Class 7 Maths

MP Board Class 7 Maths Solutions For Chapter 2 Fractions and Decimals

1. \( 5 \times 2 \frac{3}{7}\)

Solution: \( 5 \times 2 \frac{3}{7}=5 \times 12 \frac{1}{7}=\frac{85}{7}=12 \frac{1}{7}\)

2. \(1 \frac{4}{9} \times 6\)

Solution: \( 1 \frac{4}{9} \times 6=a \times 6=\frac{78}{9}=8 \frac{6}{9}\)

 what is

1. \( \frac{1}{2}\) of 10 ?

Solution:

2. \( \frac{1}{4}\) of 16 ?

Solution:

3. \( \frac{2}{5}\)

Solution: \( \frac{2}{5} \text { of } 25=\frac{2}{5} \times 25=\frac{50}{5}=10\)

Fill in these boxes:

1) \( \frac{1}{2} \times \frac{1}{7}=\frac{1 \times 1}{2 \times 7}=\frac{1}{14} \)

2) \( \frac{1}{5} \times \frac{1}{7}=\frac{1 \times 1}{5 \times 7}=\frac{1}{35} \)

3) \( \frac{1}{7} \times \frac{1}{2}=\frac{1 \times 1}{7 \times 2}=\frac{1}{14} \)

4) \( \frac{1}{7} \times \frac{1}{5}=\frac{1 \times 1}{7 \times 5}=\frac{1}{35} \)

Mp Board Class 7 Book Solutions

Find:

1) \( \frac{1}{3} \times \frac{4}{5} \)

Solution: \( \frac{1}{3} \times \frac{4}{5}=\frac{1 \times 4}{3 \times 5}=\frac{4}{15} \)

2) \( \frac{2}{3} \times \frac{1}{5} \)

Solution: \( \frac{2}{3} \times \frac{1}{5}=\frac{2 \times 1}{3 \times 5}=\frac{2}{15} \)

Solutions To Try These

1) \( \frac{8}{3} \times \frac{4}{7} \)

Solution: \( \frac{8}{3} \times \frac{4}{7}=\frac{8 \times 4}{3 \times 7}=\frac{32}{21} \)

2) \( \frac{3}{4} \times \frac{2}{3} \)

Solution: \( \frac{3}{4} \times \frac{2}{3}=\frac{3 \times 2}{4 \times 3}=\frac{6}{12}=\frac{6 \div 6}{12 \div 6}=\frac{1}{2} \)

MP Board Class 7 Maths Solutions For Chapter 2 Fractions and Decimals

1. Find

(1) \( \frac{1}{4} of \) 1) \frac{1}{4} 2) \( \frac{3}{5} \) 3) \( \frac{4}{3}\)

1) \( \frac{1}{4} \text { of } \frac{1}{4}=\frac{1}{4} \times \frac{1}{4}=\frac{1 \times 1}{4 \times 4}=\frac{1}{16} \)

2). \( \frac{1}{4} \text { of } \frac{3}{5}=\frac{1}{4} \times \frac{3}{5}=\frac{1 \times 3}{4 \times 5}=\frac{3}{20} \)

3) \( \frac{1}{4} \text { of } \frac{4}{3}=\frac{1}{4} \times \frac{4}{3}=\frac{1 \times 4}{4 \times 3}=\frac{4}{12} \)

Mp Board Class 7 Book Solutions

2) \( \frac{1}{7} \) of 1) \( \frac{2}{9} \) 2) \( \frac{6}{5} \) 3) \( \frac{3}{10} \)

Solution:

1) \( \frac{1}{7} \text { of } \frac{2}{9}=\frac{1}{7} \times \frac{2}{9}=\frac{1 \times 2}{7 \times 9}=\frac{2}{63} \)

2)\( \frac{1}{7} \text { of } \frac{6}{5}=\frac{1}{7} \times \frac{6}{5}=\frac{1 \times 6}{7 \times 5}=\frac{6}{35} \)

3)\( \frac{1}{7} \text { of } \frac{3}{10}=\frac{1}{7} \times \frac{3}{10}=\frac{1 \times 3}{7 \times 10}=\frac{3}{70} \)

2. Multiply and reduce to lowest form (if possible):

1. \( \frac{2}{3} \times 2 \frac{2}{3} \)

Solution: \( \frac{2}{3} \times 2 \frac{2}{3}=\frac{2}{3} \times \frac{8}{3}=\frac{2 \times 8}{3 \times 3}=\frac{16}{9}=1 \frac{7}{9} \)

2. \( \frac{2}{7} \times \frac{7}{9} \)

Solution:

3. \( \frac{3}{8} \times \frac{6}{4} \)

Solution:

Mp Board Class 7 Maths Solutions

4. \( \frac{9}{5} \times \frac{3}{5} \)

Solution:

5. \( \frac{1}{3} \times \frac{15}{8} \)

Solution:

6. \( \frac{11}{2} \times \frac{3}{10} \)

Solution: \( \frac{11}{2} \times \frac{3}{10}=\frac{11 \times 3}{2 \times 10}=\frac{33}{20}=1 \frac{13}{20} \)

7. \( \frac{4}{5} \times \frac{12}{7} \)

Solution: \(\frac{4}{5} \times \frac{12}{7}=\frac{4 \times 12}{5 \times 7}=\frac{48}{35}=1 \frac{13}{35} \)

3. Multiply the following fractions:

1) \( \frac{2}{5} \times 5 \frac{1}{4} \)

Solution: \( \frac{2}{5} \times 5 \frac{1}{4}=\frac{2}{5} \times \frac{21}{4} \)

Class 7 Maths Chapter 2 Solutions Mp Board

2) \( 6 \frac{2}{5} \times \frac{7}{9} \)

Solution: \( 6 \frac{2}{5} \times \frac{7}{9}=\frac{32}{5} \times \frac{7}{9} \)

3) \( \frac{3}{2} \times 5 \frac{1}{3} \)

Solution: \( \frac{3}{2} \times 5 \frac{1}{3}=\frac{3}{2} \times \frac{16}{3}=\frac{3 \times 16}{2 \times 3}=\frac{48}{6} \)

4. \( \frac{5}{6} \times 2 \frac{3}{7} \)

Solution: \( \frac{5}{6} \times 2 \frac{3}{7}=\frac{5}{6} \times \frac{17}{7}=\frac{5 \times 17}{6 \times 7}=\frac{85}{42}=2 \frac{1}{42} \)

5. \( 3 \frac{2}{5} \times \frac{4}{7} \)

Solution: \( 3 \frac{2}{5} \times \frac{4}{7}=\frac{17}{5} \times \frac{4}{7}=\frac{17 \times 4}{5 \times 7}=\frac{68}{35}=1 \frac{33}{35} \)

6. \( 2 \frac{3}{5} \times 3 \)

Solution: \( 2 \frac{3}{5} \times 3=\frac{13}{5} \times 3=\frac{13 \times 3}{5}=\frac{39}{5}=7 \frac{4}{5} \)

7.\( 3 \frac{4}{7} \times \frac{3}{5} \)

Solution: \( =\frac{75 \div 5}{35 \div 5}=\frac{15}{7}=2 \frac{1}{7} \)

Mp Board Maths Chapter 2 Solutions

4. Which is greater:

1) \( \frac{2}{7} \) of \( \frac{3}{4} \) or \( \frac{3}{5} \) of \( \frac{5}{8} \)

Solution: \( \frac{2}{7} \text { of } \frac{3}{4}=\frac{2}{7} \times \frac{3}{4}=\frac{2 \times 3}{7 \times 4} \)

LCM = 2 x 7X 4 = 56

LCM of 14, 8 is 56

∴ 21 > 12

Mp Board Maths Chapter 2 Solutions

2) \( \frac{1}{2} \text { of } \frac{6}{7} \text { or } \frac{2}{3} \text { of } \frac{3}{7} \)

Solution:

3 > 2;

5. Saili plants 4 saplings,in a row,in her garden. The distance between two adjacent saplings is \( \frac{3}{4} \) m. Find the distance between the first and the last sapling.

Solution:

Distance between two saplings = \( \frac{3}{4} \) m.

Distance between the first and the last sapling = \( 3 \times \frac{3}{4}=\frac{3 \times 3}{4}=\frac{9}{4}=2 \frac{1}{4} \mathrm{~m} \)

6. Lipika reads a book for \( 1 \frac{3}{4} \) hours everyday. She reads the entire book in 6 days. How many hours in all were required by her to read the book ?

Solution:

No. of hours taken by Lipika to read the bookin a day = \( 1 \frac{3}{4} \)

No. of days taken toread the entire book = 6 days

Time takenby Lipika to read the book

7. A car covers 16 km using 1 litre of petrol.Howmuch distance willit cover using \( 2 \frac{3}{4} \) litres of petrol ?

Solution:

Distance covered with 1 litre of petrol = 16 km

Distance covered with \( 2 \frac{3}{4} \) litres of petrol = \( 16 \times 2 \frac{3}{4} \)

Class 7 Mp Board Maths Question Answers

8.

1) 1) Provide the number in the box such that □, \( \frac{2}{3} \) × □ = \( \frac{10}{30} \)

2) The simplest form of the number obtained in □ is…………….

Solution:

1. \( \frac{2}{3}  \) x  = \( \frac{5}{10}  \) \(\frac{10}{30} \)

2. \( \frac{5}{10}=\frac{5 \div 5}{10 \div 5}=\frac{1}{2} \)

2) 1) Provide the number in the box □, such that \( \frac{3}{5}\) x □ = \( \frac{24}{75}\)

2) The simplest form of the number obtained in □ is ……….

Solution:

1) \( \frac{3}{5} \times \frac{8}{15}=\frac{24}{75} \)

2) \( \frac{8}{15} \)

Think, Discuss And Write

1) Will the reciprocal of a proper fraction be again a proper fraction?

Solution. No, the reciprocal of a proper fraction will not be a proper fraction. It will be an improper fraction.

Example: \( \frac{2}{9} \text { its reciprocal is } \frac{9}{2} \)

\( \frac{2}{9} \) is aproper fraction; \( \frac{9}{2} \) is animproper fraction.

2) Will the reciprocal of an improper fraction be again animproper fraction?

Solution: No, the reciprocal of an improper fraction will not.be an improper fraction again.It will be a proper fraction.

Example: \( \frac{5}{4} \) is animproper fraction.But its reciprocal \( \frac{4}{5} \) is a proper fraction.

MP Board Class 7 Maths Solutions For Chapter 1 Integers

1. Evaluate each of the following:

1) (-30) ÷ 10

Solution: (-30) ÷ 10 = -3

2) 50 ÷ (-5)

Solution: 50 ÷ (-5) = -10

3) (-36) ÷ (-9)

Solution: (-36) ÷ (-9) =4

4) (-49) ÷ 49

Solution: (-49) ÷ 49 =-1

5) 13 ÷ [(-2) + 1]

Solution: 13 ÷ [(-2) + 1] = 13 ÷ (-1) = -13

6) 0 ÷(-12)

Solution: 0 ÷ (-12) = 0

7) (-31) ÷ [(-30) + (-1)]

Solution: (-31)+ [(-30) +(-1)] = (-31) + (-31) = 1

8) (-36) +12) ÷ 3

Solution: [(-36) +12)] ÷ 3 =(-3) ÷ 3 =-1

Mp Board Class 7 Maths Solutions

9) [(-6) +5] ÷ [(-2)+1]

Solution: [(- 6) + 5]÷ [(-2) +1] = (-1) + (-1) =1

2) Verify that a + (b + c) (a + b) + (a + c) for each of the following values of a, b and c.

1) a = 12 ; b = -4 ; c = 2 

Solution: a ÷ (b + c) =12 ÷ [(- 4) + 2]

= 12 ÷ (-2) =- 6

[a ÷ b] +[a ÷ c] =[12 ÷ (- 4)] +[12 ÷ 2]

= (-3) +6 = 3

a ÷ (b ÷ c) ≠ (a ÷ b) + (a ÷ c)

2) a = -10 ; b =1 ; c =1

Solution: a ÷ (b + c) = (-10) ÷ (1 + 1)

= (-10) ÷ 2 = -5

[a ÷ b] + [a ÷ c] = [(-10) +1]

+ [(-10) ÷ 1] = (-10) + (-10) = -20

a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c)

Class 7 Maths Chapter 1 Solutions Mp Board

3. Fill in the blanks:

1) 369 + 1 =369

2) (-75) +75 =-1

3) (-206) + (-206) =1

4) (-87) + (-1) = 87

5) -87 + 1 = -87

7) (-48) + 48 =-1

8) 20 +(-10) = -2

9) (-12) +4 = -3 .

4. Write five pairs of integers (a, b) such that a ÷ b = -3. One such pair is (6, -2) because 6 ÷ (-2) = -3.

Solution:

Five pairs of integers (a, b) such that a ÷ b = -3 are

1) 9 ÷ (-3) =-3              ∴ [9,-3]

2) 12 ÷ (-4) =-3            ∴ [12,-4]

3) 27 ÷ (-9) = -3          ∴ [27,-9]

4) 15÷ (-5)= -3            ∴ [15,-5]

5) (-18) ÷ 6 =-3            ∴ [-18,6]

Mp Board Maths Chapter 1 Solutions

5. The temperature at 12 noon was 10°C above zero.If it decreases at the rate of 2°C. per horn: until midnight, at what time would the temperature be 8°C below zero? What would be the temperature at mid night?

Solution:

Temperature. at 12 noon = 10°C

Temperature at 1 PM = = 10°C- 2°C = 8°C

Temperature at 3 PM = 6°C – 2°C = 4°C

Temperature at 4 PM = 4°C- 2°C = 2°C

Temperature at 5 PM = 2°C- 2°C = 0°C

Temperature at 6 PM = 0°C- 2°C = – 2°C

Temperature at 7PM = -2°C- 2°C =-4°C

Temperature at 8 PM =- 4°C- 2°C =- 6°C

Temperature at 9 PM = – 6°C- 2°C =- 8°C

The temperature will be 8°C below zero at 9 PM

Temperature at 10 PM =- 8°C- 2°C =- 10°C

Temperature at11 PM =- 10°C- 2°C =- 12°C

Temperature at12 PM i.e. midnight = -12°C- 2°C =- 14°C

Temperature at midnight is 14 degrees below zero.

Mp Board Class 7 Book Solutions

6. In a class test (+3) marks are given for every correct answer and (-2) marks are given for every incorrect answer and no marks for not attempting any question.

1) Radhika scored 20 marks. If she has got 12 correct answers, how many questions has she attempted incorrectly ?

2) Mohini scores -5 marks in this test, though she has got 7 correct answers.How many questions has she attempted incorrectly?

Solution:

1) Marks given for each correct answer = 3

Marks given for 12 correct answers =12×3 = 36

Radhika’s score = 20

Marks given for incorrect answers = 20 -36 = -16

Marks given for each incorrect answer = (-2)

So number of incorrect answers = (-16) + (-2) = 8

2) Marks given for 7 correct answers = 7 x 3 = 21

Mohini’s score = -5

Marks received for incorrect answers = -5-21 = -26

Marks given for one incorrect answer = -2

Number of incorrect answers = -26/-2 =13

7. An elevator descends into a mine shaft at the rate of 6 m/min. If the descent starts from 10 m above the ground level, how long will it take to reach – 350 m ?

Solution: Total distance covered by elevator = 10 -(-350) =360m

Speed of the elevator = 6 m / min

360 m descent = -360/6 = 60 min

∴ The elevator will take 60 minutes or 1hour to reach- 350m below the ground level.

 1) (-100)÷ 5

Solution: (-100) ÷ 5 = -20

2) (-81) ÷ 9

Solution: (-81) ÷ 9 = -9

3) (-75) ÷ 5

Solution: (-75) ÷ 5 = -15

4) (-32) ÷ 2

Solution: (-32) ÷ 2 = -16

Find

1) 125 ÷ (-25)

Solution: 125 ÷ (-25) = -5

2) 80 ÷ (-5)

Solution: 80÷(-5) = -16

3) 64 ÷(-16)

Solution: 64 +(-16) =- 4

Find

1) (-36) ÷ (-4)

Solution: (-36) ÷ (-4) =9

2) (-201) ÷ (-3)

Solution: (-201) ÷ (-3) =67

3) (-325) ÷ (-13)

Solution: (-325) ÷ (-13) =25

Class 7 Mp Board Maths Question Answers

Solutions To Try These

Is

1) 1 ÷ a = 1 ?

Solution: No,1 ÷ a ≠ 1 unless a =1

2) a ÷ (-1) = -a ? for any integer a. Take different values of a and check.

Solution: Yes, a ÷ (-1) = -a

Examples:

1) Take a = 4

1. 1 ÷ 4 ≠ 1
2. 4 ÷ (-1) = -4

2) Take a = -5

1. 1 ÷ (-5) ≠ 1
2. (-5) ÷ (-1) = 5 =-(-5)

MP Board Class 7 Maths Solutions For Chapter 4 Simple Equations

1. Give the first step you will unr to separate the variable and then solve the equation :

1) x-1 -0

Solution:

Given equation isi – 1- 0

Adding 1 on both sides we get

x -1 +1-0 +1

x=1

2) x +1 =0

Solution:

Given equation is x + 1 -0

Subtracting ‘1’ from both sides we get

x+1-1 -0-1

x =-l

3) x-1 = 5

Solution:

Given equation is x-1 =5

Adding 1 on both sides we get

x-1 +1=5+1

x = 6

4) x + 6 = 2

Solution:

Given equation is x + 6 = 2

Subtracting ‘6’ from both sides we get

x+6-6= 2-6

x = – 4

Mp Board Class 7 Maths Solutions

5) y- 4 = -7

Solution:

Given equation is y – 4 = – 7

Adding 4 on both sides we get

y-4+4=-7+4

y = – 3

6) y – 4 = 4

Solution:

Given equation is y- 4 = 4

Adding 4 on both sides we get

y-4 + 4 = 4+4

y = 8

7) y + 4 = 4

Given equation is y + 4 = 4

Subtracting 4 from both sides we get

y+4-4=4-4

y=0

8) y + 4 = – 4

Solution:

Given equation is y + 4 = – 4

Subtracting 4 from both sides we get

y+4-4 = -4-4

y=-8

Class 7 Maths Chapter 4 Solutions Mp Board

2. Give the first step you will use to separate the variable and then solve the equation:

1) 3l = 42

Solution:

The given equation is 3l = 42

Dividing both sides by 3 we get

l = 14

2) \( \frac{b}{2}=6 \)

Solution: Given equation is \( \frac{b}{2}=6 \)

Multiplying both sides by 2 we get

b = 12

3) \( \frac{p}{7}=4 \)

Solution: Given equation is \( \frac{p}{7}=4 \)

Multiplying both sides by 7 we get

p = 28

Mp Board Maths Chapter 4 Solutions

4) 4x = 25

Solution: Given equation is 4x = 25

Dividing both sides by 4 we get

5) 8y = 36

Solution:

Given equation is 8y 36

Dividing both sides by 8 we get

6) \( \frac{z}{3}=\frac{5}{4} \)

Solution:

Given equation is \( \frac{z}{3}=\frac{5}{4} \)

Multiplying both sides by 3 we get

Mp Board Class 7 Book Solutions

7) \( \frac{a}{5}=\frac{7}{15}\)

Solution:

Given equation is \( \frac{a}{5}=\frac{7}{15} \)

8) 20t = – 10

Solution:

Given equation is 20t = -10

Dividing both sides by 20 we get

3. Give the steps you will use to separate the variable and then solve the equation:

1) 3n – 2 = 46

Solution: Given equation is 3n – 2 = 46

Adding 2 on both sides we get

3n -2 + 2 = 46 + 2

3n = 48

Dividing both sides by 3 we get

n = 16

2) 5m + 7 = 17

Solution:

Given equation is 5m + 7 = 17

Subtracting 7 from both sides we get

5m + 7-7 = 17-7

=> 5m = 10

Dividing both sides by 5 we get

25m = 50

m = 2

Class 7 Mp Board Maths Question Answers

3. \( \frac{20 p}{3}=40 \)

Solution:

Given equation is \( \frac{20 p}{3}=40 \)

Multiplying both sides by 3 we get

20p = 120

Dividing both sides by 20 we get

p = 6

4. \( \frac{3 p}{10}=6 \)

Solution:

Given equation is \( \frac{3 p}{10}=6 \)

Multiplying both sides by 10 we get

3p = 60

Dividing both sides by 3 we get

p = 20

4. Solve the following equations:

1)10p = 100

Solution:

Given equation is 10p = 100

Dividing both sides by 10 we get

p = 10

2) 10p + 10 = 100

Solution:

Given equation is 10p+10 = 100

Subtracting 10 from both sides we get

10p + 10 – 10 = 100 – 10

10p = 90

Dividing both sides by 10 we get

=> 100p =900

p= 9

Mp Board Class 7 Maths Solutions

3) \( \frac{p}{4}=5 \)

Solution:

Given equation is \( \frac{p}{4}=5 \)

Multiplying both sides by 4 we get

p =20

4) \( \frac{-p}{3}=5 \)

Solution:

Given equation is \( \frac{-p}{3}=5 \)

Multiplying both sides by -3 we get

p = – 15

5) \( \frac{3 p}{4}=6 \)

Solution:

Given equation is \( \frac{3 p}{4}=6 \)

Multiplying both sides by 4 we get

3p = 24

Dividing both sides by 3 we get

9p = 72

p = 8

6) 3s = – 9

Solution:

Given equation is 3s = -9

Dividing both sides by 3 we get

9s = – 27

s = – 3

Mp Board Maths Chapter 4 Solutions

7) 3s + 12 = 0

Solution:

Given equation is 3s + 12 = 0

Substracting 12 from both sides we get

3s + 12- 12 = 0 – 12

=> 3s =- 12

Dividing both sides by 3 we get

=> 9s = -36

s = – 4

8) 3s = 0

Solution:

Given equation is 3s = 0

Dividing both sides by 3 we get

=> 9s = 0

s= 0

9) 2q = 6

Solution:

Given equation is 2q =6

Dividing both sides by 2 we get

4q = 12

q = 3

10) 2q – 6 = 0

Solution:

Given equation is 2q- 6 = 0

Adding 6 on both sides we get

2q-6 + 6 = 0 + 6

2q = 6

Dividing both sides by 2 we get

=4q = 12

q= 3

11) 2q + 6 = 0

Solution:

Given equation is 2q + 6 = 0

Subtracting 6 from both sides we get

2q +6-6 = 0-6

=> 2q= – 6

Dividing both sides by 2 we get

4q = -12

q = -3

Mp Board Maths Chapter 4 Solutions

12) 2q + 6 = 12

Solution:

Given equation is 2q + 6 = 12

Subtracting 6 from both sides we get

2q + 6-6 = 12 -6

=> 2q = 6

Dividing both sides by 2 we get

4q = 12

q = 3