Rs Aggarwal 2017 for Class 8 Math Chapter 8 - Linear Equations

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Rs Aggarwal 2017 Solutions for Class 8 Math Chapter 8 Linear Equations are provided here with simple step-by-step explanations. These solutions for Linear Equations are extremely popular among Class 8 students for Math Linear Equations Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2017 Book of Class 8 Math Chapter 8 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2017 Solutions. All Rs Aggarwal 2017 Solutions for class Class 8 Math are prepared by experts and are 100% accurate.

Page No 107:

Question 1:

Solve:
8x + 3 = 27 + 2x

Answer:

$8 x + 3 = 27 + 2 x \Rightarrow 8 x - 2 x = 27 - 3 \Rightarrow 6 x = 24 \Rightarrow x = \frac{24}{6} = 4 ∴ x = 4$

Page No 107:

Question 2:

Solve:
5x + 7 = 2x − 8

Answer:

$5 x + 7 = 2 x - 8 \Rightarrow 5 x - 2 x = - 8 - 7 \Rightarrow 3 x = - 15 \Rightarrow x = \frac{- 15}{3} = - 5 ∴ x = - 5$

Page No 107:

Question 3:

Solve:
2z − 1 = 14 − z

Answer:

$2 z - 1 = 14 - z \Rightarrow 2 z + z = 14 + 1 \Rightarrow 3 z = 15 \Rightarrow z = \frac{15}{3} = 5 ∴ z = 5$

Page No 107:

Question 4:

Solve:
9x + 5 = 4(x − 2) + 8

Answer:

$9 x + 5 = 4 (x - 2) + 8 \Rightarrow 9 x + 5 = 4 x - 8 + 8 \Rightarrow 9 x + 5 = 4 x \Rightarrow 9 x - 4 x = - 5 \Rightarrow 5 x = - 5 \Rightarrow x = \frac{- 5}{5} = - 1 ∴ x = - 1$

Page No 107:

Question 5:

Solve:
$\frac{7 y}{5} = y - 4$

Answer:

$\frac{7 y}{5} = y - 4 By cross multiplication : \Rightarrow 7 y = 5 (y - 4) \Rightarrow 7 y = 5 y - 20 \Rightarrow 7 y - 5 y = - 20 \Rightarrow 2 y = - 20 \Rightarrow y = \frac{- 20}{2} = - 10 ∴ y = - 10$

Page No 107:

Question 6:

Solve:
$3 x + \frac{2}{3} = 2 x + 1$

Answer:

$3 x + \frac{2}{3} = 2 x + 1 \Rightarrow 3 x - 2 x = 1 - \frac{2}{3} \Rightarrow x = \frac{1}{1} - \frac{2}{3} (L . C . M . of 1 and 3 is 3) \Rightarrow x = \frac{3 - 2}{3} \Rightarrow x = \frac{1}{3} \Rightarrow x = \frac{1}{3} ∴ x = \frac{1}{3}$

Page No 107:

Question 7:

Solve:
15(y − 4) − 2(y − 9) + 5(y + 6) = 0

Answer:

$15 (y - 4) - 2 (y - 9) + 5 (y + 6) = 0 \Rightarrow 15 y - 60 - 2 y + 18 + 5 y + 30 = 0 \Rightarrow 15 y - 2 y + 5 y - 60 + 18 + 30 = 0 \Rightarrow 18 y - 12 = 0 \Rightarrow 18 y = 12 \Rightarrow y = \frac{12}{18} = \frac{2}{3} ∴ y = \frac{2}{3}$

Page No 107:

Question 8:

Solve:
3(5x − 7) − 2(9x − 11) = 4(8x − 13) − 17

Answer:

$3 (5 x - 7) - 2 (9 x - 11) = 4 (8 x - 13) - 17 \Rightarrow 15 x - 21 - 18 x + 22 = 32 x - 52 - 17 \Rightarrow 15 x - 18 x - 21 + 22 = 32 x - 69 \Rightarrow - 3 x + 1 = 32 x - 69 \Rightarrow 1 + 69 = 32 x + 3 x \Rightarrow 70 = 35 x \Rightarrow 35 x = 70 (b y transposition) \Rightarrow x = \frac{70}{35} = 2 ∴ x = 2$

Page No 107:

Question 9:

Solve:
$\frac{x - 5}{2} - \frac{x - 3}{5} = \frac{1}{2}$

Answer:

$\frac{x - 5}{2} - \frac{x - 3}{5} = \frac{1}{2} \Rightarrow 10 (\frac{x - 5}{2}) - 10 (\frac{x - 3}{5}) = 10 (\frac{1}{2}) (multiplying throughout by 10, which is the L . C . M . of 2, 2 and 5) \Rightarrow 5 (x - 5) - 2 (x - 3) = 5 \Rightarrow 5 x - 25 - 2 x + 6 = 5 \Rightarrow 5 x - 2 x - 25 + 6 = 5 \Rightarrow 3 x - 19 = 5 \Rightarrow 3 x = 5 + 19 \Rightarrow 3 x = 24 \Rightarrow x = \frac{24}{3} = 8 ∴ x = 8$

Page No 107:

Question 10:

Solve:
$\frac{3 t - 2}{4} - \frac{2 t + 3}{3} = \frac{2}{3} - t$

Answer:

$\frac{3 t - 2}{4} - \frac{2 t + 3}{3} = \frac{2}{3} - t \Rightarrow \frac{3 t - 2}{4} - \frac{2 t + 3}{3} = \frac{2 - 3 t}{3} (3 is the L . C . M . of 1 and 3) \Rightarrow 12 (\frac{3 t - 2}{4}) - 12 (\frac{2 t + 3}{3}) = 12 (\frac{2 - 3 t}{3}) (multiplying throughout by 12, which is the L . C . M . of 4, 3 and 3) \Rightarrow 3 (3 t - 2) - 4 (2 t + 3) = 4 (2 - 3 t) \Rightarrow 9 t - 6 - 8 t - 12 = 8 - 12 t \Rightarrow 9 t - 8 t - 6 - 12 = 8 - 12 t \Rightarrow t - 18 = 8 - 12 t \Rightarrow t + 12 t = 18 + 8 \Rightarrow 13 t = 26 \Rightarrow t = \frac{26}{13} = 2 ∴ t = 2$

Page No 107:

Question 11:

Solve:
$\frac{2 x + 7}{5} - \frac{3 x + 11}{2} = \frac{2 x + 8}{3} - 5$

Answer:

$\frac{2 x + 7}{5} - \frac{3 x + 11}{2} = \frac{2 x + 8}{3} - 5 \Rightarrow \frac{2 x + 7}{5} - \frac{3 x + 11}{2} = \frac{2 x + 8 - 15}{3} (L . C . M . of 3 and 1 is 3) \Rightarrow 30 (\frac{2 x + 7}{5}) - 30 (\frac{3 x + 11}{2}) = 30 (\frac{2 x + 8 - 15}{3}) (multiplyin g throughout by 30, which is the L . C . M . of 5, 2 and 3) \Rightarrow 6 (2 x + 7) - 15 (3 x + 11) = 10 (2 x + 8 - 15) \Rightarrow 12 x + 42 - 45 x - 165 = 20 x - 70 \Rightarrow 12 x - 45 x + 42 - 165 = 20 x - 70 \Rightarrow - 33 x - 123 = 20 x - 70 \Rightarrow - 33 x - 20 x = 123 - 70 \Rightarrow - 53 x = 53 \Rightarrow x = \frac{53}{- 53} \Rightarrow x = - 1 ∴ x = - 1$

Page No 107:

Question 12:

Solve:
$\frac{5 x - 4}{6} = 4 x + 1 - \frac{3 x + 10}{2}$

Answer:

$\frac{5 x - 4}{6} = 4 x + 1 - \frac{3 x + 10}{2} \Rightarrow \frac{5 x - 4}{6} = \frac{2 (4 x + 1) - 3 x - 10}{2} (L . C . M . of 1 and 2 is 2) \Rightarrow \frac{5 x - 4}{6} = \frac{8 x + 2 - 3 x - 10}{2} \Rightarrow \frac{5 x - 4}{6} = \frac{8 x - 3 x + 2 - 10}{2} \Rightarrow \frac{5 x - 4}{6} = \frac{5 x - 8}{2} \Rightarrow 2 (5 x - 4) = 6 (5 x - 8) \Rightarrow 10 x - 8 = 30 x - 48 \Rightarrow 10 x - 30 x = - 48 + 8 \Rightarrow - 20 x = - 40 \Rightarrow x = \frac{- 40}{- 20} = 2 ∴ x = 2$

Page No 107:

Question 13:

Solve:
$5 x - \frac{1}{3} (x + 1) = 6 (x + \frac{1}{30})$

Answer:

$5 x - \frac{1}{3} (x + 1) = 6 (x + \frac{1}{30}) \Rightarrow 5 x - \frac{1 (x + 1)}{3} = 6 (\frac{30 x + 1}{30}) (L . C . M . of 1 and 30 is 30) \Rightarrow 5 x - \frac{(x + 1)}{3} = \frac{30 x + 1}{5} \Rightarrow \frac{15 x - x - 1}{3} = \frac{30 x + 1}{5} (L . C . M . of 1 and 3 is 3) \Rightarrow \frac{14 x - 1}{3} = \frac{30 x + 1}{5} \Rightarrow 5 (14 x - 1) = 3 (30 x + 1) (by cross multiplication) \Rightarrow 70 x - 5 = 90 x + 3 \Rightarrow 70 x - 90 x = 3 + 5 \Rightarrow - 20 x = 8 \Rightarrow x = \frac{8}{- 20} = \frac{- 2}{5} ∴ x = - \frac{2}{5}$

Page No 107:

Question 14:

Solve:
$4 - \frac{2 (z - 4}{3} = \frac{1}{2} (2 z + 5)$

Answer:

$4 - \frac{2 (z - 4)}{3} = \frac{1}{2} (2 z + 5) \Rightarrow \frac{12 - 2 (z - 4)}{3} = \frac{1 (2 z + 5)}{2} (L . C . M . of 1 and 3 is 3) \Rightarrow \frac{12 - 2 z + 8}{3} = \frac{2 z + 5}{2} \Rightarrow \frac{20 - 2 z}{3} = \frac{2 z + 5}{2} \Rightarrow 2 (20 - 2 z) = 3 (2 z + 5) (by cross multiplication) \Rightarrow 40 - 4 z = 6 z + 15 \Rightarrow 40 - 15 = 6 z + 4 z \Rightarrow 25 = 10 z \Rightarrow 10 z = 25 (by transposition) \Rightarrow z = \frac{25}{10} = \frac{5}{2} ∴ z = \frac{5}{2}$

Page No 107:

Question 15:

Solve:
$\frac{3 (y - 5)}{4} - 4 y = 3 - \frac{(y - 3)}{2}$

Answer:

$\frac{3 (y - 5)}{4} - 4 y = 3 - \frac{(y - 3)}{2} \Rightarrow \frac{3 y - 15}{4} - 4 y = 3 - \frac{y - 3}{2} \Rightarrow \frac{3 y - 15 - 16 y}{4} = 3 - \frac{y - 3}{2} (L . C . M . of 4 and 1 is 4) \Rightarrow \frac{- 13 y - 15}{4} = \frac{6 - y + 3}{2} \Rightarrow \frac{- 13 y - 15}{4} = \frac{9 - y}{2} \Rightarrow 2 (- 13 y - 15) = 4 (9 - y) \Rightarrow - 26 y - 30 = 36 - 4 y \Rightarrow - 26 y + 4 y = 36 + 30 \Rightarrow - 22 y = 66 \Rightarrow 22 y = - 66 (multiplying both the sides with a - ve sign) \Rightarrow y = - \frac{66}{22} = - 3 ∴ y = - 3$

Page No 107:

Question 16:

Solve:
$\frac{8 x - 3}{3 x} = 2$

Answer:

$\frac{8 x - 3}{3 x} = 2 \Rightarrow 8 x - 3 = 2 (3 x) (by cross multiplication) \Rightarrow 8 x - 3 = 6 x \Rightarrow 8 x - 6 x = 3 \Rightarrow 2 x = 3 \Rightarrow x = \frac{3}{2} ∴ x = \frac{3}{2}$

Page No 107:

Question 17:

Solve:
$\frac{9 x}{7 - 6 x} = 15$

Answer:

$\frac{9 x}{7 - 6 x} = 15 \Rightarrow \frac{9 x}{7 - 6 x} = \frac{15}{1} \Rightarrow 1 (9 x) = 15 (7 - 6 x) (by cross multiplication) \Rightarrow 9 x = 105 - 90 x \Rightarrow 9 x + 90 x = 105 \Rightarrow 99 x = 105 \Rightarrow x = \frac{105}{99} = \frac{35}{33} ∴ x = \frac{35}{33}$

Page No 107:

Question 18:

Solve:
$\frac{3 x}{5 x + 2} = - 4$

Answer:

$\frac{3 x}{5 x + 2} = - 4 \Rightarrow \frac{3 x}{5 x + 2} = \frac{- 4}{1} \Rightarrow 1 (3 x) = - 4 (5 x + 2) (by cross multiplication) \Rightarrow 3 x = - 20 x - 8 \Rightarrow 3 x + 20 x = - 8 \Rightarrow 23 x = - 8 \Rightarrow x = \frac{- 8}{23} ∴ x = \frac{- 8}{23}$

Page No 107:

Question 19:

Solve:
$\frac{6 y - 5}{2 y} = \frac{7}{9}$

Answer:

$\frac{6 y - 5}{2 y} = \frac{7}{9} \Rightarrow 9 (6 y - 5) = 7 (2 y) (by cross multiplication) \Rightarrow 54 y - 45 = 14 y \Rightarrow 54 y - 14 y = 45 \Rightarrow 40 y = 45 \Rightarrow y = \frac{45}{40} = \frac{9}{8} ∴ y = \frac{9}{8}$

Page No 107:

Question 20:

Solve:
$\frac{2 - 9 z}{17 - 4 z} = \frac{4}{5}$

Answer:

$\frac{2 - 9 z}{17 - 4 z} = \frac{4}{5} \Rightarrow 5 (2 - 9 z) = 4 (17 - 4 z) (by cross multiplication) \Rightarrow 10 - 45 z = 68 - 16 z \Rightarrow 10 - 68 = 45 z - 16 z \Rightarrow - 58 = 29 z \Rightarrow 29 z = - 58 (by transposition) \Rightarrow z = \frac{- 58}{29} = - 2 ∴ z = - 2$

Page No 107:

Question 21:

Solve:
$\frac{4 x + 7}{9 - 3 x} = \frac{1}{4}$

Answer:

$\frac{4 x + 7}{9 - 3 x} = \frac{1}{4} \Rightarrow 4 (4 x + 7) = 1 (9 - 3 x) (by cross multiplication) \Rightarrow 16 x + 28 = 9 - 3 x \Rightarrow 16 x + 3 x = 9 - 28 \Rightarrow 19 x = - 19 \Rightarrow x = \frac{- 19}{19} = - 1 ∴ x = - 1$

Page No 107:

Question 22:

Solve:
$\frac{7 y + 4}{y + 2} = \frac{- 4}{3}$

Answer:

$\frac{7 y + 4}{y + 2} = \frac{- 4}{3} \Rightarrow 3 (7 y + 4) = - 4 (y + 2) (by cross multiplication) \Rightarrow 21 y + 12 = - 4 y - 8 \Rightarrow 21 y + 4 y = - 8 - 12 \Rightarrow 25 y = - 20 \Rightarrow y = \frac{- 20}{25} = \frac{- 4}{5} ∴ y = \frac{- 4}{5}$

Page No 107:

Question 23:

Solve:
$\frac{15 (2 - y) - 5 (y + 6)}{1 - 3 y} = 10$

Answer:

$\frac{15 (2 - y) - 5 (y + 6)}{1 - 3 y} = 10 \Rightarrow \frac{30 - 15 y - 5 y - 30}{1 - 3 y} = 10 \Rightarrow \frac{- 20 y}{1 - 3 y} = 10 \Rightarrow 1 (- 20 y) = 10 (1 - 3 y) (by cross multiplication) \Rightarrow - 20 y = 10 - 30 y \Rightarrow - 20 y + 30 y = 10 \Rightarrow 10 y = 10 \Rightarrow y = \frac{10}{10} = 1 ∴ y = 1$

Page No 107:

Question 24:

Solve:
$\frac{2 x - (7 - 5 x)}{9 x - (3 + 4 x)} = \frac{7}{6}$

Answer:

$\frac{2 x - (7 - 5 x)}{9 x - (3 + 4 x)} = \frac{7}{6} \Rightarrow \frac{2 x - 7 + 5 x}{9 x - 3 - 4 x} = \frac{7}{6} \Rightarrow \frac{7 x - 7}{5 x - 3} = \frac{7}{6} \Rightarrow 6 (7 x - 7) = 7 (5 x - 3) (by cross multiplication) \Rightarrow 42 x - 42 = 35 x - 21 \Rightarrow 42 x - 35 x = 42 - 21 \Rightarrow 7 x = 21 \Rightarrow x = \frac{21}{7} = 3 ∴ x = 3$

Page No 107:

Question 25:

Solve:
$m - \frac{(m - 1)}{2} = 1 - \frac{(m - 2)}{3}$

Answer:

$m - \frac{(m - 1)}{2} = 1 - \frac{(m - 2)}{3} \Rightarrow \frac{2 m - m + 1}{2} = 1 - \frac{(m - 2)}{3} (L . C . M . of 1 and 2 is 2) \Rightarrow \frac{m + 1}{2} = \frac{3 - m + 2}{3} (L . C . M . of 1 and 3 is 3) \Rightarrow \frac{m + 1}{2} = \frac{5 - m}{3} \Rightarrow 3 (m + 1) = 2 (5 - m) (by cross multiplication) \Rightarrow 3 m + 3 = 10 - 2 m \Rightarrow 3 m + 2 m = 10 - 3 \Rightarrow 5 m = 7 \Rightarrow m = \frac{7}{5} ∴ m = \frac{7}{5}$

Page No 107:

Question 26:

Solve:
$\frac{3 x + 5}{4 x + 2} = \frac{3 x + 4}{4 x + 7}$

Answer:

$\frac{3 x + 5}{4 x + 2} = \frac{3 x + 4}{4 x + 7} \Rightarrow (4 x + 7) (3 x + 5) = (4 x + 2) (3 x + 4) (by cross multiplication) \Rightarrow 12 x^{2} + 20 x + 21 x + 35 = 12 x^{2} + 16 x + 6 x + 8 \Rightarrow 12 x^{2} + 41 x + 35 = 12 x^{2} + 22 x + 8 \Rightarrow 12 x^{2} - 12 x^{2} + 41 x - 22 x = 8 - 35 \Rightarrow 19 x = - 27 \Rightarrow x = \frac{- 27}{19} ∴ x = \frac{- 27}{19}$

Page No 107:

Question 27:

Solve:
$\frac{9 x - 7}{3 x + 5} = \frac{3 x - 4}{x + 6}$

Answer:

$\frac{9 x - 7}{3 x + 5} = \frac{3 x - 4}{x + 6} \Rightarrow (x + 6) (9 x - 7) = (3 x + 5) (3 x - 4) (by cross multiplication) \Rightarrow 9 x^{2} - 7 x + 54 x - 42 = 9 x^{2} - 12 x + 15 x - 20 \Rightarrow 9 x^{2} + 47 x - 42 = 9 x^{2} + 3 x - 20 \Rightarrow 9 x^{2} - 9 x^{2} + 47 x - 3 x = - 20 + 42 \Rightarrow 44 x = 22 \Rightarrow x = \frac{22}{44} = \frac{1}{2} ∴ x = \frac{1}{2}$

Page No 107:

Question 28:

Solve:
$\frac{2 - 7 x}{1 - 5 x} = \frac{3 + 7 x}{4 + 5 x}$

Answer:

$\frac{2 - 7 x}{1 - 5 x} = \frac{3 + 7 x}{4 + 5 x} \Rightarrow (4 + 5 x) (2 - 7 x) = (1 - 5 x) (3 + 7 x) (by cross multiplication) \Rightarrow 8 - 28 x + 10 x - 35 x^{2} = 3 + 7 x - 15 x - 35 x^{2} \Rightarrow - 35 x^{2} - 18 x + 8 = - 35 x^{2} - 8 x + 3 \Rightarrow - 35 x^{2} + 35 x^{2} - 18 x + 8 x = - 8 + 3 \Rightarrow - 10 x = - 5 \Rightarrow x = \frac{- 5}{- 10} = \frac{1}{2} ∴ x = \frac{1}{2}$

Page No 111:

Question 1:

Two numbers are in the ratio 8 : 3. If the sum of the numbers is 143, find the numbers.

Answer:

$Let the numbers be 8 x and 3 x . 8 x + 3 x = 143 \Rightarrow 11 x = 143 \Rightarrow x = \frac{143}{11} \Rightarrow x = 13 ∴ O ne number = 8 x = 8 \times 13 = 104 Other number = 3 x = 3 \times 13 = 39$

Page No 111:

Question 2:

$\frac{2}{3}$ of a number is 20 less than the original number. Find the number.

Answer:

$Let the original number be x . \frac{2}{3} of the number is 20 less than the original number . ∴ \frac{2}{3} x = x - 20 \Rightarrow \frac{2 x}{3} = x - 20 \Rightarrow 2 x = 3 (x - 20) (by cross multiplication) \Rightarrow 2 x = 3 x - 60 \Rightarrow 2 x - 3 x = - 60 \Rightarrow - x = - 60 \Rightarrow x = 60 Therefore, the original number is 60 .$

Page No 111:

Question 3:

Four-fifths of a number is 10 more than two-thirds of the number. Find the number.

Answer:

$Let the number be x . Four fifths of the number is 10 more than two thirds of the number . ∴ \frac{4}{5} x = 10 + \frac{2}{3} x \Rightarrow \frac{4 x}{5} = 10 + \frac{2 x}{3} \Rightarrow \frac{4 x}{5} = \frac{30 + 2 x}{3} (L . C . M . of 1 and 3 is 3) \Rightarrow 3 (4 x) = 5 (30 + 2 x) (by cross multiplication) \Rightarrow 12 x = 150 + 10 x \Rightarrow 12 x - 10 x = 150 \Rightarrow 2 x = 150 \Rightarrow x = \frac{150}{2} = 75 Therefore, the number is 75 .$

Page No 111:

Question 4:

Twenty-four is divided into two parts such that 7 times the first part added to 5 times the second part makes 146. Find each part.

Answer:

$Let one part be x . 7 times the first part = 7 x Let the other part be (24 - x) . 5 times the second part = 5 (24 - x) ∴ 7 x + 5 (24 - x) = 146 \Rightarrow 7 x + 120 - 5 x = 146 \Rightarrow 7 x - 5 x = 146 - 120 \Rightarrow 2 x = 26 \Rightarrow x = \frac{26}{2} = 13 Therefore, one part is 13 . Other part = (24 - x) = (24 - 13) = 11$

Page No 111:

Question 5:

Find the number whose fifth part increased by 5 is equal to its fourth part diminished by 5.

Answer:

$Let the number be x . Fifth part increased by 5 = \frac{x}{5} + 5 Fourt part diminished by 5 = \frac{x}{4} - 5 ∴ \frac{x}{5} + 5 = \frac{x}{4} - 5 \Rightarrow 5 + 5 = \frac{x}{4} - \frac{x}{5} \Rightarrow 10 = \frac{5 x - 4 x}{20} \Rightarrow 10 = \frac{x}{20} \Rightarrow 200 = x \Rightarrow x = 200 Therefore, the number is 200 .$

Page No 111:

Question 6:

Three numbers are in the ratio of 4 : 5 : 6. If the sum of the largest and the smallest equals the sum of the third and 55, find the numbers.

Answer:

$Let the common multiple for the given three numbers be x . Then, the three numbers would be 4 x, 5 x and 6 x . ∴ 4 x + 6 x = 5 x + 55 \Rightarrow 10 x = 5 x + 55 \Rightarrow 10 x - 5 x = 55 \Rightarrow 5 x = 55 \Rightarrow x = \frac{55}{5} = 11 ∴ S mallest number = 4 x = 4 (11) = 44 L argest number is = 6 x = 6 (11) = 66 T hird number = 5 x = 5 (11) = 55 Therefore, t h e three numbers are 44, 55 and 66 .$

Page No 111:

Question 7:

If 10 be added to four times a certain number, the result is 5 less than five times the number. Find the number.

Answer:

$Let the number be x . ∴ 10 + 4 x = 5 x - 5 \Rightarrow 10 + 5 = 5 x - 4 x \Rightarrow 15 = x \Rightarrow x = 15 (by transposition) Therefore, the number is 15 .$

Page No 111:

Question 8:

Two numbers are such that the ratio between them is 3 : 5. If each is increased by 10, the ratio between the new numbers so formed is 5 : 7. Find the original numbers.

Answer:

Let us consider x as the common multiple of both the number.
Then, first number = 3x
Second number = 5x

$∴ \frac{3 x + 10}{5 x + 10} = \frac{5}{7} \Rightarrow 7 (3 x + 10) = 5 (5 x + 10) (by cross multiplication) \Rightarrow 21 x + 70 = 25 x + 50 \Rightarrow 21 x - 25 x = 50 - 70 \Rightarrow - 4 x = - 20 \Rightarrow x = \frac{- 20}{- 4} = 5$

Therefore, the common multiple of both the numbers is 5.
First number = $3 x = 3 \times 5 = 15$
Second number = $5 x = 5 \times 5 = 25$

Page No 111:

Question 9:

Find three consecutive odd numbers whose sum is 147.

Answer:

$Let the first odd number be x . Let the second odd number be (x + 2) . Let the third odd number be (x + 4) . ∴ x + (x + 2) + (x + 4) = 147 \Rightarrow x + x + 2 + x + 4 = 147 \Rightarrow 3 x + 6 = 147 \Rightarrow 3 x = 147 - 6 \Rightarrow 3 x = 141 \Rightarrow x = \frac{141}{3} = 47 Therefore, the first odd number is 47 . S econd odd number = (x + 2) = (47 + 2) = 49 T hird odd number = (x + 4) = (47 + 4) = 51$

Page No 111:

Question 10:

Find three consecutive even numbers whose sum is 234.

Answer:

$Let the first even number b e x . Let the second even number b e x + 2 . Let the third even number be x + 4 . ∴ x + x + 2 + x + 4 = 234 \Rightarrow x + x + 2 + x + 4 = 234 \Rightarrow 3 x + 6 = 234 \Rightarrow 3 x = 234 - 6 \Rightarrow 3 x = 228 \Rightarrow x = \frac{228}{3} = 76 ∴ F irst even number = x = 76 S econd even number = x + 2 = 76 + 2 = 78 T hird even number = x + 4 = 80$

Page No 111:

Question 11:

The sum of the digits of a two-digit number is 12. If the new number formed by reversing the digits is greater than the original number by 54, find the original number. Check your solution.

Answer:

$Let the digit in t h e units place be x . D igit in t h e tens place = (12 - x) ∴ Original number = 10 (12 - x) + x = 120 - 9 x On reversing the digits, we have x at the tens place and (12 - x) at the units place . ∴ New number = 10 x + 12 - x = 9 x + 12 New number - Original number = 54 \Rightarrow 9 x + 12 - (120 - 9 x) = 54 \Rightarrow 9 x + 12 - 120 + 9 x = 54 \Rightarrow 18 x - 108 = 54 \Rightarrow 18 x = 54 + 108 \Rightarrow 18 x = 162 \Rightarrow x = \frac{162}{18} = 9 Therefore, the digit in t h e units place is 9 . D igit in tens place = (12 - x) = (12 - 9) = 3 Therefore, the original number is 39 . Check : The original number is 39 . S um of the digits in the original number = (3 + 9) = 12 N ew number obtained on reversing the digit s = 93 New number - Original number = (93 - 39) = 54 Thus, both the given conditions are satisfied by 39 . Hence, the original number i s 39 .$

Page No 111:

Question 12:

The digit in the tens place of a two-digit number is three times that in the units place. If the digits are reversed, the new number will be 36 less than the original number. Find the original number. Check your solution.

Answer:

$Let the digit in t h e units place be x . D igit in t h e tens place = 3 x O riginal number = 10 (3 x) + x = 30 x + x On reversing the digits, we have x at the tens place and (3 x) at the units place . ∴ N ew number = 10 (x) + 3 x = 10 x + 3 x New number = Original number - 36 \Rightarrow 10 x + 3 x = 30 x + x - 36 \Rightarrow 13 x = 31 x - 36 \Rightarrow 36 = 31 x - 13 x \Rightarrow 36 = 18 x \Rightarrow 18 x = 36 \Rightarrow x = \frac{36}{18} = 2 Therefore, the digit in t h e units place is 2 . D igit in t h e tens place = (3 x) = 3 \times 2 = 6 Therefore, the original number is 62 . Check : New number + 36 = Original Number 26 + 36 = 62 Henc e, both the conditions are satisfied . Therefore, the original number i s 62 .$

Page No 112:

Question 13:

The denominator of a rational number is greater than its numerator by 7. If the numerator is increased by 17 and the denominator decreased by 6, the new number becomes 2. Find the original number.

Answer:

$Let the numerator be x . The denominator is greater than the numerator by 7 . ∴ (x + 7) ∴ \frac{x + 17}{(x + 7) - 6} = 2 \Rightarrow \frac{x + 17}{x + 1} = 2 \Rightarrow x + 17 = 2 (x + 1) (by cross multiplication) \Rightarrow x + 17 = 2 x + 2 \Rightarrow x - 2 x = 2 - 17 \Rightarrow - x = - 15 \Rightarrow x = 15 Therefore, the numerator is 15 . Denominator = (x + 7) = (15 + 7) = 22 ∴ Original number = \frac{15}{22}$

Page No 112:

Question 14:

In a fraction, twice the numerator is 2 more than the denominator. If 3 is added to the numerator and to the denominator, the new fraction is $\frac{2}{3}$ . Find the original fraction.

Answer:

$Denominator, d = x It is given that twice the numerator is equal to two more than the denominator . ∴ Twice of numerator, 2 n = x + 2 ∴ Numerator, n = \frac{x + 2}{2} ∴ \frac{n + 3}{d + 3} = \frac{2}{3} \Rightarrow 3 (n + 3) = 2 (d + 3) (b y c r o s s m u l t i p l i c a t i o n) \Rightarrow 3 n + 9 = 2 d + 6 \Rightarrow 3 n - 2 d = 6 - 9 \Rightarrow 3 n - 2 d = - 3 On replace d by x and n by \frac{x + 2}{2} : \Rightarrow 3 (\frac{x + 2}{2}) - 2 x = - 3 \Rightarrow \frac{3 x + 6 - 4 x}{2} = - 3 (taking the L . C . M . of 2 and 1 as 2) \Rightarrow 6 - x = - 6 (by cross multipliction) \Rightarrow - x = - 6 - 6 \Rightarrow x = 12 The denominator is 12 . ∴ Numerator = \frac{x + 2}{2} = \frac{12 + 2}{2} = \frac{14}{2} = 7 ∴ O riginal fraction = \frac{7}{12}$

Page No 112:

Question 15:

The length of a rectangle exceeds its breadth by 7 cm. If the length is decreased by 4 cm and the breadth is increased by 3 cm, the area of the new rectangle is the same as the area of the original rectangle. Find the length and the breadth of the original rectangle.

Answer:

$Let the breadth of the original rectangle be x cm . Then, its length will be (x + 7) cm . The area of the rectangle will be (x) (x + 7) {cm}^{2} . ∴ (x + 3) (x + 7 - 4) = (x) (x + 7) \Rightarrow (x + 3) (x + 3) = x^{2} + 7 x \Rightarrow x^{2} + 3 x + 3 x + 9 = x^{2} + 7 x \Rightarrow x^{2} + 6 x + 9 = x^{2} + 7 x \Rightarrow 9 = x^{2} - x^{2} + 7 x - 6 x \Rightarrow 9 = x \Rightarrow x = 9 (by transposition) Breadth of the original rectangle = 9 cm Length of the original rectangle = (x + 7) = (9 + 7) = 16 cm$

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Question 16:

The width of a rectangle is two-thirds its length. If the perimeter is 180 metres, find the dimensions of the rectangle.

Answer:

$Let the width of the rectangle be x cm . It is \frac{2}{3} of the length of the rectangle . This means that the length of the rectangle will be \frac{3}{2} x . Perimeter of the rectangle = 2 (x) + 2 (\frac{3}{2}) x = 180 m ∴ 2 x + \frac{6 x}{2} = 180 \Rightarrow \frac{4 x + 6 x}{2} = 180 (taking the L . C . M . of 1 on the L . H . S . of the equation) \Rightarrow 10 x = 2 \times 180 (by cross multiplication) \Rightarrow 10 x = 360 \Rightarrow x = \frac{360}{10} = 36 Therefore, the width of the rectangle is 36 m . Length of the rectangle will be = \frac{3}{2} x = \frac{3}{2} (36) = 54 m$

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Question 17:

An altitude of a triangle is five-thirds thelength of its corresponding base. If the altitude be increased by 4 cm and the base decreased by 2 cm, the area of the triangle remains the same. Find the base and the altitude of the triangle.

Answer:

$Let the length of the base of the triangle be x cm . Then, its altitude will be \frac{5}{3} x cm . Area of the triangle = \frac{1}{2} (x) (\frac{5}{3} x) = \frac{5}{6} x^{2} ∴ \frac{1}{2} (x - 2) (\frac{5}{3} x + 4) = \frac{5}{6} x^{2} \Rightarrow (\frac{x - 2}{2}) (\frac{5 x + 12}{3}) = \frac{5 x^{2}}{6} \Rightarrow \frac{(x - 2) (5 x + 12)}{6} = \frac{5 x^{2}}{6} \Rightarrow \frac{5 x^{2} + 12 x - 10 x - 24}{6} = \frac{5 x^{2}}{6} \Rightarrow 5 x^{2} + 2 x - 24 = 5 x^{2} (cancelling the denominators from both the sides since they are same) \Rightarrow 5 x^{2} - 5 x^{2} + 2 x = 24 \Rightarrow 2 x = 24 \Rightarrow x = \frac{24}{2} = 12 m Therefore, the bas e of the triangle is 12 m . Altitude of the triangle = \frac{5}{3} x = \frac{5}{3} (12) = 20 m$

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Question 18:

Two angles of a triangle are in the ratio 4 : 5. If the sum of these angles is equal to the third angle, find the angles of the triangle.

Answer:

$Let the common multiple of all the three angles be x . Then, the first angle will be 4 x . And the second angle w i l l be 5 x . In a triangle, sum of all the three angles will be equal to 180 ° . ∴ T hird angle = 180 - (4 x + 5 x) = 180 - 9 x ∴ 4 x + 5 x = 180 - 9 x \Rightarrow 9 x = 180 - 9 x \Rightarrow 9 x + 9 x = 180 \Rightarrow 18 x = 180 \Rightarrow x = \frac{180}{18} = 10 First angle = 4 x = 4 \times 10 = 40^{°} Second angle = 5 x = 5 \times 10 = 50 ° Third angle = 4 x + 5 x = 9 x = 9 \times 10 = 90 °$

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Question 19:

A steamer goes downstream from one port to another in 9 hours. It covers the same distance upstream in 10 hours. If the speed of the stream be 1 km/h, find the speed of the steamer in still water and the distance between the ports.

Answer:

$Let the speed of the steamer in still water be x km / h . Speed (downstream) = (x + 1) km / h Speed (upstream) = (x - 1) km / h Distance covered in 9 hours while going downstream = 9 (x + 1) km Distance covered in 10 hours while going upstream = 10 (x - 1) km But both of these distances will be same . 9 (x + 1) = 10 (x - 1) \Rightarrow 9 x + 9 = 10 x - 10 \Rightarrow 9 + 10 = 10 x - 9 x \Rightarrow 19 = x \Rightarrow x = 19 Therefore, the speed of the steamer in still water is 19 km / h . Distance between the ports = 9 (x + 1) = 9 (19 + 1) = 9 \times 20 = 180 km$

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Question 20:

The distance between two stations is 300 km. Two motorcyclists start simultaneously from these stations and move towards each other. The speed of one of them is 7 km/h more than that of the other. If the distance between them after 2 hours of their start is 34 km, find the speed of each motorcyclist. Check your solution.

Answer:

$Let the speed of one motorcyclist be x km / h . So, the speed of the other motorcyclist will be (x + 7) km / h . D istance travelled by the first motorcyclist in 2 hours = 2 x km D istance travelled by the second motorcyclist in 2 hours = 2 (x + 7) km Therefore, 300 - (2 x + (2 x + 14)) = 34 \Rightarrow 300 - (2 x + 2 x + 14) = 34 \Rightarrow 300 - 4 x - 14 = 34 \Rightarrow 286 - 4 x = 34 \Rightarrow 286 - 34 = 4 x \Rightarrow 252 = 4 x \Rightarrow x = \frac{252}{4} = 63 Therefore, the speed of the first motorcyclist is 63 km / h . The speed of the second motorcyclist is (x + 7) = (63 + 7) = 70 km / h . Check : The distance covered by the first motorcyclist in 2 hours = 63 \times 2 = 126 km The distance covered by the second motorcyclist in 2 hours = 70 \times 2 = 140 km The distance between the motorcyclists after 2 hours = 300 - (126 + 140) = 34 km (which is the same as given) Therefore, the speeds of the motorcyclists are 63 km / h and 70 km / h, respectively .$

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Question 21:

Divide 150 into three parts such that the second number is five-sixths the first and the third number is four-fifths the second.

Answer:

$Let the first number be x . Then, the second number will be \frac{5}{6} x . Third numbe r = \frac{4}{5} (\frac{5}{6} x) = \frac{2}{3} x ∴ x + \frac{5 x}{6} + \frac{2 x}{3} = 150 \Rightarrow \frac{6 x + 5 x + 4 x}{6} = 150 (multiplying the L . H . S . by 6, which is the L . C . M . of 1, 6 and 3) \Rightarrow 15 x = 150 \times 6 (by cross multiplication) \Rightarrow 15 x = 900 \Rightarrow x = \frac{900}{15} = 60 Therefore, the first number is 60 . Second number = \frac{5}{6} x = \frac{5}{6} (60) = 50 Third number = \frac{2}{3} x = \frac{2}{3} (60) = 40$

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Question 22:

Divide 4500 into two parts such that 5% of the first part is equal to 10% of the second part.

Answer:

$Let the first part be x . Let the second part be (4500 - x) . ∴ 5 % o f x = 10 % o f (4500 - x) \Rightarrow (\frac{5}{100}) x = (\frac{10}{100}) (4500 - x) \Rightarrow \frac{5 x}{100} = \frac{45000 - 10 x}{100} \Rightarrow 5 x = 45000 - 10 x (by cancellation of same denominators from both the sides) \Rightarrow 5 x + 10 x = 45000 \Rightarrow 15 x = 45000 \Rightarrow x = \frac{45000}{15} = 3000 Therefore, the first part is 3000 . Second part = (4500 - x) = (4500 - 3000) = 1500$

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Question 23:

Rakhi's mother is four times as old as Rakhi. After 5 years, her mother will be three times as old as she will be then. Find their present ages.

Answer:

$Let the present age of Rakhi be x . Then, the present age of Rakhi' s mother will be 4 x . After five years, Rakhi' s age will be (x + 5) . After five years, her mother' s age will be (4 x + 5) . 4 x + 5 = 3 (x + 5) \Rightarrow 4 x + 5 = 3 x + 15 \Rightarrow 4 x - 3 x = 15 - 5 \Rightarrow x = 10 Present age of Rakhi = 10 years Present age of Rakhi' s mother = 4 (x) = 4 \times 10 = 40 years$

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Question 24:

Monu's father is 26 years younger than Monu's grandfather and 29 years older than Monu. The sum of the ages of all the three is 135 years. What is the age of each one of them?

Answer:

$Let the age of Monu' s father be x years . The age of Monu' s grandfather will be (x + 26) . Then, the age of Monu will be (x - 29) . ∴ x + (x + 26) + (x - 29) = 135 \Rightarrow x + x + 26 + x - 29 = 135 \Rightarrow 3 x - 3 = 135 \Rightarrow 3 x = 135 + 3 \Rightarrow 3 x = 138 \Rightarrow x = \frac{138}{3} = 46 ∴ A ge of Monu' s father = 46 years Age of Monu' s grandfather = (x + 26) = (46 + 26) = 72 years Age of Monu = (x - 29) = 46 - 29 = 17 years$

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Question 25:

A man is 10 times older than his grandson. He is also 54 years older than him. Find their present ages.

Answer:

$Let the age of the grandson be x years . Then, his grandfather' s age will be 10 x . Also, the grandfather is 54 years older than his grandson . ∴ Age of the grandson = x + 54 10 x = x + 54 \Rightarrow 10 x - x = 54 \Rightarrow 9 x = 54 \Rightarrow x = \frac{54}{9} = 6 Therefore, the grandson' s age is 6 years . Grandfather' s age = 10 (x) = 10 \times 6 = 60 years$

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Question 26:

The difference between the ages of two cousins is 10 years. 15 years ago, if the elder one was twice as old as the younger one, find their present ages.

Answer:

$Let the age of the younger cousin be x . Then, the age of the elder cousin will be (x + 10) . 15 years ago : Age of the younger cousin = (x - 15) Ag e of elder cousin = (x + 10 - 15) = (x - 5) ∴ (x - 5) = 2 (x - 15) \Rightarrow x - 5 = 2 x - 30 \Rightarrow x - 2 x = - 30 + 5 \Rightarrow - x = - 25 \Rightarrow x = 25 Therefore, the present age of the younger cousin is 25 years . Present age of elder cousin = (x + 10) = (25 + 10) = 35 years$

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Question 27:

Half of a herd of deer are grazing in the field and three-fourths of the remaining are playing nearby. The rest 9 are drinking water from the pond. Find thenumber of deer in the herd.

Answer:

$Let the number of deer in the herd be x . The number of deer grazing in the field is (\frac{1}{2}) x . Remaining deer = x - \frac{x}{2} = \frac{x}{2} N umber of deer playing nearby = \frac{3}{4} (\frac{x}{2}) = \frac{3}{8} x The number of deer drinking water from the pond is 9 . ∴ 9 + \frac{3}{8} x + \frac{1}{2} x = x \Rightarrow \frac{72 + 3 x + 4 x}{8} = x (multiplying the L . H . S . by 8, which is the L . C . M . of 1, 8 and 2) \Rightarrow 72 + 7 x = 8 x (by cross multiplication) \Rightarrow 72 = 8 x - 7 x \Rightarrow 72 = x \Rightarrow x = 72 T otal number of deer in the herd = 72$

Page No 113:

Question 1:

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

Answer:

Page No 113:

Question 2:

Tick (âœ“) 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:

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

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Question 3:

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

Answer:

(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 (âœ“) 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:

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

Answer:

(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:

Tick (âœ“) the correct answer:
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:

(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:

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

Answer:

(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:

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

Answer:

(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 (âœ“) 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:

(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:

Tick (âœ“) the correct answer:
If $\frac{4 x + 8}{5 x + 8} = \frac{5}{6},$ then x = ?
(a) 4
(b) 6
(c) 8
(d) 12

Answer:

(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:

Tick (âœ“) the correct answer:
If $\frac{n}{n + 15} = \frac{4}{9}$ , then n = ?
(a) 4
(b) 6
(c) 9
(d) 12

Answer:

(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:

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

Answer:

(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:

Tick (âœ“) the correct answer:
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:

(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:

Tick (âœ“) the correct answer:
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:

(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:

Tick (âœ“) the correct answer:
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:

(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:

Tick (âœ“) the correct answer:
Sum of three consecutive integers is 51. The middle one is
(a) 14
(b) 15
(c) 16
(d) 17

Answer:

(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:

Tick (âœ“) the correct answer:
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:

(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 (âœ“) 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:

(c) 48

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

Page No 115:

Question 1:

Subtract 4a² + 5b² − 6c² + 8 from 2a² − 3b² − 4c² − 5.

Answer:

$2 a^{2} - 3 b^{2} - 4 c^{2} - 5 - (4 a^{2} + 5 b^{2} - 6 c^{2} + 8) = 2 a^{2} - 3 b^{2} - 4 c^{2} - 5 - 4 a^{2} - 5 b^{2} + 6 c^{2} - 8 = 2 a^{2} - 4 a^{2} - 3 b^{2} - 5 b^{2} - 4 c^{2} + 6 c^{2} - 5 - 8 = - 2 a^{2} - 8 b^{2} + 2 c^{2} - 13$

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Question 2:

Find each of the following products:
(i) (4a + 5b) × (5a − 6b)
(ii) (6x² − x + 8) × (x² − 3)

Answer:

$(i) (4 a + 5 b) \times (5 a - 6 b) = 20 a^{2} - 24 a b + 25 a b - 30 b^{2} = 20 a^{2} + a b - 30 b^{2} (i i) (6 x^{2} - x + 8) \times (x^{2} - 3) = 6 x^{4} - 18 x^{2} - x^{3} + 3 x + 8 x^{2} - 24 = 6 x^{4} - x^{3} - 18 x^{2} + 8 x^{2} + 3 x - 24 = 6 x^{4} - x^{3} - 10 x^{2} + 3 x - 24$

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Question 3:

Divide (5a³ − 4a² + 3a + 18) by (a² − 2a + 3).

Answer:

Therefore, $(5 a + 6)$ is the quotient.

Page No 115:

Question 4:

If $(x - \frac{1}{x}) = 4$ , find the value of
(i) $(x^{2} + \frac{1}{x^{2}}) .$
(ii) $(x^{4} + \frac{1}{x^{4}}) .$

Answer:

$(i) Given, (x - \frac{1}{x}) = 4 Squaring both the sides : {(x - \frac{1}{x})}^{2} = 4^{2} \Rightarrow {(x)}^{2} - 2 \times (x) \times (\frac{1}{x}) + {(\frac{1}{x})}^{2} = 16 \Rightarrow x^{2} - 2 + \frac{1}{x^{2}} = 16 \Rightarrow x^{2} + \frac{1}{x^{2}} = 16 + 2 ∴ (x^{2} + \frac{1}{x^{2}}) = 18$ ² + 2⇒ 4² + 2⇒ 16 + 2⇒ 18

^{$(i i) From the first part : (x^{2} + \frac{1}{x^{2}}) = 18 Squaring both the sides : {(x^{2} + \frac{1}{x^{2}})}^{2} = 18^{2} \Rightarrow (x^{2})^{2} + 2 \times (x^{2}) \times (\frac{1}{x^{2}}) + {(\frac{1}{x^{2}})}^{2} = 324 \Rightarrow x^{4} + 2 + \frac{1}{x^{4}} = 324 \Rightarrow x^{4} + \frac{1}{x^{4}} = 324 - 2 ∴ x^{4} + \frac{1}{x^{4}} = 322$}

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Question 5:

Evaluate {(83)² − (17)²}.

Answer:

${(83)}^{2} - {(17)}^{2} = (83 + 17) (83 - 17) [according to the formula {(a)}^{2} - {(b)}^{2} = (a + b) (a - b)] = (100) (66) = 6600$

Page No 115:

Question 6:

Factorise:
(i) x³ − 3x² + x − 3
(ii) 63x²y² − 7
(iii) 1 − 6x + 9x²
(iv) 7x² − 19x − 6

Answer:

$(i) x^{3} - 3 x^{2} + x - 3 = x^{2} (x - 3) + 1 (x - 3) = (x^{2} + 1) (x - 3) (ii) 63 x^{2} y^{2} - 7 = 7 (9 x^{2} y^{2} - 1) = 7 ((3 x y)^{2} - (1)^{2}) [according to the formula a^{2} - b^{2} = (a + b) (a - b)] = 7 (3 x y + 1) (3 x y - 1) (iii) 1 - 6 x + 9 x^{2} = 9 x^{2} - 6 x + 1 = 9 x^{2} - 3 x - 3 x + 1 = 3 x (3 x - 1) - 1 (3 x - 1) = (3 x - 1) (3 x - 1) = {(3 x - 1)}^{2} (iv) 7 x^{2} - 19x - 6 = 7 x^{2} - 21 x + 2 x - 6 = 7 x (x - 3) + 2 (x - 3) = (7 x + 2) (x - 3)$

Page No 115:

Question 7:

Solve: $\frac{2 x + 7}{3 x + 5} = \frac{15}{17} .$

Answer:

$\frac{2 x + 7}{3 x + 5} = \frac{15}{17} \Rightarrow 17 (2 x + 7) = 15 (3 x + 5) \Rightarrow 34 x + 119 = 45 x + 75 \Rightarrow 119 - 75 = 45 x - 34 x \Rightarrow 44 = 11 x \Rightarrow x = \frac{44}{11} = 4 ∴ x = 4$

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Question 8:

5 years ago a man was 7 times as old as his son. After 5 years he will be thrice as old as his son. Find their present ages.

Answer:

$Let the present age of the son be x years and that of the father be f years . 5 years back, the father was 7 times as old as his son . ∴ (f - 5) = 7 (x - 5) f = 7 x - 35 + 5 f = 7 x - 30 . . . (1) After 5 years, ages of the father and son will be (f + 5) and (x + 5), respectively . After 5 years, the father will be three times older than his son . ∴ (f + 5) = 3 (x + 5) 7 x - 30 + 5 = 3 x + 15 [inserting the value of f from equation (1)] 7 x - 25 = 3 x + 15 7 x - 3 x = 25 + 15 4 x = 40 x = \frac{40}{4} = 10 Therefore, the present age of the son is 10 years . Father' s present age = (7 x - 30) = 7 (10) - 30 = 70 - 30 = 40 years$

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Question 9:

Mark (âœ“) against the correct answer:
ab − a − b + 1 = ?
(a) (1 − a)(1 − b)
(b) (1 − a)(b − 1)
(c) (a − 1)(b − 1)
(d) (a − 1)(1 − b)

Answer:

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Question 10:

Mark (âœ“) against the correct answer:
3 + 23x − 8x² = ?
(a) (1 − 8x)(3 + x)
(b) (1 + 8x)(3 − x)
(c) (1 − 8x)(3 − x)
(d) none of these

Answer:

(b) $(1 + 8 x) (3 - x)$

$3 + 23 x - 8 x^{2} = 3 + 24 x - x - 8 x^{2} = 3 (1 + 8 x) - x (1 + 8 x) = (1 + 8 x) (3 - x)$

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Question 11:

Mark (âœ“) against the correct answer:
7x² − 19x − 6 = ?
(a) (x − 3)(7x + 2)
(b) (x + 3)(7x − 2)
(c) (x − 3)(7x − 2)
(d) (7x − 3)(x + 2)

Answer:

(a) $(x - 3) (7 x + 2)$

$7 x^{2} - 19 x - 6 = 7 x^{2} - 21 x + 2 x - 6 = 7 x (x - 3) + 2 (x - 3) = (x - 3) (7 x + 2)$

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Question 12:

Mark (âœ“) against the correct answer:
12x² + 60x + 75 = ?
(a) (2x + 5)(6x + 5)
(b) (3x + 5)²
(c) 3(2x + 5)²
(d) none of these

Answer:

(c) $3 {(2 x + 5)}^{2}$

$12 x^{2} + 60 x + 75 = 3 (4 x^{2} + 20 x + 25) = 3 ({(2 x)}^{2} + 2 \times 2 x \times 5 + 5^{2}) = 3 {(2 x + 5)}^{2}$

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Question 13:

Mark (âœ“) against the correct answer:
10p² + 11p + 3 = ?
(a) (2p + 3)(5p + 1)
(b) (5p + 3)(2p + 1)
(c) (5p − 3)(2p − 1)
(d) none of these

Answer:

(b) $(5 p + 3) (2 p + 1)$

$10 p^{2} + 11 p + 3 = 10 p^{2} + 5 p + 6 p + 3 = 5 p (2 p + 1) + 3 (2 p + 1) = (5 p + 3) (2 p + 1)$

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Question 14:

Mark (âœ“) against the correct answer:
8x³ − 2x = ?
(a) (4x − 1)(2x − 1)x
(b) (2x² + 1)(2x − 1)
(c) 2x(2x − 1)(2x + 1)
(d) none of these

Answer:

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Question 15:

Mark (âœ“) against the correct answer:
$\frac{x + 5}{2} + \frac{x - 5}{3} = \frac{25}{6} gives$
(a) x = 3
(b) x = 4
(c) x = 5
(d) x = 2

Answer:

(b) $x = 4$

$\frac{x + 5}{2} + \frac{x - 5}{3} = \frac{25}{6} \Rightarrow \frac{3 (x + 5) + 2 (x - 5)}{6} = \frac{25}{6} (because L . C . M . of 2 & 3 is 6) \Rightarrow \frac{3 x + 15 + 2 x - 10}{6} = \frac{25}{6} \Rightarrow \frac{5 x + 5}{6} = \frac{25}{6} \Rightarrow 5 x + 5 = 25 (cancelling 6 from the denominator on both the sides) \Rightarrow 5 x = 25 - 5 \Rightarrow 5 x = 20 \Rightarrow x = \frac{20}{5} \Rightarrow x = 4$

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Question 16:

Fill in the blanks.
(i) x²− 18x + 81 = (......)
(ii) 4 − 36x² = (......)(......)(......)
(iii) x² − 14x + 13 = (......)(......)
(iv) 9z² − x² − 4y² + 4xy = (......)(......)
(v) abc − ab − c + 1 = (......)(......)

Answer:

(i) ${(x - 9)}^{2}$

$x^{2} - 18 x + 81 = x^{2} - 2 \times x \times 9 + 9^{2} = {(x - 9)}^{2}$

(ii) $(4) (1 - 3 x) (1 + 3 x)$

$4 - 36 x^{2} = 4 (1 - 9 x^{2}) = 4 (1^{2} - {(3 x)}^{2}) = (4) (1 - 3 x) (1 + 3 x)$

(iii) $(x - 13) (x - 1)$

$x^{2} - 14 x + 13 = x^{2} - 13 x - x + 13 = x (x - 13) - 1 (x - 13) = (x - 13) (x - 1)$

(iv) $(3 z + x - 2 y) (3 z - x + 2 y)$

$9 z^{2} - x^{2} - 4 y^{2} + 4 x y = 9 z^{2} - (x^{2} - 4 x y + 4 y^{2}) = 9 z^{2} - (x^{2} - 2 \times x \times 2 y + {(2 y)}^{2}) = 9 z^{2} - {(x - 2 y)}^{2} = {(3 z)}^{2} - {(x - 2 y)}^{2} = (3 z + (x - 2 y)) (3 z - (x - 2 y)) = (3 z + x - 2 y) (3 z - x + 2 y)$

(v) $(c - 1) (a b - 1)$

$a b c - a b - c + 1 = a b (c - 1) - 1 (c - 1) = (c - 1) (a b - 1)$

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Question 17:

Write 'T' for true and 'F' for false for each of the following:
(i) (5 − 3x²) is a binomial.
(ii) −8 is a monomial.
(iii) (5a − 9b) − (−6a + 2b) = (−a − 7b).
(iv) When x = 2 and y = 1, the value of $\frac{- 8}{7} x^{3} y^{4} is \frac{- 64}{7} .$
(v) $\frac{x}{4} + \frac{x}{6} - \frac{x}{2} = \frac{3}{4} \Rightarrow x = - 9 .$
(vi) $2 x - 5 = 0 \Rightarrow x = \frac{2}{5} .$

Answer:

(i) T
Binomial expression is an expression that shows the sum or the difference of two unlike terms. The above expression has two unlike terms, i.e. 5 and 3x².

(ii) T
Any expression that contains only one term is called a monomial. It can either be a constant or a variable.

(iii) F

$L . H . S . = (5 a - 9 b) - (- 6 a + 2 b) = 5 a - 9 b + 6 a - 2 b = (- a - 7 b) = 5 a + 6 a - 9 b - 2 b = (- a - 7 b) = (11 a - 11 b) This is not equal to (- a - 7 b) . Thus, the L . H . S . is not equal to the R . H . S .$

(iv) T

$\frac{- 8}{7} x^{3} y^{4} Given : x = 2 y = 1 Given expression = \frac{- 8}{7} {(2)}^{3} {(1)}^{4} = \frac{- 8}{7} \times 8 \times 1 = \frac{- 64}{7}$

(v) T

$\frac{x}{4} + \frac{x}{6} - \frac{x}{2} = \frac{3}{4} \Rightarrow \frac{3 x + 2 x - 6 x}{12} = \frac{3}{4} \Rightarrow \frac{- x}{12} = \frac{3}{4} \Rightarrow - 4 x = 36 \Rightarrow x = - \frac{36}{4} \Rightarrow x = - 9$

(vi) F

$2 x - 5 = 0 \Rightarrow 2 x = 5 \Rightarrow x = \frac{5}{2}$

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