Maths Ace Class 7 Solutions Chapter 7

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Maths Ace Class 7 Solutions Chapter 7 Simple Linear Equations

Welcome to NCTB Solutions. Here with this post we are going to help 7th class students for the Solutions of Maths Ace Prime Class 7 Math Book, Chapter 7, Simple Linear Equations. Here students can easily find step by step solutions of all the problems for Simple Linear Equations, Exercise 7.1, 7.2 and 7.3 Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 7 solutions.

Simple Linear Equations Exercise 7.1 Solution :

Question no – (1) 

Solution :

(a) The product of 5 and x is 20.

= 5x = 20

or, x = 20/5 = 4

(b) 3 taken away from twice of y is 7.

= 2y – 3 = 7

or, 2y = 7 + 3 = 10

or, y = 10/2 = 5

(c) Two-fifths of a number p added to 7 gives 30.

= 2p/5 + 7 = 30 or, 2p + 35/5 = 30

or, 2p + 35 = 150

or, 2p = 150 – 35

or, p = 115/2

or, p = 57.5

(d) A number a added to its one-third is equal to 32.

= a/3 + a = 32

or, a + 3a/3 = 32

or, 4a = 96

or, a = 96/4

= 24

Question no – (2)

Solution :

(a) p – 10 = 50

or, p = 50 + 10

or, p = 60

(b) m/3 + 2 = 14

or, m + 6/3 = 14

or, m = (14 × 3) – 6

= 42 – 6

= 36

(c) 6x + 3 = 21

= 6x = 21 – 3 = 18

or, x = 18/6

= 3

(d) t/5 = 6

or, t = 30

Question no – (3)

Solution :

Rajinis mother’s age = 36 years

Let, Rajini’s age = x

According to question,

= 3x + 6 = 36

or, 3x = 36 – 6 = 30

or, x = 30/3

= 10 years

Therefore, Rajini’s age is 10 years old.

Question no – (4)

Solution :

Let, smaller number = p

greater number = P + 15

According to question’s,

= p + (p + 15) = 65

or, 2p = 65 – 15 = 50

or, p = 50/2

= 25

So, The smaller number is 25

Question no – (5) 

Solution :

(a) y + 3 = 10 ∴ y + 3 – 3 = 10 – 3

(b) x – 2 = 5 ∴ x – 2 + 5 = 5 + 5

(c) a/2 = 9 ∴ a/2 × 2 = 9 × 2

(d) 3k = 27 ∴ 3k ÷ 3 = 27 ÷ 3

Simple Linear Equations Exercise 7.2 Solution :

Question no – (1)

Solution :

(a) 7p + 5 = 19 [p = – 2]

L.H.S

= 7p + 5

= 7 (- 2) + 5

= – 14 + 5 = – 9

∴ No

(b) 4m – 3 = – 5 [m = – 3]

L.H.S /4m – 3

= 4 (- 3) =- 3

= 12 – 3 = – 15

∴ No

(c) 15 = 3 – 4x [x = – 2]

R.H.S

= 3 – 4x

= 2 – 4 (- 2)

= 3 + 8 = 11

∴ No

(d) t/3 = 6 [t = 18]

= L.H.S /t/3 = 18/3

∴ Yes

Question no – (2)

Solution :

(a) 2x + 10 = 18

= 2.1 + 10 ≠ 18 [x = 1]

= 2.2 + 10 ≠ 18 [x = 2]

= 2.3 + 10 ≠ 18 [x = 3]

= 2.4 + 10 = 18 [∵ x = 4]

(b) 3 (x – 1) = 6

= 3 (1 – 1) ≠ [x = 1]

= 3 (2 – 1) ≠ 6 [x = 2]

= 3 (3 – 1) = 6 [x = 3]

(c) 5a – 12 = 23

= 5.1 – 12 ≠ 23 [∵ x = 1]

= 5.2 – 12 ≠ 23 [x = 2]

= 5.3 – 12 ≠ 23 [x = 3]

= 5.4 – 12 ≠ 23 [x = 4]

= 5.5 – 12 ≠ 23 [x = 5]

= 5.6 – 12 ≠ 23 [x = 6]

= 5.7 – 12 = 23 [x = 7]

(d) 5x = – 75

or, 5(-10) ≠ – 75 [x = – 10]

or, 5 (-11) ≠ – 75 [x = – 11]

or, 5(- 12) ≠ – 75 [x = – 12]

or, 5(- 13) ≠ – 75 [x = 13]

or, 5 (-14) ≠ – 75 [x = – 14]

or, 5(- 15) – 75 [x = – 15]

Question no – (3)

Solution :

(a) x + 7 = 10

= or, 3 + 7 = 10 [∴ x = 3]

(b) 3x + 2 = 8

= 3.2 + 2 = 8 [∴ x = 2]

(c) 5x – 9 = 1

or, 5.2 – 9 = 1 [∴ x = 2]

(d) 7 (x – 1) = 20

or, 7 (27/7 – 1) = 20 [∴ x = 27/7]

(e) 6x = 54

or, 6 × 9 = 54 [x = 9]

(f) x – 4/2 = 1

or, 6 – 4/2 = 1 [∴ x = 6]

Question no – (4)

Solution :

(a) 6(t – 2) = 18

= or, t – 2 = 18/6 = 3

or, t = 3 + 2 = 5

(b) 3/2 p – 4 = 20

or, 3/2 p = 20 + 4 = 24

or, p = 24 × 2/3 = 16

(c) x/2 + 3/4 = 7/8

or, x/2 = 7/8 – 3/4

= 7 – 6/8 = 1/8

or, x = x/8 = 1/4

(d) 2 (x + 4) – 10 = 44

or, 2 (x + 4) = 444 + 10 = 54

or, x + 4 = 54/2 = 27

or, x = 27 – 4 = 23

(e) 4 – x/6 = 7

or, 4 – x = 42

or, – x = 42 – 4 = 38

or, x = – 38

(f) 2y + 7/2 = 39/2

or, 2y = 39/2 – 7/2 = 39 – 7/2

or, y = 32 × 2/2 = 32

Question no – (5) 

Solution :

(i) 2m + 1 = – 9

(ii) m + 2 = – 3

(iii) 2(m + 6) = 2

Question no – (6) 

Solution :

(i) 2 ++ 3 = 3

(ii) + + 1 = 3/2

(iii) 2 + 1 = 1

Simple Linear Equations Exercise 7.3 Solution :

Question no – (1) 

Solution :

(a) Three-fifth of a number is 21.

Let, the number = x

or, x = 21 × 5/3

= 35

(b) Four times a number is 28.

= 4x = 28

or, x = 28/4

= 7

(c) When you add 5 to three times a number you get 7.

= 3x + 5 = 7

or, 3x = 7 – 5 = 2

or, x = 2/3

(d) Riya thinks of a number. She takes away 6 from it and then multiply the result by 5 to get zero.

= 5 (x – 6) = 0

or, x – 6 = 0

or, x = 6

Question no – (2)

Solution :

According to question,

= x + x + 1 = 101

or, 2x = 101 – 1 = 100

or, x = 100/2 = 50

∴ One number = 50

another number = 50 + 1

= 51

So, the number is 51.

Question no – (3)

Solution :

Let, Soni’s age = x

According to question,

= 4x/5 + 8 = 32

or, 4/5 x = 32 – 8 = 24

or, x = 24 × 5/4

= 30 years

So, Soni’s age is 30 years old.

Question no – (4)

Solution :

Let, Neena has x toffes

According to question,

4x – 2 = 14

or, 4x = 14 + 2 = 16

or, x = 16/4

= 4

Question no – (5)

Solution :

Sum of both’s score = 100 + 11

Let, Rajesh score = 2x

Rajesh = 2x

According to question,

= 2x + x = 111

or, 3x = 111

or, x = 111/3 = 37

∴ Rajesh score = 37

Rajesh score = 37 × 2

= 74

Therefore, each score is 74.

Question no – (6)

Solution :

Let, no of red balls = x

According to question,

= 4x + 4 = 24

or, 4x = 24 – 4 = 20

or, x = 20/4

= 5

So, the number of red balls are 5.

Question no – (7)

Solution :

According to the question,

Rahim gave the shopkeeper = 100 Rs

Shopkeeper returned him = 10 Rs

Now,

100 – 10

= 90

= 90/2

= 45

Therefore, Price of two each chocolate is 45 rupee.

Question no – (8)

Solution :

Let, breadth = x

Length = 3x + 6

According to question,

= 2 (x + 3x + 6) = 100

or, 4x + 6 = 100/x = 50

or, 4x = 50 – 6 = 44

or, x = 44/4

= 11 m

So, the dimensions of the garden is 11m.

Question no – (9)

Solution :

Let, Shreya’s present age = x years

According to question,

= 2x + 10 = 40

or, 2x = 40 – 10 = 30

or, x = 30/2

= 15 years

So, Shreya’s present age is 15 years.

Question no – (10)

Solution :

Let, no of cows = x

According to question,

= x/3 + 7 = 21

or, x/3 = 21 – 7 = 14

or, x = 14 × 3 = 42

∴ Total animals = 42 + 21

= 63

Therefore, the total number of animals was 63.

Question no – (11)

Solution :

Let, Ashish’s money = x

Suresh money = 3x

According to question,

= 3x + x = 700

or, 4x = 700

or, x = 700/4

= 175

∴ Ashish’s money = 175

Suresh’s money = 175 × 3

= 525

So, they have total 525 rupees.

Next Chapter Solution :  

👉 Chapter 8 👈