Rs Aggarwal 2020 2021 for Class 7 Maths Chapter 7 - Linear Equations In One Variable

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Rs Aggarwal 2020 2021 Solutions for Class 7 Maths 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 Maths 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 2020 2021 Book of Class 7 Maths Chapter 7 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 7 Maths are prepared by experts and are 100% accurate.

Page No 111:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

Answer:

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

Page No 112:

Answer:

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

Page No 112:

Answer:

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

Page No 112:

Answer:

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

Page No 112:

Answer:

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

Page No 112:

Answer:

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

Page No 112:

Answer:

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

Page No 112:

Answer:

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

Page No 114:

Answer:

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

Page No 114:

Answer:

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

Page No 114:

Answer:

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

Page No 114:

Answer:

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

Page No 114:

Answer:

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

Page No 114:

Answer:

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

Page No 114:

Answer:

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

Page No 115:

Answer:

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

Page No 115:

Answer:

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

Page No 115:

Answer:

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

Page No 115:

Answer:

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

Page No 115:

Answer:

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

Page No 115:

Answer:

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

Page No 115:

Answer:

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

Page No 115:

Answer:

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

Page No 115:

Answer:

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

Page No 115:

Answer:

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

Page No 115:

Answer:

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

Page No 115:

Answer:

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

Page No 115:

Answer:

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

Page No 115:

Answer:

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

Page No 115:

Answer:

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

Page No 115:

Answer:

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

Page No 115:

Answer:

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

Page No 115:

Answer:

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

Page No 115:

Answer:

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

Page No 115:

Answer:

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

Page No 115:

Answer:

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

Page No 115:

Answer:

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

Page No 115:

Answer:

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

Page No 115:

Answer:

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

Page No 115:

Answer:

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

Page No 116:

Answer:

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

Page No 116:

Answer:

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

Page No 116:

Answer:

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

Page No 116:

Answer:

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

Page No 116:

Answer:

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

Page No 116:

Answer:

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

Page No 116:

Answer:

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

Page No 116:

Answer:

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

Page No 116:

Answer:

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

Page No 116:

Answer:

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

Page No 116:

Answer:

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

Page No 117:

Answer:

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

Page No 117:

Answer:

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

Page No 117:

Answer:

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

Page No 117:

Answer:

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

Page No 117:

Answer:

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

Page No 117:

Answer:

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

Page No 117:

Answer:

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

Page No 117:

Answer:

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

Page No 117:

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:

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:

Answer:

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

Page No 118:

Answer:

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

Page No 118:

Answer:

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

Page No 118:

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:

Answer:

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

Page No 118:

Answer:

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

Page No 118:

Answer:

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

Page No 118:

Answer:

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

Page No 118:

Answer:

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

Page No 118:

Answer:

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

Page No 118:

Answer:

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

Page No 118:

Answer:

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

Page No 118:

Answer:

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

Page No 118:

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:

Answer:

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

Page No 118:

Answer:

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

Page No 118:

Answer:

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

Page No 118:

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.

Page No 118:

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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