RD Sharma solution class 7 chapter 8 Linear Equations In One Variables Objective Type Question

Objective Type Question

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Question 1:

Mark the correct alternative in the following question:

The zero of 3x + 2 is

$\left(\mathrm{a}\right) \frac{2}{3} \left(\mathrm{b}\right) \frac{3}{2} \left(\mathrm{c}\right) -\frac{2}{3} \left(\mathrm{d}\right) \frac{-3}{2}$

Answer 1:

$$\begin{gathered} \mathrm{If} 3x+2=0, \mathrm{then} \\ 3x=-2 \left(\mathrm{Transposing} +2 \mathrm{to} \mathrm{RHS}\right) \\ \Rightarrow x=-\frac{2}{3} \end{gathered}$$

So, the zero of 3x + 2 is $-\frac{2}{3}$.

Note: A zero is that number, when put in place of the variable, makes the expression equal to zero.

Hence, the correct alternative is option (c).

Question 2:

Mark the correct alternative in the following question:

$$\begin{gathered} \mathrm{If} 2x-\frac{3}{2}=5x+\frac{3}{4}, \mathrm{then} x= \\ \left(\mathrm{a}\right) \frac{3}{4} \left(\mathrm{b}\right) -\frac{3}{4} \left(\mathrm{c}\right) \frac{4}{3} \left(\mathrm{d}\right) -\frac{4}{3} \end{gathered}$$

Answer 2:

$$\begin{gathered} \mathrm{As}, 2x-\frac{3}{2}=5x+\frac{3}{4} \\ \Rightarrow 2x-5x=\frac{3}{2}+\frac{3}{4} \left(\mathrm{By} \mathrm{transposing} -\frac{3}{2} \mathrm{to} \mathrm{RHS} \mathrm{and} 5x \mathrm{to} \mathrm{LHS}\right) \\ \Rightarrow -3x=\frac{6}{4}+\frac{3}{4} \\ \Rightarrow -3x=\frac{6+3}{4} \\ \Rightarrow x=\frac{9}{4\times \left(-3\right)} \left(\mathrm{By} \mathrm{transposing} -3 \mathrm{to} \mathrm{RHS}\right) \\ \Rightarrow x=\frac{3}{4\times \left(-1\right)} \\ \Rightarrow x=\frac{3}{-4} \\ \therefore x=-\frac{3}{4} \end{gathered}$$

Hence, the correct alternative is option (b).

Question 3:

Mark the correct alternative in the following question:

$$\begin{gathered} \mathrm{If} \frac{x}{2}-4=\frac{x}{3}-1, \mathrm{then} x= \\ \left(\mathrm{a}\right) 3 \left(\mathrm{b}\right) 6 \left(\mathrm{c}\right) 18 \left(\mathrm{d}\right) 2 \end{gathered}$$

Answer 3:

$$\begin{gathered} \mathrm{As}, \frac{x}{2}-4=\frac{x}{3}-1 \\ \Rightarrow \frac{x}{2}-\frac{x}{3}=4-1 \left(\mathrm{By} \mathrm{transposing} \frac{x}{3} \mathrm{to} \mathrm{LHS} \mathrm{and} -4 \mathrm{to} \mathrm{RHS}\right) \\ \Rightarrow \frac{3x}{6}-\frac{2x}{6}=3 \\ \Rightarrow \frac{3x-2x}{6}=3 \\ \Rightarrow \frac{x}{6}=3 \\ \Rightarrow x=3\times 6 \left(\mathrm{By} \mathrm{transposing} 6 \mathrm{to} \mathrm{RHS}\right) \\ \therefore x=18 \end{gathered}$$

Hence, the correct alternative is option (c).

Question 4:

Mark the correct alternative in the following question:

$$\begin{gathered} \mathrm{If} \frac{x+2}{x-2}=\frac{2}{3}, \mathrm{then} x= \\ \left(\mathrm{a}\right) -10 \left(\mathrm{b}\right) 10 \left(\mathrm{c}\right) \frac{4}{3} \left(\mathrm{d}\right) -\frac{4}{3} \end{gathered}$$

Answer 4:

$$\begin{gathered} \mathrm{As}, \frac{x+2}{x-2}=\frac{2}{3} \\ \Rightarrow 3\left(x+2\right)=2\left(x-2\right) \left(\mathrm{By} \mathrm{cross} \mathrm{multiplication}\right) \\ \Rightarrow 3x+6=2x-4 \\ \Rightarrow 3x-2x=-6+4 \left(\mathrm{By} \mathrm{transposing} 2x \mathrm{to} \mathrm{LHS} \mathrm{and} 6 \mathrm{to} \mathrm{RHS}\right) \\ \therefore x=-10 \end{gathered}$$

Hence, the correct alternative is option (a).

Question 5:

Mark the correct alternative in the following question:

$$\begin{gathered} \mathrm{If} \frac{x}{6}+\frac{x}{4}=\frac{x}{2}+\frac{3}{4}, \mathrm{then} x= \\ \left(\mathrm{a}\right) 9 \left(\mathrm{b}\right) 6 \left(\mathrm{c}\right) -9 \left(\mathrm{d}\right) 4 \end{gathered}$$

Answer 5:

$$\begin{gathered} \mathrm{As}, \frac{x}{6}+\frac{x}{4}=\frac{x}{2}+\frac{3}{4} \\ \Rightarrow \frac{x}{6}+\frac{x}{4}-\frac{x}{2}=\frac{3}{4} \left(\mathrm{By} \mathrm{transposing} \frac{x}{2} \mathrm{to} \mathrm{LHS}\right) \\ \Rightarrow \frac{2x}{12}+\frac{3x}{12}-\frac{6x}{12}=\frac{3}{4} \\ \Rightarrow \frac{2x+3x-6x}{12}=\frac{3}{4} \\ \Rightarrow \frac{-x}{12}=\frac{3}{4} \\ \Rightarrow -x\times 4=3\times 12 \left(\mathrm{By} \mathrm{cross} \mathrm{multiplication}\right) \\ \Rightarrow -4x=36 \\ \Rightarrow x=\frac{36}{-4} \\ \therefore x=-9 \end{gathered}$$

Hence, the correct alternative is option (c).

Question 6:

Mark the correct alternative in the following question:

$$\begin{gathered} \mathrm{If} 2x+\frac{5}{3}=\frac{1}{4}x+4, \mathrm{then} x= \\ \left(\mathrm{a}\right) 3 \left(\mathrm{b}\right) 4 \left(\mathrm{c}\right) \frac{3}{4} \left(\mathrm{d}\right) \frac{4}{3} \end{gathered}$$

Answer 6:

$$\begin{gathered} \mathrm{As}, 2x+\frac{5}{3}=\frac{1}{4}x+4 \\ \Rightarrow 2x-\frac{1}{4}x=4-\frac{5}{3} \left(\mathrm{By} \mathrm{transposing} \frac{5}{3} \mathrm{to} \mathrm{RHS} \mathrm{and} \frac{1}{4}x \mathrm{to} \mathrm{LHS}\right) \\ \Rightarrow \frac{2x}{1}-\frac{x}{4}=\frac{4}{1}-\frac{5}{3} \\ \Rightarrow \frac{8x}{4}-\frac{x}{4}=\frac{12}{3}-\frac{5}{3} \\ \Rightarrow \frac{8x-x}{4}=\frac{12-5}{3} \\ \Rightarrow \frac{7x}{4}=\frac{7}{3} \\ \Rightarrow 7x\times 3=4\times 7 \left(\mathrm{By} \mathrm{cross} \mathrm{multiplication}\right) \\ \Rightarrow 21x=28 \\ \Rightarrow x=\frac{28}{21} \\ \therefore x=\frac{4}{3} \end{gathered}$$

Hence, the correct alternative is option (d).

Question 7:

Mark the correct alternative in the following question:

$$\begin{gathered} \mathrm{If} \frac{x}{2}-\frac{x}{3}=5, \mathrm{then} x= \\ \left(\mathrm{a}\right) 8 \left(\mathrm{b}\right) 16 \left(\mathrm{c}\right) 24 \left(\mathrm{d}\right) 30 \end{gathered}$$

Answer 7:

$$\begin{gathered} \mathrm{As}, \frac{x}{2}-\frac{x}{3}=5 \\ \Rightarrow \frac{3x}{6}-\frac{2x}{6}=5 \\ \Rightarrow \frac{3x-2x}{6}=5 \\ \Rightarrow \frac{x}{6}=5 \\ \Rightarrow x=5\times 6 \left(\mathrm{By} \mathrm{transposing} 6 \mathrm{to} \mathrm{RHS}\right) \\ \therefore x=30 \end{gathered}$$

Hence, the correct alternative is option (d).

Question 8:

Mark the correct alternative in the following question:

$$\begin{gathered} \mathrm{If} \frac{x-2}{3}=\frac{2x-1}{3}-1, \mathrm{then} x= \\ \left(\mathrm{a}\right) 2 \left(\mathrm{b}\right) 4 \left(\mathrm{c}\right) 6 \left(\mathrm{d}\right) 8 \end{gathered}$$

Answer 8:

$$\begin{gathered} \mathrm{As}, \frac{x-2}{3}=\frac{2x-1}{3}-1 \\ \Rightarrow \frac{x-2}{3}-\frac{2x-1}{3}=-1 \left(\mathrm{By} \mathrm{transposing} \frac{2x-1}{3} \mathrm{to} \mathrm{LHS}\right) \\ \Rightarrow \frac{\left(x-2\right)-\left(2x-1\right)}{3}=-1 \\ \Rightarrow \frac{x-2-2x+1}{3}=-1 \\ \Rightarrow \frac{-x-1}{3}=-1 \\ \Rightarrow -x-1=-1\times 3 \left(\mathrm{By} \mathrm{transposing} 3 \mathrm{to} \mathrm{RHS}\right) \\ \Rightarrow -x-1=-3 \\ \Rightarrow -x=-3+1 \left(\mathrm{By} \mathrm{transposing} -1 \mathrm{to} \mathrm{RHS}\right) \\ \Rightarrow -x=-2 \\ \therefore x=2 \end{gathered}$$

Hence, the correct alternative is option (a).

Question 9:

Mark the correct alternative in the following question:

The sum of two consecutive whole numbers is 43. The smaller number is

(a) 21                                 (b) 22                                 (c) 23                                 (d) 24

Answer 9:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{two} \mathrm{consecutive} \mathrm{whole} \mathrm{numbers} \mathrm{be} x \mathrm{and} x+1. \\ \mathrm{As}, \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{the} \mathrm{two} \mathrm{consecutive} \mathrm{whole} \mathrm{numbers} \mathrm{is} 43. \\ \Rightarrow x+\left(x+1\right)=43 \\ \Rightarrow 2x+1=43 \\ \Rightarrow 2x=43-1 \left(\mathrm{By} \mathrm{transposing} 1 \mathrm{to} \mathrm{RHS}\right) \\ \Rightarrow 2x=42 \\ \Rightarrow x=\frac{42}{2} \left(\mathrm{By} \mathrm{transposing} 2 \mathrm{to} \mathrm{RHS}\right) \\ \therefore x=21 \end{gathered}$$

So, the smaller number is 21.

Hence, the correct alternative is option (a).

Question 10:

Mark the correct alternative in the following question:

The sum of two consecutive odd numbers is 36. The larger number is

(a) 17                                  (b) 15                                  (c) 19                                  (d) 21

Answer 10:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{two} \mathrm{consecutive} \mathrm{odd} \mathrm{numbers} \mathrm{be} x \mathrm{and} x+2. \\ \mathrm{As}, \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{the} \mathrm{two} \mathrm{consecutive} \mathrm{odd} \mathrm{numbers} \mathrm{is} 36. \\ \Rightarrow x+\left(x+2\right)=36 \\ \Rightarrow 2x+2=36 \\ \Rightarrow 2x=36-2 \\ \Rightarrow 2x=34 \\ \Rightarrow x=\frac{34}{2} \\ \Rightarrow x=17 \\ \therefore x+2=17+2=19 \end{gathered}$$

So, the larger number is 19.

Hence, the correct alternative is option (c).

Question 11:

Mark the correct alternative in the following question:

Twice a number when increased by 7 gives 25. The number is

(a) 7                                   (b) 9                                   (c) 10                                   (d) 8

Answer 11:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{number} \mathrm{be} x. \\ \mathrm{As}, \mathrm{twice} \mathrm{the} \mathrm{number} \mathrm{when} \mathrm{increased} \mathrm{by} 7 \mathrm{gives} 25. \\ \Rightarrow 2x+7=25 \\ \Rightarrow 2x=25-7 \left(\mathrm{By} \mathrm{transposing} 7 \mathrm{to} \mathrm{RHS}\right) \\ \Rightarrow 2x=18 \\ \Rightarrow x=\frac{18}{2} \left(\mathrm{By} \mathrm{transposing} 2 \mathrm{to} \mathrm{RHS}\right) \\ \therefore x=9 \end{gathered}$$

So, the number is 9.

Hence, the correct alternative is option (b).

Question 12:

Mark the correct alternative in the following question:

The length of a rectangle is three times its width and its perimeter 56 m. The length is

(a) 7 m                               (b) 14 m                               (c) 21 m                               (d) 28 m

Answer 12:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{width} \mathrm{of} \mathrm{the} \mathrm{rectangle} \mathrm{be} x. \mathrm{Then}, \\ \mathrm{the} \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{rectangle}=3x \\ \mathrm{As}, \mathrm{perimeter} \mathrm{of} \mathrm{the} \mathrm{rectangle}=56 \mathrm{m} \\ \Rightarrow 2\times \left(\mathrm{Length}+\mathrm{Breadth}\right)=56 \\ \Rightarrow 2\times \left(3x+x\right)=56 \\ \Rightarrow 2\times 4x=56 \\ \Rightarrow 8x=56 \\ \Rightarrow x=\frac{56}{8} \\ \therefore x=7 \\ \mathrm{So}, \mathrm{the} \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{rectangle}=3x=3\times 7=21 \mathrm{m} \end{gathered}$$

Hence, the correct alternative is option (c).

Question 13:

Mark the correct alternative in the following question:

Two-third of a number is greater than one-third of the number by 5. The number is

(a) 10                                (b) 5                                (c) 15                                (d) 12

Answer 13:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{number} \mathrm{be} x. \\ \mathrm{As}, \mathrm{two}-\mathrm{third} \mathrm{of} \mathrm{a} \mathrm{number} \mathrm{is} \mathrm{greater} \mathrm{than} \mathrm{one}-\mathrm{third} \mathrm{of} \mathrm{the} \mathrm{number} \mathrm{by} 5. \\ \Rightarrow \frac{2}{3}x-\frac{1}{3}x=5 \\ \Rightarrow \frac{2x-x}{3}=5 \\ \Rightarrow \frac{x}{3}=5 \\ \Rightarrow x=5\times 3 \\ \therefore x=15 \end{gathered}$$

So, the number is 15.

Hence, the correct alternative is option (c).

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Question 14:

Mark the correct alternative in the following question:

If the sum of a number and its two-fifth is 70. The number is

(a) 70                                 (b) 50                                 (c) 60                                 (d) 90

Answer 14:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{number} \mathrm{be} x. \\ \mathrm{As}, \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{a} \mathrm{number} \mathrm{and} \mathrm{its} \mathrm{two}-\mathrm{fifth} \mathrm{is} 70. \\ \Rightarrow x+\frac{2}{5}x=70 \\ \Rightarrow \frac{x}{1}+\frac{2x}{5}=70 \\ \Rightarrow \frac{5x}{5}+\frac{2x}{5}=70 \\ \Rightarrow \frac{5x+2x}{5}=70 \\ \Rightarrow \frac{7x}{5}=70 \\ \Rightarrow 7x=70\times 5 \left(\mathrm{By} \mathrm{transposing} 5 \mathrm{to} \mathrm{RHS}\right) \\ \Rightarrow 7x=350 \\ \Rightarrow x=\frac{350}{7} \left(\mathrm{By} \mathrm{transposing} 7 \mathrm{to} \mathrm{RHS}\right) \\ \therefore x=50 \end{gathered}$$

So, the number is 50.

Hence, the correct alternative is option (b).

Question 15:

Mark the correct alternative in the following question:

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

(a) 30                                  (b) 40                                  (c) 50                                  (d) 60

Answer 15:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{number} \mathrm{be} x. \\ \mathrm{As}, \frac{2}{3} \mathrm{of} \mathrm{the} \mathrm{number} \mathrm{is} \mathrm{less} \mathrm{than} \mathrm{the} \mathrm{original} \mathrm{number} \mathrm{by} 20. \\ \Rightarrow x-\frac{2}{3}x=20 \\ \Rightarrow \frac{x}{1}-\frac{2x}{3}=20 \\ \Rightarrow \frac{3x}{3}-\frac{2x}{3}=20 \\ \Rightarrow \frac{3x-2x}{3}=20 \\ \Rightarrow \frac{x}{3}=20 \\ \Rightarrow x=20\times 3 \left(\mathrm{By} \mathrm{transposing} 3 \mathrm{to} \mathrm{RHS}\right) \\ \therefore x=60 \end{gathered}$$

So, the number is 60.

Hence, the correct alternative is option (d).

Question 16:

Mark the correct alternative in the following question:

A number is as much greater than 31 as it is less than 81. The number is

(a) 46                                   (b) 56                                   (c) 66                                   (d) 76

Answer 16:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{number} \mathrm{be} x. \\ \mathrm{As}, \mathrm{the} \mathrm{number} \mathrm{is} \mathrm{as} \mathrm{much} \mathrm{greater} \mathrm{than} 31 \mathrm{as} \mathrm{it} \mathrm{is} \mathrm{less} \mathrm{than} 81. \\ \Rightarrow x-31=81-x \\ \Rightarrow x+x=81+31 \left(\mathrm{By} \mathrm{transposing} -x \mathrm{to} \mathrm{LHS} \mathrm{and} -31 \mathrm{to} \mathrm{RHS}\right) \\ \Rightarrow 2x=112 \\ \Rightarrow x=\frac{112}{2} \left(\mathrm{By} \mathrm{transposing} 2 \mathrm{to} \mathrm{RHS}\right) \\ \therefore x=56 \end{gathered}$$

So, the number is 56.

Hence, the correct alternative is option (b).

Question 17:

Mark the correct alternative in the following question:

Two complementary angles differ by 20°. The smaller angle is

(a) 55°                                    (b) 25°                                    (c) 65°                                    (d) 35°

Answer 17:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{smaller} \mathrm{angle} \mathrm{be} x. \mathrm{Then}, \\ \mathrm{The} \mathrm{larger} \mathrm{angle}=\left(x+20°\right) \\ \mathrm{As}, \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{the} \mathrm{two} \mathrm{complementary} \mathrm{angles} \mathrm{is} \mathrm{always} 90°. \\ \Rightarrow x+\left(x+20°\right)=90° \\ \Rightarrow 2x+20°=90° \\ \Rightarrow 2x=90°-20° \\ \Rightarrow 2x=70° \\ \Rightarrow x=\frac{70°}{2} \left(\mathrm{By} \mathrm{transposing} 2 \mathrm{to} \mathrm{RHS}\right) \\ \therefore x=35° \end{gathered}$$

So, the smaller angle is 35°.

Hence, the correct alternative is option (d).

Question 18:

Mark the correct alternative in the following question:

Two supplementary angles differ by 40°. The measure of the larger angle is

(a) 70°                                     (b) 80°                                     (c) 110°                                     (d) 100°

Answer 18:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{larger} \mathrm{angle} \mathrm{be} x. \mathrm{Then}, \\ \mathrm{The} \mathrm{smaller} \mathrm{angle}=\left(x-40°\right) \\ \mathrm{As}, \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{the} \mathrm{two} \mathrm{supplementary} \mathrm{angles} \mathrm{is} \mathrm{always} 180°. \\ \Rightarrow x+\left(x-40°\right)=180° \\ \Rightarrow 2x-40°=180° \\ \Rightarrow 2x=180°+40° \\ \Rightarrow 2x=220° \\ \Rightarrow x=\frac{220°}{2} \left(\mathrm{By} \mathrm{transposing} 2 \mathrm{to} \mathrm{RHS}\right) \\ \therefore x=110° \end{gathered}$$

So, the measure of the larger angle is 110°.

Hence, the correct alternative is option (c).

Question 19:

Mark the correct alternative in the following question:

The sum of three consecutive odd numbers is 81. The middle number is

(a) 25                                     (b) 27                                     (c) 31                                     (d) 29

Answer 19:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{three} \mathrm{consecutive} \mathrm{odd} \mathrm{numbers} \mathrm{be} x, x+2 \mathrm{and} x+4. \\ \mathrm{As}, \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{the} \mathrm{three} \mathrm{consecutive} \mathrm{numbers} \mathrm{is} 81. \\ \Rightarrow x+\left(x+2\right)+\left(x+4\right)=81 \\ \Rightarrow 3x+6=81 \\ \Rightarrow 3x=81-6 \left(\mathrm{By} \mathrm{transposing} 6 \mathrm{to} \mathrm{RHS}\right) \\ \Rightarrow 3x=75 \\ \Rightarrow x=\frac{75}{3} \left(\mathrm{By} \mathrm{transposing} 3 \mathrm{to} \mathrm{RHS}\right) \\ \Rightarrow x=25 \\ \therefore x+2=25+2=27 \end{gathered}$$

So, the middle number is 27.

Hence, the correct alternative is option (b).

Question 20:

If 2(2n + 5) = 3(3n $-$ 10), then n =

(a) 5                                       (b) 3                                      (c) 7                                      (d) 8

Answer 20:

$$\begin{gathered} \mathrm{As}, 2\left(2n+5\right)=3\left(3n-10\right) \\ \Rightarrow 4n+10=9n-30 \\ \Rightarrow 4n-9n=-10-30 \left(\mathrm{By} \mathrm{transposing} 10 \mathrm{to} \mathrm{RHS} \mathrm{and} 9n \mathrm{to} \mathrm{LHS}\right) \\ \Rightarrow -5n=-40 \\ \Rightarrow n=\frac{-40}{-5} \left(\mathrm{By} \mathrm{transposing} -5 \mathrm{to} \mathrm{RHS}\right) \\ \therefore n=8 \end{gathered}$$

Hence, the correct alternative is option (d).