## Maharashtra Board Practice Set 18 Class 7 Maths Solutions Chapter 4 Angles and Pairs of Angles

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 18 Answers Solutions Chapter 4 Angles and Pairs of Angles.

## Angles and Pairs of Angles Class 7 Practice Set 18 Answers Solutions Chapter 4

Question 1.
Name the pairs of opposite rays in the figure alongside.

Solution:

1. Ray PL and ray PM
2. Ray PN and ray PT

Question 2.
Are the ray PM and PT opposite rays? Give reasons for your answer.

Solution:
No.
Ray PM and Ray PT do not form a straight line and hence are not opposite rays.

Maharashtra Board Class 7 Maths Chapter 4 Angles and Pairs of Angles Practice Set 18 Intext Questions and Activities

Question 1.
Observe the adjacent figure and answer the following questions. (Textbook pg. no. 28)

1. Name the rays in the figure alongside.
2. Name the origin of the rays
3. Name the angle in the given figure

Solution:

1. Ray BA and ray BC
2. Point B
3. ∠ABC or ∠CBA

Question 2.
Observe the adjacent figure and answer the following questions. (Textbook pg. no. 28)

1. Name the angle in the figure alongside.
2. Name the rays whose origin is point B

Solution:

1. ∠ABC or ∠CBA
2. Ray BA and ray BC

## Maharashtra Board Practice Set 28 Class 7 Maths Solutions Chapter 6 Indices

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 28 Answers Solutions Chapter 6 Indices.

## Indices Class 7 Practice Set 28 Answers Solutions Chapter 6

Question 1.
Simplify:
i. a6 ÷ a4
ii. m5 ÷ m8
iii. p3 ÷ p13
iv. x10 ÷ x10
Solution:
i. a6 ÷ a4
= a6-4
= a2

ii. m5 ÷ m8
= m5-8
= m-3

iii. p3 ÷ p13
= p3-13
= p-10

iv. x10 ÷ x10
= x10-10
= x0
= 1

Question 2.
Find the value of:
i. (-7)12 ÷ (-7)12
ii. 75 ÷ 73
iii. $$\left(\frac{4}{5}\right)^{3} \div\left(\frac{4}{5}\right)^{2}$$
iv. 47 ÷ 45
Solution:
i. (-7)12 ÷ (-7)12
= (-7)12-12
= (-7)0
= 1

ii. 75 ÷ 73
= 75-3
= 72
= 49

iii. $$\left(\frac{4}{5}\right)^{3} \div\left(\frac{4}{5}\right)^{2}$$
$$=\left(\frac{4}{5}\right)^{3-2}=\frac{4}{5}$$

iv. 4 ÷ 4
= 47-5
= 42
= 16

## Maharashtra Board Practice Set 27 Class 7 Maths Solutions Chapter 6 Indices

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 27 Answers Solutions Chapter 6 Indices.

## Indices Class 7 Practice Set 27 Answers Solutions Chapter 6

Question 1.
Simplify:
i. 74 × 72
ii. (-11)5 × (-11)2
iii. $$\left(\frac{6}{7}\right)^{3} \times\left(\frac{6}{7}\right)^{5}$$
iv. $$\left(-\frac{3}{2}\right)^{5} \times\left(-\frac{3}{2}\right)^{3}$$
v. a16 × a7
vi. $$\left(\frac{\mathrm{P}}{5}\right)^{3} \times\left(\frac{\mathrm{P}}{5}\right)^{7}$$
Solution:
i. 74 × 72
= 74+2
= 76

ii. (-11)5 × (-11)2
= (-11)5+2
= (-11)7

iii. $$\left(\frac{6}{7}\right)^{3} \times\left(\frac{6}{7}\right)^{5}$$
$$=\left(\frac{6}{7}\right)^{3+5}=\left(\frac{6}{7}\right)^{8}$$

iv. $$\left(-\frac{3}{2}\right)^{5} \times\left(-\frac{3}{2}\right)^{3}$$
$$=\left(-\frac{3}{2}\right)^{5+3}=\left(-\frac{3}{2}\right)^{8}$$

v. a16 × a7
= a16+7
= a23

vi. $$\left(\frac{\mathrm{P}}{5}\right)^{3} \times\left(\frac{\mathrm{P}}{5}\right)^{7}$$
$$=\left(\frac{\mathrm{P}}{5}\right)^{3+7}=\left(\frac{\mathrm{P}}{5}\right)^{10}$$

## Maharashtra Board Practice Set 26 Class 7 Maths Solutions Chapter 6 Indices

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 26 Answers Solutions Chapter 6 Indices.

## Indices Class 7 Practice Set 26 Answers Solutions Chapter 6

Question 1.
Complete the table below:

 Sr. No. Indices (Numbers in index form) Base Index Multiplication form Value i. 34 3 4 3 x 3 x 3 x 3 81 ii. 163 iii. (-8) 2 iv. $$\frac{3}{7} \times \frac{3}{7} \times \frac{3}{7} \times \frac{3}{7}$$ $$\frac { 81 }{ 2401 }$$ v. (-13)4

Solution:

 Sr. No. Indices (Numbers in index form) Base Index Multiplication form Value i. 34 3 4 3 x 3 x 3 x 3 81 ii. 163 16 3 16 x 16 x 16 4096 iii. (-8)² (-8) 2 -8 x -8 64 iv. $$\left(\frac{3}{7}\right)^{4}$$ $$\frac { 7 }{ 7 }$$ 4 $$\frac{3}{7} \times \frac{3}{7} \times \frac{3}{7} \times \frac{3}{7}$$ $$\frac { 81 }{ 2401 }$$ v. (-13)4 -13 4 (-13) x (-13) x (-13) x (-13) 28561

Question 2.
Find the value of.
i. 210
ii. 53
iii. (-7)4
iv. (-6)3
v. 93
vi. 81
vii. $$\left(\frac{4}{5}\right)^{3}$$
viii. $$\left(-\frac{1}{2}\right)^{4}$$
Solution:
i. 210
= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= 1024

ii. 53
= 5 × 5 × 5
= 125

iii. (-7)4
= (-7) × (-7) × (-7) × (-7)
= 2401

iv. (-6)3
= (-6) × (-6) × (-6)
= -216

v. 93
= 9 × 9 × 9
= 729

vi. 81
= 8

vii. $$\left(\frac{4}{5}\right)^{3}$$
$$=\frac{4}{5} \times \frac{4}{5} \times \frac{4}{5}=\frac{64}{125}$$

viii. $$\left(-\frac{1}{2}\right)^{4}$$
$$=\left(-\frac{1}{2}\right) \times\left(-\frac{1}{2}\right) \times\left(-\frac{1}{2}\right) \times\left(-\frac{1}{2}\right)=\frac{1}{16}$$

## Maharashtra Board Practice Set 13 Class 7 Maths Solutions Chapter 3 HCF and LCM

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 13 Answers Solutions Chapter 3 HCF and LCM.

## HCF and LCM Class 7 Practice Set 13 Answers Solutions Chapter 3

Question 1.
Find the LCM:
i. 12, 15
ii. 6, 8, 10
iii. 18, 32
iv. 10, 15, 20
v. 45, 86
vi. 15, 30, 90
vii. 105, 195
viii. 12,15,45
ix. 63,81
x. 18, 36, 27
Solution:
i. 12, 15

∴ LCM of 12 and 15 = 3 x 2 x 2 x 5
= 60

ii. 6, 8, 10

∴ LCM of 6, 8 and 10 = 2 x 2 x 3 x 2 x 5
= 120

iii. 18, 32

∴ LCM of 18 and 32 = 2 x 2 x 2 x 2 x 3 x 3 x 2
= 288

iv. 10, 15, 20

∴ LCM of 10, 15 and 20 = 5 x 2 x 3 x 2
= 60

v. 45, 86

∴ LCM of 45 and 86 = 2 x 3 x 3 x 5 x 43
= 3870

vi. 15, 30, 90

∴ LCM of 15,30 and 90 = 3 x 5 x 2 x 3
= 90

vii. 105, 195

∴ LCM of 105 and 195 = 5 x 3 x 7 x 13
= 1365

viii. 12, 15, 45

∴ LCM of 12, 15 and 45 = 3 x 3 x 2 x 5 x 2
= 180

ix. 63, 81

∴ LCM of 63 and 81 = 3 x 3 x 3 x 7 x 3
= 567

x. 18, 36, 27

∴ LCM of 18, 36 and 27 = 3 x 3 x 2 x 2 x 3
= 108

Question 2.
Find the HCF and LCM of the numbers given below. Verify that their product is equal to the product of the given numbers:
i. 32, 37
ii. 46, 51
iii. 15, 60
iv. 18, 63
v. 78, 104
Solution:
i. 32 = 2 x 16
= 2 x 2 x 8
= 2 x 2 x 2 x 4
= 2 x 2 x 2 x 2 x 2 x 1
37 = 37 x 1
∴ HCF of 32 and 37 =1
LCM of 32 and 37 = 2 x 2 x 2 x 2 x 2 x 37
= 1184
HCF x LCM = 1 x 1184
= 1184
Product of the given numbers = 32 x 37
= 1184
∴ HCF x LCM = Product of the given numbers.

ii. 46 = 2 x 23 x 1
51 = 3 x 17 x 1
∴ HCF of 46 and 51 = 1
LCM of 46 and 51 = 2 x 23 x 3 x 17
= 2346
HCF x LCM = 1 x 2346
= 2346
Product of the given numbers = 46 x 51
= 2346
∴ HCF x LCM = Product of the given numbers.

iii. 15 = 3 x 5
60 = 2 x 30
= 2 x 2 x 15
= 2 x 2 x 3 x 5
∴ HCF of 15 and 60 = 3 x 5
= 15
LCM of 15 and 60 = 3 x 5 x 2 x 2
= 60
HCF x LCM = 15 x 60
= 900
Product of the given numbers = 15 x 60
= 900
∴ HCF x LCM = Product of the given numbers.

iv. 18 = 2 x 9
= 2 x 3 x 3
63 = 3 x 21
= 3 x 3 x 7
∴ HCF of 18 and 63 = 3 x 3
= 9
LCM of 18 and 63 = 3 x 3 x 2 x 7
= 126
HCF x LCM = 9 x 126
= 1134
Product of the given numbers = 18 x 63
= 1134
∴ HCF x LCM = Product of the given numbers.

v. 78 = 2 x 39
= 2 x 3 x 13
104 = 2 x 52
= 2 x 2 x 26
= 2 x 2 x 2 x 13
∴ HCF of 78 and 104 = 2 x 13
= 26
LCM of 78 and 104 = 2 x 13 x 3 x 2 x 2
= 312
HCF x LCM = 26 x 312
= 8112
Product of the given numbers = 78 x 104
= 8112
∴ HCF x LCM = Product of the given numbers.

Maharashtra Board Class 7 Maths Chapter 3 HCF and LCM Practice Set 13 Intext Questions and Activities

Question 1.
Write the tables of the given numbers and find their LCM. (Textbook pg. no. 19)
i. 6, 7
ii. 8, 12
iii. 5, 6, 15
Solution:
i. Multiples of 6 : 6, 12, 18, 24, 30, 36, 42
Multiples of 7 : 7, 14, 21, 28, 35, 42, 49
∴ LCM of 6 and 7 = 42

ii. Multiples of 8 : 8, 16, 24, 32, 40
Multiples of 12 : 12, 24, 36, 48
∴ LCM of 8 and 12 = 24

iii. Multiples of 5 : 5, 10, 15, 20, 25, 30, 35
Multiples of 6 : 6, 12, 18, 24, 30, 36
Multiples of 15 : 15, 30, 45, 60
∴ LCM of 5,6 and 15 = 30

## Maharashtra Board Practice Set 24 Class 7 Maths Solutions Chapter 5 Operations on Rational Numbers

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 24 Answers Solutions Chapter 5 Operations on Rational Numbers.

## Operations on Rational Numbers Class 7 Practice Set 24 Answers Solutions Chapter 5

Question 1.
Write the following rational numbers in decimal form.
i. $$\frac { 13 }{ 4 }$$
ii. $$\frac { -7 }{ 8 }$$
iii. $$7\frac { 3 }{ 5 }$$
iv. $$\frac { 5 }{ 12 }$$
v. $$\frac { 22 }{ 7 }$$
vi. $$\frac { 4 }{ 3 }$$
vii. $$\frac { 7 }{ 9 }$$
Solution:
i. $$\frac { 13 }{ 4 }$$

ii. $$\frac { -7 }{ 8 }$$

iii. $$7\frac { 3 }{ 5 }$$

iv. $$\frac { 5 }{ 12 }$$

v. $$\frac { 22 }{ 7 }$$

vi. $$\frac { 4 }{ 3 }$$

vii. $$\frac { 7 }{ 9 }$$

Maharashtra Board Class 7 Maths Chapter 5 Operations on Rational Numbers Practice Set 24 Intext Questions and Activities

Question 1.
Without using division, can we tell from the denominator of a fraction, whether the decimal form of the fraction will be a terminating decimal? Find out. (Textbook pg. no. 40)
Solution:
If the prime factorization of the denominator of a fraction has only factors as 2 or 5 or a combination of 2 and 5 then the decimal form of that fractional will be a terminating decimal form.
Consider the fractions $$\frac { 17 }{ 20 }$$ and $$\frac { 19 }{ 6 }$$
Now, 20 = 2 x 2 x 5, and 6 = 2 x 3
∴ $$\frac { 17 }{ 20 }$$ is terminating decimal form while $$\frac { 19 }{ 6 }$$ is recurring decimal form.

## Maharashtra Board Practice Set 23 Class 7 Maths Solutions Chapter 5 Operations on Rational Numbers

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 23 Answers Solutions Chapter 5 Operations on Rational Numbers.

## Operations on Rational Numbers Class 7 Practice Set 23 Answers Solutions Chapter 5

Question 1.
Write three rational numbers that lie between the two given numbers.
i. $$\frac{2}{7}, \frac{6}{7}$$
ii. $$\frac{4}{5}, \frac{2}{3}$$
iii. $$-\frac{2}{3}, \frac{4}{5}$$
iv. $$\frac{7}{9},-\frac{5}{9}$$
v. $$\frac{-3}{4}, \frac{+5}{4}$$
vi. $$\frac{7}{8}, \frac{-5}{3}$$
vii. $$\frac{5}{7}, \frac{11}{7}$$
viii. $$0, \frac{-3}{4}$$
Solution:
i. $$\frac{2}{7}, \frac{6}{7}$$
The three numbers lying between $$\frac { 2 }{ 7 }$$ and $$\frac { 6 }{ 7 }$$ are $$\frac{3}{7}, \frac{4}{7}, \frac{5}{7}$$

ii. $$\frac{4}{5}, \frac{2}{3}$$
$$\frac{4}{5}=\frac{24}{30}, \frac{2}{3}=\frac{20}{30}$$
The three numbers between $$\frac { 4 }{ 5 }$$ and $$\frac { 2 }{ 3 }$$ are $$\frac{21}{30}, \frac{22}{30}, \frac{23}{30}$$

iii. $$-\frac{2}{3}, \frac{4}{5}$$
$$\frac{-2}{3}=\frac{-10}{15}, \frac{4}{5}=\frac{12}{15}$$
The three numbers between $$\frac { -2 }{ 3 }$$ and $$\frac { 4 }{ 5 }$$ are $$\frac{-9}{15}, \frac{-7}{15}, \frac{4}{15}$$

iv. $$\frac{7}{9},-\frac{5}{9}$$
The three numbers between $$\frac { 7 }{ 9 }$$ and $$\frac { -5 }{ 9 }$$ are $$\frac{6}{9}, 0, \frac{-4}{9}$$

v. $$\frac{-3}{4}, \frac{+5}{4}$$
The three numbers between $$\frac { -3 }{ 4 }$$ and $$\frac { +5 }{ 4 }$$ are $$\frac{-2}{4}, \frac{-1}{4}, \frac{3}{4}$$

vi. $$\frac{7}{8}, \frac{-5}{3}$$
$$\frac{7}{8}=\frac{21}{24}, \frac{-5}{3}=\frac{-40}{24}$$
The three numbers between $$\frac { 7 }{ 8 }$$ and $$\frac { -5 }{ 3 }$$ are $$\frac{17}{24}, \frac{11}{24}, \frac{-13}{24}$$

vii. $$\frac{5}{7}, \frac{11}{7}$$
The three numbers between $$\frac { 5 }{ 7 }$$ and $$\frac { 11 }{ 7 }$$ are $$\frac{6}{7}, \frac{8}{7}, \frac{9}{7}$$

viii. $$0, \frac{-3}{4}$$
The three numbers between 0 and $$\frac { -3 }{ 4 }$$ are $$\frac{-1}{8}, \frac{-2}{8}, \frac{-5}{8}$$

Maharashtra Board Class 7 Maths Chapter 5 Operations on Rational Numbers Practice Set 23 Intext Questions and Activities

Question 1.
Answer the following questions: (Textbook pg. no. 36)

1. Write all the natural numbers between 2 and 9.
2. Write all the integers between -4, and 5.
3. Which rational numbers are there between $$\frac { 1 }{ 2 }$$ and $$\frac { 3 }{ 4 }$$ ?

Solution:

1. 3, 4, 5, 6, 7, 8
2. -3, -2, -1, 0, 1, 2, 3, 4
3. $$\frac{1}{2}=\frac{1 \times 2}{2 \times 2}=\frac{2}{4}=\frac{2 \times 10}{4 \times 10}=\frac{20}{40}$$
$$\frac{3}{4}=\frac{3 \times 10}{4 \times 10}=\frac{30}{40}$$
∴ The rational numbers between $$\frac { 1 }{ 2 }$$ and $$\frac { 3 }{ 4 }$$ are $$\frac{21}{40}, \frac{22}{40}, \frac{25}{40}, \frac{27}{40}$$ etc.

## Maharashtra Board Practice Set 21 Class 7 Maths Solutions Chapter 4 Angles and Pairs of Angles

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 21 Answers Solutions Chapter 4 Angles and Pairs of Angles.

## Angles and Pairs of Angles Class 7 Practice Set 21 Answers Solutions Chapter 4

Question 1.
∠ACD is an exterior angle of ∆ABC. The measures of ∠A and ∠B are equal. If m∠ACD = 140°, find the measures of the angles ∠A and ∠B.

Solution:
Let the measures of ∠A be x°.
m∠A = m∠B = x°
∠ACD is the exterior angle of ∆ABC
∴ m∠ACD = m∠A + m∠B
∴ 140 = x + x
∴ 140 = 2x
∴ 2x = 140
∴ x = $$\frac { 140 }{ 2 }$$
= 70
∴ The measures of the angles ∠A and ∠B is 70° each.

Question 2.
Using the measures of the angles given in the figure alongside, find the measures of the remaining three angles.

Solution:
m∠EOD = m∠AOB = 8y ….(vertically opposite angles)
∠FOL, ∠EOD and ∠COD form a straight angle.
∴ m∠FOE + m∠EOD + m∠COD = 180°
∴ 4y + 8y + 6y = 180
∴ 18y = 180
∴ y = $$\frac { 180 }{ 18 }$$
∴ y = 10
m∠EOD = 8y = 8 x 10 = 80°
m∠AOF = m∠COD ….(Vertically opposite angles)
= 6y = 6 x 10 = 60°
m∠BOC = m∠FOE ….(Vertically opposite angles)
= 4y = 4 x 10 = 40°
∴ The measures of ∠EOD, ∠AOF and ∠BOC are 80°, 60° and 40° respectively.

Question 3.
In the isosceles triangle ABC, ∠A and ∠B are equal. ∠ACD is an exterior angle of ∆ABC. The measures of ∠ACB and ∠ACD are (3x – 17)° and (8x + 10)° respectively. Find the measures of ∠ACB and ∠ACD. Also find the measures of ∠A and ∠B.

Solution:
Let the measure of ∠A be y°. A
∴ m∠A = m∠B = y°
∠ACB and ∠ACD form a pair of linear angles.
∴ m∠ACB + m∠ACD = 180°
∴ (3x – 17) + (8x + 10) = 180
∴ 3x + 8x – 17 + 10 = 180
∴ 11x – 7 = 180
∴ 11x – 7 + 7 = 180 + 7 …(Adding 7 on both sides.)
∴ 11x = 187
∴ x = $$\frac { 187 }{ 11 }$$ = 17
m∠ACB = 3x – 17 = (3 x 17) – 17 = 51 – 17 = 34°
m∠ACD = 8x + 10 = 8 x 17 + 10 = 136 + 10 = 146°
Here ∠ACD is the exterior angle of ∆ABC and ∠A and ∠B are its remote interior angles.
∴ m∠ACD = m∠A + m∠B
∴ 146 = y + y
∴ 146 = 2y
∴ 2y = 146
∴ y = $$\frac { 146 }{ 2 }$$ = 73
∴ The measures of ∠ACB, ∠ACD, ∠A and ∠B are 34°, 146°, 73° and 73° respectively.

Maharashtra Board Class 7 Maths Chapter 4 Angles and Pairs of Angles Practice Set 21 Intext Questions and Activities

Question 1.
Use straws or sticks to make all the kinds of angles that you have learnt about. (Textbook pg. no. 29)
Solution:
(Student should attempt the activity on their own)

Question 2.
Observe the table given below and draw your conclusions (Textbook pg. no. 31)

Solution:
i. 180°
ii. 360°
iii. 540°
iv. 720°
v. 180° x 5 = 900°
vi.  , 180° x 6 = 1080°

## Maharashtra Board Practice Set 22 Class 7 Maths Solutions Chapter 5 Operations on Rational Numbers

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 22 Answers Solutions Chapter 5 Operations on Rational Numbers.

## Operations on Rational Numbers Class 7 Practice Set 22 Answers Solutions Chapter 5

Question 1.
Carry out the following additions of rational numbers:
i. $$\frac{5}{36}+\frac{6}{42}$$
ii. $$1 \frac{2}{3}+2 \frac{4}{5}$$
iii. $$\frac{11}{17}+\frac{13}{19}$$
iv. $$2 \frac{3}{11}+1 \frac{3}{77}$$
Solution:
i. $$\frac{5}{36}+\frac{6}{42}$$

ii. $$1 \frac{2}{3}+2 \frac{4}{5}$$

iii. $$\frac{11}{17}+\frac{13}{19}$$

iv. $$2 \frac{3}{11}+1 \frac{3}{77}$$

Question 2.
Carry out the following subtractions involving rational numbers.
i. $$\frac{7}{11}-\frac{3}{7}$$
ii. $$\frac{13}{36}-\frac{2}{40}$$
iii. $$1 \frac{2}{3}-3 \frac{5}{6}$$
iv. $$4 \frac{1}{2}-3 \frac{1}{3}$$
Solution:
i. $$\frac{7}{11}-\frac{3}{7}$$

ii. $$\frac{13}{36}-\frac{2}{40}$$

iii. $$1 \frac{2}{3}-3 \frac{5}{6}$$

iv. $$4 \frac{1}{2}-3 \frac{1}{3}$$

Question 3.
Multiply the following rational numbers.
i. $$\frac{3}{11} \times \frac{2}{5}$$
ii. $$\frac{12}{5} \times \frac{4}{15}$$
iii. $$\frac{(-8)}{9} \times \frac{3}{4}$$
iv. $$\frac{0}{6} \times \frac{3}{4}$$
Solution:
i. $$\frac{3}{11} \times \frac{2}{5}$$
$$=\frac{3 \times 2}{11 \times 5}=\frac{6}{55}$$

ii. $$\frac{12}{5} \times \frac{4}{15}$$
$$=\frac{4}{5} \times \frac{4}{5}=\frac{4 \times 4}{5 \times 5}=\frac{16}{25}$$

iii. $$\frac{(-8)}{9} \times \frac{3}{4}$$
$$=\frac{(-2)}{3} \times \frac{1}{1}=\frac{-2}{3}$$

iv. $$\frac{0}{6} \times \frac{3}{4}$$
$$=0 \times \frac{3}{4}=0$$

Question 4.
Write the multiplicative inverse of.
i. $$\frac{2}{5}$$
ii. $$\frac{-3}{8}$$
iii. $$\frac{-17}{39}$$
iv. 7
v. $$-7 \frac{1}{3}$$
Solution:
i. $$\frac{5}{2}$$
ii. $$\frac{-8}{3}$$
iii. $$\frac{-39}{17}$$
iv. $$\frac {1}{7}$$
v. $$\frac {-3}{22}$$

Question 5.
Carry out the divisions of rational numbers:
i. $$\frac{40}{12} \div \frac{10}{4}$$
ii. $$\frac{-10}{11} \div \frac{-11}{10}$$
iii. $$\frac{-7}{8} \div \frac{-3}{6}$$
iv. $$\frac{2}{3} \div(-4)$$
v. $$2 \frac{1}{5} \div 5 \frac{3}{6}$$
vi. $$\frac{-5}{13} \div \frac{7}{26}$$
vii. $$\frac{9}{11} \div(-8)$$
viii. $$5 \div \frac{2}{5}$$
Solution:
i. $$\frac{40}{12} \div \frac{10}{4}$$
$$=\frac{40}{12} \times \frac{4}{10}=\frac{4}{3}$$

ii. $$\frac{-10}{11} \div \frac{-11}{10}$$
$$=\frac{-10}{11} \times \frac{-10}{11}=\frac{100}{121}$$

iii. $$\frac{-7}{8} \div \frac{-3}{6}$$
$$=\frac{-7}{8} \times \frac{-6}{3}=\frac{-7}{4} \times \frac{-3}{3}=\frac{7}{4}$$

iv. $$\frac{2}{3} \div(-4)$$
$$=\frac{2}{3} \times \frac{-1}{4}=\frac{1}{3} \times \frac{-1}{2}=\frac{-1}{6}$$

v. $$2 \frac{1}{5} \div 5 \frac{3}{6}$$
$$=\frac{11}{5} \div \frac{33}{6}=\frac{11}{5} \times \frac{6}{33}=\frac{1}{5} \times \frac{6}{3}=\frac{2}{5}$$

vi. $$\frac{-5}{13} \div \frac{7}{26}$$
$$=\frac{-5}{13} \times \frac{26}{7}=\frac{-10}{7}$$

vii. $$\frac{9}{11} \div(-8)$$
$$=\frac{9}{11} \times \frac{-1}{8}=\frac{-9}{88}$$

viii. $$5 \div \frac{2}{5}$$
$$=\frac{5}{1} \times \frac{5}{2}=\frac{25}{2}$$

Maharashtra Board Class 7 Maths Chapter 5 Operations on Rational Numbers Practice Set 22 Intext Questions and Activities

Question 1.
Complete the table given below. (Textbook pg. no. 34)

 -3 $$\frac {3}{5}$$ -17 $$\frac { -5 }{ 11 }$$ 5 Natural Numbers x ✓ Integers ✓ Rational Numbers ✓

Solution:

 -3 $$\frac {3}{5}$$ -17 $$\frac { -5 }{ 11 }$$ 5 Natural Numbers x x x x ✓ Integers ✓ x ✓ x ✓ Rational Numbers ✓ ✓ ✓ ✓ ✓

Question 2.
Discuss the characteristics of various groups of numbers in class and complete the table below. In front of each group, write the inference you make after carrying out the operations of addition, subtraction, multiplication and division, using a (✓) or a (x).
Remember that you cannot divide by zero. (Textbook pg. no. 35)

 Group of Numbers Addition Subtraction Multiplication Division Natural Numbers ✓ x (7- 10 =-3) ✓ x (3÷5=$$\frac { 3 }{ 5 }$$) Integers Rational Numbers

Solution:

 Group of Numbers Addition Subtraction Multiplication Division Natural Numbers ✓ x (7- 10 =-3) ✓ x (3÷5=$$\frac { 3 }{ 5 }$$) Integers ✓ ✓ ✓ x (4÷9=$$\frac { 4 }{ 9 }$$) Rational Numbers ✓ ✓ ✓ ✓

## Maharashtra Board Practice Set 20 Class 7 Maths Solutions Chapter 4 Angles and Pairs of Angles

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 20 Answers Solutions Chapter 4 Angles and Pairs of Angles.

## Angles and Pairs of Angles Class 7 Practice Set 20 Answers Solutions Chapter 4

Question 1.
Lines AC and BD intersect at point P. m∠APD = 47° Find the measures of ∠APB, ∠BPC, ∠CPD.

Solution:
∠APD and ∠APB are angles in a linear pair.
∴m∠APD + m∠APB = 180°
∴47 + m∠APB = 180
∴47 + m∠APB – 47 = 180 – 47 ….(Subtracting 47 from both sides)
∴m∠APB = 133°
m∠CPD = m∠APB = 133° … .(Vertically opposite angles)
m∠BPC = m∠APD = 47° … .(Vertically opposite angles)
∴The measures of ∠APB, ∠BPC and ∠CPD are 133°, 47° and 133° respectively.

Question 2.
Lines PQ and RS intersect at point M. m∠PMR = x°.What are the measures of ∠PMS, ∠SMQ and ∠QMR?
Solution:
∠PMR and ∠PMS are angles in a linear pair.
∴ m∠PMR + m∠PMS = 180°
∴ x + m∠PMS = 180
∴ m∠PMS = (180-x)°
m∠QMR = m∠PMS = (180 – x)° … .(Vertically opposite angles)
m∠SMQ = m∠PMR = x° …. (Vertically opposite angles)
∴The measures of ∠PMS, ∠SMQ and ∠QMR are (180 – x)°, x° and (180 – x)° respectively.