MCQ Questions for Class 8 Maths Linear Equations In One Variable Quiz 3

Solve the equation:

$\displaystyle\frac { x-5 }{ 7 } =\displaystyle\frac { x+3 }{ 2 }$ .

0%
$x=-\displaystyle\frac { 31 }{ 5 }$
0%
$x=-\displaystyle\frac { 11 }{ 5 }$
0%
$x= \displaystyle\frac { 31 }{ 9 }$
0%
$x=-\displaystyle\frac { 26 }{ 5 }$

Explanation

Given,

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

$\Rightarrow \displaystyle 2(x-5)=7(x+3)$ .....(By cross multiplication of denominators)

$\Rightarrow2x-10=7x+21$ ...(By Distribution Law).

Transposing

$x$ terms to one side, we get,

$\Rightarrow 7x-2x=-10-21$

$\Rightarrow 5x=-31$

$\Rightarrow x=\cfrac{-31}{5}$ .

Hence, option

$A$ is correct.

Lalit's brother is thrice as old as Lalit. If brother's age is 60 years then Lalit's age is

0%
30 yrs
0%
10 yrs
0%
20 yrs
0%
60 yrs

Explanation

Let Lalit brothers age be 3x and Lalit's age be x.
3x = 60
x =

$\displaystyle{\frac{60}{3}}$ = 20

Linear equation in one variable has

0%
Only constant term.
0%
Only one variable with power $1$ .
0%
Only one term with a variable.
0%
Only one variable with any power.

Explanation

${\textbf{Step-1: Use definition of linear equation in one variable }}$

$\text{Linear equation is in the form ax+b=0 where x is variable and a and b are constants}$

${\text{Example, 5x-10=5 has only one variable.}}$

${\textbf{Hence, answer is an equation one variable raised to the power 1.}}$

$x-4=5$ , then the value of

$x$ is-

0%
$-1$
0%
$9$
0%
$-9$
0%
$1$

Explanation

$x-5=4$

$\Rightarrow$

$x=5+4=9$

Find

$x$ , if

$12-(9+3x)=18x$ .

0%
$\dfrac{-1}{7}$
0%
$\dfrac{7}{5}$
0%
$\dfrac{1}{7}$
0%
$-7$

Explanation

Given,

$\displaystyle 12-(9+3x)=18x$ .

$\Longrightarrow \displaystyle 12-9-3x=18x$ ...[By Distribution Law]

$\Longrightarrow\displaystyle -3x+3=18x$ ...[On simplifying].

Transposing

$x$ terms to one side, we get,

$\Longrightarrow\displaystyle 18x+3x=3$

$\Longrightarrow 21x=3$

$\Longrightarrow x=\dfrac{3}{21}$

$\Longrightarrow x=\dfrac{1}{7}$ .

Hence, option

$C$ is correct.

Solve the equation:

$\dfrac{7x - 3}{3x}=2$

0%
$x=3$
0%
$x=2$
0%
$x=1$
0%
$x=-1$

Explanation

$\dfrac{7x - 3}{3x}=2$

$7x - 3 = 6x$
On transposing

$6x$ to the L.H.S and

$3$ to the R.H.S we obtain

$7x - 6x = 3$

$x = 3$

Find the value of

$p$ in the linear equation:

$4p + 2 = 6p + 10$

0%
$p=-4$
0%
$p=-3$
0%
$p=-2$
0%
$p=-1$

Explanation

$4p + 2 = 6p + 10$

$6p - 4p =2-10$

$2p = -8$

$p = -4$

$1$ is subtracted from a number it becomes

$\dfrac{1}{8}$ . Find the original number.

0%
$\dfrac{8}{9}$
0%
$\dfrac{-9}{8}$
0%
$\dfrac{9}{8}$
0%
None of the above.

Explanation

Let original number be

$x$ .

$\displaystyle \therefore x-1=\dfrac { 1 }{ 8 }$

$\therefore x=1+\dfrac { 1 }{ 8 }$

$\therefore x=\dfrac { 9 }{ 8 }$

Reduce the following linear equation:

$2x + 5 = 3$

0%
$x=1$
0%
$x=-1$
0%
$x=0$
0%
$x=2$

Explanation

Given,

$2x + 5 = 3$

$2x = 3 - 5$

$2x = -2$

$x = -1$

$\dfrac{3}{5}$ th of a number subtracted from

$\dfrac{3}{4}$ th of that number gives

$18$ . Find the number.

0%
$110$
0%
$100$
0%
$120$
0%
None of the above.

Explanation

Let the required number be

$x$ .

According to the question:

$\displaystyle \frac { 3 }{ 4 } x-\frac { 3 }{ 5 } x=18$

$\Longrightarrow\displaystyle \frac { 3x\times 5-3x\times 4 }{ 20 } =18$

$\Longrightarrow\displaystyle 15x-12x=18\times 20$

$\Longrightarrow\displaystyle 3x=18\times 20$

$\Longrightarrow\displaystyle x=6\times 20$

$\therefore x=120$

When a number is divided by

$25$ , the result is

$-25$ . The number is:

0%
$625$
0%
$-625$
0%
$15625$
0%
$3125$

Explanation

Let the required number be

$x$ .

According to the question,

$\displaystyle \dfrac { x }{ 25 } =-25$

$\displaystyle x=-25\times 25$ .......Multiply both sides by

$25$

$\displaystyle x=-625$

Find

$v$ , if

$\displaystyle v+11=77-9v$ .

0%
$\dfrac{33}{5}$
0%
$\dfrac{5}{33}$
0%
$-\dfrac{5}{33}$
0%
$\dfrac{15}{33}$

Explanation

Given,

$\displaystyle v+11=77-9v$ .

Transposing

$v$ terms to one side, we get,

$\Longrightarrow\displaystyle v+9v=77-11$

$\Longrightarrow 10v=66$

$\Longrightarrow\displaystyle v=\dfrac{66}{10}$

$\therefore v=\dfrac{33}{5}$ .

Hence, option

$A$ is correct.

$\displaystyle 4(3-n)=3n$ , then find the value of

$n$ .

0%
$\dfrac{7}{12}$
0%
$-\dfrac{12}{7}$
0%
$\dfrac{12}{7}$
0%
$-\dfrac{7}{12}$

Explanation

Given,

$\displaystyle 4(3-n)=3n$

$\Longrightarrow\displaystyle 12-4n=3n$ ...[By Distribution Law].

Transposing

$n$ terms to one side, we get,

$\Longrightarrow\displaystyle -4n-3n=-12$

$\Longrightarrow\displaystyle -7n=-12$

$\Longrightarrow\displaystyle n=\dfrac{12}{7}$ .

Hence, option

$C$ is correct.

If the sum of three consecutive integers is

$63$ . Find the smallest number.

0%
$11$
0%
$15$
0%
$23$
0%
$20$

Explanation

Let the smallest integer be

$x$ .

The three consecutive numbers

$= x, x+1, x+2$

Given,

$x + x+1 + x +2 = 63$

$3x + 3 = 63$

$3x = 60$

$\therefore x =20$

The sum of four consecutive even integers is

$92$ . Find the largest number.

0%
$20$
0%
$22$
0%
$24$
0%
$26$

Explanation

Let four consecutive even numbers be:

$\displaystyle x,x+2,x+4,x+6$

According to the question,

$\displaystyle x+x+2+x+4+x+6=92$

$\Longrightarrow\displaystyle 4x+12=92$

$\displaystyle x=\dfrac{80}{4}=20$

$\therefore$ the largest number among them is

$x+6$ =

$20+6$ =

$26$

$10$ is subtracted from the product of

$x$ and

$9$ , the result is

$80$ . The number is

0%
$20$
0%
$10$
0%
$1$
0%
$9$

Explanation

According to the question,

$\displaystyle 9x-10=80$

$\Longrightarrow\displaystyle 9x=80+10$ (on transposing)

$\Longrightarrow\displaystyle 9x=90$

$\Longrightarrow\displaystyle x=\frac { 90 }{ 9 } =10$

Calculate the value of

$x$ , when

$2x+3=3x-4$ .

0%
$-7$
0%
$-\cfrac{1}{5}$
0%
$1$
0%
$\cfrac{1}{5}$
0%
$7$

Explanation

Given,

$\ 2x+3=3x-4$ .

Transposing

$x$ terms to one side, we get,

$\Rightarrow 2x-3x=-4-3$

$\Rightarrow -x=-7$

$\Rightarrow x=7$ .

Hence, option

$E$ is correct.

Solve the equation:

$\dfrac{x+2}{2x}=1$

0%
$x=2$
0%
$x=3$
0%
$x=4$
0%
$x=5$

Explanation

Given,

$\dfrac{x+2}{2x}=1$
Multiplying both sides by

$2x$ , we obtain

$x + 2 = 2x$

$2x - x = 2$

$x = 2$

$x + 1 = 23$ , calculate the value of

$3x + 3$ .

0%
$22$
0%
$46$
0%
$66$
0%
$69$

Explanation

Given that,

$x+1=23$

To find out: The value of

$3x+3$

$x+1=23$

$\therefore \ x=23-1=22$

Hence,

$3x+3 =3\left( 22 \right) +3\quad \quad [\because \ x=22]$

$\Rightarrow 66+3=69$

Hence, if

$x+1=23$ , then the value of

$3x+3$ is

$69$ .

Find the value of

$x$ which satisfies

$\displaystyle 5x+3=10x-17$ .

0%
$1$
0%
$2$
0%
$3$
0%
$4$
0%
$5$

Explanation

Given,

$5x+3=10x-17$ .

Transposing

$x$ terms to one side, we get,

$\Rightarrow 5x-10x=-17-3$

$\Rightarrow -5x=-20$

$\Rightarrow x=\dfrac{-20}{-5}$

$\Rightarrow x=4$ .

Hence, option

$D$ is correct.

$\displaystyle \frac{a-b}{b}=\frac{3}{7}$ , which of the following must also be true?

0%
$\displaystyle \frac{a}{b}=-\frac{4}{7}$
0%
$\displaystyle \frac{a}{b}=\frac{10}{7}$
0%
$\displaystyle \frac{a+b}{b}=\frac{10}{7}$
0%
$\displaystyle \frac{a-2b}{b}=-\frac{11}{7}$

Explanation

Given:

$\displaystyle \frac {a-b}{b}=\frac 37$

Separating denominators,

$\Rightarrow \displaystyle \frac ab -1=\frac 37$

$\Rightarrow \displaystyle \frac ab=\frac 37+1=\frac {10}{7}$

$\Rightarrow \dfrac {a}{b}=\dfrac {10}{7}$

Therefore, option B is correct.

Solve for

$x$ :

$\dfrac {7}{3x + 4} = \dfrac {7}{6x - 2}$

0%
$-2$
0%
$-1$
0%
$0$
0%
$2$

Explanation

Given,

$\dfrac { 7 }{ 3x+4 } =\dfrac { 7 }{ 6x-2 }$

$\Rightarrow 42x-14=21x+28$

$\Rightarrow 42x-21x=28+14$

$\Rightarrow 21x=42$

$\Rightarrow x=2$

Solve the equation:

$\dfrac{2z}{1-z}=6$

0%
$z = \dfrac{1}{4}$
0%
$z = \dfrac{3}{4}$
0%
$z = \dfrac{2}{4}$
0%
$z = \dfrac{3}{7}$

Explanation

Given,

$\dfrac{2z}{1-z}=6$
On multiplying both sides by

$1 - z$ , we get

$2z = 6(1 - z)$

$2z = 6 - 6z$

$8z = 6$

$z = \dfrac{3}{4}$

Solve the equation:

$-(x-5)=x-2$ .

0%
$x=\dfrac { 7 }{ 2 }$
0%
$x=-5$
0%
$x=3$
0%
$x=\dfrac{1}{2}$

Explanation

Given,

$-\left( x-5 \right) =x-2$

$\implies$

$-x+5=x-2$ ...[By Distribution Law].

Transposing

$x$ terms to one side, we get,

$\implies$

$x+x=5+2$

$\implies$

$2x=7$

$\implies$

$x=\dfrac { 7 }{ 2 }$ .

Hence, option

$A$ is correct.

Solve the equation:

$-2x+6=4x-2$ .

0%
no solutions
0%
$0$
0%
$3$
0%
$\dfrac {1}{2}$
0%
$\dfrac {4}{3}$

Explanation

Given,

$-2x+6=4x-2$ .

Transposing

$x$ terms to one side, we get,

$4x+2x=6+2$

$\Rightarrow 6x=8$

$\Rightarrow x=\dfrac { 8 }{ 6 }$

$\Rightarrow x=\dfrac { 4 }{ 3 }$ .

Hence, option

$E$ is correct.

Three more than twice a integer is equal to

$4$ . What is the integer?

0%
$0.2$
0%
$0.3$
0%
$0.4$
0%
$0.5$

Explanation

Let the number be

$'n'$

Twice the number gives

$'2n'$

Three more than twice the number is

$2n+3$

Given that three more than twice the number

$=$

$4$

$\Rightarrow 2n$

$+$

$3$

$=$

$4$

$\Rightarrow 2$

$\times$

$n$

$=$

$4$

$-$

$3$

$\Rightarrow n$

$=$

$\dfrac {1}{2}$

$\Rightarrow n$

$=$

$0.5$

Therefore, the number is

$0.5$

$a\neq 0$ and

$\dfrac{5}{x}=\dfrac{5+a}{x+a}$ , what is the value of

$x$ ?

0%
$-5$
0%
$-1$
0%
$5$
0%
$2$

Explanation

Given:

$a$

$\neq$

$0$ and

$\dfrac {5}{x}$

$=$

$\dfrac {5 \space + \space a}{x \space + \space a}$

To find the value of

$x$ ,

$\Rightarrow \dfrac {5}{x}$

$=$

$\dfrac {5 \space + \space a}{x \space + \space a}$

$\Rightarrow 5$

$\times$

$(x$

$+$

$a)$

$=$

$x$

$\times$

$(5$

$+$

$a)$

$\Rightarrow 5x$

$+$

$5a$

$=$

$5x$

$+$

$xa$

Get the co-efficients of

$'x'$ on one-side,

$\Rightarrow 5x$

$+$

$xa$

$-$

$5x$

$=$

$5a$

$\Rightarrow xa$

$=$

$5a$

$(a$

$\neq$

$0)$ ,

$x$

$=$

$5$

Therefore, the value of

$'x'$ is

$'5'$ .

$3$ times a number is equal to

$\dfrac{3}{2}$ , what is the number?

0%
$\dfrac{1}{3}$
0%
$\dfrac{1}{2}$
0%
$\dfrac{2}{3}$
0%
$2$

Explanation

Let the number be

$'n'$

Given that

$3$ times the number

$'n'$ is equal to

$\dfrac {3}{2}$

$\Rightarrow 3$

$\times$

$n$

$=$

$\dfrac {3}{2}$

$\Rightarrow n$

$=$

$\dfrac{\frac {3}{2}}{3}$

$\Rightarrow n$

$=$

$\dfrac {3}{6}$

$\Rightarrow n$

$=$

$\dfrac {1}{2}$

Therefore, the number is

$'$

$\dfrac {1}{2}$

$'$ .

$\dfrac { 1 }{ 7 } +\dfrac { x }{ 7 } =3$ , calculate the value of

$x$ .

0%
$20$
0%
$7$
0%
$3$
0%
$1$

Explanation

Given :-

$\frac{1}{7} + \frac{x}{7} = 3$
multiplying

$7$ to both sides

$\Rightarrow 1+ x = 21$

$\Rightarrow x = 21-1$

$\Rightarrow x = 20$

Which of the following equations does not have a solution in integers?

0%
$x + 1 = 1$
0%
$x - 1 = 3$
0%
$2x + 1 = 6$
0%
$1 - x = 5$

Explanation

Let us try and solve each of the equations given using the trial method.

(A)

$x+1=1$
We can see that for

$x= 0$

$0+1= 1$
Therefore, the value of

$x= 0$ satisfies equation in option A and

$\textbf{it is an Integer}$ .
So option A is not correct

(B)

$x-1=3$
We can see that for

$x= 4$

$4-1= 3$
Therefore, the value of

$x = 4$ satisfies equation B and

$\textbf{it is an Integer}$ .
So option B is not correct

(C)

$2 x+1=6$
Now, lets assume

$x= 1$
So,

$(2 \times 1) + 1=3$ which is not equal to

$6$
Now, lets assume

$x= 2$
So,

$(2 \times 2) + 1=5$ which is not equal to

$6$
Now, lets assume

$x= 3$
So,

$(2 \times 3) + 1=7$ which is not equal to

$6$

Here for

$x=3$ the value of LHS

$(7)$ is more than the value of RHS

$(6)$ and for any value more than

$3$ the LHS will be bigger than LHS.
Also, there is no integer between

$2$ and

$3$ which gives a value of

$2 x+1=6$
Hence we can say that

$2 x+1=6$ does not have a solution which is an Integer.
So option C is correct

(D) $1-x=5$

We can see that for $x= -4$

$1-x= 1-(-4)=1+4$

$=5$

Therefore, the value of $x= -4$ satisfies equation D and $\textbf{it is an Integer}$ .

So option D is not correct

$\textbf{Hence, Option C is the correct answer.}$

CBSE Questions for Class 8 Maths Linear Equations In One Variable Quiz 3 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

CBSE Questions for Class 8 Maths Linear Equations In One Variable Quiz 3 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

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