecutive odd numbers (ii) 4
(c) The difference between twin prime number (iii) product of the number
(d) Number of factors of a prime number (iv) 60
(e) Lowest composite number (v)2
(f) LCM of 12 and 5 (vi) 4587594
(g) The smallest prime number (vii) 1
(h) Product of HCF and LCM is equal to (viii) 2
Solution:
(a) → (vi)
(b) → (vii)
(c) → (viii)
(d) → (i)
(e) → (ii)
(f) → (iv)
(g) → (v)
(h) → (iii)

Playing With Numbers Class 6 Extra Questions Higher Order Thinking Skills (HOTS)

Question 26.
Ill cows, 185 sheep and 296 goats are to be taken across a river. There is only one boat and the boatsman says; he will take the same number and same kind of animals in each trip. Find the largest number of animals in each trip and the number of trips he will have to make.
Solution:
We have
Number of cows = 111
Number of sheep = 185
Number of goats = 296
According to the condition of the boatsman, we need to find HCF of 111, 185 and 296
111 = 3 x 37;
185 = 5 x 37;
296 = 2 x 2 x 2 x 37
∴ the HCF = 37
So, the number of animals of same kind = 37.
Number of trips
= \(\frac { 111 }{ 37 }\) + \(\frac { 185 }{ 37 }\) + \(\frac { 296 }{ 37 }\)
= 3 + 5 + 8 = 16
Hence, the number of animals in each trip = 37
and the number of trips = 16.

Question 27.
In a seminar, the number of participants in Mathematics, Physics and Chemistry are 60, 96 and 144 respectively. Find the number of rooms required if in each room, the same number of ‘ participants are to be seated and all of them are to be in the same subject.
Solution:
The number of participants in each room must be the HCF of 60, 96 and 144.
∴ 60 = 2 x 2 x 3 x 5
96 = 2 x 2 x 2 x 2 x 2 x 3
144 = 2 x 2 x 2 x 2 x 3 x 3
HCF = 2 x 2 x 3 = 12
Number of rooms required
= \(\frac { 60 }{ 12 }\) + \(\frac { 96 }{ 12 }\) + \(\frac { 144 }{ 12 }\)
= 5 + 8 + 12 = 25
Hence, number of participants = 12
in each room and number of rooms required = 25.

Class 6 Maths Extra Questions

Whole Numbers Class 6 Extra Questions and Answers CBSE

September 22, 2024

NCERT Class 6 Maths Chapter 2 Extra Questions and Answers

Extra Questions for Class 6 Maths Chapter 2 Whole Numbers

Whole Numbers Class 6 Extra Questions Very Short Answer Type

Class 6 Maths Chapter 2 Extra Questions Question 1.
Write the smallest whole number.
Solution:
0 is the smallest whole number.

Whole Numbers Class 6 Extra Questions Question 2.
What is the predecessor of whole number 0?
Solution:
Whole number 0 has no predecessor.

Whole Numbers Class 6 Questions And Answers Question 3.
Which property do the following statements hold?
(a) 6 + 4 = 4 + 6
(b) 3 + 2 = whole number
Solution:
(a) 6 + 4 = 4 + 6 holds commutative property of addition
(b) 3 + 2 = whole number holds closure property.

Extra Questions For Class 6 Maths Chapter 2 Question 4.
Add the following in three ways. Indicate the property used.
(a) 25 + 36 + 15
(b) 30 + 18 + 22
Solution:
(a) 25 + 36 + 15
Way I: 25 + (36 + 15) = 25 + 51 = 76
Way II: (25 + 36) + 15 = 61 + 15 = 76
Way III: (25 + 15) + 36 = 40 + 36 = 76
Here, we have used associative property.

(b) 30 + 18 + 22
Way I: 30 + (18 + 22) = 30 + 40 = 70
Way II: (30 + 18) + 22 = 48 + 22 = 70
Way III: (30 + 22) + 18 = 52 + 18 = 70
Here, we have used associative property.

Whole Numbers Class 6 Worksheet Question 5.
Using distributive property, solve the following:
(а) 360 x 102
(b) 35 x 98
Solution:
(a) 36 x 102 = 36 x (100 + 2)
= 36 x 100 + 36 x 2
= -36000 + 72 = 36072
(b) 35 x 98 = 35 x (100 – 2) = 35 x 100 – 35 x 2
= 3500 – 70 = 3430

Class 6 Maths Chapter 2 Worksheet With Answers Question 6.
Find the product of the greatest 3-digit number and the smallest 2-digit number.
Solution:
The greatest 3-digit number = 999
The smallest 2-digit number = 10
∴ Product = 999 x 10 = 9990

Questions On Whole Numbers For Class 6 Question 7.
Write any two numbers which can be shown as rectangles.
Solution:
Class 6 Whole Numbers Extra Questions

Chapter 2 Maths Class 6 Extra Questions Question 8.
Write the predecessor of the smallest 4-digit number.
Solution:
The smallest 4-digit number = 1000
∴ The predecessor of 1000 = 1000 – 1 = 999

Ncert Class 6 Maths Chapter 2 Extra Questions Question 9.
For n = 5, verify the given statement 10 x n + 1 = n1
Solution:
Given statement is
10 x n + 1 = n1
Put n = 5, 10 x 5 + 1 = 51
⇒ 50 + 1 = 51
⇒ 51 = 51. Hence, verified.

Extra Questions On Whole Numbers For Class 6 Question 10.
Write the next two steps:
Class 6 Whole Numbers Worksheet
Solution:
Next two steps are 123 x 9 + 4 = 1111 and 1234 x 9 + 5 = 11111.

Whole Numbers Class 6 Extra Questions Short Answer Type

Word Problems On Whole Numbers For Class 6 Question 11.
Using the properties, find the values of each of the following:
(a) 736 x 102
(b) 8165 x 169 – 8165 x 69
Solution:
(a) 736 x 102 = 736 x (100 + 2)
= 736 x 100 + 736 x 2 [Using distributive property]
= 73600 + 1472 = 75072

(b) 8165 x 169 – 8165 x 69 = 8165 x (169 – 69) [Using distributive property]
= 8165 x 100 = 816500

Class 6 Maths Ch 2 Extra Questions Question 12.
Observe the following patterns and extend them by two more terms.
Class 6 Maths Chapter 2 Worksheet
Solution:
Next two terms are
1010101 × 1010101= 1020304030201
101010101 × 101010101=10203040504030201

Whole Numbers Questions And Answers Question 13.
Observe the following patterns and extend them by two more terms:
Maths Class 6 Chapter 2 Extra Questions
Solution:
Next two terms are
15873 x 7 x 3 = 333333
15873 x 7 x 4 = 444444

Whole Numbers Extra Questions Question 14.
Write the three whole numbers which can be arranged as squares.
Solution:
The required number are 4, 9, 16.
Whole Numbers Worksheet Class 6

Chapter 2 Class 6 Maths Extra Questions Question 15.
Using the properties of whole numbers, find the value of the following in suitable way:
(a) 945 x 4 x 25
(b) 40 x 328 x 25
Solution:
(a) 945 x 4 x 25 = 945 x (4 x 25)
= 945 x 100 = 94500
(b) 40 x 328 x 25 = 328 x (40 x 25)
= 328 x 1000 = 328000

Whole Numbers Class 6 Worksheet With Answers Question 16.
Represent the following on number line:
(a) 3 + 4
(b) 6 – 2
(c) 2 × 4
Solution:
(a) 3 + 4
Whole Numbers Questions For Class 6
(b) 6 – 2
Whole Numbers Class 6 Extra Questions Maths Chapter 2
(c) 2 × 4

Question 17.
Give one example for each of the following properties for whole numbers.
(a) Closure property
(b) Commutative property
(c) Associative property
(d) Distributive property
Solution:
(a) 3 + 4 = 7 (whole number) closure property
(b) 4 + 5 = 5 + 4 Commutative property
(c) 3 + (5 + 7) = (3 + 5) + 7 Associative property
(d) 6 x (8 + 3) = 6 x 8 + 6 x 3 Distributive property.

Question 18.
A dealer purchased 124 LED sets. If the cost of one set is ₹38,540, determine their total cost.
Solution:
Total cost of 124 LED sets = ₹(38,540 x 124)
= ₹ [38,540 x (100 + 20 + 4)]
= ₹ [38,540 x 100 + 38,540 x 20 + 38,540 x 4]
= ₹ [38,54,000 + 7,70,800 + 1,54,160]
= ₹ 47,789,60

Question 19.
Find the product of the greatest 3-digit number and the greatest 2-digit number.
Solution:
Greatest 3-digit number = 999
Greatest 2-digit number = 99
∴ Product = 999 x 99 = 999 x (100 – 1)
= 999 x 100 – 999 x 1
= 99900 – 999 = 98901

Question 20.
Write 10 such numbers which can be shown only as line.
Solution:
2, 5, 7, 11, 13, 17, 19, 23, 29 and 31 are such numbers which can be shown only as line.
123 x 9 + 4 = 1111.

Whole Numbers Class 6 Extra Questions Long Answer Type

Question 21.
320 km distance is to be covered partially by bus and partially by train. Bus covers 180 km distance with a speed of 40 km/h and the rest of the distance is covered by the train at a speed of 70 km/h. Find the time taken by a passenger to cover the whole distance.
Solution:
Total distance = 320 km
Distance covered by the bus = 180 km
Speed of the bus = 40 km/h
Whole Numbers Class 6 Extra Questions Maths Chapter 2
Distance covered by the train = 320 – 180 = 140 km.
Speed of the train = 70 km/h
∴ Time taken by the train

Hence, the total time taken by the passenger
= \(\frac { 9 }{ 2 }\) hours + 2 hours
= 4 hours 30 min + 2 hours
= 6 hours 30 min

Question 22.
Solve the following and establish a pattern:
(a) 84 x 9
(b) 84 x 99
(c) 84 x 999
(d) 84 x 9999
Solution:
(a) 84 x 9 = 84 x (10 – 1) = 84 x 10 – 84 x 1 = 840 – 84 = 756
(b) 84 x 99 = 84 x (100 – 1) = 84 x 100 – 84 x 1 = 8400 – 84 = 8316
(c) 84 x 999 = 84 x (1000 – 1) = 84 x 1000 – 84 x 1 = 84000 – 84 = 83916
(d) 84 x 9999 = 84 x (10000 – 1) = 84 x 10000 – 84 x 1 = 840000 – 84 = 839916

Question 23.
Solve the following with suitable and short-cut method: .
(a) 86 x 5
(b) 86 x 15
(c) 86 x 25
(d) 86 x 35
(e) 86 x 50
(f) 96 x 125
(g) 96 x 250
(h) 112 x 625
Solution:
Whole Numbers Class 6 Extra Questions Maths Chapter 2

Question 24.
Ramesh buys 10 containers of juice from one shop and 18 containers of the same juice from another shop. If the capacity of each container is same and the cost of each of the container is ₹150, find the total money spend by Ramesh.
Solution:
Ramesh buys 10 containers from one shop Cost of 1 container = ₹150
He buys 18 containers of the same capacity from another shop.
Cost of 1 container = ₹150
∴ Total money spent by Ramesh
= ₹ [10 x 150 + 18 x 150]
= ₹150 x (10 + 18)
= ₹150 x 28
= ₹4200

Question 25.
Fill in the blanks.
(а) The smallest whole number is ……… .
(b) The smallest natural number is ……… .
(c) Difference between 5-digit smallest number and 4-digit largest number is ……… .
(d) Any number divided by 0 is not ……… .
(e) The property used in 84 x 25 = 25 x 84 is ……… .
(f) The property used in 80 x (60 + 3) = 80 x 60 + 80 x 3 is ……… .
(g) The smallest number which can be shown by two doted rectangles is ……… .
(h) Every whole number except ……… is a natural number.
(i) When any counting number is multiplied by zero, the product is ……… .
(j) When zero is divided by any non-zero whole number, the quotient is ……… .
Solution:
(a) The smallest whole number is 0.
(b) The smallest natural number is 1.
(c) Difference between 5-digit smallest number and 4-digit largest number is 1.
(d) Any number divided by 0 is not defined.
(e) The property used in 84 x 25 = 25 x 84 is commutative property.
(f) The property used in 80 x (60 + 3) = 80 x 60 + 80 x 3 is distributive property.
(g) The smallest number which can be shown by two doted rectangles is 6.
(h) Every whole number except 0 is a natural number.
(i) When any counting number is multiplied by zero, the product is 0.
(j) When zero is divided by any non-zero whole number, the product is 0.

Question 26.
Which of the following statements are true (T) and which are false (F)?
(a) The sum of two whole numbers is always less than their product.
(b) There is only one whole number n such that n x n – n
(c) For two non-zero whole numbers a and b, a ÷ b = b ÷ a.
(d) The sum of two odd whole numbers is an even number.
(e) There jloes not exist any whole number m for which m ÷ m = m.
(f) (16 ÷ 4) ÷ 2 = 16 ÷ (4 ÷ 2)
(g) 7 – 8 = whole number
(h) If 1 is added to a number, we get its successor.
(i) The whole number 15 lies between 14 and 21.
(j) 84 x (10 + 5) = 84 x 10 + 84 x 5 represents distributive property.
Solution:
(a) False statement
(b) False statement
(c) False statement
(d) True statement
(e) False statement
(f) False statement
(g) False statement
(h) True statement
(i) True statement
(j) True statement.

Question 27.
The value of 27 ÷ (9 ÷ 3) is
(a) 3
(b) 6
(c) 9
(d) 27
Solution:
27 ÷ (9 ÷ 3) = 27 ÷ (\(\frac { 9 }{ 3 }\)) = 27 ÷ 3 = 9
Hence, the correct option is (c).

Question 28.
The whole number 7 can be arranged as,
(a) line
(b) square
(c) rectangle
(d) triangle
Solution:
7 can be arranged as line.
Hence, the correct option is (a).

Whole Numbers Class 6 Extra Questions Higher Order Thinking Skills (HOTS)

Question 29.
A housing complex built by DLF consists of 25 large buildings and 40 small buildings. Each large building has 15 floors with 4 apartments on each floor and each small building has 9 floors with 3 apartments on each floor. How many apartments are there in all?
Solution:
Number of large buildings = 25
Number of floors = 15
Number of apartments on each floor = 4
∴ Total number of apartments in large buildings = 25 x 15 x 4
Number of small building = 40
Number of floors = 9
Number of apartments on each floor = 3
∴ Total number of apartments in small buildings = 40 x 9 x 3
Hence, the number of apartments in all = 25 x 15 x 4 + 40 x 9 x 3 = 1500 + 1080 = 2580.

Question 30.
A school principal places orders for 85 chairs and 25 tables with a dealer. Each chair cost ₹180 and each table cost ₹140. If the principal has given ₹2500 to the dealer as an advance money, then what amount to be given to the dealer now?
Solution:
Number of chairs = 85
Cost of one chair = ₹ 180
Cost of 85 chairs = ₹ (85 x 180)
Number of tables = 25
Cost of one table = ₹ 140
Cost of 25 tables = ₹ (25 x 140)
Total cost of all chairs and tables = ₹(85 x 180 + 25 x 140)
= ₹ (15300 + 3500) = ₹ 18800
Money given in advance = ₹ 2500
∴ Balance money to be paid to the dealer = ₹ 18800 – ₹ 2500 = ₹ 16300

Class 6 Maths Extra Questions

Ratio and Proportion Class 6 Extra Questions Maths Chapter 12

Ratio and Proportion Class 6 Extra Questions and Answers CBSE

September 22, 2024

NCERT Class 6 Maths Chapter 12 Extra Questions and Answers

Extra Questions for Class 6 Maths Chapter 12 Ratio and Proportion

Ratio and Proportion Class 6 Extra Questions Very Short Answer Type

Ratio And Proportion Class 6 Extra Questions Question 1.
Find the ratio of 75 cm to 1.5 m.
Solution:
The given numbers are not in the same units. So, converting them into same units.
1.5 m = 1.5 x 100 cm = 150 cm
[∵ 1 m = 100 cm]
∴ The required ratio is 75 cm : 150 cm.

∴ Required ratio = 1 : 2

Ratio Questions For Class 6 Question 2.
Give two equivalent ratios of 3 : 5.
Solution:

Thus, 9 : 15 and 6 : 10 are the two equivalent ratios of 3 : 5.

Ratio And Proportion Questions And Answers For Class 6 Pdf Question 3.
Fill in the blank box.

Solution:

Questions On Ratio And Proportion For Class 6 Question 4.
Check whether the given ratios are equivalent or not. \(\frac { 2 }{ 7 }\), \(\frac { 6 }{ 21 }\)
Solution:

∴ They are equivalent ratios.

Ratio And Proportion Questions For Class 6 Pdf Question 5.
Divide 60 in the ratio of 2 : 3.
Solution:
Sum = 2 + 3 = 5
∴ First part = \(\frac { 2 }{ 5 }\) x 60 =24 5
∴ Second part = \(\frac { 3 }{ 5 }\) x 60 =36 5
Thus, the required two parts = 24 and 36.

Extra Questions Of Ratio And Proportion For Class 6 Question 6.
Find the ratio of the following:
(a) 56 to 63.
(b) 55 to 120.
Solution:

Class 6 Ratio And Proportion Questions Question 7.
Ramesh deposited ₹ 2050 in a bank and in the month of January he withdrew ₹ 410 from his account on the last date of the month. Find the ratio of
(a) Money withdrawn to the total money deposited.
(b) Money withdrawn to the remaining amount in the bank.
Solution:
Total money deposited = ₹ 2050
Amount of money withdrawn = ₹ 410
Amount of money left in the bank = ₹ 2050 – ₹ 410 = ₹ 1640
(a) Ratio of money withdrawn to the total money deposited

∴ Required ratio = 1 : 5
(b) Ratio of money withdrawn to the money left in the bank

∴ Required ratio = 1 : 4

Ratio And Proportion Questions Class 6 Question 8.
There are 180 students in a class. Number of girls are 75. Find the ratio of the girls to the number of boys.
Solution:
Total number of students = 180
Number of girls = 75
Number of boys = 180 – 75 = 105
∴ Ratio of number of girls to the number of boys

Required ratio = 5 : 7

Ratio and Proportion Class 6 Extra Questions Short Answer Type

Ratio And Proportion Class 6 Worksheet Question 9.
Green paint is made by mixing blue, yellow and white paints in the ratio 2 : 7 : 1. How much blue paint is needed to make 64 litres of green paint?
Solution:
Here, sum of ratios = 2 + 7 + 1 = 10
∴ Total quantity of green paint = 64 litres
Quantity of blue paint = \(\frac { 2 }{ 10 }\) x 64 = 12.8 litres
Therefore, the required blue paint = 12.8 litres.

Ratio And Proportion Class 9 Extra Questions Question 10.
From the figure, find the ratio of
(a) The number of squares to the number of triangles.
(b) The number of circles to the number of rectangles.

Solution:
(a) Number of squares = 2
Number of triangles = 3 2
∴ Ratio = \(\frac { 2 }{ 3 }\) or 2 : 3

(b) Number of circles = 3
Number of rectangles = 3
∴ Ratio = \(\frac { 3 }{ 3 }\) or 1 : 1

Ratio And Proportion Extra Questions Class 6 Question 11.
In each of the following figures, find the ratio of the shaded region to the unshaded region.

Solution:
(a) Number of shaded parts = 4
Number of unshaded parts = 12

Required ratio = 1 : 3

(b) Number of shaded parts = 2
Number of unshaded parts = 4

Required ratio = 1 : 2

Proportion Questions For Class 6 Question 12.
Are 20, 25, 12, 15 in proportion?
Solution:
We have 20, 25, 12, 15
Product of extremes = 20 x 15 = 300
Product of middles = 25 x 12 = 300
Since both the products are same.
∴ The four numbers 20, 25, 12, 15 are in proportion.

Question 13.
The first, second and fourth terms in a proportion are 32, 112, 217 respectively. Find the third term.
Solution:
Let the third term be x.
∴ 32, 112, x and 217 are in proportion.
∴ 32 : 112 :: x : 217

Thus, the third term = 62.

Question 14.
Find the value of x, if
(а) 8, x, x, 50 are in proportion.
(b) 36, 90, 90, x are in proportion.
Solution:
(a) Since 8, x, x, 50, are in proportion.
∴ x × x = 8 × 50
⇒ x² = 400
∴ x = 20

(b) Since 36, 90, 90, x are in proportion.
∴ 36 × x = 90 × 90
⇒ x = \(\frac { 90\times 90 }{ 36 }\) = 225
∴ x = 225

Question 15.
The cost of 10 tables is ₹ 7500. Find the number of tables that can be purchased with ₹ 9000.
Solution:
Number of tables purchased in ₹ 7500 = 10
Number of tables purchased in ₹ 1 = \(\frac { 10 }{ 7500 }\)
∴ Number of tables purchased in ₹ 9000
= \(\frac { 10\times 9000 }{ 7500 }\) = 12

Question 16.
39 packets of 12 pens each costs ₹ 374.40. Find the cost of 52 packets of 10 pens each.
Solution:
Number of pens in 1 packet = 12
Number of pens in 39 packets = 12 x 39 = 468
Number of pens in 1 packet = 10
Number of pens in 52 packets = 10 x 52 = 520
Now cost of 468 pen = ₹ 374.40

Class 6 Maths Extra Questions

Mensuration Class 6 Extra Questions Maths Chapter 10

Mensuration Class 6 Extra Questions and Answers CBSE

September 22, 2024

NCERT Class 6 Maths Chapter 10 Extra Questions and Answers

Extra Questions for Class 6 Maths Chapter 10 Mensuration

Mensuration Class 6 Extra Questions Very Short Answer Type

Mensuration Class 6 Extra Questions Question 1.
The perimeter of a square is 64 cm. Find the length of each side.
Solution:
Perimeter of the square = 64 cm

Mensuration Questions For Class 6 Question 2.
Length and breadth of a rectangular table-top are 36 cm and 24 cm respectively. Find its perimeter.
Solution:
Length of the rectangular table-top = 36 cm
and its breadth = 24 cm.
∴ Perimeter of the table-top = 2 [length + breadth]
= 2 [36 cm + 24 cm]
= 2 x 60 cm = 120 cm.

Class 6 Maths Chapter 10 Extra Questions Question 3.
Which of the following figure has greater perimeter?

Solution:
Fig. (i) Perimeter of the square = 4 x side
= 4 x 4 cm = 16 cm
Fig. (ii) Perimeter of the rectangle
= 2 [length + breadth]
= 2[8 cm + 3 cm]
= 2 x 11 cm = 22 cm
Since 22 cm > 16 cm
∴ Rectangle has greater perimeter than the square.

Mensuration Class 6 Questions Question 4.
How much distance will you have to travel in going around each of the following figures?

Solution:
Distance travelled in going around Fig. (i)
= 12 cm + 3 cm + 12 cm + 3 cm = 30 cm
Distance travelled in going around Fig. (ii)
= 6 cm + 4 cm + 4 cm + 4 cm = 18 cm

Class 6 Maths Mensuration Extra Questions Question 5.
Find the perimeter of a square whose side is 15 cm.
Solution:
Side of the square = 15 cm
∴ Perimeter of the square = 15 cm x 4 = 60 cm

Mensuration Class 6 Extra Questions With Answers Question 6.
Find the cost of fencing a rectangular park 300 m long and 200 m wide at the rate of ₹4 per metre.
Solution:
Length of the park = 300 m
Breadth = 200 m
∴ Perimeter of the park = 2 [length + breadth]
= 2 [300 m + 200 m]
= 2 x 500 m = 1000 m.
Cost of fencing the rectangular park = 1000 x 4 = ₹4000

Class 6 Maths Ch 10 Extra Questions Question 7.
Find the area of a square field whose each side is 150 m.
Solution:
Side of the square field = 150 m
∴ Area of the square field = Side x Side
= 150 m x 150 m
= 22500 sq m.

Mensuration Class 6 Worksheet Question 8.
Length and breadth of a rectangular paper are 22 cm and 10 cm respectively. Find the area of the paper.
Solution:
Length of the rectangular paper = 22 cm
Breadth = 10 cm
∴ Area of the rectangular paper = length x breadth
= 22 cm x 10 cm
= 220 sq cm.

Mensuration Class 6 Extra Questions Short Answer Type

Questions On Mensuration For Class 6 Question 9.
Find the length of a rectangle given that its perimeter is 880 m and breadth is 88 m.
Solution:
Perimeter of the rectangle = 2 [length + breadth]
∴ 2 [length + breadth] = 880
length + breadth = 880 ÷ 2 = 440
∵ Breadth = 88 m
∴ Length = 440 m – 88 m = 352 m
Hence, the required length = 352 m.

Class 6 Mensuration Questions Question 10.
How many trees can be planted at a distance of 6 metres each around a rectangular plot whose length is 120 m and breadth is 90 m?
Solution:
Length of the rectangular plot = 120 m
Breadth = 90 m
∴ Perimeter of the rectangular plot
= 2 [length + breadth]
= 2 [120 m + 90 m]
= 2 x 210 m = 420 m
Now distance between two trees = 6 m
∴ Number of trees around the rectangular plot = 420 m ÷ 6 m = 70

Extra Questions Of Mensuration Class 6 Question 11.
A rectangular park is 30 metres long and 20 metres broad. A steel wire fence is put up all around it. Find the cost of putting the fence at the rate of ₹15 per metre.
Solution:
Length of the rectangular park = 30 m
Breadth = 20 m
∴ Perimeter of the rectangular park = 2(length + breadth)
= 2 [30 + 20] = 2 x 50 m = 100 m
∴ Cost of fencing all around the park = ₹15 x 100 = ₹1500

Class 6 Mensuration Questions Pdf Question 12.
Find the area of the figures A, B, C and D drawn on a squared paper in the following figure by counting squares.

Solution:
(A) Counting the squares, we have 8 squares
∴ Area = 8 sq units
(B) Counting the squares, we have 4 squares
∴ Area = 4 sq units
(C) Counting the squares, we have 5 squares
∴ Area = 5 sq units
(D) Counting the squares, we have 7 squares
∴ Area = 7 sq units

Mensuration Class 6 Extra Questions Higher Order Thinking Skills (HOTS)

Extra Questions On Mensuration For Class 6 Question 13.
A rectangle and a square have the same perimeter 100 cm. Find the side of the square. If the rectangle has a breadth 2 cm less than that of the square. Find the breadth, length and area of the rectangle.
Solution:
Perimeter of the square = 100 cm
Perimeter 100

= 25 cm.
∴ Breadth of the rectangle = 25 cm – 2 cm = 23 cm
Now perimeter of the rectangle = 100 cm
∴ 2 [length + breadth] = 100
length + breadth = 100 ÷ 2 = 50 cm
But breadth = 23 cm
∴ Length = 50 cm – 23 cm = 27 cm
Now, Area of the rectangle
= length x breadth = 27 cm x 23 cm
= 621 sq cm.

Mensuration Class 6 Question Bank Question 14.
Fencing the compound of a house costs ₹5452. If the rate is ₹94 per metre, find the perimeter of the compound. If the breadth is 10 m, find its length.
Solution:
Cost of fencing the compound = ₹5452
and the rate of fencing = ₹94 per metre
∴ Perimeter of the compound = 5452 ÷ 94 = 58 metres
Now breadth of the compound = 10 m.
2 [length + breadth] = 58 m
∴ length + breadth = 58 + 2 m = 29 m
∴ Length of the compound = 29 m – 10 m = 19 m.

Class 6 Maths Extra Questions

Lines and Angles Class 7 Extra Questions Maths Chapter 5

Lines and Angles Class 7 Extra Questions with Answers CBSE

September 22, 2024

Extra Questions for Class 7 Maths PDF are provided here. Students can download the pdf of these solutions from the given links. Lines and Angles Class 7 Extra Questions Maths Chapter 5 provided in accordance with the latest syllabus of CBSE which, in turn, help the students to build a strong foundation and secure excellent marks in their board exams. https://meritbatch.com/lines-and-angles-class-7-extra-questions/

NCERT Class 7 Maths Chapter 5 Extra Questions and Answers

Extra Questions for Class 7 Maths Chapter 5 Lines and Angles

Lines and Angles Class 7 Extra Questions Very Short Answer Type

Lines And Angles Class 7 Extra Questions Question 1.
Find the angles which is \(\frac { 1 }{ 5 }\) of its complement.
Solution:
Let the required angle be x°
its complement = (90 – x)°
As per condition, we get

Class 7 Maths Chapter 5 Extra Questions Question 2.
Find the angles which is \(\frac { 2 }{ 3 }\) of its supplement.
Solution:
Let the required angle be x°.
its supplement = (180 – x)°
As per the condition, we get
\(\frac { 2 }{ 3 }\) of (180 – x)° = x°

Lines And Angles Class 7 Questions And Answers Question 3.
Find the value of x in the given figure.

Solution:
∠POR + ∠QOR = 180° (Angles of linear pair)
⇒ (2x + 60°) + (3x – 40)° = 180°
⇒ 2x + 60 + 3x – 40 = 180°
⇒ 5x + 20 = 180°
⇒ 5x = 180 – 20 = 160
⇒ x = 32
Thus, the value of x = 32.

Lines And Angles Class 7 Worksheet With Answers Pdf Question 4.
In the given figure, find the value of y.

Solution:
Let the angle opposite to 90° be z.
z = 90° (Vertically opposite angle)
3y + z + 30° = 180° (Sum of adjacent angles on a straight line)
⇒ 3y + 90° + 30° = 180°
⇒ 3y + 120° = 180°
⇒ 3y = 180° – 120° = 60°
⇒ y = 20°
Thus the value of y = 20°.

Lines And Angles Extra Questions Class 7 Question 5.
Find the supplements of each of the following:
(i) 30°
(ii) 79°
(iii) 179°
(iv) x°
(v) \(\frac { 2 }{ 5 }\) of right angle
Solution:
(i) Supplement of 30° = 180° – 30° = 150°
(ii) Supplement of 79° = 180° – 79° = 101°
(iii) Supplement of 179° = 180° – 179° = 1°
(iv) Supplement of x° = (180 – x)°
(v) Supplement of \(\frac { 2 }{ 5 }\) of right angle
= 180° – \(\frac { 2 }{ 5 }\) × 90° = 180° – 36° = 144°

Class 7 Maths Lines And Angles Extra Questions Question 6.
If the angles (4x + 4)° and (6x – 4)° are the supplementary angles, find the value of x.
Solution:
(4x + 4)° + (6x – 4)° = 180° (∵ Sum of the supplementary angle is 180°)
⇒ 4x + 4 + 6x – 4 = 180°
⇒ 10x = 180°
⇒ x = 18°
Thus, x = 18°

Class 7 Maths Chapter 5 Worksheet With Answers Question 7.
Find the value of x.

Solution:
(6x – 40)° + (5x + 9)° + (3x + 15) ° = 180° (∵ Sum of adjacent angles on straight line)
⇒ 6x – 40 + 5x + 9 + 3x + 15 = 180°
⇒ 14x – 16 = 180°
⇒ 14x = 180 + 16 = 196
⇒ x = 14
Thus, x = 14

Extra Questions On Lines And Angles Class 7 Question 8.
Find the value of y.

Solution:
l || m, and t is a transversal.
y + 135° = 180° (Sum of interior angles on the same side of transversal is 180°)
⇒ y = 180° – 135° = 45°
Thus, y = 45°

Lines and Angles Class 7 Extra Questions Short Answer Type

Ncert Class 7 Maths Chapter 5 Extra Questions Question 9.
Find the value ofy in the following figures:

Solution:
(i) y + 15° = 360° (Sum of complete angles round at a point)
⇒ y = 360° – 15° = 345°
Thus, y = 345°
(ii) (2y + 10)° + 50° + 40° + 130° = 360° (Sum of angles round at a point)
⇒ 2y + 10 + 220 = 360
⇒ 2y + 230 = 360
⇒ 2y = 360 – 230
⇒ 2y = 130
⇒ y = 65
Thus, y = 65°
(iii) y + 90° = 180° (Angles of linear pair)
⇒ y = 180° – 90° = 90°
[40° + 140° = 180°, which shows that l is a straight line]

Ch 5 Maths Class 7 Extra Questions Question 10.
In the following figures, find the lettered angles.

Solution:
(i) Let a be represented by ∠1 and ∠2
∠a = ∠1 + ∠2
∠1 = 35° (Alternate interior angles)
∠2 = 55° (Alternate interior angles)
∠1 + ∠2 = 35° + 55°
∠a = 90°
Thus, ∠a = 90°

Extra Questions For Class 7 Maths Lines And Angles Question 11.
In the given figure, prove that AB || CD.

Solution:
∠CEF = 30° + 50° = 80°
∠DCE = 80° (Given)
∠CEF = ∠DCE
But these are alternate interior angle.
CD || EF ……(i)
Now ∠EAB = 130° (Given)
∠AEF = 50° (Given)
∠EAB + ∠AEF = 130° + 50° = 180°
But these are co-interior angles.
AB || EF …(ii)
From eq. (i) and (ii), we get
AB || CD || EF
Hence, AB || CD
Co-interior angles/Allied angles: Sum of interior angles on the same side of transversal is 180°.

Lines And Angles Class 7 Questions Question 12.
In the given figure l || m. Find the values of a, b and c.

Solution:
(i) We have l || m
∠b = 40° (Alternate interior angles)
∠c = 120° (Alternate interior angles)
∠a + ∠b + ∠c = 180° (Sum of adjacent angles on straight angle)
⇒ ∠a + 40° + 120° = 180°
⇒ ∠a + 160° = 180°
⇒ ∠a = 180° – 160° = 20°
Thus, ∠a = 20°, ∠b = 40° and ∠c = 120°.
(ii) We have l || m
∠a = 45° (Alternate interior angles)
∠c = 55° (Alternate interior angles)
∠a + ∠b + ∠c = 180° (Sum of adjacent angles on straight line)
⇒ 45 + ∠b + 55 = 180°
⇒ ∠b + 100 = 180°
⇒ ∠b = 180° – 100°
⇒ ∠b = 80°

Question 13.
In the adjoining figure if x : y : z = 2 : 3 : 4, then find the value of z.

Solution:
Let x = 2s°
y = 3s°
and z = 4s°
∠x + ∠y + ∠z = 180° (Sum of adjacent angles on straight line)
2s° + 3s° + 4s° = 180°
⇒ 9s° = 180°
⇒ s° = 20°
Thus x = 2 × 20° = 40°, y = 3 × 20° = 60° and z = 4 × 20° = 80°

Question 14.
In the following figure, find the value of ∠BOC, if points A, O and B are collinear. (NCERT Exemplar)

Solution:
We have A, O and B are collinear.
∠AOD + ∠DOC + ∠COB = 180° (Sum of adjacent angles on straight line)
(x – 10)° + (4x – 25)° + (x + 5)° = 180°
⇒ x – 10 + 4x – 25 + x + 5 = 180°
⇒ 6x – 10 – 25 + 5 = 180°
⇒ 6x – 30 = 180°
⇒ 6x = 180 + 30 = 210
⇒ x = 35
So, ∠BOC = (x + 5)° = (35 + 5)° = 40°

Question 15.
In given figure, PQ, RS and UT are parallel lines.
(i) If c = 57° and a = \(\frac { c }{ 3 }\), find the value of d.
(ii) If c = 75° and a = \(\frac { 2 }{ 5 }\)c , find b.

Solution:
(i) We have ∠c = 57° and ∠a = \(\frac { \angle c }{ 3 }\)
∠a = \(\frac { 57 }{ 3 }\) = 19°
PQ || UT (given)
∠a + ∠b = ∠c (Alternate interior angles)
19° + ∠b = 57°
∠b = 57° – 19° = 38°
PQ || RS (given)
∠b + ∠d = 180° (Co-interior angles)
38° + ∠d = 180°
∠d = 180° – 38° = 142°
Thus, ∠d = 142°
(ii) We have ∠c = 75° and ∠a = \(\frac { 2 }{ 5 }\) ∠c
∠a = \(\frac { 2 }{ 5 }\) × 75° = 30°
PQ || UT (given)
∠a + ∠b = ∠c
30° + ∠b = 75°
∠b = 75° – 30° = 45°
Thus, ∠b = 45°

NCERT Solutions for Class 7 Maths

Visualising Solid Shapes Class 7 Extra Questions Maths Chapter 15

Visualising Solid Shapes Class 7 Extra Questions with Answers CBSE

September 22, 2024

Extra Questions for Class 7 Maths PDF are provided here. Students can download the pdf of these solutions from the given links. Visualising Solid Shapes Class 7 Extra Questions Maths Chapter 15 provided in accordance with the latest syllabus of CBSE which, in turn, help the students to build a strong foundation and secure excellent marks in their board exams. https://meritbatch.com/visualising-solid-shapes-class-7-extra-questions/

NCERT Class 7 Maths Chapter 15 Extra Questions and Answers

Extra Questions for Class 7 Maths Chapter 15 Visualising Solid Shapes

Visualising Solid Shapes Class 7 Extra Questions Very Short Answer Type

Class 7 Maths Chapter 15 Extra Questions Question 1.
If three cubes of dimensions 2 cm × 2 cm × 2 cm are placed end to end, what would be the dimension of the resulting cuboid?
Solution:
Length of the resulting cuboid = 2 cm + 2 cm + 2 cm = 6 cm
Breadth = 2 cm
Height = 2 cm

Hence the required dimensions = 6 cm × 2 cm × 2 cm.

Visualising Solid Shapes Class 7 Worksheets Pdf Question 2.
Answer the following:
(i) Why a cone is not a pyramid?
(ii) How many dimension a solid have?
(iii) Name the solid having one curved and two flat faces but no vertex.
Solution:
(i) Cone is not a pyramid because its base is not a polygon.
(ii) Three.
(iii) Cylinder

Visualising Solid Shapes Class 7 Worksheets With Answers Question 3.
Write down the number of edges on each of the following solid figures:
(i) Cube
(ii) Tetrahedron
(iii) Sphere
(iv) Triangular prism
Solution:
(i) 12
(ii) 6
(iii) 0
(iv) 9

Worksheet On Visualising Solid Shapes Class 7 Question 4.
What cross-section do you get when you give a horizontal cut to an ice cream cone?
Solution:
Circle.

Questions On Visualising Solid Shapes For Class 7 Question 5.
Determine the number of edges, vertices and faces in the given figure.

Solution:
Edges = 8
Vertices = 5
Faces = 5

Visualizing Solid Shapes Class 7 Questions Question 6.
Draw the sketch of two figure that has no edge.
Solution:

Extra Questions On Visualising Solid Shapes For Class 7 Question 7.
Draw the sketches of two figures that have no vertex.
Solution:

Visualising Solid Shapes Class 7 Worksheets Question 8.
Name any three objects which resemble a sphere and cone.
Solution:
Sphere: Football, Earth, Round table
Cone: Conical funnel, ice cream cone, conical cracker

Questions On Visualising Solid Shapes Question 9.
What shape would we get from the given figure?

Solution:
From the given net, we get a rectangular pyramid.

Visualising Solid Shapes Class 7 Extra Questions Short Answer Type

Question 10.
For the solids given below sketch the front, side and top view

Solution:

Question 11.
Match the following:

Solution:
(i) → (e)
(ii) → (a)
(iii) → (b)
(iv) → (c)
(v) → (d)

Question 12.
Complete the following table:

Solution:

Question 13.
Draw a plan, front and side elevations of the following solids.

Solution:

Question 14.
Name the solid that would be formed by each net:

Solution:
(i) Triangular pyramid
(ii) Square pyramid
(iii) Hexagonal pyramid

Question 15.
Name the solids that have:
(i) 1 curved surface
(ii) 4 faces
(iii) 6 faces
(iv) 5 faces and 5 vertices
(v) 8 triangular faces
(vi) 6 triangular faces and 2 hexagonal faces.
Solution:
(i) Cylinder
(ii) Tetrahedron
(iii) Cube and cuboid
(iv) Square pyramid or rectangular pyramid
(v) Regular octahedron
(vi) Hexagonal prism.

Question 16.
Draw the top, side and front views of the given solids:

Solution:

Question 17.
Draw the net of a cuboid having same breadth and height, but length double the breadth.
Solution:

Question 18.
Draw the nets of the following:
(i) Triangular prisms
(ii) Tetrahedron
(iii) Cuboid.
Solution:

NCERT Solutions for Class 7 Maths

Comparing Quantities Class 7 Extra Questions Maths Chapter 8

Comparing Quantities Class 7 Extra Questions with Answers CBSE

September 22, 2024

Extra Questions for Class 7 Maths PDF are provided here. Students can download the pdf of these solutions from the given links. Comparing Quantities Class 7 Extra Questions Maths Chapter 8 provided in accordance with the latest syllabus of CBSE which, in turn, help the students to build a strong foundation and secure excellent marks in their board exams. https://meritbatch.com/comparing-quantities-class-7-extra-questions/

NCERT Class 7 Maths Chapter 8 Extra Questions and Answers

Extra Questions for Class 7 Maths Chapter 8 Comparing Quantities

Comparing Quantities Class 7 Extra Questions Very Short Answer Type

Comparing Quantities Class 7 Extra Questions Question 1.
Find the ratio of:
(a) 5 km to 400 m
(b) 2 hours to 160 minutes
Solution:
(a) 5 km = 5 × 1000 = 5000 m
Ratio of 5 km to 400 m
= 5000 m : 400 m
= 25 : 2
Required ratio = 25 : 2
(b) 2 hours = 2 × 60 = 120 minutes
Ratio of 2 hours to 160 minutes
= 120 : 160
= 3 : 4
Required ratio = 3 : 4

Comparing Quantities Class 7 Worksheet Question 2.
State whether the following ratios are equivalent or not?
(a) 2 : 3 and 4 : 5
(b) 1 : 3 and 2 : 6
Solution:
(a) Given ratios = 2 : 3 and 4 : 5

Hence 2 : 3 and 4 : 5 are not equivalent ratios.
(b) Given ratios = 1 : 3 and 2 : 6
LCM of 3 and 6 = 6

Hence, 1 : 3 and 2 : 6 are equivalent ratios.

Comparing Quantities Class 7 Worksheets With Answers Pdf Question 3.
Express the following ratios in simplest form:
(a) 6\(\frac { 1 }{ 5 }\) : 2\(\frac { 1 }{ 3 }\)
(b) 42 : 56
Solution:

Class 7 Maths Comparing Quantities Extra Questions Question 4.
Compare the following ratios:
3 : 4, 5 : 6 and 3 : 8
Solution:
Given: 3 : 4, 5 : 6 and 3 : 8
or \(\frac { 3 }{ 4 }\) , \(\frac { 5 }{ 6 }\) and \(\frac { 3 }{ 8 }\)
LCM of 4, 6 and 8 = 24

Hence, 3 : 8 < 3 : 4 < 5 : 6

Class 7 Maths Ch 8 Extra Questions Question 5.
State whether the following ratios are proportional or not:
(i) 20 : 45 and 4 : 9
(ii) 9 : 27 and 33 : 11
Solution:
(i) 20 : 45 and 4 : 9
Product of extremes = 20 × 9 = 180
Product of means = 45 × 4 = 180
Here, the product of extremes = Product of means
Hence, the given ratios are in proportion.
(ii) 9 : 27 and 33 : 11
Product of extremes = 9 × 11 = 99
Product of means = 27 × 33 = 891
Here, the product of extremes ≠ Product of means
Hence, the given ratios are not in proportion.

Ratio And Proportion Class 7 Extra Questions Question 6.
24, 36, x are in continued proportion, find the value of x.
Solution:
Since, 24, 36, x are in continued proportion.
24 : 36 :: 36 : x
⇒ 24 × x = 36 × 36
⇒ x = 54
Hence, the value of x = 54.

Comparing Quantities Extra Questions Class 7 Question 7.
Find the mean proportional between 9 and 16.
Solution:
Let x be the mean proportional between 9 and 16.
9 : x :: x : 16
⇒ x × x = 9 × 16
⇒ x² = 144
⇒ x = √144 = 12
Hence, the required mean proportional = 12.

Extra Questions On Comparing Quantities For Class 7 Question 8.
Find:
(i) 36% of 400
(ii) 16\(\frac { 2 }{ 3 }\)% of 32
Solution:

Class 7 Comparing Quantities Worksheet Question 9.
Find a number whose 6\(\frac { 1 }{ 4 }\)% is 12.
Solution:
Let the required number be x.

Hence, the required number = 192.

Class 7 Chapter 8 Maths Extra Questions Question 10.
What per cent of 40 kg is 440 g?
Solution:
Let x% of 40 kg = 440 g

Hence, the required Percentage = 1.1%

Comparing Quantities Class 7 Extra Questions Short Answer Type

Questions On Comparing Quantities For Class 7 Question 11.
Convert each of the following into the decimal form:
(а) 25.2%
(b) 0.15%
(c) 25%
Solution:

Chapter 8 Class 7 Maths Extra Questions Question 12.
What per cent of
(a) 64 is 148.48?
(b) 75 is 1225?
Solution:

Class 7 Maths Chapter 8 Worksheet With Answers Question 13.
A machine costs ₹ 7500. Its value decreases by 5% every year due to usage. What will be its price after one year?
Solution:
The cost price of the machine = ₹ 7500
Decrease in price = 5%
Decreased price after one year

= 75 × 95
= ₹ 7125
Hence, the required price = ₹ 7125.

Question 14.
What sum of money lent out at 12 per cent p.a. simple interest would produce ₹ 9000 as interest in 2 years?
Solution:
Here, Interest = ₹ 9000
Rate = 12% p.a.
Time = 2 years
Principal = ?

Hence, the required principal amount = ₹ 37500.

Question 15.
Rashmi obtains 480 marks out of 600. Rajan obtains 560 marks out of 700. Whose performance is better?
Solution:
Rashmi obtains 480 marks out of 600
Marks Percentage = \(\frac { 480 }{ 600 }\) × 100 = 80%
Rajan obtains 560 marks out of 700
Marks Percentage = \(\frac { 560 }{ 700 }\) × 100 = 80%
Since, both of them obtained the same per cent of marks i.e. 80%.
So, their performance cannot be compared.

Question 16.
₹ 9000 becomes ₹ 18000 at simple interest in 8 years. Find the rate per cent per annum.
Solution:
Here, Principal = ₹ 9000
Amount = ₹ 18000
Interest = Amount – Principal = ₹ 18000 – ₹ 9000 = ₹ 9000

Hence, the required rate of interest = 12\(\frac { 1 }{ 2 }\)%.

Question 17.
The cost of an object is increased by 12%. If the current cost is ₹ 896, what was its original cost?
Solution:
Here, rate of increase in cost = 12%
Increased Cost = ₹ 896
Original Cost = ?
Let the Original Cost be ₹ x

Hence, the required cost = ₹ 800.

Comparing Quantities Class 7 Extra Questions Long Answer Type

Question 18.
Radhika borrowed ₹ 12000 from her friends. Out of which ₹ 4000 were borrowed at 18% and the remaining at 15% rate of interest per annum. What is the total interest after 3 years? (NCERT Exemplar)
Solution:
Total amount borrowed by Radhika = ₹ 12,000
The amount borrowed by her at 18% p.a. = ₹ 4000

Total interest = ₹ 2160 + ₹ 3600 = ₹ 5760
Hence, the total interest = ₹ 5760.

Question 19.
Bhavya earns ₹ 50,000 per month and spends 80% of it. Due to pay revision, her monthly income increases by 20% but due to price rise, she has to spend 20% more. Find her new savings. (NCERT Exemplar)
Solution:
Monthly income of Bhavya = ₹ 50,000
Money spent by her = 80% of ₹ 50,000
= \(\frac { 80 }{ 100 }\) × 50,000 = ₹ 40,000
Due to pay revision, income is increased by 20%

So, the new savings = ₹ 60,000 – ₹ 48,000 = ₹ 12,000

Question 20.
The simple interest on a certain sum at 5% per annum for 3 years and 4 years differ by ₹ 82. Find the sum.
Solution:
Let the required sum be ₹ P.
Simple interest for 3 years

Alternate Method
Simple Interest gained from 3rd to 4th year = ₹ 82
Time (4th year – 3rd year) = 1 year

Required sum = ₹ 1640

Comparing Quantities Class 7 Extra Questions Higher Order Thinking Skills (HOTS) Type

Question 21.
Rajan’s monthly income is 20% more than the monthly income of Sarita. What per cent of Sarita’s income is less than Rajan’s monthly income?
Solution:
Let the monthly income of Sarita be ₹ 100.
Rajan’s monthly income

Now, Sarita’s monthly income is less than the monthly income of Raj an by = ₹ 120 – ₹ 100 = ₹ 20
Per cent of less in Rajan’s monthly income
= \(\frac { 20\times 100 }{ 120 }\) = \(\frac { 50 }{ 3 }\)% = 16\(\frac { 2 }{ 3 }\)%
Hence, the required per cent = 16\(\frac { 2 }{ 3 }\)%

Question 22.
If 10 apples are bought for ₹ 11 and sold at the rate of 11 apples for ₹ 10. Find the overall gain or loss per cent in these transactions.
Solution:
CP of 10 apples = ₹ 11
CP of 1 apple = ₹ \(\frac { 11 }{ 10 }\)
SP of 11 apples = ₹ 10
SP of 1 apple = ₹ \(\frac { 10 }{ 11 }\)

Question 23.
If 25 men can do a work in 36 hours, find the number of men required to do the same work in 108 hours.
Solution:
Let the number of men required to be x.
Men : Hours :: Men : Hours
25 : 36 :: x : 108
Product of extremes = 25 × 108
Product of means = 36 × x
Product of means = Product of extremes
36 × x = 25 × 108
⇒ x = 25 × 3 = 75
Hence, the required number of men = 75.

Question 24.
A machine is sold by A to B at a profit of 10% and then B sold it to C at a profit of 20%. If C paid ₹ 1200 for the machine, what amount was paid by A to purchase the machine?
Solution:
Cost price of machine for C = Selling price of the machine for B = ₹ 1200

Hence, the required cost price = ₹ 909\(\frac { 10 }{ 11 }\) or ₹ 909.09 (approx)

Category: CBSE

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