MCQ Questions for Class 8 Maths Linear Equations In One Variable Quiz 1

Two numbers are in the ratio of

$6:3$ . If they differ by

$18$ , what are the numbers?

0%
$17, 26$
0%
$42, 76$
0%
$15, 13$
0%
$36, 18$

Explanation

Let the two numbers be

$6x$ and

$3x$ .

They are in the ratio of

$6:3$ .

By the given condition, we have

$6x -3x =18$

$\therefore 3x = 18$

$\therefore x =6$

Therefore, the numbers would be

$6x=6\times 6=36$ and

$3x=3\times 6=18$ .

So, the numbers are

$36$ and

$18$ .

Solve the following equation:

$\cfrac{3y+4}{2-6y}=\cfrac{-2}{5}$

0%
$2$
0%
$-4$
0%
$4$
0%
$-8$

Explanation

Given,

$\dfrac{3y+4}{2-6y}=\dfrac{-2}{5}$

On cross multiplying, we get

$5(3y+4)=-2(2-6y)$

$\Rightarrow 15y+20=-4+12y$

$\Rightarrow 3y=-24$

$\Rightarrow y=-8$

Solve the following equation:

$\cfrac{8x-3}{3x}=2$

0%
$x = \cfrac{5}{2}$
0%
$x = \cfrac{5}{6}$
0%
$x = \cfrac{9}{8}$
0%
$x = \cfrac{3}{2}$

Explanation

Given,

$\dfrac{8x-3}{3x}=2$

Multiply

$3x$ on both the sides, we get

$8x-3=6x$

$8x-6x=3$

$2x=3$

$\therefore x=\dfrac{3}{2}$

Solve the following linear equation:

$m-\cfrac{m-1}{2} = 1-\cfrac{m-2}{3}$

0%
$m = \cfrac{7}{5}$
0%
$m = \cfrac{2}{3}$
0%
$m = \cfrac{10}{3}$
0%
$m = \cfrac{3}{8}$

Explanation

Given,

$m - \cfrac { (m-1) }{ 2 } = 1-\cfrac { (m-2) }{ 3 }$

L.C.M. of

$2$ and

$3$ is

$6$

Multiplying with 6 on both sides of the equation, we get

$6m -3(m-1) = 6 - 2(m-2)$

$6m - 3m +3 = 6 - 2m +4$

$5m = 7$

$m = \cfrac { 7 }{ 5 }$

Solve the following equation:

$\cfrac{z}{z+15}=\cfrac{4}{9}$

0%
$2$
0%
$9$
0%
$12$
0%
$16$

Explanation

Given,

$\dfrac{z}{z+15}=\dfrac{4}{9}$

$9z=4(z+15)$

$9z=4z+60$

Subtract

$4z$ from both the sides, we get

$9z-4z=4z+60-4z$

$5z=60$

$\therefore z=12$

Which of the following is the solution of the equation

$\displaystyle \frac{7y+4}{y+2}=\frac{-4}{3}$ ?

0%
$\displaystyle y = -\frac{4}{5}$
0%
$\displaystyle y = \frac{4}{5}$
0%
$\displaystyle y = -\frac{5}{4}$
0%
$\displaystyle y = \frac{5}{4}$

Explanation

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

$21y+12=-4y-8$

$25y=-20$

$y=\dfrac{-20}{25} = \dfrac{-4}{5}$

Solve the following equation:

$\cfrac{9x}{7-6x}=15$

0%
$x = \cfrac{25}{72}$
0%
$x = \cfrac{35}{33}$
0%
$x = \cfrac{45}{39}$
0%
$x = \cfrac{22}{45}$

Explanation

Given,

$\dfrac{9x}{7-6x}=15$

$9x=105-90x$

Add

$90x$ on both the sides, we get

$9x+90x=105-90x+90x$

$99x=105$

$\therefore x=\dfrac{105}{99}=\dfrac{35}{33}$

Solve the following linear equation. If

$\cfrac{3t-2}{4}-\cfrac{2t+3}{3} = \cfrac{2}{3}-t$ , then

$t$ is equal to

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

Explanation

Given

$\cfrac { 3t-2 }{ 4 } -\cfrac { (2t+3) }{ 3 } = \cfrac { 2 }{ 3 } - t$

L.C.M. of

$4,3,3$ is

$12$

Multiplying with 12 on both sides of the equation, we get

$\Rightarrow 3(3t-2) - 4(2t+3) = 8 - 12t$

$\Rightarrow 9t - 6 -(8t +12) = 8 -12t$

$\Rightarrow t -18 = 8 -12t$

$\Rightarrow 13 t = 26$

$\Rightarrow t = 2$

Linear equation in one variable is :

0%
$2x = y$
0%
$y^2 = 3y + 5$
0%
$4x - y = 5$
0%
$3t + 5 = 9t- 7$

Explanation

${\textbf{Step - 1 : State the definition of linear equation with one variable}}{\text{.}}$

${\text{As per the definition , Linear equation with one variable is that equation which have,}}$

${\text{literally one variable with highest power of 1}}{\text{.}}$

${\textbf{Step - 2 : Choose the correct option using the definition}}{\text{.}}$

${\text{The first equation is 2x = y}}{\text{.Where x and y are variables}}{\text{. So there are two variable}}{\text{.}}$

$\Rightarrow {\text{This is not a linear equation in one variable}}.$

${\text{The second option is, }}{{\text{y}}^2}{\text{ = 3y + 5}}{\text{. Here there is only one variable.}}\\{\text{But the highest power is 2}}{\text{. }}$

$\Rightarrow {\text{This is not a linear equation in one variable}}.$

${\text{In third option 4x}} - {\text{y = 5}}{\text{. Here there are two variables , x and y}}{\text{.}}$

$\Rightarrow {\text{This is not a linear equation in one variable}}.$

${\text{In fourth option 3t + 5 = 9t}} - 7.{\text{ Here there is only one variable and the $\\$highest power is also 1}}{\text{. }}$

$\Rightarrow {\text{This is a linear equation in one variable}}.$

${\textbf{Thus, the linear equation with one variable is option (D) 3t + 5 = 9t}} - 7.$

Three consecutive integers add up to

$51$ . What are these integers?

0%
$15, 16, 17$
0%
$16, 17, 18$
0%
$10, 11, 12$
0%
$18, 19, 20$

Explanation

Let the three consecutive numbers be

$x, x+1, x+2$

Sum of the numbers is

$51$ .

$\therefore x + x+ 1+ x+2 = 51$

$3x + 3 = 51$

$3x = 48$

$x = 16$

The numbers would be

$x=16$ ,

$x+1=16+1=17$ ,

$x+2=16+2=18$

So, the numbers are

$16, 17, 18$ .

A student divided a number by two when he was required to multiply it by

$2$ . The answer he got was

$2$ . The correct answer should have been

0%
$8$
0%
$1$ 2
0%
$6$
0%
$4$

Explanation

Let the number be

$x$

As per the given condition, we have:

$\cfrac { x }{ 2 } =2$

$\Rightarrow x=4$

$\therefore$ the number is

$4$ .

This number was supposed to be multiplied by

$2$ .

Hence, the correct answer is

$4\times 2=8$ .

Solve the following:

$3\left ( 3x-4 \right )-2\:\left ( 4x-5 \right )=\:6$

0%
$4$
0%
$8$
0%
$12$
0%
$6$

Explanation

Given,

$3(3x-4)-2(4x-5)=6$

$\Rightarrow 9x-12-8x+10=6$

$\Rightarrow x-2=6$

Add

$2$ on both the sides, we get

$x-2+2=6+2$

$\Rightarrow x=8$

$\displaystyle \frac{1}{3}\, =\, \frac {\sqrt{x}}{2}$ , then value of

$x$ is

0%
$\displaystyle \frac{6}{7}$
0%
$\displaystyle \frac{7}{2}$
0%
$\displaystyle \frac{4}{9}$
0%
$\displaystyle \frac{12}{7}$

Explanation

$\dfrac { 1 }{ 3 } = \dfrac { \sqrt { x } }{ 2 } \\ \sqrt { x } =\dfrac { 2 }{ 3 }$

On squaring both the sides.

$\Rightarrow x = \dfrac { 4 }{ 9 }$

Solve for

$x$ :

$\sqrt[3]{x}\,- 4\, =\, 0$

0%
$64$
0%
$36$
0%
$81$
0%
$49$

Explanation

$\sqrt [ 3 ]{ x } - 4 = 0\\ \sqrt [ 3 ]{ x } = 4$

Cubing on both sides

$x = { 4 }^{ 3 } = 64$

Solve for

$x$ :

$5\, -\, \sqrt{x}\, =\, 0$

0%
$25$
0%
$54$
0%
$15$
0%
$20$

Explanation

Given,

$5 - \sqrt { x } = 0$

$\sqrt { x } = 5$

Squaring on both sides, we get

$x = { 5 }^{ 2 } = 25$

Twice a number decreased by

$15$ , equals

$25$ . Find the number.

0%
$25$
0%
$20$
0%
$18$
0%
$21$

Explanation

Let the number be

$x$

In the question, it is given that twice a number, i.e.

$2x$ and decreased by

$15$ .

That means,

$2x-15$ . This is further equals to

$2x-15=25$

$2x=40$

$x=20$

Therefore, the number is

$20$ .

The sum of seven consecutive odd integer is

$133$ . The least integer is

0%
$17$
0%
$19$
0%
$11$
0%
$13$

Explanation

Let the seven consecutive odd integers be

$(2x+1),\:(2x+3),\:(2x+5),\:(2x+7),\:(2x+9),\:(2x+11),\:(2x+13)$
Then,

$(2x+1)+(2x+3)+(2x+5)+(2x+7)+(2x+9)+(2x+11)+(2x+13)=133$

$\Rightarrow 14x+49=133$

$\Rightarrow \displaystyle 14x=133-49=84$

$\Rightarrow x=\dfrac{84}{14}=6$

Hence, the least odd integer is

$=2x+1=2\times 6+1=13$ .

Mohan gets

$3$ marks for each correct sum and loses

$2$ marks for each wrong sum. He attempts

$30$ sums and obtains

$40$ marks. The number of sums solved correctly is

0%
$10$
0%
$15$
0%
$20$
0%
$25$

Explanation

It is given that, Mohan gets

$3$ marks for each correct answer and loses

$2$ marks for each wrong answer.

He attempted

$30$ sums and got

$40$ marks.

Let the number of sums solved correctly be

$n$ .
Then number of wrong sums

$= 30 - n$

$\therefore 3n-2.(30-n) = 40$

$\Rightarrow 3n-60+2n=40$

$\Rightarrow 5n=100$

$\Rightarrow n=20$

Two distinct natural numbers are such that the one number is twice the other number and their sum is

$6$ . The two numbers are ____

0%
$8$ and $3$
0%
$2$ and $7$
0%
$4$ and $2$
0%
$6$ and $3$

Explanation

Let the number be

$x$ .
Then, according to question, we have

$x+2x=6$

$\Rightarrow 3x=6$

$\Rightarrow x=2$

$\Rightarrow 2x=4$
Therefore, the two numbers are

$2$ and

$4$ .

You take

$5$ less than thrice a number and add

$7$ . The result is

$14$ . The number is

0%
$5$
0%
$4$
0%
$6$
0%
$2$

Explanation

Let the number be

$x$ .

Then according to the given condition,

$(3x-5)+7=14$

$\Rightarrow 3x+2=14$

$\Rightarrow \displaystyle 3x=12$

$\Rightarrow x=\dfrac{12}{3}$

$\Rightarrow x=4$
Hence, the required number is

$4$ .

The difference between

$\displaystyle \frac{3}{4}$ of a line and

$\displaystyle \frac{2}{5}$ of the same line is

$28$ cm. Find the length of the line.

0%
$60$ cm
0%
$90$ cm
0%
$20$ cm
0%
$80$ cm

Explanation

Let the length of the line be

$x$ .

$\cfrac { 3 }{ 4 } x - \cfrac { 2 }{ 5 } x = 28$

L.C.M. of

$4$ and

$5$ is

$20$ .

$\therefore 15x - 8x = 28(20)$

$\therefore 7x = 28(20)$

$\therefore x = 80$

Therefore, length of the line is

$80$ cm.

On a three day hiking trip you hiked the same distance (in km) on both first and second days.You hiked

$5$ km on the third day. If you hiked a total of

$19$ km, how many kilometer were left to hike after the first day?

0%
$11$ km
0%
$12$ km
0%
$9$ km
0%
$10$ km

Explanation

Let the distance hiked on first two days be

$x$ km each.

Then,

$x+x+5=19$

$\Rightarrow 2x+5=19$

$\Rightarrow \displaystyle 2x=19-5=14$

$\Rightarrow x=\dfrac{14}{2}=7$

$\therefore$ distance left to be hiked after the first day

$=(7+5)=12$ km.

Find two consecutive even whole numbers, whose sum is

$38$ .

0%
$10$ and $14$
0%
$18$ and $20$
0%
$16$ and $34$
0%
$12$ and $29$

Explanation

Let the two consecutive numbers be

$x$ and

$x+2$ .

Given their sum is

$38$ .

Therefore,

$x+x+2 = 38$

$\Rightarrow 2x = 36$

$\Rightarrow x = 18$

Other number is

$x+2=18+2=20$ .

So, the numbers are

$18$ and

$20$ .

Solve:

$\displaystyle \frac{7x\, -\, 1}{4}\, - \, \frac{1}{3}(2x\, -\, \frac{1\, -\,x}{2})\, =\, {6}\frac{1}{3}$ .

0%
7
0%
5
0%
-7
0%
3

Explanation

$\displaystyle \frac{7x\, -\, 1}{4}\, - \, \frac{1}{3}(2x\, -\, \frac{1\, -\,x}{2})\, =\, {6}\frac{1}{3}$

$=>\dfrac{7x-1}{4}-\dfrac{2x}{3}+\dfrac{1-x}{6}-\dfrac{19}{3}=0$

$=>\dfrac{21x-3-8x+2-2x-76}{12}=0$

$=>11x-77=0$

$=>x=\dfrac{77}{11}$

$=>x=7$

You are decorating a gift pack with

$15$ flowers. You want an equal number of flowers in each of the

$3$ rows on the gift pack. Which equation would you use to find the number of flowers,

$r$ in each row?

0%
$r+3=15$
0%
$15-r=3$
0%
$3r=15$
0%
$\displaystyle \frac{3}{r}=15$

Explanation

We have, number of rows

$\times$ Number of flowers in each row

$=$ Total number of flowers.

$\Rightarrow 3\times r=15$

$\Rightarrow 3r=15$

Hence, option C is correct.

If a number with

$12$ has the same ratio as

$8$ having with

$10$ , then the number is :

0%
$15$
0%
$9.6$
0%
$7.5$
0%
$10$

Explanation

Let the required number be

$x$ .

$\therefore$

$\displaystyle{\frac{x}{12} = \frac{8}{10} \Rightarrow x = 9.6}$
Hence, option 'B' is correct.

Which of the following equation is not a linear equation?

0%
$2x + 3 = 7x - 2$
0%
$\displaystyle \frac{2}{3}x + 5 = 3x - 4$
0%
$\displaystyle x^2 + 3 = 5x - 3$
0%
$\displaystyle (x - 2)^2 = x^2 + 8$

Explanation

An algebraic equation is said to be linear if the variable or variables in the equation are of the first degree.

In option

$A$ , all the

$x$ terms have a degree

$1$ . So, it is a linear equation.

In option

$B$ , all the

$x$ terms have a degree

$1$ . So, it is a linear equation.

In option

$C$ ,

$x^{2}+3=5x-3, x$ is raised to the power

$2$ .
So,

$x^{2}+3=5x-3$ is not a linear equation.

In option

$D$ ,

$(x-2)^2=x^2+8$

$\Rightarrow x^2-4x+4=x^2+8$

$\Rightarrow -4x=4$
i.e, the

$x$ term has a degree

$1$ . So, it is a linear equation.

Hence, Option

$C$ is the correct option.

The three even consecutive integers whose sum is 90 are

0%
$26 , 30, 34$
0%
$24, 32, 34$
0%
$24, 28, 38$
0%
$28, 30, 32$

Explanation

Let, the three even consecutive integers be

$x, x+2, x+4$

A.T.Q,

$x+x+2+x+4=90$

$\implies 3x+6=90$

$\implies 3x=90-6=84$

$\implies x=\dfrac{84}{3}=28$

$\therefore$ The three even consecutive integers are

$28,30, 32$

A is 2 years older than B who is twice as old as C If the total of the ages of A, B and C be 27 years then how old is B?

0%
7 years
0%
8 years
0%
9 years
0%
10 years

Explanation

Let C' age

$= x$ years Then
B's age

$= 2x$ years
As age

$(2 x + 2)$ years
Given

$(2x + 2 )$ years

$\displaystyle \Rightarrow 5x+2=27\Rightarrow 5x=25\Rightarrow x=5$

$\displaystyle \therefore$ B's age

$= (2 \times 5)$ years

$=10$ years.

The sum of the ages of 5 children born at intervals of 3 years each is 50 years What is the age of the youngest child?

0%
4 years
0%
8 years
0%
10 years
0%
6 years

Explanation

Let the age of the youngest child be

$x$ years
Then

$x+(x+3)+(x+6)+(x+9)+(x+12)=50$

$\displaystyle \Rightarrow 5x+30=50\Rightarrow 5x=20\Rightarrow x=4$

$\displaystyle \therefore$ Age of the youngest child is

$4$ years.

CBSE Questions for Class 8 Maths Linear Equations In One Variable Quiz 1 - 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 1 - MCQExams.com

0:0:1

Practice Class 8 Maths Quiz Questions and Answers

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