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NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.2 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.2.

• Fractions Class 6 Ex 7.1
• Fractions Class 6 Ex 7.3
• Fractions Class 6 Ex 7.4
• Fractions Class 6 Ex 7.5
• Fractions Class 6 Ex 7.6

NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.2

Question 1.
Draw number lines and locate the points on them:

Solution :
(a)

(b)

(c)

Question 2.
Express the following as mixed fractions:
(a) \(\frac { 20 }{ 3 }\)
(b) \(\frac { 11 }{ 5 }\)
(c) \(\frac { 17 }{ 7 }\)
(d) \(\frac { 28 }{ 5 }\)
(e) \(\frac { 19 }{ 6 }\)
(f) \(\frac { 35 }{ 9 }\).
Solution :

Question 3.
Express the following as improper fractions:

Solution :

We hope the NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.2 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.3 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.3.

• Integers Class 6 Ex 6.1
• Integers Class 6 Ex 6.2

NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.3

Question 1.
Find:
(a) 35 – (20)
(b) 72 – (90)
(c) (-15) – (-18)
(d) (-20) – (13)
(e) 23 -(-12)
(f) (-32) -(- 40).
Solution.
(a) 35-(20)
35 – (20)
= 35 + (additive inverse of 20)
= 35 + (- 20)
= 15 + 20 + (- 20)
= 15 + 0= 15

(b) 72-90
72-90
= 12 + (additive inverse of 90)
= 72 + (- 90)
= 72 + (- 72) + (- 18)
= 0 + (- 18) = – 18

(c) (- 15) – (- 18)
(-15)-(-18)
= (- 15) + (additive inverse of -18)
= (- 15) + (18)
= (- 15) + (15) + (3)
= 0 + (3) = 3

(d) (- 20) – (13)
= (-20)-(13)
= (- 20) + (additive inverse of 13)
= (- 20) + (- 13) = – 33

(e) 23 – (- 12)
23 – (- 12)
= 23 + (additive inverse of – 12)
= 23+12 = 35

(f) (- 32) – (- 40)
(-32)-(-40)
= (- 32) + (additive inverse of – 40)
= (-32)+ (+40)
= (- 32) + (+ 32) + (+ 8)
= 0 + (+ 8) = 8.

Question 2.
Fill in the blanks with >, < or = sign :
(a) (- 3) + (- 6)……… (-3)-(-6)
(b) (- 21) – (- 10)….. (-31)+ (-11)
(c) 45 -(- 11)……. 57 +(-4)
(d) (- 25) – (- 42)……. (-42)-(-25).
Solution.
(a) L.H.S. = (- 3) + (- 6) = – 9
R.H.S. = (- 3) – (- 6)
= (- 3) + (additive inverse of – 6)
= (- 3) + 6
= (- 3) + 3 + 3
=0+3=3
∴ (- 3) + (- 6) < (- 3) – (- 6)

(b) L.H.S. = (- 21) – (- 10)
= (- 21) + (additive inverse of -10)
= (- 21) + 10
= (- ll) + (- 10)+ 10
= (- 11) + 0 = – 11
R.H.S. = (-31)+ (-11)
= -42
∴ (-21)-(-10) >(-31)+ (-11)

(c) L.H.S. =45-(-11)
= 45 + (additive inverse of – 11)
= 45 + 11 = 56
R.H.S.;= 57 + (-4)
= 53 + 4 + (- 4)
= 53 + 0 = 53
∴ 45-(-11) >57 +(-4)

(d) L.H.S. = (-25) – (-42)
= (- 25) + (additive inverse of – 42)
= (-25)+ (+42)
= (- 25) + (+ 25) + (+ 17)
= 0 + (+ 17)= 17
R.H.S. = (- 42) – (- 25)
= (- 42) + (additive inverse of – 25)
= (- 42) + (+ 25)
= (- 17) + (- 25) + (+ 25)
= (- 17) + 0 = – 17
∴ (-25)-(-42) >(-42)-(-25).

Question 3.
Fill in the blanks: 
(a) (-8) + =0
(b) 13 + = 0
(c) 12 + (-12) –
(d) (-4) + =-72
(e) – 75 = -10.
Solution.
(a) (-8)+ 8 = 0
(b) 13 + (-13) = 0
(c) 12 + (- 12) = 0.
(d)(-4) + (-8) = -12
(e) (+5)-15 =-10.

Question 4.
Find:
(a) (-7) -8 -(-25)
(b) (-13)+ 32-8-1
(c) (-7) + (-8) + (-90)
(d) 50-(-40)-(-2)
Solution.
(a) (- 7) – 8 – (- 25)
(- 7) – 8 – (- 25)
= (- 7) + (additive inverse of 8) – (- 25)
= (- 7) + (- 8) – (- 25)
= – 15-(-25)
= – 15 + (additive inverse of – 25)
= -15+ (+25)
= – 15 + (+ 15) + (+ 10)
= 0 + (+ 10) = 10

(b) (- 13) + 32 – 8 – 1
(-13)+ 32-8-1
= (-13)+ 32-9
= (- 13) + 32 + (additive inverse of 9)
= (- 13) + 32 + (- 9)
= (- 13) + 23 + 9 + (- 9)
= (-13)+ 23 + 0
= (- 13)+ 23
= (-13)+13+10
= 0+ 10 = 10

(c) (- 7) + (- 8) + (- 90)
(-7) +(-8)+ (-90)
= (-15)+ (-90)
= -105

(d) 50-(-40)-(-2) 
= 50 – (- 40) – (- 2)
= 50 + (additive inverse of – 40) – (- 2)
= 50+ (40)-(-2)
= 90-(-2)
= 90 + (additive inverse of – 2)
= 90 + 2 = 92.

We hope the NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.3 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.2 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.2.

• Integers Class 6 Ex 6.1
• Integers Class 6 Ex 6.3

NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.2

Question 1.
Using number line write the integer which is:
(a) 3 more than 5
(b) 5 more than – 5
(c) 6 less than 2
(d) 3 less than – 2.
Solution.
(a) 3 more than 5
We start from 5 and proceed 3 steps to the right of 5 to reach 8 as shown below:

Therefore, 3 more than 5 is 8.

(b) 5 more than – 5
We start from – 5 and move to the right by 5 steps and obtain 0 as shown below:

Therefore, 5 more than – 5 is 0.

(c) 6 less than 2
We start from 2 and proceed 6 steps to the left of 2 to reach – 4 as shown below:

(d) 3 less than – 2
We start from – 2 and move to the left by 3 steps and obtain – 5 as shown below:

Therefore, 3 less than – 2 is – 5.

Question 2.
Use number line and add the following integers:
(a) 9 +(-6)
(b) 5+ (-11)
(c) (- 1) +(- 7)
(d) (-5)+ 10
(e) (-1) +(-2) + (-3)
(f) (-2) + 8 + (-4).
Solution.
(a) 9 +(-6)
On the number line, we first move 9 steps to the right from 0 reaching 9 and then we move 6 steps to the left of 9 and reach 3.
Thus, 9 + (- 6) = 3.

(b) 5 + (-11)
On the number line, we first move 5 steps to the right from 0 reaching 5 and then we move 11 steps to the left of 5 and reach – 6.
Thus, 5 + (- 11) = – 6.

(c) (-1) + (-7)
On the number line we first move 1 step to the left of 0 reaching – 1, then we move 7 steps to the left of – 1 and reach – 8.
Thus, (- 1) + (- 7) = – 8.

(d) (- 5) + 10
First, we move 5 steps to the left of 0 reaching – 5, then from this point, we move 10 steps to the right. We reach the point + 5.
Thus, (- 5) + 10 = 5.

(e) (-l) + (- 2) + (-3)
First, we move 1 step to the left of 0 reaching – 1, then from this point, we move 2 steps to the left to reach – 3 and finally from – 3, we move 3 steps to the left. We reach the point – 6.
Thus, (- 1) + (- 2) + (- 3) = – 6.

(f) (- 2) + 8 + (- 4)
First, we move 2 steps to the left of 0 reaching – 2, then from this point, we move 8 steps to the right to reach + 6 and finally from + 6 we move 4 steps to the left. We reach the point 2.
Thus, (- 2) + 8 + (- 4) = 2.

Question 3.
Add without using number line:
(a) 11 + (- 7)
(b) (- 13) + (+ 18)
(c) (-10) + (+ 19)
(d) (-250) + (+ 150)
(e) (- 380) + (- 270)
(f) (-217) + (-100).
Solution.
(a) 11 + (- 7)
11+(-7)
= 4 + 7 + (- 7)
=4+0=4

(b) (- 13) + (+ 18)
(-13)+ (+18)
= (- 13)+ (+.13) +(+5)
= 0 + (+5) = 5

(c) (-10) + (+ 19)
(-10)+ (+19)
= (- 10) + (+ 10) + (+ 9)
= 0 + (+ 9) = 9=

(d) (- 250) + (+ 150)
(- 250) + (+ 150)
= (- 100) + (- 150) + (+ 150)
= (- 100) + 0 = – 100

(e) (- 380) + (- 270)
(- 380) + (- 270)
= -650 (/) (- 217) + (-100)
= -317.

Question 4.
Find the sum of:
(a) 137 and – 354
(b) – 52 and 52
(c) – 312, 39 and 192
(d) – 50, – 200 and 300.
Solution.
(a) 137 and – 354
137+ (-354)
= 137+ (- 137)+ (-217)
= 0 +(-217) =-217

(b) – 52 and 52
– 52 + (52)
= 0

(c) – 312,39 and 192
(-312)+ (39)+ (192)
= (-312)+ (231)
= (-81)+ (-231)+ (231)
= (- 81) + 0 = – 81

(d) – 50, – 200 and 300
(-50)+ (-200)+ (300)
= (- 250) + (300)
= (- 250) + (250) + (50)
= 0 + (50) = 50.

Question 5.
Find the value of:
(a) (-7) +(-9)+ 4 + 16
(b) (37) +(-2) +(-65) +(-8).
Solution.
(a) (- 7) + (- 9) + 4 + 16
(- 7) + (- 9) + 4 + 16
= (-16)+ 20 = (-16) +16 + 4 =0+4=4

(b) (37) +(-2) +(-65) +(-8)
= (37)+ (75)
= (37)+ (-37)+ (-38) = 0 + (- 38)
= -38.

We hope the NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.2 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.9 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.9.

• Understanding Elementary Shapes Class 6 Ex 5.1
• Understanding Elementary Shapes Class 6 Ex 5.2
• Understanding Elementary Shapes Class 6 Ex 5.3
• Understanding Elementary Shapes Class 6 Ex 5.4
• Understanding Elementary Shapes Class 6 Ex 5.5
• Understanding Elementary Shapes Class 6 Ex 5.6
• Understanding Elementary Shapes Class 6 Ex 5.7
• Understanding Elementary Shapes Class 6 Ex 5.8

NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.9

Chapter 5 Understanding Elementary Shapes Ex 5.9

Question 1.
Match the following :
(a) Cone

(b) Sphere

(c) Cylinder

(d) Cuboid

(e) Pyramid

Give two new examples of each shape.
Solution :
(a) ↔ (ii)
(b) ↔ (iv)
(c) ↔ (v)
(d) ↔ (iii)
(e) ↔ (i).
(i) Birthday cap, ice-cream cone
(ii) Tennis Ball, laddu
(iii) Road-roller, gas cylinder
(iv) Brick, book
(v) Pyramids of Egypt, right pyramid.

Question 2.
What shape is
(a) Your installments box?
(b) A brick?
(c) A matchbox?
(d) A road-roller?
(e) A sweet laddu?
Solution :
(a) Cuboid
(b) Cuboid
(c) Cuboid
(d) Cylinder
(e) Sphere.

We hope the NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.9 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.9, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.8 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.8.

• Understanding Elementary Shapes Class 6 Ex 5.1
• Understanding Elementary Shapes Class 6 Ex 5.2
• Understanding Elementary Shapes Class 6 Ex 5.3
• Understanding Elementary Shapes Class 6 Ex 5.4
• Understanding Elementary Shapes Class 6 Ex 5.5
• Understanding Elementary Shapes Class 6 Ex 5.6
• Understanding Elementary Shapes Class 6 Ex 5.7
• Understanding Elementary Shapes Class 6 Ex 5.9

NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.8

Question 1.
Examine whether the following are polygons. If anyone among them is not, say why?

Solution :
(a) is not a closed figure and hence is not a polygon.
(b) is a polygon of six sides.
(c) is not a polygon since it is not made of line segments.
(d) is not a polygon since it is not made of line segments.

Question 2.
Name each polygon:

Make two more examples of each of these.
Solution :
(a) A Quadrilateral
(b) A Triangle
(c) A Pentagon (5-sided)
(d) An Octagon (8-sided).

Question 3.
Draw a rough sketch of a regular hexagon. Connecting any three of its vertices, draw a triangle. Identify the type of the triangle you have drawn.
Solution :

The triangle ABC drawn is an obtuse-angled isosceles triangle.

Question 4.
Draw a rough sketch of a regular octagon. (Use squared paper if you wish). Draw a rectangle by joining exactly four of the vertices of the octagon.
Solution :

Question 5.
A diagonal is a line segment that joins any two vertices of the polygon and is not a side of the polygon. Draw a rough sketch of a pentagon and draw its diagonals.
Solution :

We hope the NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.8 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.8, drop a comment below and we will get back to you at the earliest.