Rs Aggarwal 2018 for Class 7 Math Chapter 7 - Linear Equations In One Variable

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Rs Aggarwal 2018 Solutions for Class 7 Math Chapter 7 Linear Equations In One Variable are provided here with simple step-by-step explanations. These solutions for Linear Equations In One Variable are extremely popular among Class 7 students for Math Linear Equations In One Variable Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2018 Book of Class 7 Math Chapter 7 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2018 Solutions. All Rs Aggarwal 2018 Solutions for class Class 7 Math are prepared by experts and are 100% accurate.

Page No 111:

Question 1:

3x − 5 = 0

Answer:

$3 x - 5 = 0 \Rightarrow 3 x = 5 (Transposing - 5 to RHS) \Rightarrow x = \frac{5}{3} CHECK : By substituting x = \frac{5}{3} in the given equation, we get : LHS = 3 (\frac{5}{3}) - 5 = 5 - 5 = 0 RHS = 0 ∴ LHS = RHS Hence checked .$

Page No 111:

Question 2:

8x − 3 = 9 − 2x

Answer:

8x − 3 = 9 − 2x
$\Rightarrow$ 8x + 2x = 9 + 3 (By transposition)
$\Rightarrow$ 10x = 12
$\Rightarrow x = \frac{12}{10} = \frac{6}{5} CHECK : By substituting x = \frac{6}{5} in the given equation, we get : LHS : 8 (\frac{6}{5}) - 3 = \frac{48}{5} - 3 = \frac{48 - 15}{5} = \frac{33}{5} RHS : 9 - 2 (\frac{6}{5}) = 9 - \frac{12}{5} = \frac{45 - 12}{5} = \frac{33}{5} ∴ LHS = RHS Hence checked .$

Page No 111:

Question 3:

7 − 5x = 5 − 7x

Answer:

$\begin{array}{l} We have: \\ 7 - 5 x = 5 - 7 x \\ \Rightarrow - 5 x + 7 x = 5 - 7 [transposing -7 x to LHS and 7 to RHS] \\ \Rightarrow 2 x = - 2 \\ \Rightarrow x = \frac{- 2^{- 1}}{2^{1}} \\ \Rightarrow x = - 1 \\ Thus, x = - 1 is a solution to the given equation. \\ CHECK: Substituting x = - 1 in the given equation, we get: \\ LHS: = 7 - 5 x \\ = 7 - 5 \times (- 1) \\ = 7 + 5 \\ = 12 \\ RHS: \\ = 5 - 7 x \\ =5 - 7 \times (- 1) \\ = 5 + 7 \\ =12 \\ ∴ LHS = RHS \\ Hence, x = - 1 is a solution of the given equation. \end{array}$

Page No 111:

Question 4:

3 + 2x = 1 − x

Answer:

$\begin{array}{l} We have: \\ 3 + 2 x = 1 – x \\ ⇒ 2 x + x + 3 – 1 = 0 (By transposition) \\ \Rightarrow 3 x + 2 = 0 \\ \Rightarrow x = – \frac{2}{3} \\ CHECK: Substituting x = - \frac{2}{3} in the given equation, we get: \\ LHS: 3+2 x \\ =3+2 \times (- \frac{2}{3}) \\ =3 - \frac{4}{3} \\ = \frac{9 - 4}{3} \\ = \frac{5}{3} \\ RHS: 1 - x \\ =1 - (\frac{- 2}{3}) \\ =1+ \frac{2}{3} \\ = \frac{3 + 2}{3} \\ = \frac{5}{3} \\ ∴ LHS = RHS \\ Hence, x = - \frac{2}{3} is a solution of the given equation . \end{array}$

Page No 111:

Question 5:

2(x − 2) +3(4x − 1) = 0

Answer:

$\begin{array}{l} We have: \\ 2 (x - 2) + 3 (4 x - 1) = 0 \\ \Rightarrow 2 x - 4 + 12 x - 3 =0 \\ \Rightarrow 14 x - 7 =0 \\ \Rightarrow 14 x = 7 (By transposition) \\ \Rightarrow x = \frac{1}{2} \\ CHECK: Substituting x = \frac{1}{2} in the given equation, we get: \\ LHS: 2 (x - 2) + 3 (4 x - 1) \\ =2 x - 4 + 12 x - 3 \\ =2 \times \frac{1}{2} - 4 + 12 \times \frac{1}{2} - 3 \\ =1-4+6-3 \\ = - 7 + 7 \\ =0 \\ RHS: 0 \\ ∴ LHS= RHS \\ Hence, x = \frac{1}{2} is a solution of the given equation . \end{array}$

Page No 111:

Question 6:

5(2x − 3) − 3(3x − 7) = 5

Answer:

$\begin{array}{l} We have: \\ 5 (2 x - 3) - 3 (3 x - 7) = 5 \\ \Rightarrow 10 x - 15 - 9 x + 21 = 5 \\ \Rightarrow 10 x - 9 x =5+15 - 21 (By transposition) \\ \Rightarrow x = 20 - 21 \\ \Rightarrow x = - 1 \\ CHECK: Substituting x = - 1 in the given equation, we get: \\ LHS: 5 (2 x - 3) - 3 (3 x - 7) \\ =10 x - 15 - 9 x + 21 \\ =10 \times (- 1) - 15 - 9 \times (- 1) + 21 \\ = - 10 - 15 + 9 + 21 \\ = - 25 + 30 \\ =5 \\ RHS: 5 \\ ∴ LHS = RHS \\ Hence, x = - 1 is a solution of the given equation . \end{array}$

Page No 111:

Question 7:

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

Answer:

$\begin{array}{l} We have: \\ 2 x - \frac{1}{3} = \frac{1}{5} - x \\ \Rightarrow 2 x + x = \frac{1}{5} + \frac{1}{3} \\ \Rightarrow 3 x = \frac{3 \times 1 + 5 \times 1}{15} \\ \Rightarrow 3 x = \frac{3 + 5}{15} \\ \Rightarrow 3 x = \frac{8}{15} \\ \Rightarrow x = \frac{8}{15 \times 3} \\ \Rightarrow x = \frac{8}{45} \\ CHECK: Substituting x = \frac{8}{45} in the given equation, we get: \\ LHS: 2 x - \frac{1}{3} \\ =2 \times \frac{8}{45} - \frac{1}{3} \\ = \frac{16}{45} - \frac{1}{3} \\ = \frac{16 \times 1 - 15 \times 1}{45} \\ = \frac{16 - 15}{45} \\ = \frac{1}{45} \\ RHS : \frac{1}{5} - x \\ = \frac{1}{5} - \frac{8}{45} \\ = \frac{1 \times 9 - 1 \times 8}{45} \\ = \frac{9 - 8}{45} \\ = \frac{1}{45} \\ ∴ LHS=RHS \end{array} Hence, x = \frac{8}{45} is a solution of the given equation .$

Page No 111:

Question 8:

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

Answer:

$\begin{array}{l} We have: \\ \frac{1}{2} x - 3 = 5 + \frac{1}{3} x \\ \Rightarrow \frac{1}{2} x - \frac{1}{3} x = 5 + 3 (transposing \frac{1}{3} x to LHS and - 3 to RHS) \\ \Rightarrow (\frac{1 \times 3 - 1 \times 2}{6}) x = 8 \\ \Rightarrow (\frac{3 - 2}{6}) x = 8 \\ \Rightarrow \frac{1}{6} x = 8 \\ \Rightarrow x = 8 \times 6 \\ \Rightarrow x = 48 \\ CHECK: Substituting x =48 in the given equation, we get: \\ LHS: \frac{1}{2} x - 3 \\ = \frac{1}{2^{1}} \times 4 8^{24} - 3 \\ =24 - 3 \\ =21 \\ RHS: 5 + \frac{1}{3} x \\ =5+ \frac{1}{3^{1}} \times 4 8^{16} \\ =5+16 \\ =21 \\ ∴ LHS=RHS \\ Hence, x =48 is a solution of the given equation . \end{array}$

Page No 111:

Question 9:

$\frac{x}{2} + \frac{x}{4} = \frac{1}{8}$

Answer:

$\begin{array}{l} \frac{x}{2} + \frac{x}{4} = \frac{1}{8} \\ \Rightarrow \frac{x \times 2 + x \times 1}{4} = \frac{1}{8} \\ \Rightarrow \frac{2 x + x}{4} = \frac{1}{8} \\ \Rightarrow \frac{3 x}{4} = \frac{1}{8} \\ \Rightarrow 3 x = \frac{1}{8^{2}} \times 4^{1} \\ \Rightarrow 3 x = \frac{1}{2} \\ \Rightarrow x = \frac{1}{6} \\ CHECK: Substituting x = \frac{1}{6} in the given equation, we get: \\ LHS: \frac{x}{2} + \frac{x}{4} \\ = \frac{x \times 2 + x \times 1}{4} \\ = \frac{2 x + x}{4} \\ = \frac{3 x}{4} \\ = \frac{3^{1}}{4} \times \frac{1}{6^{2}} \\ = \frac{1}{8} \\ RHS: \frac{1}{8} \\ ∴ LHS = RHS \\ Hence, x = \frac{1}{3} is a solution of the given equation . \end{array}$

Page No 111:

Question 10:

3x + 2(x + 2) = 20 − (2x − 5)

Answer:

$\begin{array}{l} We have: \\ 3 x + 2 (x + 2) = 20 - (2 x - 5) \\ \Rightarrow 3 x + 2 x + 4 = 20 - 2 x + 5 \\ \Rightarrow 3 x + 2 x + 2 x = 20 + 5 - 4 (T ransposing - 2x to LHS and 4 to RHS) \\ \Rightarrow 7 x =21 \\ \Rightarrow x = \frac{2 1^{3}}{7^{1}} \\ \Rightarrow x = 3 \\ CHECK: Substituting x =3 in the given equation, we get: \\ LHS= 3 x + 2 (x + 2) \\ =3 x + 2 x + 4 \\ =5 x + 4 \\ =5 \times 3+4 \\ =15+4 \\ =19 \\ RHS= 20 - (2 x - 5) \\ =20 - 2 x + 5 \\ =25 - 2 \times 3 \\ =25 - 6 \\ =19 \\ ∴ LHS = RHS \\ Hence, x =3 is a solution of the given equation . \end{array}$

Page No 111:

Question 11:

13(y − 4) − 3(y − 9) − 5(y + 4) = 0

Answer:

$\begin{array}{l} We have: \\ 13 (y - 4) - 3 (y - 9) - 5 (y + 4) = 0 \\ \Rightarrow 13y - 52 - 3 y + 27 - 5 y - 20 = 0 \\ \Rightarrow 13y - 3 y - 5 y = 52 + 20 - 27 (T ransposing - 52, - 20 and 27 to RHS) \\ \Rightarrow 5y =45 \\ \Rightarrow y = \frac{4 5^{9}}{5^{1}} \\ \Rightarrow y = 9 \\ CHECK: Substituting x =9 in the given equation, we get: \\ LHS= 13 (y - 4) - 3 (y - 9) - 5 (y + 4) \\ =13y - 52 - 3 y + 27 - 5 y - 20 \\ =13y - 3 y - 5 y - 52 + 27 - 20 \\ =5y - 45 \\ =5 \times 9 - 45 \\ =45 - 45 \\ =0 \\ RHS=0 \\ ∴ LHS=RHS \\ Hence, x =9 is a solution of the given equation . \end{array}$

Page No 111:

Question 12:

$\frac{2 m + 5}{3} = 3 m - 10$

Answer:

$\begin{array}{l} We have, \\ \frac{2 m + 5}{3} = 3 m - 10 \\ \Rightarrow 2 m + 5 = 3 (3 m - 10) \\ \Rightarrow 2 m + 5 = 9 m - 30 \\ \Rightarrow 2 m - 9 m = - 30 - 5 (T ransposing 9 m to LHS and 5 to RHS) \\ \Rightarrow - 7 m = - 35 \\ \Rightarrow m = \frac{- 3 5^{5}}{- 7^{1}} \\ \Rightarrow m = 5 \\ CHECK: Substituting m = 5 in the given equation, we get: \\ LHS= \frac{2 m + 5}{3} \\ = \frac{2 \times 5 + 5}{3} \\ = \frac{10 + 5}{3} \\ = \frac{1 5^{5}}{3^{1}} \\ =5 \\ RHS= 3 m - 10 \\ =3 \times 5 - 10 \\ =15 - 10 \\ =5 \\ ∴ LHS=RHS \\ Hence, x =5 is a solution of the given equation . \end{array}$

Page No 111:

Question 13:

6(3 x + 2) − 5(6x − 1) = 3(x − 8) − 5(7x − 6) + 9x

Answer:

$\begin{array}{l} We have: \\ 6 (3 x + 2) - 5 (6 x - 1) = 3 (x - 8) - 5 (7 x - 6) + 9 x \\ ⇒ 18 x +1 2 - 30 x + 5 =3x - 24 - 35 x + 30 + 9 x \\ \Rightarrow 18 x - 30 x - 3 x + 35 x - 9 x = - 24 + 30 - 12 - 5 (T ransposing 3 x, 9 x and - 35 x to LHS and 12 and 5 to RHS) \\ \Rightarrow 53 x - 42 x =30 - 41 \\ \Rightarrow 11 x = - 11 \\ \Rightarrow x = \frac{- 1 1^{1}}{1 1^{1}} \\ \Rightarrow x = - 1 \\ CHECK: Substituting x = - 1 in the given equation, we get: \\ LHS= 6 (3 x + 2) - 5 (6 x - 1) \\ =18 x +1 2 - 30 x + 5 \\ = - 12 x + 17 \\ = - 12 \times (- 1) + 17 \\ =12+17 \\ =29 \\ RHS= 3 (x - 8) - 5 (7 x - 6) + 9 x \\ =3 x - 24 - 35 x + 30 + 9 x \\ =12 x - 35 x - 24 + 30 \\ = - 23 x + 6 \\ = - 23 \times (- 1) + 6 \\ = 23+6 \\ =29 \\ ∴ LHS=RHS \\ Hence, x = - 1 is a solution of the given equation . \end{array}$

Page No 111:

Question 14:

t − (2t + 5) − 5(1 − 2t) = 2(3 + 4t) −3(t − 4)

Answer:

$\begin{array}{l} We have: \\ t - (2 t + 5) - 5 (1 - 2 t) = 2 (3 + 4 t) - 3 (t - 4) \\ \Rightarrow t - 2 t - 5 - 5 + 10 t =6+8t - 3 t + 12 \\ \Rightarrow t - 2 t + 10 t - 8 t + 3 t = 6 + 12 + 5 + 5 (By transposition) \\ \Rightarrow 14 t - 10 t =28 \\ \Rightarrow 4 t = 28 \\ \Rightarrow x = \frac{2 8^{7}}{4^{1}} \\ \Rightarrow x = 7 \\ CHECK: Substituting x =7 in the given equation, we get: \\ LHS= t - (2 t + 5) - 5 (1 - 2 t) \\ = t - 2 t - 5 - 5 + 10 t \\ =11 t - 2 t - 10 \\ =9 t - 10 \\ =9 \times 7 - 10 \\ =63 - 10 \\ =53 \\ RHS= 2 (3 + 4 t) - 3 (t - 4) \\ =6+8 t - 3 t + 12 \\ =5 t + 18 \\ =5 \times 7 + 18 \\ =35 + 18 \\ = 53 \\ ∴ LHS=RHS \\ Hence, x =7 is a solution of the given equation . \end{array}$

Page No 111:

Question 15:

$\frac{2}{3} x = \frac{3}{8} x + \frac{7}{12}$

Answer:

$\begin{array}{l} We have: \\ \frac{2}{3} x = \frac{3}{8} x + \frac{7}{12} \\ \Rightarrow \frac{2}{3} x - \frac{3}{8} x = \frac{7}{12} (T ransposing \frac{3}{8} x to LHS) \\ \Rightarrow (\frac{2 \times 8 - 3 \times 3}{24}) x = \frac{7}{12} \\ \Rightarrow (\frac{16 - 9}{24}) x = \frac{7}{12} \\ \Rightarrow \frac{7}{24} x = \frac{7}{12} \\ \Rightarrow x = \frac{7^{1}}{1 2^{1}} \times \frac{2 4^{2}}{7^{1}} \\ \Rightarrow x = 2 \\ CHECK: Substituting x =2 in the given equation, we get: \\ LHS= \frac{2}{3} x \\ = \frac{2}{3} \times 2 \\ = \frac{4}{3} \\ RHS= \frac{3}{8} x + \frac{7}{12} \\ = \frac{3}{8} \times 2 + \frac{7}{12} \\ = \frac{6}{8} + \frac{7}{12} \\ = \frac{6 \times 3 + 7 \times 2}{24} \\ = \frac{18 + 14}{24} \\ = \frac{3 2^{4}}{2 4^{3}} \\ = \frac{4}{3} \\ \begin{array}{l} ∴ LHS=RHS \\ Hence, x =2 is a solution of the given equation . \end{array} \end{array}$

Page No 111:

Question 16:

$\frac{3 x - 1}{5} - \frac{x}{7} = 3$

Answer:

$\begin{array}{l} We have: \\ \frac{3 x - 1}{5} - \frac{x}{7} = 3 \\ \Rightarrow \frac{7 (3 x - 1) - 5 \times x}{35} = 3 \\ \Rightarrow (\frac{21 x - 7 - 5 x}{35}) = 3 \\ \Rightarrow (\frac{16 x - 7}{35}) = 3 \\ \Rightarrow 16 x - 7 = 3 \times 35 (T ransposing 35 to RHS) \\ \Rightarrow 16 x - 7 = 105 \\ \Rightarrow 16 x = 105 + 7 \\ \Rightarrow 16 x = 112 \\ \Rightarrow x = \frac{11 2^{7}}{1 6^{1}} \\ \Rightarrow x = 7 \\ CHECK: Substituting x =7 in the given equation, we get: \\ LHS= \frac{3 x - 1}{5} - \frac{x}{7} \\ = \frac{7 (3 x - 1) - 5 \times x}{35} \\ = (\frac{21 x - 7 - 5 x}{35}) \\ = (\frac{16 x - 7}{35}) \\ = (\frac{16 \times 7 - 7}{35}) \\ = \frac{112 - 7}{35} \\ = \frac{10 5^{3}}{3 5^{1}} \\ =3 \\ RHS=3 \\ ∴ LHS=RHS \\ Hence, x =3 is a solution of the given equation . \end{array}$

Page No 111:

Question 17:

$2 x - 3 = \frac{3}{10} (5 x - 12)$

Answer:

$\begin{array}{l} We have: \\ 2 x - 3 = \frac{3}{10} (5 x - 12) \\ \Rightarrow 10 (2 x - 3) = 3 (5 x - 12) \\ \Rightarrow 20 x - 30 = 15 x - 36 \\ \Rightarrow 20 x - 15 x = - 36 + 30 (T ransposing 15 x to LHS and - 30 to RHS) \\ \Rightarrow 5 x = - 6 \\ \Rightarrow x = \frac{- 6}{5} \\ CHECK: Substituting x = \frac{- 6}{5} in the given equation, we get: \\ LHS= 2 x - 3 \\ =2 \times (\frac{- 6}{5}) - 3 \\ = \frac{- 12}{5} - 3 \\ = \frac{- 12 - (3 \times 5)}{5} \\ = \frac{- 12 - 15}{5} \\ = \frac{- 27}{5} \\ RHS= \frac{3}{10} (5 x - 12) \\ = \frac{3}{10} (5^{1} \times \frac{- 6}{5^{1}} - 12) \\ = \frac{3}{10} \times (- 18) \\ = \frac{3}{1 0^{5}} \times - 1 8^{9} \\ = \frac{- 27}{5} \\ ∴ LHS=RHS \\ Hence, x = \frac{- 6}{5} is a solution of the given equation . \end{array}$

Page No 111:

Question 18:

$\frac{y - 1}{3} - \frac{y - 2}{4} = 1$

Answer:

$\begin{array}{l} We have: \\ \frac{y - 1}{3} - \frac{y - 2}{4} = 1 \\ \Rightarrow \frac{4 (y - 1) - 3 (y - 2)}{12} = 1 \\ \Rightarrow (\frac{4 y - 4 - 3 y + 6}{12}) = 1 \\ \Rightarrow (\frac{y + 2}{12}) = 1 \\ \Rightarrow y + 2 = 1 \times 12 \\ \Rightarrow y = 12 - 2 \\ \Rightarrow y = 10 \\ CHECK: Substituting y=10 in the given equation, we get: \\ LHS= \frac{y - 1}{3} - \frac{y - 2}{4} \\ = \frac{4 (y - 1) - 3 (y - 2)}{12} \\ = (\frac{y + 2}{12}) \\ = (\frac{10 + 2}{12}) \\ = \frac{1 2^{1}}{1 2^{1}} \\ =1 \\ RHS=1 \\ ∴ LHS=RHS \\ Hence, y=10 is a solution of the given equation . \end{array}$

Page No 112:

Question 19:

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

Answer:

$\begin{array}{l} We have: \\ \frac{x - 2}{4} + \frac{1}{3} = x - \frac{2 x - 1}{3} \\ \Rightarrow \frac{x - 2}{4} + \frac{2 x - 1}{3} - x = - \frac{1}{3} (T ransposing - \frac{2 x - 1}{3} to LHS and \frac{1}{3} to RHS) \\ \Rightarrow (\frac{3 (x - 2) + 4 (2 x - 1) - 12 x}{12}) = - \frac{1}{3} \\ \Rightarrow (\frac{3 x - 6 + 8 x - 4 - 12 x}{12}) = - \frac{1}{3} \\ \Rightarrow 11 x - 12 x - 10 = - \frac{1}{3^{1}} \times 1 2^{4} \\ \Rightarrow - x = - 4 + 10 \\ \Rightarrow - x = 6 \\ \Rightarrow x = - 6 \\ CHECK: Substituting x = - 6 in the given equation, we get: \\ LHS= \frac{x - 2}{4} + \frac{1}{3} \\ = \frac{- 6 - 2}{4} + \frac{1}{3} \\ = - 2 + \frac{1}{3} \\ = \frac{- 5}{3} \\ RHS= x - \frac{2 x - 1}{3} \\ = - 6 - \frac{2 \times (- 6) - 1}{3} \\ = - 6 - \frac{(- 13)}{3} \\ = - 6 + \frac{13}{3} \\ = \frac{- 5}{3} \\ ∴ LHS=RHS \\ Hence, y=10 is a solution of the given equation . \end{array}$

Page No 112:

Question 20:

$\frac{2 x - 1}{3} - \frac{6 x - 2}{5} = \frac{1}{3}$

Answer:

$\begin{array}{l} We have: \\ \frac{2 x - 1}{3} - \frac{6 x - 2}{5} = \frac{1}{3} \\ \Rightarrow \frac{5 (2 x - 1) - 3 (6 x - 2)}{15} = \frac{1}{3} \\ \Rightarrow \frac{10 x - 5 - 18 x + 6}{15} = \frac{1}{3} \\ \Rightarrow \frac{- 8 x + 1}{15} = \frac{1}{3} \\ \Rightarrow - 8 x + 1 = \frac{1}{3} \times 15 \\ \Rightarrow - 8 x = 5 - 1 \\ \Rightarrow - x = \frac{4}{8} \\ \Rightarrow x = - \frac{2}{4} = \frac{- 1}{2} \\ CHECK: Substituting x = - \frac{1}{2} in the given equation, we get: \\ LHS= \frac{2 x - 1}{3} - \frac{6 x - 2}{5} \\ = \frac{- 8 x + 1}{15} \\ = \frac{- 8 \times (- \frac{1}{2}) + 1}{15} \\ = \frac{5}{15} \\ = \frac{1}{3} \\ RHS= \frac{1}{3} \\ \begin{array}{l} ∴ LHS=RHS \\ Hence, y= - \frac{1}{2} is a solution of the given equation . \end{array} \end{array}$

Page No 112:

Question 21:

$\frac{y + 7}{3} = 1 + \frac{3 y - 2}{5}$

Answer:

$\begin{array}{l} We have: \\ \frac{y + 7}{3} = 1 + \frac{3 y - 2}{5} \\ \Rightarrow \frac{y + 7}{3} = \frac{5 \times 1 + 3 y - 2}{5} \\ \Rightarrow 5 (y + 7) = 3 (3 + 3 y) \\ \Rightarrow 5 y + 35 = 9 + 9 y \\ \Rightarrow 9 y - 5 y = 35 - 9 \\ \Rightarrow 4 y = 26 \\ \Rightarrow y = \frac{13}{2} \\ CHECK: Substituting x = \frac{13}{2} in the given equation, we get: \\ LHS= \frac{y + 7}{3} \\ = \frac{\frac{13}{2} + 7}{3} \\ = \frac{1 \times 13 + 2 \times 7}{2} \\ = \frac{13 + 14}{6} \\ = \frac{27}{6} \\ = \frac{9}{2} \\ RHS=1+ \frac{3 \times \frac{13}{2} - 2}{5} \\ = 1 + \frac{\frac{39 - 2 \times 2}{2}}{5} \\ = 1 + \frac{35}{10} \\ = \frac{45}{10} \\ = \frac{9}{2} \\ ∴ LHS=RHS \\ Hence, y= \frac{13}{2} is a solution of the given equation . \end{array}$

Page No 112:

Question 22:

$\frac{2}{7} (x - 9) + \frac{x}{3} = 3$

Answer:

$\begin{array}{l} We have: \\ \Rightarrow \frac{2}{7} (x - 9) + \frac{x}{3} = 3 \\ \Rightarrow \frac{2 \times 3 (x - 9) + 7 x}{21} = 3 \\ \Rightarrow 6 (x - 9) + 7 x = 3 \times 21 \\ \Rightarrow 6 x - 54 + 7 x = 63 \\ \Rightarrow 13 x = 63 + 54 \\ \Rightarrow 13 x = 117 \\ \Rightarrow x = 9 \\ CHECK: Substituting x =9 in the given equation we get . \\ LHS= \frac{2}{7} (x - 9) + \frac{x}{3} \\ = \frac{2}{7} (9 - 9) + \frac{x}{3} \\ =0+ \frac{9}{3} \\ = \frac{9}{3} \\ =3 \\ RHS=3 \\ ∴ LHS = RHS \\ Hence, x =9 is a solution of the given equation . \end{array}$

Page No 112:

Question 23:

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

Answer:

$\begin{array}{l} We have: \\ \Rightarrow \frac{2 x - 3}{5} + \frac{x + 3}{4} = \frac{4 x + 1}{7} \\ \Rightarrow \frac{4 (2 x - 3) + 5 (x + 3)}{20} = \frac{4 x + 1}{7} \\ \Rightarrow \frac{8 x - 12 + 5 x + 15}{20} = \frac{4 x + 1}{7} \\ \Rightarrow \frac{13 x + 3}{20} = \frac{4 x + 1}{7} \\ \Rightarrow 7 (13 x + 3) = 20 (4 x + 1) \\ \Rightarrow 91 x + 21 = 80 x + 20 \\ \Rightarrow 91 x - 80 x = 20 - 21 \\ \Rightarrow 11 x = - 1 \\ \Rightarrow x = \frac{- 1}{11} \\ CHECK: Substituting x = - \frac{1}{11} in the given equation, we get: \\ LHS: \\ LHS= \frac{2 x - 3}{5} + \frac{x + 3}{4} \\ = \frac{2 \times \frac{1}{11} - 3}{5} + \frac{- \frac{1}{11} + 3}{4} \\ = \frac{- 2 - 33}{55} + \frac{33 - 1}{44} \\ = - \frac{35}{55} + \frac{32}{44} \\ = \frac{- 140 + 160}{220} \\ = \frac{20}{220} = \frac{1}{11} \\ RHS= \frac{4 x + 1}{7} \\ = \frac{4 \times (- \frac{1}{11}) + 1}{7} \\ = \frac{- 4 + 11}{7 \times 11} \\ = \frac{7}{77} \\ = \frac{1}{11} \\ ∴ LHS = RHS \\ Hence, x = \frac{- 1}{11} is a solution of the given equation . \end{array}$

Page No 112:

Question 24:

$\frac{3}{4} (7 x - 1) - (2 x - \frac{1 - x}{2}) = x + \frac{3}{2}$

Answer:

$\begin{array}{l} We have: \\ \frac{3}{4} (7 x - 1) - (2 x - \frac{1 - x}{2}) = x + \frac{3}{2} \\ \Rightarrow \frac{3}{4} (7 x - 1) - 2 x + \frac{1 - x}{2} - x = \frac{3}{2} \\ \Rightarrow \frac{3 \times 7}{4} x - \frac{3}{4} - 2 x + \frac{1}{2} - \frac{x}{2} - x = \frac{3}{2} \\ \Rightarrow \frac{21}{4} x - 2 x - \frac{x}{2} - x = \frac{3}{2} + \frac{3}{4} - \frac{1}{2} (By transposition) \\ \Rightarrow \frac{21 x - 8 x - 2 \times x - 4 x}{4} = 1 + \frac{3}{4} \\ \Rightarrow \frac{21 x - 14 x}{4} = \frac{7}{4} \\ \Rightarrow \frac{7 x}{4} = \frac{7}{4} \\ \Rightarrow x = 1 \\ CHECK: Substituting x =1 in the given equation, we get: \\ LHS= \frac{3}{4} (7 x - 1) - (2 x - \frac{1 - x}{2}) \\ = \frac{3}{4} (7 \times 1 - 1) - (2 \times 1 - \frac{1 - 1}{2}) \\ = \frac{3}{4} \times 6 - 2 \\ = \frac{9}{2} - 2 \\ = \frac{9 - 4}{2} \\ = \frac{5}{2} \\ RHS= x + \frac{3}{2} \\ = 1 + \frac{3}{2} \\ = \frac{2 + 3}{2} \\ = \frac{5}{2} \\ \begin{array}{l} ∴ LHS = RHS \\ Hence, x =1 is a solution of the given equation . \end{array} \end{array}$

Page No 112:

Question 25:

$\frac{x + 2}{6} - (\frac{11 - x}{3} - \frac{1}{4}) = \frac{3 x - 4}{12}$

Answer:

$\begin{array}{l} We have : \\ \frac{x + 2}{6} - (\frac{11 - x}{3} - \frac{1}{4}) = \frac{3 x - 4}{12} \\ \Rightarrow \frac{x + 2}{6} - (\frac{11 - x}{3}) + \frac{1}{4} = \frac{3 x - 4}{12} \\ \Rightarrow \frac{x + 2}{6} - (\frac{11 - x}{3}) - \frac{3 x - 4}{12} = - \frac{1}{4} (By transposition) \\ \Rightarrow \frac{2 (x + 2) - 4 (11 - x) - 1 (3 x - 4)}{12} = - \frac{1}{4} \\ \Rightarrow \frac{2 x + 4 - 44 + 4 x - 3 x + 4}{12} = - \frac{1}{4} \\ \Rightarrow 3 x - 36 = - \frac{1}{4} \times 12 \\ \Rightarrow 3 x = - 3 + 36 \\ \Rightarrow x = \frac{33}{3} \\ \Rightarrow x = 11 \\ CHECK: Substituting x = 11 in the given equation, we get: \\ LHS= \frac{x + 2}{6} - (\frac{11 - x}{3} - \frac{1}{4}) \\ = \frac{11 + 2}{6} - (\frac{11 - 11}{3} - \frac{1}{4}) \\ = \frac{13}{6} - (- \frac{1}{4}) \\ = \frac{13}{6} + \frac{1}{4} \\ = \frac{13 \times 2 + 3}{12} \\ = \frac{29}{12} \\ RHS= \frac{3 x - 4}{12} \\ = \frac{3 \times 11 - 4}{12} \\ = \frac{33 - 4}{12} \\ = \frac{29}{2} \\ ∴ LHS=RHS \\ Hence, x = 11 is a solution of the given equation . \\ Verified. \end{array}$

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Question 26:

$\frac{9 x + 7}{2} - (x - \frac{x - 2}{7}) = 36$

Answer:

$\begin{array}{l} We have: \\ \frac{9 x + 7}{2} - (x - \frac{x - 2}{7}) = 36 \\ \Rightarrow \frac{9 x + 7}{2} - x + \frac{x - 2}{7} = 36 \\ \Rightarrow \frac{7 (9 x + 7) - 14 \times x + 2 \times (x - 2)}{14} = 36 \\ \Rightarrow \frac{63 x + 49 - 14 x + 2 x - 4}{14} = 36 \\ \Rightarrow 51 x + 45 = 36 \times 14 \\ \Rightarrow 51 x = 504 - 45 \\ \Rightarrow x = \frac{459}{51} \\ \Rightarrow x = 9 \\ \Rightarrow x = 9 \\ CHECK: Substituting x = 9 in the given equation, we get: \\ LHS= \frac{9 x + 7}{2} - (x - \frac{x - 2}{7}) \\ = \frac{9 \times 9 + 7}{2} - (9 - \frac{9 - 2}{7}) \\ = \frac{88}{2} - 9 + \frac{7}{7} \\ =44 - 9 + 1 \\ =36 \\ RHS =36 \\ ∴ LHS=RHS \\ Hence, x = 11 is a solution of the given equation . \\ Verified. \end{array}$

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Question 27:

$0.5 x + \frac{x}{3} = 0.25 x + 7$

Answer:

$\begin{array}{l} We have: \\ 0.5 x + \frac{x}{3} = 0.25 x + 7 \\ \Rightarrow \frac{1}{2} x + \frac{x}{3} = \frac{x}{4} + 7 \\ \Rightarrow \frac{x}{2} + \frac{x}{3} - \frac{x}{4} = 7 \\ \Rightarrow \frac{6 x + 4 x - 3 x}{12} = 7 \\ \Rightarrow \frac{7 x}{12} = 7 \\ \Rightarrow x = 12 \\ CHECK: Substituting x = 9 in the given equation, we get: \\ LHS= 0.5 x + \frac{x}{3} \\ = 0.5 \times 12 + \frac{12}{3} \\ = \frac{1}{2} \times 12 + 4 \\ =6+4 \\ =10 \\ RHS= 0.25 x + 7 \\ = 0.25 \times 12 + 7 \\ = 3 + 7 \\ = 10 \\ ∴ LHS=RHS \\ Hence, x = 12 is a solution of the given equation . \\ Verified. \end{array}$

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Question 28:

0.18(5x − 4) = 0.5x + 0.8

Answer:

$\begin{array}{l} We have: \\ 0.18 (5 x - 4) = 0.5 x + 0.8 \\ \Rightarrow 100 \times 0.18 (5 x - 4) = 100 (0.5 x + 0.8) (M ultipling both sides by 100) \\ \Rightarrow 18 (5 x - 4) = 100 \times 0.5 x + 100 \times 0.8 \\ \Rightarrow 90 x - 72 = 50 x + 80 \\ \Rightarrow 90 x - 50 x = 80 + 72 \\ \Rightarrow 40 x = 152 \\ \Rightarrow x = \frac{152}{40} \\ \Rightarrow x= \frac{19}{5} =3.8 \\ CHECK: Substituting x = 3.8 in the given equation, we get: \\ LHS= 0.18 (5 x - 4) \\ = 0.18 (5 \times 3.8 - 4) \\ =0 .18 \times 15 \\ =2 .7 \\ RHS= 0.5 x + 0.8 \\ = 0.5 \times 3.8 + 0.8 \\ = 1.9 + 0.8 \\ = 2.7 \\ ∴ LHS=RHS \\ Hence, x = 3.8 is a solution of the given equation . \\ Verified. \end{array}$

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Question 29:

2.4(3 − x) − 0.6(2x − 3) = 0

Answer:

$\begin{array}{l} We have: \\ \Rightarrow 2.4 (3 - x) - 0.6 (2 x - 3) = 0 \\ \Rightarrow 10 \times 2.4 (3 - x) - 10 \times 0.6 (2 x - 3) =0 (M ultiplying both sides by 10 to remove decimals) \\ \Rightarrow 24 (3 - x) - 6 (2 x - 3) = 0 \\ \Rightarrow 6 [4 (3 - x) - (2 x - 3)] = 0 \\ \Rightarrow 4 (3 - x) - (2 x - 3) = 0 \\ \Rightarrow 12 - 4 x - 2 x + 3 = 0 \\ \Rightarrow 15 - 6 x = 0 \\ \Rightarrow - 6x= - 15 \\ \Rightarrow x = \frac{15}{6} \\ \Rightarrow x= \frac{5}{2} =2.5 \\ CHECK: Substituting x = 2.5 in the given equation, we get: \\ LHS= 2.4 (3 - x) - 0.6 (2 x - 3) \\ = 2.4 (3 - 2.5) - 0.6 (2 \times 2.5 - 3) \\ =2.4 \times 0.5 - 0.6 \times 2 \\ =1 .2-1 .2 \\ = 0 \\ RHS= 0 \\ ∴ LHS = RHS \\ Hence, x = \frac{19}{5} is a solution of the given equation . \\ Verified. \end{array}$

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Question 30:

0.5x −(0.8 − 0.2x) = 0.2 − 0.3x

Answer:

$\begin{array}{l} We have: \\ 0.5 x - (0.8 - 0.2 x) = 0.2 - 0.3 x \\ \Rightarrow 0.5 x + 0.3 x - 0.8 + 0.2 x =0 .2 (By transposition) \\ \Rightarrow (0.5 + 0.3 + 0.2) x = 0.2 + 0.8 \\ \Rightarrow 1 x = 1 \\ \Rightarrow x = 1 \\ CHECK: Substituting x = 1 in the given equation, we get: \\ LHS= 0.5 x - (0.8 - 0.2 x) \\ = 0.5 \times 1 - (0.8 - 0.2 \times 1) \\ =0 .5 - 0.8 + 0.2 \\ = - 0 .1 \\ RHS= 0.2 - 0.3 x \\ = 0.2 - 0.3 \times 1 \\ = - 0.1 \\ ∴ LHS=RHS \\ Hence, x = 1 is a solution of the given equation . \\ Verified. \end{array}$

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Question 31:

$\frac{x + 2}{x - 2} = \frac{7}{3}$

Answer:

$\begin{array}{l} We have: \\ \frac{x + 2}{x - 2} = \frac{7}{3} \\ \Rightarrow (x + 2) \times 3=7 \times (x - 2) (C ross multiplication) \\ \Rightarrow 3 x + 6 = 7 x - 14 \\ \Rightarrow 4 x = 20 \\ \Rightarrow x = \frac{20}{4} \\ \Rightarrow x = 5 \\ CHECK: Substituting x = 5 in the given equation, we get . \\ LHS= \frac{x + 2}{x - 2} \\ = \frac{5 + 2}{5 - 2} \\ = \frac{7}{3} \\ RHS= \frac{7}{3} \\ ∴ LHS=RHS \\ Hence, x = 5 is a solution of the given equation . \\ Verified. \end{array}$

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Question 32:

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

Answer:

$\begin{array}{l} We have: \\ \frac{2 x + 5}{3 x + 4} = 3 \\ \Rightarrow \frac{2 x + 5}{3 x + 4} = \frac{3}{1} \\ \Rightarrow 1 \times (2 x + 5) = 3 \times (3 x + 4) \\ \Rightarrow 2 x + 5 = 9 x + 12 \\ \Rightarrow 7 x = - 7 \\ \Rightarrow x = - 1 \\ CHECK: Substituting x = - 1 in the given equation, we get: \\ LHS : \frac{2 x + 5}{3 x + 4} \\ = \frac{2 \times (- 1) + 5}{3 \times (- 1) + 4} \\ = \frac{- 2 + 5}{- 3 + 4} \\ = \frac{3}{1} \\ RHS =3 \\ ∴ LHS = RHS \\ Hence, x = 5 is a solution of the given equation . \\ Verified. \end{array}$

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Question 1:

Twice a number when decreased by 7 gives 45. Find the number.

Answer:

$\begin{array}{l} Let the number be x . \\ Then, we have : \\ \Rightarrow 2 x - 7 = 45 \\ \Rightarrow 2 x = 45 + 7 \\ \Rightarrow x = \frac{45 + 7}{2} \\ \Rightarrow x = \frac{5 2^{26}}{2_{1}} \\ \Rightarrow x = 26 \\ ∴ The required number is 26 . \end{array}$

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Question 2:

Thrice a number when increased by 5 gives 44. Find the number.

Answer:

$\begin{array}{l} Let the number be x . \\ Then, we have: \\ \Rightarrow 3 x + 5 = 44 \\ \Rightarrow 3 x = 44 - 5 \\ \Rightarrow x = \frac{44 - 5}{3} \\ \Rightarrow x = \frac{3 9^{13}}{3_{1}} \\ \Rightarrow x = 13 \\ ∴ The required number is 13 \end{array}$

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Question 3:

Four added to twice a number yields $\frac{26}{5}$ . Find the fractions.

Answer:

$\begin{array}{l} Let the number be x . \\ Then, we have: \\ \Rightarrow 2 x + 4 = \frac{26}{5} \\ \Rightarrow 2 x = \frac{26}{5} - 4 \\ \Rightarrow 2 x = \frac{26 - 20}{5} \\ \Rightarrow x = \frac{6^{3}}{1 0_{5}} \\ \Rightarrow x = \frac{3}{5} \\ ∴ The required fraction is \frac{3}{5} . \end{array}$

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Question 4:

A number when added to its half gives 72. Find the number.

Answer:

$\begin{array}{l} Let the required number be x . \\ Then, we have: \\ \Rightarrow x + \frac{x}{2} = 72 \\ \Rightarrow \frac{2 x + x}{2} = 72 \\ \Rightarrow \frac{3 x}{2} = 72 \\ \Rightarrow 3 x = 72 \times 2 \\ \Rightarrow x = \frac{7 2^{24} \times 2}{\begin{array}{l} 3_{1} \\ = 48 \end{array}} \\ ∴ The required number is 48. \end{array}$

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Question 5:

A number added to its two-thirds is equal to 55. Find the number.

Answer:

$\begin{array}{l} Let the required number be x. \\ Then, we have: \\ \Rightarrow x + \frac{2 x}{3} = 55 \\ \Rightarrow \frac{3 x + 2 x}{3} = 55 \\ \Rightarrow 5 x = 55 \times 3 \\ \Rightarrow x = \frac{5 5^{11} \times 3}{\begin{array}{l} 5_{1} \\ = 33 \end{array}} \\ ∴ The required number is 33. \end{array}$

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Question 6:

A number when multiplied by 4, exceeds itself by 45. Find the number.

Answer:

$\begin{array}{l} Let the required number be x . \\ Then, we have: \\ \Rightarrow 4 x - x = 45 \\ \Rightarrow 3 x = \frac{45}{3} \\ \Rightarrow x = 15 \\ ∴ The required number is 15 . \end{array}$

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Question 7:

A number is as much greater than 21 as it is less than 71. Find the number.

Answer:

$\begin{array}{l} Let the number be x . \\ Then, we have: \\ (x - 21) = (71 - x) \\ \Rightarrow x + x = 71 + 21 \\ \Rightarrow 2 x = 92 \\ \Rightarrow x = \frac{9 2^{46}}{2_{1}} \\ \Rightarrow x = 46 \\ ∴ The required number is 46. \end{array}$

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Question 8:

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

Answer:

$\begin{array}{l} Let the original number be x . \\ Then, we have: \\ \Rightarrow \frac{2}{3} x = x - 20 \\ \Rightarrow \frac{2 x}{3} - x = - 20 \\ \Rightarrow \frac{2 x - 3 x}{3} = - 20 \\ \Rightarrow - x = - 20 \times 3 \\ \Rightarrow x = 60 \\ ∴ The original number is 6 0 . \end{array}$

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Question 9:

A number is $\frac{2}{5}$ times another number. If their sum is 70, find the numbers.

Answer:

$\begin{array}{l} Let the number be x . \\ Then, the other number will be \frac{2 x}{5} . \\ \begin{array}{l} Now, we have: \\ \Rightarrow x + \frac{2 x}{5} = 70 \\ \Rightarrow \frac{5 x + 2 x}{5} = 70 \\ \Rightarrow \frac{7 x}{5} = 70 \\ \Rightarrow x = \frac{7 0^{10} \times 5}{7_{1}} \\ ∴ Other number = 50 \times \frac{2}{5} = 20 \\ Hence, the numbers are 50 and 20 . \end{array} \end{array}$

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Question 10:

Two-thirds of a number is greater than one-third of the number by 3. Find the number.

Answer:

$\begin{array}{l} Let the number be x . \\ Then, we have: \\ \frac{2}{3} x = \frac{1}{3} x + 3 \\ \Rightarrow \frac{1}{3} x = \frac{2 x}{3} - 3 \\ \Rightarrow \frac{x}{3} - \frac{2 x}{3} = - 3 \\ \Rightarrow \frac{x - 2 x}{3} = - 3 \\ \Rightarrow x - 2 x = 3 \times (- 3) \\ \Rightarrow - x = - 9 \\ ∴ The required number is 9. \end{array}$

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Question 11:

The fifth part of a number when increased by 5 equals its fourth part decreased by 5. Find the number.

Answer:

$\begin{array}{l} Let the number be x . \\ Then, we have: \\ \Rightarrow \frac{x}{5} + 5 = \frac{x}{4} - 5 \\ \Rightarrow \frac{x}{5} - \frac{x}{4} = - 5 - 5 \\ \Rightarrow \frac{- x}{20} = - 10 \\ \Rightarrow x = 200 \\ ∴ The required number is 200. \end{array}$

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Question 12:

Find two consecutive natural numbers whose sum is 63.

Answer:

$\begin{array}{l} Let the two consecutive natural number be x and (x + 1) . \\ Then, we have: \\ x + (x + 1) = 63 \\ \Rightarrow x + x + 1 = 63 \\ \Rightarrow 2 x = 63 - 1 \\ \Rightarrow x = \frac{6 2^{31}}{2_{1}} \\ \Rightarrow x = 31 \\ ∴ The required numbers are 31 and 32 (i.e., 31+1). \end{array}$

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Question 13:

Find two consecutive positive odd integers whose sum is 76.

Answer:

$\begin{array}{l} Let the two consecutive odd integers whose sum is 76 be x and (x + 2) . \\ Then, x + x + 2 = 76 \\ \Rightarrow 2x + 2 = 76 \\ \Rightarrow 2x = 76 - 2 \\ \Rightarrow x = 74 \div 2 \\ \Rightarrow x = 37 \\ ∴ The required integers are 37 and 39 (i . e ., 37 + 2) . \end{array}$

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Question 14:

Find two consecutive positive even integers whose sum is 90.

Answer:

$\begin{array}{l} Let the three consecutive positive even integers be x, (x + 2) and (x + 4) . \\ Let x be the even number. \\ Then, x + x + 2 + x + 4 = 90 \\ \Rightarrow 3 x = 90 - 6 \\ \Rightarrow 3 x = 84 \\ \Rightarrow x = \frac{84}{3} = 28 \\ ∴ The r equired numbers are 28, 30 and 32. \end{array}$

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Question 15:

Divide 184 into two parts such that one-third of one part may exceed one-seventh of the other part by 8.

Answer:

$\begin{array}{l} Let the two parts be x and (184 - x) . \\ Then, we have: \\ \frac{1}{3} x = \frac{1}{7} (184 - x)+8 \\ \begin{array}{l} ⇒ \frac{1}{3} x - \frac{1}{7} (184 - x) = 8 \\ ⇒ \frac{1}{3} x - \frac{184}{7} + \frac{x}{7} = 8 \\ ⇒ \frac{1}{3} x + \frac{1}{7} x = \frac{184}{7} + 8 \\ ⇒ \frac{7 x + 3 x}{21} = 8 + \frac{184}{7} \\ ⇒ \frac{10 x}{21} = \frac{56 + 184}{7} \\ ⇒ \frac{10 x}{21} = \frac{240}{7} \\ ⇒ x = \frac{240 \times 21}{7 \times 10} \\ = 72 \\ Now, ot her part =184 - 72 = 112 \end{array} \end{array} ∴ The two parts are 72 and 112 .$

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Question 16:

A sum of 500 is in the form of denominations of 5 and 10. If the total number of notes is 90, find the number of notes of each type.

Answer:

$\begin{array}{l} Let the number of five rupee notes be x . \\ Then, the number of ten rupee notes will be (90 - x) . \\ According to the question, we have : \\ \begin{array}{l} 5 x + 10 (90 - x) = 500 \\ ⇒5 x + 900 - 10 x = 500 \\ \Rightarrow - 5 x = - 400 \\ \Rightarrow x = 80 \\ Number of ten rupee notes = 90 - 80 = 10 \\ ∴ There are 80 five rupee notes and 10 ten rupee notes . \end{array} \end{array}$

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Question 17:

Sumitra has 34 in 50-paise and 25-paise coins. If the number of 25-paise coins is twice the number of 50-paise coins, how many coins of each kind does she have?

Answer:

$\begin{array}{l} Let the numbers of 5 0 paise coins and 25 paise coins be x and 2 x, respectively. \\ Then, we have : \\ 50 x + 25 \times 2 x = 3400 \\ \Rightarrow 50 x + 50 x = 3400 \\ \Rightarrow 100 x = 3400 \\ \Rightarrow x = 34 \\ ∴ Number of 50 paise coins = 34 \\ and number of 25 paise coins = 68 \end{array}$

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Question 18:

Raju is 19 years younger than his cousin. After 5 years, their ages will be in the ratio 2 : 3. Find their present ages.

Answer:

$\begin{array}{l} Let the present ages of Raju and his cousin be (x -19) yrs and x yrs . \\ According to the question, we have : \\ \begin{array}{l} \frac{(x - 19) + 5}{x + 5} = \frac{2}{3} \\ ⇒3 (x - 14) = 2 x + 10 \\ \Rightarrow 3 x - 42 = 2 x + 10 \\ \Rightarrow x = 52 \\ ∴ Age of Raju' s c ousin = 52 yrs \\ and age of Raju = 52 - 19 = 33 yrs \end{array} \end{array}$

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Question 19:

A father is 30 years older than his son. In 12 years, the man will be three times as old as his son. Find their present ages.

Answer:

$\begin{array}{l} Let the age of the son and the father be x yrs and (x + 30) yrs, respectively . \\ According to the question, we have : \\ \begin{array}{l} 3 \times (x + 12) = x + 30 + 12 \\ ⇒3 x + 36 = x + 42 \\ \Rightarrow 3 x - x = 42 - 36 \\ \Rightarrow 2 x = 6 \\ \Rightarrow x = 3 \\ ∴ Son' s a ge = 3 yrs \\ Father's age = (x + 30) yrs = (3 + 30) yrs = 33 yrs \end{array} \end{array}$

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Question 20:

The ages of Sonal and Manoj are in the ratio 7 : 5. Ten years hence, the ratio of their ages will be 9 : 7. Find their present ages.

Answer:

$\begin{array}{l} Given ratio of Sonal's and Manoj's ages = 7 : 5 \\ Let the ages of Sonal and Manoj be 7 x yrs and 5 x yrs . \\ According to the question, we have : \\ \begin{array}{l} \frac{7 x + 10}{5 x + 10} = \frac{9}{7} \\ \Rightarrow 7 (7 x + 10) = 9 (5 x + 10) \\ \Rightarrow 49 x + 70 = 45 x + 90 \\ \Rightarrow 49 x - 45 x = 90 - 70 \\ \Rightarrow 4 x = 20 \\ \Rightarrow x = 5 \\ ∴ Sonal's present age is 7 \times 5=35 yrs \end{array} \\ Manoj's present age is 5 \times 5=25 yrs \end{array}$

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Question 21:

Five years ago a man was seven times as old as his son. Five years hence, the father will be three times as old as his son. Find their present ages.

Answer:

$\begin{array}{l} Let x yrs be the present age of son. \\ Then, the age of the son 5 years ago would be (x - 5) y r s \end{array} Then, Age of father = 7 (x - 5) yrs After 5 yrs, the age of the son will be (x + 5) y r s Then, Age of father = 3 (x + 5) yrs \begin{array}{l} Now, we have 3 (x + 5) = 7 (x - 5) + 10 \\ ⇒ 3 x + 15 = 7 x - 35 + 10 \\ \Rightarrow 4 x = 40 \\ \Rightarrow x = 10 \\ ∴ P resent age of the father is = 3(x +5)-5 \end{array} = 3 (10 + 5) - 5 = 40 yrs$

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Question 22:

After 12 years Manoj will be 3 times as old as he was 4 years ago. Find his present age.

Answer:

$\begin{array}{l} Let x be the present age of Manoj. \end{array} According to the question, we have : \begin{array}{l} \Rightarrow x + 12 = 3 (x - 4) \\ \Rightarrow x + 12 = 3 x - 12 \\ \Rightarrow 2 x = 24 \\ \Rightarrow x = 12 \\ ∴ Manoj's present age is 12 years . \end{array}$

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Question 23:

In an examination, a student requires 40% of the total marks to pass. If Rupa gets 185 marks and fails by 15 marks, find the total marks.

Answer:

$\begin{array}{l} Let x be the total marks. \\ According to the question, we have: \\ 40 % of x = 185 + 15 \\ \Rightarrow \frac{40 x}{100} = 200 \\ \Rightarrow 40 x = 200 \times 100 \\ \Rightarrow 40 x = 20000 \\ \Rightarrow x = 500 \\ ∴ Total marks = 500 \end{array}$

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Question 24:

A number consists of two digits whose sum is 8. If 18 is added to the number its digits are reversed. Find the number.

Answer:

$\begin{array}{l} Let x be the digit in the units place. \\ Sum of the units and tens digits = 8 \\ Then, tens digit = (8 - x) \\ ∴ T he number is 10 (8 - x) + x . \\ Now, 10 (8 - x) + x + 18 = 10 x + (8 - x) \\ \Rightarrow 80 - 10 x + x + 18 = 10 x + 8 - x \\ \Rightarrow 98 - 9 x = 9 x + 8 \\ \Rightarrow 18 x = 90 \\ \Rightarrow x = 5 \\ i.e., tens digit= (8 - 5) =3 \\ ∴ Required number=10 (8 - 5) +5=10 \times 3+5=35 \end{array}$

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Question 25:

The total cost of 3 tables and 2 chairs is 1850. If a table costs 75 more than a chair, find the price of each.

Answer:

$\begin{array}{l} Let Rs x be the cost of the chair. \\ Then, the cost of the table is Rs (x + 75) . \\ Now, 3 (x + 75) + 2 x = 1850 \\ \Rightarrow 3 x + 225 + 2 x = 1850 \\ \Rightarrow 5 x = 1625 \\ \Rightarrow x = \frac{1625}{5} = 325 \\ ∴ Cost of the chair = Rs 325; cost of the table = (325+75)=Rs 400 \end{array}$

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Question 26:

A man sold an article for 495 and gained 10% on it. Find the cost price of the article.

Answer:

$\begin{array}{l} Let the cost price of the article be Rs x . \\ According to the question, we have: \\ SP = Rs 495 \end{array} ∴ Gain %= \frac{Gain}{CP} ×100 \Rightarrow 10 = \frac{Gain}{x} \times 100 \Rightarrow Gain = \frac{10 x}{100} = Rs \frac{x}{10} \begin{array}{l} Now, CP + Gain = SP \\ \Rightarrow x + \frac{x}{10} = 495 \\ \Rightarrow \frac{x + 10 x}{10} = 495 \\ \Rightarrow 11 x = 495 \times 10 \\ \Rightarrow x = \frac{495 \times 10}{11} \\ \Rightarrow x = \frac{4950}{11} \\ \Rightarrow x = 450 \\ ∴ CP = Rs 45 0 \end{array}$

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Question 27:

The length of a rectangular field is twice its breadth. If the perimeter of the field is 150 metres, find its length and breadth.

Answer:

$\begin{array}{l} Let the length and breadth of the rectangular field be l m and b m, respectively. \\ According to the question, we have : \\ 2 (l + b) = 150 . . . (i) \\ \Rightarrow l + b = 75 \\ Given that l = 2 b ... (ii) \\ Using (ii) in (i), we have: \\ 2 b + b = 75 \\ \Rightarrow 3 b = 75 \\ \Rightarrow b = 25 \\ ∴ l = 50 m and b = 25 m \end{array}$

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Question 28:

Two equal sides of a triangle are each 5 metres less than twice the third side. If the perimeter of the triangle is 55 metres, find the lengths of its sides.

Answer:

$\begin{array}{l} Let the length of third side be x m . Then, the length of the two equal sides will be (2 x - 5) m . \\ ∴ (2 x - 5) + (2 x - 5) + x = 55 \\ \Rightarrow 2 x - 5 + 2 x - 5 + x = 55 \\ \Rightarrow 5 x - 10 = 55 \\ \Rightarrow 5 x = 65 \\ \Rightarrow x = \frac{65}{5} = 13 \\ ∴ Length of the third side=13 m \\ And length of the other two equal sides= (2 \times 13) - 5 = 21 m \end{array}$

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Question 29:

Two complementary angles differ by 8⁰. Find the angles.

Answer:

$\begin{array}{l} Let the two complementary angles be x ° and (90 - x) ° . \\ According to the question, we have : \\ x - (90 - x) = 8 \\ \Rightarrow x - 90 + x = 8 \\ \Rightarrow 2 x = 98 \\ \Rightarrow x = 49 \\ ∴ The measures of the complementary angles are 49 ° and (90 - 49) ° = 41 ° . \end{array}$

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Question 30:

Two supplementary angles differ by 44⁰. Find the angles.

Answer:

$\begin{array}{l} Let the two supplementary angles be x ° and (180 - x) ° . \\ ∴ x - (180 - x) = 44 \\ ⇒ x - 180 + x = 44^{0} \\ \Rightarrow 2 x = 224 \\ \Rightarrow x = 112 \\ ∴ The measures of the supplementary angles are 112 ° and (180 - 112) °, i . e ., 68 ° . \end{array}$

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Question 31:

In an isosceles triangle the base angles are equal and the vertex angle is twice of each base angle. Find the measures of the angles of the triangle.

Answer:

$\begin{array}{l} Let the base angles of the isosceles triangle be x ° each . \\ Then, the measure the vertex angle will be (2x)°. \\ According to the question, we have : \\ x + x + 2 x = 180 (S um of three sides of a triangle) \\ \Rightarrow 4 x = 180 \\ \Rightarrow x = \frac{180}{4} \\ \Rightarrow x = 45 \\ ∴ Each base angle measures 45 ° and the vertex angle measures (2 \times 45) °, i . e ., 90 ° . \end{array}$

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Question 32:

A man travelled $\frac{3}{5}$ of his journey by rail, $\frac{1}{4}$ by a taxi, $\frac{1}{8}$ by a bus and the remaining 2 km on foot. What is the length of his total journey?

Answer:

$\begin{array}{l} Let the length of the total journey be x km. \\ According to the question, we have: \\ \frac{3}{5} x + \frac{1}{4} x + \frac{1}{8} x + 2 = x \\ \Rightarrow \frac{24 x + 10 x + 5 x + 80}{40} = x \\ \Rightarrow 39 x + 80 = 40 x \\ \Rightarrow x = 80 \\ ∴ The length of his total journey is 80 km. \end{array}$

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Question 33:

A labourer is engaged for 20 days on the condition that he will receive 120 for each day he works and will be fined 10 for each day he is absent. If he receives 1880 in all, for how many days did he remain absent?

Answer:

$\begin{array}{l} Let x be the number of days of his absence. \\ ∴ Number of days of his presence = (20 - x) \\ Now, (20 - x) 120 - 10 x = 1880 \\ \Rightarrow 2400 - 120 x - 10 x = 1880 \\ \Rightarrow 2400 - 1880 = 130 x \\ \Rightarrow 130 x = 520 \\ \Rightarrow x = 4 \\ ∴ Number of days of his absence = 4 \end{array}$

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Question 34:

Hari Babu left one-third of his property to his son, one-fourth to his daughter and the remainder to his wife. If his wife's share is 18000, what was the worth of his total property?

Answer:

$\begin{array}{l} Let the worth of Hari Babu's property be Rs x . \\ According to the question, we have: \\ Son's share = \frac{1}{4} x \\ Daughter's share= \frac{1}{3} x \\ Wife's share = {x - (\frac{1}{4} x + \frac{1}{3} x)} \\ It is given that his wife's share is Rs 18000 . \\ i.e., x - (\frac{1}{4} x + \frac{1}{3} x) = 18000 \\ \Rightarrow x - (\frac{1}{3} x + \frac{1}{4} x) = 18000 \\ \Rightarrow x - \frac{7 x}{12} = 18000 \\ \Rightarrow \frac{5 x}{12} = 18000 \\ \Rightarrow x = \frac{1800 0^{3600} \times 12}{5} \\ \Rightarrow x = 43200 \\ ∴ Hari Babu ’ s total property is worth Rs 432 00 . \end{array}$

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Question 35:

How much pure alcohol must be added to 400 mL of a 15% solution to make its strength 32%.

Answer:

$\begin{array}{l} Let the volume of the pure alcohol be x ml . \\ Initial concentration= 15 % \\ So, initial amount of alcohol in the solution will be = \frac{15}{100} \times 400= 60 ml \\ To make the strength of the solution 32%, we will keep the amount of water constant a nd add x volume of pure alcohol . \\ On adding pure alcohol, the volume of the solution increases to 400 + x . \\ According to the question, we have : \\ \frac{x + 60}{400 + x} = \frac{32}{100} \\ ⇒100 x + 6000 = 12800 + 32 x \\ \Rightarrow 100 x - 32 x = 12800 - 6000 \\ \Rightarrow 68 x = 6800 \\ \Rightarrow x = 100 \\ So, amount of pure alcohol to be added=100 ml \end{array}$

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Question 1:

Mark (âœ“) 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:

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

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Question 2:

Mark (âœ“) 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:

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

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Question 3:

Mark (âœ“) 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:

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

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Question 4:

Mark (âœ“) against the correct answer

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

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

Answer:

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

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Question 5:

Mark (âœ“) 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:

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

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Question 6:

Mark (âœ“) against the correct answer

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

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

Answer:

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

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Question 7:

Mark (âœ“) 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:

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

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Question 8:

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

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

Answer:

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

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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:

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

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Question 10:

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

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

Answer:

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

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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:

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

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Question 12:

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

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

Answer:

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

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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:

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

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Question 14:

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

(a) 60°
(b) 50°
(c) 64°
(d) 54°

Answer:

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

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Question 15:

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

(a) 60°
(b) 80°
(c) 100°
(d) 120°

Answer:

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

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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:

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

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Question 17:

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

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

Answer:

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

Page No 117:

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:

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

Page No 118:

Question 1:

Evaluate x³ + y³ + z³ −3xyz when x = −2, y = −1 and z = 3.

Answer:

$\begin{array}{l} We have: \\ x^{3} + y^{3} + z^{3} - 3 x y z \\ = (- 2)^{3} + (- 1)^{3} + (3)^{3} - 3 \times (- 2) \times (- 1) \times 3 \\ = - 8 - 1 + 27 - 18 \\ = - 27 + 27 \\ = 0 \end{array}$

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Question 2:

Write the coefficient of x in each of the following:

(i) −5xy
(ii) 2xy²z
(iii) $- \frac{3}{2} a b c$

Answer:

$\begin{array}{l} Coefficient of x in the given numbers are \\ (i) - 5y \\ (ii) {2y}^{2} z \\ (iii) - \frac{3}{2} a b \end{array}$

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Question 3:

Subtract x²− 2xy + 5y² − 4 from 4xy − 5x² − y² + 6.

Answer:

$\begin{array}{l} We have: \\ (4 x y - 5 x^{2} - y^{2} + 6) - (x^{2} - 2 x y + 5 y^{2} - 4) \\ = 4 x y - 5 x^{2} - y^{2} + 6 - x^{2} + 2 x y - 5 y^{2} + 4 \\ = - 6 x^{2} - 6 y^{2} + 6 x y + 10 \\ = - 2 (3 x^{2} + 3 y^{2} - 3 x y - 5) \end{array}$

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Question 4:

How much less is x²− 2xy + 3y²than 2x² − 3y² + xy?

Answer:

$\begin{array}{l} We have: \\ (2 x^{2} - 3 y^{2} + x y) - (x^{2} - 2 x y + 3 y^{2}) \end{array} = 2 x^{2} - 3 y^{2} + x y - x^{2} + 2 x y - 3 y^{2}) \begin{array}{l} = 2 x^{2} - x^{2} - 3 y^{2} - 3 y^{2} + x y + 2 x y \\ = x^{2} - 6 y^{2} + 3 x y \end{array} ∴ x^{2} - 2 x y + 3 y^{2} is less than 2 x^{2} - 3 y^{2} + x b y x^{2} - 6 y^{2} + 3 x y .$

Page No 118:

Question 5:

Find the product $(\frac{3}{5} a b c^{3}) \times (\frac{- 25}{12} a^{3} b^{2}) \times (- 8 b^{3} c) .$

Answer:

$\begin{array}{l} We have: \\ \frac{3}{5} a b c^{3} \times \frac{(- 25)}{12} a^{3} b^{2} \times (- 8 b^{3} c) \\ = \frac{3^{1}}{5_{1}} a b c^{3} \times \frac{(- 2 5^{5})}{1 2_{4_{1}}} a^{3} b^{2} \times (- 8^{2} b^{3} c) \\ = a b c^{3} \times (- 5 a^{3} b^{2}) \times (- 2 b^{3} c) \\ = 10 a^{4} b^{6} c^{4} \end{array}$

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Question 6:

Simplify:

(3a + 4)(2a − 3) + (5a − 4)(a + 2)

Answer:

$\begin{array}{l} We have: \\ (3 a + 4) (2 a - 3) + (5 a - 4) (a + 2) \\ = {3 a (2 a - 3) + 4 (2 a - 3)} + {5 a (a + 2) - 4 (a + 2)} \\ = (6 a^{2} - 9 a + 8 a - 12) + (5 a^{2} + 10 a - 4 a - 8) \\ = (6 a^{2} - a - 12) + (5 a^{2} + 6 a - 8) \\ = (11 a^{2} + 5 a - 20) \end{array}$

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Question 7:

Solve: $\frac{3 x}{10} + \frac{2 x}{5} = \frac{7 x}{25} + \frac{29}{25} .$

Answer:

$\begin{array}{l} We have: \\ \frac{3 x}{10} + \frac{2 x}{5} = \frac{7 x}{25} + \frac{29}{25} \\ \Rightarrow \frac{3 x + 4 x}{10} = \frac{7 x + 29}{25} \\ \Rightarrow \frac{3 x + 4 x}{10} = \frac{7 x + 29}{25} \\ \Rightarrow \frac{7 x}{10} = \frac{7 x + 29}{25} \\ \Rightarrow 175 x = 70 x + 290 \\ \Rightarrow 105 x = 290 \\ \Rightarrow x = \frac{29 0^{58}}{10 5^{21}} \\ \Rightarrow x = \frac{58}{21} \end{array}$

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Question 8:

Solve: $0.5 x + \frac{x}{3} = 0.25 x + 7 .$

Answer:

$\begin{array}{l} We have: \\ 0.5 x + \frac{x}{3} = 0.25 x + 7 \\ \Rightarrow \frac{1.5 x + x}{3} = 0.25 x + 7 \\ \Rightarrow 1.5 x + x = 3 (0.25 x + 7) \\ \Rightarrow 2.5 x = 0.75 x + 21 \\ \Rightarrow 2.5 x - 0.75 x = 21 \\ \Rightarrow 1.75 x = 21 \\ \Rightarrow x = \frac{21}{1.75} \\ \Rightarrow x = 12 \end{array}$

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Question 9:

The sum of two consecutive odd numbers is 68. Find the numbers.

Answer:

$\begin{array}{l} Let the consecutive odd numbers be x and (x + 2) . \\ x + (x + 2) = 68 \\ \Rightarrow 2 x + 2 = 68 \\ \Rightarrow 2 x = 68 - 2 \\ \Rightarrow x = \frac{6 6^{33}}{2_{1}} \\ \Rightarrow x = 33 \\ ∴ The required numbers are 33 and (33 + 2), i . e ., 35. \end{array}$

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Question 10:

Reenu's father is thrice as old as Reenu. After 12 years he will be just twice his daughter. Find their present ages.

Answer:

$\begin{array}{l} Let Reenu's present age be x . \\ Then, her father's present age will be 3 x . \\ Reenu's age after 12 years = (x + 12) \\ Her father's age after 12 years = (3 x + 12) \\ Now, (3 x + 12) = 2 (x + 12) \\ \Rightarrow 3 x + 12 = 2 x + 24 \\ \Rightarrow x = 12 \\ ∴ Reenu's present age = 12 yrs \\ And her father's age = (3 \times 12) = 36 yrs \end{array}$

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Question 11:

Mark (âœ“) against the correct answer
If $2 x + \frac{5}{3} = \frac{1}{4} x + 4,$ then x = ?
(a) 3
(b) 4
(c) $\frac{3}{4}$
(d) $\frac{4}{3}$

Answer:

$(d) \frac{4}{3} \begin{array}{l} 2 x + \frac{5}{3} = \frac{1}{4} x + 4 \\ \Rightarrow 2 x - \frac{1}{4} x = 4 - \frac{5}{3} \\ \Rightarrow \frac{8 x - 1 x}{4} = \frac{12 - 5}{3} \\ \Rightarrow \frac{7 x}{4} = \frac{7}{3} \\ \Rightarrow 21 x = 28 \\ \Rightarrow x = \frac{2 8^{4}}{2 1_{3}} = \frac{4}{3} \end{array}$

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Question 12:

Mark (âœ“) against the correct answer
If $\frac{x}{2} - \frac{x}{3} = 5$ then x = ?
(a) 8
(b) 16
(c) 24
(d) 30

Answer:

$\begin{array}{l} (d) 30 \\ \frac{x}{2} - \frac{x}{3} =5 \\ \Rightarrow \frac{3 x - 2 x}{6} = 5 \\ \Rightarrow x = 30 \end{array}$

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Question 13:

Mark (âœ“) against the correct answer
If $\frac{x - 2}{3} = \frac{2 x - 1}{3} - 1,$ then x = ?
(a) 2
(b) 4
(c) 6
(d) 8

Answer:

$\begin{array}{l} (a) 2 \\ \frac{x - 2}{3} = \frac{2 x - 1}{3} - 1 \\ \Rightarrow \frac{x - 2}{3} = \frac{2 x - 1 - 3}{3} \\ \Rightarrow x - 2 = 2 x - 4 \\ \Rightarrow x-2x=-4+2 \\ \Rightarrow - x = - 2 \end{array} \Rightarrow x = 2$

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Question 14:

Mark (âœ“) against the correct answer
A number when multiplied by 4 is increased by 54. The number is

(a) 21
(b) 16
(c) 18
(d) 19

Answer:

$(c) 18 \begin{array}{l} Let the number be x . \\ According to the question, we have: \\ 4 x = x + 54 \\ \Rightarrow 3 x = 54 \\ \Rightarrow x = 18 \end{array}$

Page No 118:

Question 15:

Two complementary angles differ by 14°. The larger angle is

(a) 50°
(b) 52°
(c) 54°
(d) 56°

Answer:

$(b) 52 ° \begin{array}{l} Let the two complementary angles be x ° and (90 - x) ° . \\ According to the question, we have: \\ x - (90 - x) = 14 \\ ⇒2 x = 104 \\ \Rightarrow x = 52 \\ \Rightarrow (90 - x) ° = 90 ° - 52 ° = 38 ° \\ ∴ The larger angle is 52 ° . \end{array}$

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Question 16:

The length of a rectangle is twice its breadth and its perimeter is 96 m. The length of the rectangle is

(a) 28 m
(b) 30 m
(c) 32 m
(d) 36 m

Answer:

$(c) 32 m \begin{array}{l} Let the length and breadth of the rectangle be l m and b m, respectively. \\ According to the question, we have : \\ l = 2 b ... (i) \\ 2 (l + b) = 96 ... (ii) \\ Now, 2 (2 b + b) = 96 \\ \Rightarrow 6 b = 96 \\ \Rightarrow b = 16 \\ ∴ Length = 16 \times 2 m = 32 m \end{array}$

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Question 17:

The ages of A and B are in the ratio 4 : 3. After 6 years, their ages will be in the ratio 11 : 9. A's present age is

(a) 12 years
(b) 16 years
(c) 20 years
(d) 24 years

Answer:

$(b) 12 years Let the ages of A and B be x and y years, respectively . \begin{array}{l} Now, \frac{x}{y} = \frac{4}{3} \\ \Rightarrow 3 x = 4 y \\ \Rightarrow x = \frac{4}{3} y \\ After 6 years, we have: \\ \frac{x + 6}{y + 6} = \frac{11}{9} \\ \Rightarrow \frac{\frac{4}{3} y + 6}{y + 6} = \frac{11}{9} \\ \Rightarrow \frac{4 y + 18}{3(y + 6)} = \frac{11}{9} \\ \Rightarrow 36 y + 162 = 33 y + 198 \\ \Rightarrow 3 y = 36 \\ \Rightarrow y = 12 \\ ∴ x = \frac{4}{3} \times 1 2^{4} = 16 \\ Hence, A's present age is 16 years. \end{array} \begin{array}{l} \end{array}$

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Question 18:

Fill in the blanks

(i) −2a²b is a ...... .
(ii) (a² − 2b²) is a ...... .
(iii) (a + 2b − 3c) is a ...... .
(iv) In −5ab, the coefficient of a is ...... .
(v) In x² + 2x − 5, the ...... term is −5.

Answer:

(i) −2a²b is a monomial.
(ii) (a² − 2b²) is a binomial.
(iii) (a + 2b − 3c) is a trinomial.
(iv) In −5ab, the coefficient of a is -5b.
(v) In x² + 2x − 5, the constant term is −5.

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Question 19:

Write 'T' for true and 'F' for false

(i) In −x, the constant term is −1.
(ii) The coeffiecient of x in x² − 3x + 5 is 3.
(iii) (5x − 7) − (3x − 5) = 2x − 12.
(iv) (3x + 5y)(3x − 5y) = (9x² − 25y²).
(v) If a = 2 and b = $\frac{1}{2},$ then the value of ab (a² + b²) is $4 \frac{1}{4}$ .

Answer:

$(i) F The coefficient of x is - 1 . (i i) F The coefficient of x is - 3 . (i i i) F LHS = (5 x - 7) - (3 x - 5) = 5 x - 7 - 3 x + 5 = 2 x - 2 (i v) T LHS = (3 x + 5 y) (3 x - 5 y) = 3 x (3 x - 5 y) + 5 y (3 x - 5 y) = 9 x^{2} - 15 x y + 15 x y - 25 y^{2} = 9 x^{2} - 25 y^{2} (v) T (a^{2} + b^{2}) = [2^{2} + {(\frac{1}{2})}^{2}] = 4 + \frac{1}{4} = 4 \frac{1}{4}$

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