Rs Aggarwal 2019 Solutions for Class 7 MATH Chapter 7

Question 1:

Answer:

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

Question 2:

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

Answer:

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

Question 3:

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

Answer:

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

Question 4:

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

Answer:

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

Question 5:

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

Answer:

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

Question 6:

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

Answer:

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

Question 7:

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

Answer:

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

Question 8:

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

Answer:

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

Question 9:

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

Answer:

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

Question 10:

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

Answer:

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

Question 11:

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

Answer:

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

Question 12:

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

Answer:

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

Question 13:

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

Answer:

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

Question 14:

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

Answer:

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

Question 15:

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

Answer:

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

Question 16:

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

Answer:

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

Question 17:

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

Answer:

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

Question 18:

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

Answer:

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

Question 19:

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

Answer:

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

Question 20:

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

Answer:

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

Question 21:

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

Answer:

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

Question 22:

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

Answer:

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

Question 23:

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

Answer:

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

Question 24:

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

Answer:

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

Question 25:

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

Answer:

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

Question 26:

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

Answer:

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

Question 27:

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

Answer:

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

Question 28:

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

Answer:

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

Question 29:

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

Answer:

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

Question 30:

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

Answer:

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

Question 31:

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

Answer:

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

Question 32:

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

Answer:

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

Question 1:

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

Answer:

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

Question 2:

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

Answer:

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

Question 3:

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

Answer:

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

Question 4:

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

Answer:

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

Question 5:

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

Answer:

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

Question 6:

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

Answer:

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

Question 7:

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

Answer:

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

Question 8:

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

Answer:

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

Question 9:

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

Answer:

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

Question 10:

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

Answer:

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

Question 11:

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

Answer:

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

Question 12:

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

Answer:

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

Question 13:

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

Answer:

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

Question 14:

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

Answer:

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

Question 15:

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

Answer:

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

Question 16:

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

Answer:

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

Question 17:

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

Answer:

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

Question 18:

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

Answer:

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

Question 19:

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

Answer:

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

Question 20:

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

Answer:

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

Question 21:

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

Answer:

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

Question 22:

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

Answer:

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

Question 23:

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

Answer:

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

Question 24:

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

Answer:

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

Question 25:

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

Answer:

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

Question 26:

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

Answer:

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

Question 27:

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

Answer:

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

Question 28:

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

Answer:

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

Question 29:

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

Answer:

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

Question 30:

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

Answer:

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

Question 31:

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

Answer:

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

Question 32:

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

Answer:

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

Question 33:

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

Answer:

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

Question 34:

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

Answer:

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

Question 35:

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

Answer:

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

Question 1:

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

Answer:

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

Question 2:

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

Answer:

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

Question 3:

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

Answer:

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

Question 4:

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

Answer:

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

Question 5:

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

Answer:

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

Question 6:

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

Answer:

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

Question 7:

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

Answer:

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

Question 8:

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

Answer:

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

Question 9:

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

Answer:

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

Question 10:

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

Answer:

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

Question 11:

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

Answer:

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

Question 12:

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

Answer:

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

Question 13:

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

Answer:

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

Question 14:

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

Answer:

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

Question 15:

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

Answer:

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

Question 16:

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

Answer:

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

Question 17:

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

Answer:

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

Question 18:

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

Answer:

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

Question 1:

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

Answer:

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

Question 2:

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

Answer:

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

Question 3:

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

Answer:

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

Question 4:

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

Answer:

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

Question 5:

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

Answer:

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

Question 6:

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

Answer:

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

Question 7:

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

Answer:

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

Question 8:

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

Answer:

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

Question 9:

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

Answer:

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

Question 10:

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

Answer:

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

Question 11:

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

Answer:

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

Question 12:

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

Answer:

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

Question 13:

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

Answer:

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

Question 14:

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

Answer:

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

Question 15:

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

Answer:

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

Question 16:

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

Answer:

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

Question 17:

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

Answer:

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

Question 18:

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

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.

Question 19:

(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.

Answer:

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

RS Aggarwal 2019 Solutions for Class 7 Math Chapter 7 - Linear Equations In One Variable

Page No 111:

Question 1:

Answer:

Page No 111:

Question 2:

Answer:

Page No 111:

Question 3:

Answer:

Page No 111:

Question 4:

Answer:

Page No 111:

Question 5:

Answer:

Page No 111:

Question 6:

Answer:

Page No 111:

Question 7:

Answer:

Page No 111:

Question 8:

Answer:

Page No 111:

Question 9:

Answer:

Page No 111:

Question 10:

Answer:

Page No 111:

Question 11:

Answer:

Page No 111:

Question 12:

Answer:

Page No 111:

Question 13:

Answer:

Page No 111:

Question 14:

Answer:

Page No 111:

Question 15:

Answer:

Page No 111:

Question 16:

Answer:

Page No 111:

Question 17:

Answer:

Page No 111:

Question 18:

Answer:

Page No 112:

Question 19:

Answer:

Page No 112:

Question 20:

Answer:

Page No 112:

Question 21:

Answer:

Page No 112:

Question 22:

Answer:

Page No 112:

Question 23:

Answer:

Page No 112:

Question 24:

Answer:

Page No 112:

Question 25:

Answer:

Page No 112:

Question 26:

Answer:

Page No 112: