myhelper: RS Aggarwal 2019,2020 solution class 7 chapter 7 Linear Equation In One Variable Exercise 7C

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RS Aggarwal 2019,2020 solution class 7 chapter 7 Linear Equation In One Variable Exercise 7C

Exercise 7C

Page-116

Question 1:

Mark (✓) against the correct answer

If

$5 x - \frac{3}{4} = 2 x - \frac{2}{3},$ then x = ?

(a)

$\frac{1}{12}$
(b)

$\frac{1}{4}$
(c) 36
(d)

$\frac{1}{36}$

Answer 1:

$\begin{array}{l} (d) \frac{1}{36} \\ We have: \\ 5 x - \frac{3}{4} = 2 x - \frac{2}{3} \\ \Rightarrow 5 x - 2 x = \frac{- 2}{3} + \frac{3}{4} \\ \Rightarrow 3 x = \frac{- 8 + 9}{12} \\ \Rightarrow x = \frac{1}{12 \times 3} \\ \Rightarrow x = \frac{1}{36} \end{array}$

Question 2:

Mark (✓) against the correct answer

If

$2 z + \frac{8}{3} = \frac{1}{4} z + 5,$ then z = ?

(a) 3
(b) 4
(c)

$\frac{3}{4}$
(d)

$\frac{4}{3}$

Answer 2:

$\begin{array}{l} (d) \frac{4}{3} \\ We have: \\ 2 z + \frac{8}{3} = \frac{1}{4} z + 5 \\ \Rightarrow 2 z - \frac{1}{4} z = 5 - \frac{8}{3} \\ \Rightarrow \frac{8 z - z}{4} = \frac{15 - 8}{3} \end{array} \Rightarrow \frac{7 z}{4} = \frac{7}{3} \begin{array}{l} \Rightarrow z = \frac{7^{1} \times 4}{3 \times 7_{1}} \\ \Rightarrow z = \frac{4}{3} \end{array}$

Question 3:

Mark (✓) against the correct answer

If (2n + 5) = 3(3n − 10), then n = ?

(a) 5
(b) 3
(c)

$\frac{2}{5}$
(d)

$\frac{2}{3}$

Answer 3:

$\begin{array}{l} (a) 5 \\ We have: \\ (2 n + 5) = 3 (3 n - 10) \\ \Rightarrow 2 n + 5 = 9 n - 30 \\ \Rightarrow 2 n - 9 n = - 30 - 5 \\ \Rightarrow - 7 n = - 35 \\ \Rightarrow n = \frac{3 5^{5}}{7_{1}} \\ \Rightarrow n = 5 \end{array}$

Question 4:

Mark (✓) against the correct answer

If

$\frac{x - 1}{x + 1} = \frac{7}{9},$ then x = ?

(a) 6
(b) 7
(c) 8
(d) 10

Answer 4:

$\begin{array}{l} (c) 8 \\ We have: \\ \frac{x - 1}{x + 1} = \frac{7}{9} \\ \Rightarrow 9 (x - 1) = 7 (x + 1) \end{array} \Rightarrow 9 x - 9 = 7 x + 7 \begin{array}{l} \Rightarrow 9 x - 7 x = 7 + 9 \\ \Rightarrow 2 x = 16 \\ \Rightarrow x = \frac{1 6^{8}}{2_{1}} \\ \Rightarrow x = 8 \end{array}$

Question 5:

Mark (✓) against the correct answer

If 8(2x − 5) − 6(3x − 7) = 1, then x = ?

(a) 2
(b) 3
(c)

$\frac{1}{2}$
(d)

$\frac{1}{3}$

Answer 5:

$\begin{array}{l} (c) \frac{1}{2} \\ We have: \\ 8 (2 x - 5) - 6 (3 x - 7) = 1 \\ \Rightarrow 16 x - 40 - 18 x + 42 = 1 \\ \Rightarrow - 2 x + 2 = 1 \\ \Rightarrow - 2 x = 1 - 2 \\ \Rightarrow - x = - \frac{1}{2} \\ \Rightarrow x = \frac{1}{2} \end{array}$

Question 6:

Mark (✓) against the correct answer

If

$\frac{x}{2} - 1 = \frac{x}{3} + 4,$ then x = ?

(a) 8
(b) 16
(c) 24
(d) 30

Answer 6:

$\begin{array}{l} (d) 30 \\ We have: \\ \frac{x}{2} - 1 = \frac{x}{3} + 4 \\ \Rightarrow \frac{x - 2}{2} = \frac{x + 12}{3} \\ \Rightarrow 3 (x - 2) = 2 (x + 12) \\ \Rightarrow 3 x - 6 = 2 x + 24 \\ \Rightarrow 3 x - 2 x = 24 + 6 \\ \Rightarrow x = 30 \end{array}$

Question 7:

Mark (✓) against the correct answer

If

$\frac{2 x - 1}{3} = \frac{x - 2}{3} + 1,$ then x = ?

(a) 2
(b) 4
(c) 6
(d) 8

Answer 7:

$\begin{array}{l} (a) 2 \\ We have: \\ \frac{2 x - 1}{3} = \frac{x - 2}{3} + 1 \\ \Rightarrow \frac{2 x - 1}{3} = \frac{(x - 2) + 3}{3} \\ \Rightarrow 3 (2 x - 1) = 3 (x + 1) \\ \Rightarrow 6 x - 3 = 3 x + 3 \\ \Rightarrow 6 x - 3 x = 3 + 3 \\ \Rightarrow 3 x = 6 \\ \Rightarrow x = \frac{6^{2}}{\begin{array}{l} 3_{1} \\ = 2 \end{array}} \end{array}$

Question 8:

The sum of two consecutive whole numbers is 53. The smaller number is

(a) 25
(b) 26
(c) 29
(d) 23

Answer 8:

$\begin{array}{l} (b) 26 \\ Let the consecutive whole numbers be x and (x + 1) . \\ Then, x + (x + 1) = 53 \\ \Rightarrow 2 x + 1 = 53 \\ \Rightarrow 2 x = 53 - 1 \\ \Rightarrow x = \frac{5 2^{26}}{2_{1}} \\ \Rightarrow x = 26 \end{array}$

Page-117

Question 9:

The sum of two consecutive even numbers is 86. The larger of the two is

(a) 46
(b) 36
(c) 38
(d) 44

Answer 9:

$\begin{array}{l} (d) 44 \\ Let the two consecutive even numbers be x and (x + 2) . \\ Then, x + (x + 2) = 86 \\ \Rightarrow 2 x + 2 = 86 \\ \Rightarrow 2 x = 86 - 2 \\ \Rightarrow x = \frac{8 4^{42}}{2_{1}} \\ \Rightarrow x = 42 \\ ∴ The required numbers are 42 and (42 + 2), i . e ., 44. \end{array}$

Question 10:

The sum of two consecutive odd numbers is 36. The smaller one is

(a) 15
(b) 17
(c) 19
(d) 13

Answer 10:

$\begin{array}{l} (b) 17 \\ Let the two consecutive odd numbers be (x + 1) and (x + 3) . \\ Then, (x + 1) + (x + 3) = 36 \\ \Rightarrow 2 x + 4 = 36 \\ \Rightarrow 2 x = 36 - 4 \\ \Rightarrow x = \frac{3 2^{16}}{2_{1}} \\ \Rightarrow x = 16 \\ ∴ The smaller number is 17. \end{array}$

Question 11:

on adding 9 to the twice of a whole number gives 31. The whole number is

(a) 21
(b) 16
(c) 17
(d) 11

Answer 11:

$\begin{array}{l} (d) 11 \\ Let the whole number be x . \\ Then, 2 x + 9 = 31 \\ \Rightarrow 2 x = 31 - 9 \\ \Rightarrow 2 x = 22 \\ \Rightarrow x = \frac{2 2^{11}}{2_{1}} \\ \Rightarrow x = 11 \end{array}$

Question 12:

Thrice a number when increased by 6 gives 24. The number is

(a) 6
(b) 7
(c) 8
(d) 11

Answer 12:

$\begin{array}{l} (a) 6 \\ Let the whole number be x . \\ Then, 3 x + 6 = 24 \\ \Rightarrow 3 x = 24 - 6 \\ \Rightarrow 3 x = 18 \\ \Rightarrow x = \frac{1 8^{6}}{3_{1}} \\ \Rightarrow x = 6 \end{array}$

Question 13:

$\frac{2}{3}$ of a number is less than the original number by 10. The original number is

(a) 30
(b) 36
(c) 45
(d) 60

Answer 13:

$\begin{array}{l} (a) 30 \\ Let the original number be x . \\ Then, \frac{2}{3} x = x - 10 \\ \Rightarrow 2 x = 3 x - 30 \\ \Rightarrow 2 x - 3 x = - 30 \\ \Rightarrow - x = - 30 \\ \Rightarrow x = 30 \\ ∴ The required number is 3 0 . \end{array}$

Question 14:

Two complementary angles differ by 10°. The larger angle is

(a) 60°
(b) 50°
(c) 64°
(d) 54°

Answer 14:

$(b) 50 ° Let the angle be x ° . Then, complementary of x = 90 ° - x ° According to the question, we have : x - 90 - x = 10 \Rightarrow 2 x = 90 + 10 \Rightarrow 2 x = 100 \Rightarrow x = 50 So, the larger angle is 50 ° .$

Question 15:

Two supplementary angles differ by 20°. The smaller of the two measures

(a) 60°
(b) 80°
(c) 100°
(d) 120°

Answer 15:

$\begin{array}{l} (b) 80^{0} \\ Let the angle be x ° . \end{array} Then, complementary angle of x = 180 ° - x ° According to the question, we have : x - (180 - x) = 20 \Rightarrow x - 180 + x = 20 \Rightarrow 2 x = 10 + 180 \Rightarrow 2 x = 200 \Rightarrow x = 100 Hence, the smaller angle is 80 ° .$

Question 16:

The ages of A and B are in the ratio 5 : 3. After 6 years, their ages will be in the ratio 7 : 5. The present age of A is

(a) 5 years
(b) 10 years
(c) 15 years
(d) 20 years

Answer 16:

$\begin{array}{l} (c) 15 years \\ Let the present ages of A and B be 5 x and 3 x, respectively . \\ According to the question, we have : \\ \frac{5 x + 6}{3 x + 6} = \frac{7}{5} \\ \Rightarrow 25 x + 30 = 21 x + 42 \\ \Rightarrow 25 x - 21 x = 42 - 30 \\ \Rightarrow 4 x = 12 \\ \Rightarrow x = \frac{1 2^{3}}{4} \\ \Rightarrow x = 3 \\ ∴ A ’ s present age=5 \times 3 years=15 years \end{array}$

Question 17:

A number when multiplied by 5 is increased by 80. The number is

(a) 15
(b) 20
(c) 25
(d) 30

Answer 17:

$\begin{array}{l} (b) 20 \\ Let the number be x . \\ Then, 5 x = x + 80 \\ \Rightarrow 5 x - x = 80 \\ \Rightarrow 4 x = 80 \\ \Rightarrow x = \frac{8 0^{20}}{4_{1}} \\ \Rightarrow x = 20 \\ ∴ The required number is 2 0 . \end{array}$

Question 18:

The length of a rectangle is three times its width and its perimeter is 96 m. The length is

(a) 12 m
(b) 24 m
(c) 36 m
(d) 48 m

Answer 18:

$\begin{array}{l} (c) 32 m \\ Let the width of the rectangle be x . Then, its length will be 3 x . \\ Perimeter of the rectangle = 96 m \\ Now, 2 (l + b) = 96 \\ \Rightarrow 2 (3 x + x) = 96 \\ \Rightarrow 2 \times 4 x = 96 \\ \Rightarrow 8 x = 96 \\ \Rightarrow x = \frac{9 6^{12}}{8_{1}} \\ \Rightarrow x = 12 \\ ∴ Length of the rectangle = 3 \times 12 m = 36 m \end{array}$

RS Aggarwal 2019,2020 solution class 7 chapter 7 Linear Equation In One Variable Exercise 7C

Exercise 7C

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