myhelper: RS Aggarwal solution class 8 chapter 8 Linear Equations Exercise 8C

Exercise 8C

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Q1 | Ex-8C | Class 8 | RS AGGARWAL | Chapter 8 | Linear Equations | myhelper

Question 1:

Tick (✓) the correct answer:
If 2x − 3 = x + 2, then x = ?
(a) 1
(b) 3
(c) 5
(d) 7

Answer 1:

Q2 | Ex-8C | Class 8 | RS AGGARWAL | chapter 8 | Linear Equations | myhelper

Question 2:

Tick (✓) the correct answer:
If $5 x + \frac{7}{1} = \frac{3}{2} x - 14,$ then x = ?
(a) 5
(b) −5
(c) 6
(d) −6

Answer 2:

$(b) - 5 5 x + \frac{7}{2} = \frac{3}{2} x - 14 \Rightarrow \frac{10 x + 7}{2} = \frac{3 x - 28}{2} \Rightarrow 10 x + 7 = 3 x - 28 \Rightarrow 10 x - 3 x = - 28 - 7 \Rightarrow 7 x = - 35 \Rightarrow x = \frac{- 35}{7} = - 5$

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Question 3:

Tick (✓) the correct answer:
If $z = \frac{4}{5} (z + 10), then z = ?$
(a) 40
(b) 20
(c) 10
(d) 60

Answer 3:

(a) 40

$z = \frac{4}{5} (z + 10) \Rightarrow 5 z = 4 (z + 10) \Rightarrow 5 z = 4 z + 40 \Rightarrow 5 z - 4 z = 40 \Rightarrow z = 40$

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Question 4:

Tick (✓) the correct answer:
If 3m = 5m $- \frac{8}{5}$ , then m = ?
(a) $\frac{2}{5}$
(b) $\frac{3}{5}$
(c) $\frac{4}{5}$
(d) $\frac{1}{5}$

Answer 4:

(c) $\frac{4}{5}$

$3 m = 5 m - \frac{8}{5} \Rightarrow 3 m = \frac{25 m - 8}{5} \Rightarrow 15 m = 25 m - 8 \Rightarrow 15 m - 25 m = - 8 \Rightarrow - 10 m = - 8 \Rightarrow m = \frac{- 8}{- 10} = \frac{4}{5}$

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Question 5:

Tick (✓) the correct answer:
If 5t − 3 = 3t − 5, then t = ?
(a) 1
(b) −1
(c) 2
(d) −2

Answer 5:

(b) -1

$5 t - 3 = 3 t - 5 \Rightarrow 5 t - 3 t = 3 - 5 \Rightarrow 2 t = - 2 \Rightarrow t = \frac{- 2}{2} = - 1$

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Question 6:

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If $2 y + \frac{5}{3} = \frac{26}{3} - y$ , then y = ?
(a) 1
(b) $\frac{2}{3}$
(c) $\frac{6}{5}$
(d) $\frac{7}{3}$

Answer 6:

(d) $\frac{7}{3}$

$2 y + \frac{5}{3} = \frac{26}{3} - y \Rightarrow \frac{6 y + 5}{3} = \frac{26 - 3 y}{3} \Rightarrow 6 y + 5 = 26 - 3 y \Rightarrow 6 y + 3 y = 26 - 5 \Rightarrow 9 y = 21 \Rightarrow y = \frac{21}{9} = \frac{7}{3}$

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Question 7:

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If $\frac{6 x + 1}{3} + 1 = \frac{x - 3}{6}$ , then x = ?
(a) 1
(b) −1
(c) 3
(d) −3

Answer 7:

(b) -1

$\frac{6 x + 1}{3} + 1 = \frac{x - 3}{6} \Rightarrow \frac{6 x + 1 + 3}{3} = \frac{x - 3}{6} \Rightarrow 6 (6 x + 4) = 3 (x - 3) \Rightarrow 36 x + 24 = 3 x - 9 \Rightarrow 36 x - 3 x = - 24 - 9 \Rightarrow 33 x = - 33 \Rightarrow x = \frac{- 33}{33} = - 1$

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Question 8:

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If $\frac{n}{2} - \frac{3 n}{4} + \frac{5 n}{6} = 21,$ then n = ?
(a) 30
(b) 42
(c) 36
(d) 28

Answer 8:

(c) 36

$\frac{n}{2} - \frac{3 n}{4} + \frac{5 n}{6} = 21 \Rightarrow \frac{6 n - 9 n + 10 n}{12} = 21 \Rightarrow 7 n = 21 \times 12 \Rightarrow 7 n = 252 \Rightarrow n = \frac{252}{7} = 36$

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Question 9:

Tick (✓) the correct answer:
If $\frac{x + 1}{2 x + 3} = \frac{3}{8}$ , then x = ?
(a) $\frac{1}{4}$
(b) $\frac{1}{3}$
(c) $\frac{1}{6}$
(d) $\frac{1}{2}$

Answer 9:

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

$\frac{x + 1}{2 x + 3} = \frac{3}{8} \Rightarrow 8 (x + 1) = 3 (2 x + 3) \Rightarrow 8 x + 8 = 6 x + 9 \Rightarrow 8 x - 6 x = 9 - 8 \Rightarrow 2 x = 1 \Rightarrow x = \frac{1}{2}$

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Question 10:

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If $\frac{4 x + 8}{5 x + 8} = \frac{5}{6},$ then x = ?
(a) 4
(b) 6
(c) 8
(d) 12

Answer 10:

(c) 8

$\frac{4 x + 8}{5 x + 8} = \frac{5}{6} \Rightarrow 6 (4 x + 8) = 5 (5 x + 8) \Rightarrow 24 x + 48 = 25 x + 40 \Rightarrow 24 x - 25 x = - 48 + 40 \Rightarrow - x = - 8 \Rightarrow x = 8$

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Question 11:

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If $\frac{n}{n + 15} = \frac{4}{9}$ , then n = ?
(a) 4
(b) 6
(c) 9
(d) 12

Answer 11:

(d) 12

$\frac{n}{n + 15} = \frac{4}{9} \Rightarrow 9 n = 4 (n + 15) \Rightarrow 9 n = 4 n + 60 \Rightarrow 9 n - 4 n = 60 \Rightarrow 5 n = 60 \Rightarrow n = \frac{60}{5} = 12$

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Question 12:

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If 3(t − 3) = 5(2t + 1), then t = ?
(a) −2
(b) 2
(c) −3
(d) 3

Answer 12:

(a) -2

$3 (t - 3) = 5 (2 t + 1) \Rightarrow 3 t - 9 = 10 t + 5 \Rightarrow 3 t - 10 t = 9 + 5 \Rightarrow - 7 t = 14 \Rightarrow - t = \frac{14}{7} = 2 \Rightarrow t = - 2$

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Question 13:

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Four-fifths of a number is greater than three-fourths of the number by 4. The number is
(a) 12
(b) 64
(c) 80
(d) 102

Answer 13:

(c) 80

$Let the number be x . ∴ \frac{4}{5} x = \frac{3}{4} x + 4 \Rightarrow \frac{4 x}{5} = \frac{3 x + 16}{4} \Rightarrow 16 x = 15 x + 80 \Rightarrow 16 x - 15 x = 80 \Rightarrow x = 80$

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Question 14:

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The ages of A and B are in the ratio 5 : 7. Four years from now the ratio of their ages will be 3 : 4. The present age of B is
(a) 20 years
(b) 28 years
(c) 15 years
(d) 21 years

Answer 14:

(b) 28 years

$Let x be the common multiple of the ages of A and B . Then . the age s of A and B would be 5 x and 7 x, respectively . ∴ \frac{5 x + 4}{7 x + 4} = \frac{3}{4} \Rightarrow 4 (5 x + 4) = 3 (7 x + 4) \Rightarrow 20 x + 16 = 21 x + 12 \Rightarrow 16 - 12 = 21 x - 20 x \Rightarrow 4 = x \Rightarrow x = 4 ∴ Age of B = 7 (x) = 7 \times 4 = 28 years$

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Question 15:

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The base of an isosceles triangle is 6 cm and its perimeter is 16 cm. Length of each of the equal sides is
(a) 4 cm
(b) 5 cm
(c) 3 cm
(d) 6 cm

Answer 15:

(b) 5 cm

$Let the equal side of the isosceles triangle be x . Then, the perimeter of the triangle would be (x + x + 6) . ∴ 2 x + 6 = 16 \Rightarrow 2 x = 16 - 6 \Rightarrow 2 x = 10 \Rightarrow x = \frac{10}{2} = 5 ∴ Length of each equal side = 5 cm$

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Question 16:

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Sum of three consecutive integers is 51. The middle one is
(a) 14
(b) 15
(c) 16
(d) 17

Answer 16:

(d) 17

$Let the three consecutive integers be x, x + 1 and x + 2 . Equation = x + x + 1 + x + 2 = 51 \Rightarrow 3 x + 3 = 51 \Rightarrow 3 x = 51 - 3 \Rightarrow 3 x = 48 \Rightarrow x = \frac{48}{3} = 16 Middle integer = x + 1 = 16 + 1 = 17$

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Question 17:

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The sum of two numbers is 95. If one exceeds the other by 15, then the smaller of the two is
(a) 40
(b) 35
(c) 45
(d) 55

Answer 17:

(a) 40

$Let the numbers be x and x + 15 . ∴ x + x + 15 = 95 \Rightarrow 2 x + 15 = 95 \Rightarrow 2 x = 95 - 15 \Rightarrow 2 x = 80 \Rightarrow x = 40 The smaller number is 40 .$

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Question 18:

Tick (✓) the correct answer:
Number of boys and girls in a class are in the ratio 7 : 5. The number of boys is 8 more than the number of girls. The total class strength is
(a) 56
(b) 52
(c) 48
(d) 36

Answer 18:

(c) 48

$Let the number of boys in the class be x . Then, the number of girls will be (x - 8) . The equation becomes : \frac{x}{x - 8} = \frac{7}{5} \Rightarrow 5 x = 7 x - 56 \Rightarrow 5 x - 7 x = - 56 \Rightarrow - 2 x = - 56 \Rightarrow x = \frac{- 56}{- 2} = 28 Therefore, the number of boys is 28 . Number of girls = (x - 8) = 28 - 8 = 20 Total strength of the class = 28 + 20 = 48$

RS Aggarwal solution class 8 chapter 8 Linear Equations Exercise 8C

Exercise 8C

Page-113

Q1 | Ex-8C | Class 8 | RS AGGARWAL | Chapter 8 | Linear Equations | myhelper

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Answer 1:

Q2 | Ex-8C | Class 8 | RS AGGARWAL | chapter 8 | Linear Equations | myhelper

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Q3 | Ex-8C | Class 8 | RS AGGARWAL | chapter 8 | Linear Equations | myhelper

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