Rs Aggarwal 2020 2021 for Class 8 Maths Chapter 8 - Linear Equations

Share with your friends

Rs Aggarwal 2020 2021 Solutions for Class 8 Maths 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 Maths Linear Equations Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2020 2021 Book of Class 8 Maths Chapter 8 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2020 2021 Solutions. All Rs Aggarwal 2020 2021 Solutions for class Class 8 Maths are prepared by experts and are 100% accurate.

Page No 107:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

Answer:

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

Page No 112:

Answer:

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

Page No 112:

Answer:

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

Page No 112:

Answer:

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

Page No 112:

Answer:

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

Page No 112:

Answer:

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

Page No 112:

Answer:

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

Page No 112:

Answer:

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

Page No 112:

Answer:

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

Page No 112:

Answer:

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

Page No 112:

Answer:

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

Page No 112:

Answer:

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

Page No 112:

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:

Answer:

Page No 113:

Answer:

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

Page No 113:

Answer:

(a) 40

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

Page No 113:

Answer:

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

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

Page No 113:

Answer:

(b) -1

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

Page No 113:

Answer:

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

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

Page No 113:

Answer:

(b) -1

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

Page No 113:

Answer:

(c) 36

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

Page No 113:

Answer:

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

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

Page No 113:

Answer:

(c) 8

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

Page No 113:

Answer:

(d) 12

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

Page No 113:

Answer:

(a) -2

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

Page No 113:

Answer:

(c) 80

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

Page No 114:

Answer:

(b) 28 years

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

Page No 114:

Answer:

(b) 5 cm

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

Page No 114:

Answer:

(d) 17

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

Page No 114:

Answer:

(a) 40

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

Page No 114:

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:

Answer:

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

Page No 115:

Answer:

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

Page No 115:

Answer:

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

Page No 115:

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

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

Page No 115:

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:

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:

Answer:

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

Page No 115:

Answer:

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

Page No 115:

Answer:

Page No 115:

Answer:

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

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

Page No 115:

Answer:

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

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

Page No 115:

Answer:

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

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

Page No 115:

Answer:

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

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

Page No 115:

Answer:

Page No 115:

Answer:

(b) $x = 4$

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

Page No 115:

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

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

Page No 116:

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

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

View NCERT Solutions for all chapters of Class 8