MCQ Questions for Class 8 Maths Linear Equations In One Variable Quiz 10

The height of pole

$A$ is

$\dfrac { 5 }{ 6 }$ of pole

$B$ , and the height of pole

$B$ is

$\dfrac { 1 }{ 2 }$ of pole

$C$ . If the height of pole

$C$ is

$72 m$ , how tall is pole

$A$ ?

0%
$60 m$
0%
$30 m$
0%
$18 m$
0%
$12 m$

Explanation

Given height of pole C is 72 m

And given height of pole B is

$\dfrac{1}{2}$ of pole C

Then height of pole B

$=\dfrac{1}{2}\times 72=36$ m

And given height of pole A is

$\dfrac{5}{6}$ of pole B

Then height of pole A

$=\dfrac{5}{6}\times 36=30$ m

$p$ is added to

$-5$ , the result is

$7$ . Hence

$3p=$ ________

0%
$6$
0%
$18$
0%
$36$
0%
$4$

Explanation

Given number is

$p$

Adding

$-5,$ we get number

$p-5$

But given that this is equal to

$7$

$\therefore p-5=7$

Add

$5$ on both sides, we get

$p-5+5=7+5$

$\Rightarrow p=12$

$\therefore 3p=3\times 12=36$

Kishore has Rs.

$p$ . Hari has Rs.

$3$ more than Kishore and Ashok has Rs.

$7$ less than Hari. Altogether the three boys have Rs.

$20$ . How much money does Kishore have?

0%
Rs. $30$
0%
Rs. $10$
0%
Rs. $7$
0%
Rs. $3$

Explanation

Given : Kishore have Rs

$p$

And Hari have Rs

$3$ more than Kishore

Then, Hari have

$p+3$ Rs

Ashok have Rs

$7$ less than Hari

Then, Ashok have

$p+3-7$ Rs

But three boys all together have Rs

$20$

$\therefore p+(p+3)+(p+3-7)=20$

$\Rightarrow p+p+3+p+3-7=20$

$\Rightarrow p+p+p=20-3-3+7$

$\Rightarrow 3p=21$

$\Rightarrow p=7$

If one-third of a two digit number exceeds its one-fourth by

$8$ , then what is the sum of the digits of the number?

0%
$6$
0%
$13$
0%
$15$
0%
$17$

Explanation

Let the two digit number be

$xy$ =

$10x+y$ .

So, accoding to the problem.

$(\dfrac {1}{3}(10x + y)$

$-(10x + y)\dfrac {1}{4}) = 8$

$(10x + y)\left (\dfrac {1}{3} - \dfrac {1}{4}\right ) = 8$

$\Rightarrow 10x + y = 96$

$\Rightarrow x = 9, y = 6$

$\therefore x + y = 15$

$1$ is subtracted from the numerator of a fraction it becomes

$\dfrac 13$ , and if

$5$ added to the denominator, the fraction becomes

$\dfrac 14$ . Which fraction shall result if

$1$ is subtracted from the numerator and

$5$ is added to the denominator?

0%
$\dfrac 5{12}$
0%
$\dfrac 7{12}$
0%
$\dfrac 1{8}$
0%
$\dfrac 23$

Explanation

Let the fraction be

$\dfrac {x}{y}$

$\dfrac {x - 1}{y} = \dfrac {1}{3}$ and

$\dfrac {x}{y + 5} = \dfrac {1}{4}$

$\Rightarrow 3(x-1)=y\; and\; 4x=y+5$ ......eqn(1)

$\Rightarrow 3x-3=y\; and\; 4x=y+5$ .

$\Rightarrow 3x-3=4x-5$ .

$\Rightarrow x=-3+5$ .

$\Rightarrow x = 2$ and susbtitute in eqn (1) we get

$y = 3$ . So, the fraction is

$\dfrac {2}{3}$ .

New fraction

$= \dfrac {2 - 1}{3 + 5} = \dfrac {1}{8}$ .

$14$ is taken away from one - fifth of a number, the result is

$20$ . The equation expressing this statement is:

0%
$\dfrac{x}{5}-14=20$
0%
$x-\left (\dfrac{14}{5}\right)=\left (\dfrac{20}{5}\right)$
0%
$x-14=\left (\dfrac{20}{5}\right)$
0%
$x+4\left (\dfrac{14}{5}\right)=20$

Explanation

Let the number be

$x$ .

As per the question, we can write,

One- fifth of a number, i.e.

$\dfrac {1}{5}x$ .

Now

$14$ is taken away from this means

$14$ is subtracted i.e.,

$\dfrac {x}{5}-14$ .

The result is equal to

$20$ .

Thus all this can be written as ,

$\dfrac {x}{5}-14=20$

If the sum of two numbers is

$84$ and their difference is

$30$ , the numbers are:

0%
$-57$ and $27$
0%
$57$ and $27$
0%
$57$ and $-27$
0%
$-57$ and $-27$

Explanation

Given, the sum of two numbers is

$84$ .

Let the smaller number be

$x$ then, the larger number will be

$84-x$ .

Given, their difference is

$30$ .

$84-x-x=30$

$\Rightarrow 84-2x=30$

$\Rightarrow 84-30=2x$

$\Rightarrow 54=2x$

$\Rightarrow x=27$

Smaller number is

$27$ .

Now, larger number is,

$84-x=84-27$

$=57$

If five times a number increased by 8 is 83, then the number is:

0%
$12$
0%
$15$
0%
$11$
0%
$2$

Explanation

Given that, five times a number increased by

$8$ is

$83$ .

To find out: The number.

Let the number be

$x$ .

As per the question, we have

$5x+8=83$

$\Rightarrow 5x=83-8$

$\Rightarrow 5x=75$

$\Rightarrow x=15$

Hence, the number is

$15$ .

$\cfrac{(x + 0.7x)}{2} = 0.85$ , then the value of

$x$ is:

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

Explanation

Given,

$\dfrac {x+0.7x}{2}=0.85$

Multiply

$2$ on both the sides,

$\Rightarrow x+0.7x=1.7$

$\Rightarrow 1.7x=1.7$

Divide both sides by

$1.7$ , we get

$\Rightarrow x=1$

The numerator of a fraction is

$6$ less than the denominator. If

$3$ is added to the numerator, the fraction is equal to

$\cfrac{2}{3}$ , find the original fraction

0%
$\dfrac 39$
0%
$\dfrac 69$
0%
$\dfrac 37$
0%
None of these

Explanation

Let denominator

$=x$

thus numerator

$=x-6$

By condition

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

$3(x-3)=2x$

$3x-9=2x$

$x=9$

So denominator

$=x=9$

thus numerator

$\Rightarrow x-6=9-6=3$

Original fraction

$=\dfrac{3}{9}$

Tank

$A$ contains

$5$ times as much water as Tank

$B$ . How much water must be transfer from Tank

$A$ to Tank

$B$ so that each tank contains

$45$ liters of water?

0%
$30\ litres$
0%
$60\ litres$
0%
$75\ litres$
0%
$45\ litres$

Explanation

Let quantity of water in tank

$B$ be

$x$ litres.
Then quantity of water in the tank

$A = 5x$ litres
Total water in both the tanks

$= x + 5x$

$= 90\ liters$

$\Rightarrow 6x = 90 \Rightarrow x = 15$

$\therefore$ Water in tank

$B = 15\ litres$

$\Rightarrow$ Water in tank

$A = 15\times 5 = 75\ litres$
So, quantity of water to be transferred from tank

$A$ to tank

$B$ , so that each tank contains

$45\ litres = 75 - 30 = 45\ litres$
Therefore,

$30\ liters$ of water should be transferred.

Half of a number is added to

$18$ , then the sum is

$46$ . The number is ______ .

0%
$92$
0%
$56$
0%
$65$
0%
$0$

Explanation

$\Rightarrow$ Let the required number be

$x$ .

$\therefore$ According to given condition,

$\Rightarrow$

$\dfrac{x}{2}+18=46$

$\Rightarrow$

$\dfrac{x}{2}=46-18$

$\Rightarrow$

$\dfrac{x}{2}=28$

$\Rightarrow$

$x=28\times 2$

$\therefore$

$x=56.$

$\therefore$ The Required number is

$56$

The total cost of

$3$ tables and

$2$ chairs is Rs.

$745$ . If one table costs Rs.

$40$ more than that of one chair, find the price of each table and chair respectively.

0%
Rs. $160$ , Rs. $120$
0%
Rs. $165$ , Rs. $125$
0%
Rs. $200$ , Rs. $160$
0%
RS. $120$ , RS. $80$

Explanation

Let cost of a chair be

$x$ , then cost of table will be

$40+x$ .

It is given that,

$3(40+x)+2x=745$

$\Rightarrow 120+5x=745$

$\Rightarrow x=\dfrac{745-120}{5}=\dfrac{625}{5}=125$

$=$ cost of chair,

Hence, cost of table

$= 40+x=40+125=165$ .

Hence, B is the right option.

$6(2a-1)+8=14$ , the value of '

$a$ ' is ______ .

0%
$-1$
0%
$3\cfrac { 1 }{ 12 }$
0%
$\cfrac{4}{10}$
0%
$1$

Explanation

Give equation is

$6(2a-1)+8=14$

To find out: The value of

$a$

Solving the given equation, we get:

$6(2a-1)+8=14$

$\Rightarrow$

$12a-6+8=14$

$\Rightarrow$

$12a+2=14$

$\Rightarrow$

$12a=14-2$

$\Rightarrow$

$12a=12$

$\Rightarrow$

$a=\dfrac{12}{12}$

$\therefore$

$a=1.$

Hence, the value of

$a$ is

$1$ .

Value of

$x$ in

$\cfrac { x }{ 4 } +\cfrac { 1 }{ 2 } =4$ is ______ .

0%
$28$
0%
$-28$
0%
$14$
0%
$-14$

Explanation

$\Rightarrow$

$\dfrac{x}{4}+\dfrac{1}{2}=4$

$\Rightarrow$

$\dfrac{x}{4}+\dfrac{1\times 2}{2\times 2}=4$

$\Rightarrow$

$\dfrac{x}{4}+\dfrac{2}{4}=4$

$\Rightarrow$

$\dfrac{x+2}{4}=4$

$\Rightarrow$

$x+2=16$

$\therefore$

$x=14$

The perimeter of a triangle is

$45\ cm$ . Length of the second side is twice the length of first side. The third side is

$5$ more than the first side. Find the length of each sides.

0%
$11,19,15$
0%
$10,20,15$
0%
$10,16,19$
0%
$13,15,17$

Explanation

Let the length of first side

$=x$

Length of second side

$=2x$

Length of third side

$=x+5$

Perimeter

$=45cm$

$x+2x+x+5=45\\ 4x=45-5\\ 4x=40\\ x=10$

So the sides are

$x=10$

$2x=2\times 10=20\\ x+5=10+5=15$

Option

$B$ is correct.

A drum of kerosene oil is

$\cfrac{3}{4}$ full. When

$15$ liters of oil is drawn from it, it is

$\cfrac{7}{12}$ full. The capacity of the drum is _____.

0%
$45$ litres
0%
$90$ litres
0%
$60$ litres
0%
Can't be determined

Explanation

Let the capacity of drum $= x$ liters.

Initially, the drum was $\dfrac{3x}{4}$ part filled. When $15$ liters oil is drawn then it becomes $\dfrac{7x}{12}$ part.

Now, according to the question,

$\dfrac{3x}{4}$ $-\dfrac{7x}{12}=15$

$\dfrac{9x - 7x}{12} = 15$ { $LCM(4,12)=12$ }

$2x = 180$

so, $x = 90$ litres. Hence, B is right option.

Form an equation of the form

$ax + b = c$ , where

$a$ ,

$b$ and

$c$ are constants, such that the solution of the equation is

$x = 4$ .

0%
2x + 5 = 15
0%
7x + 2 = 10
0%
5x + 4 = 16
0%
3x + 4 = 16

Explanation

$(A)\,\,\, 2x + 5 = 15$

$\Rightarrow$

$2x = 10$

$\Rightarrow$

$x = 5$

$(B)\,\,\, 7x + 2 = 10$

$\Rightarrow$

$7x = 8$

$\Rightarrow$

$x = \dfrac{8}{7}$

$(C)\,\,\, 5x + 4 = 16$

$\Rightarrow$

$5x = 12$

$\Rightarrow$

$x = \dfrac{12}{5}$

$(D)\,\,\, 3x + 4 = 16$

$\Rightarrow$

$3x = 12$

$\Rightarrow$

$x = 4$

$\therefore$ In option

$D$ we get,

$x$ as

$4$ so,correct answer is option

$D$

Divide Rs.

$3010$ among A, B, C respectively in such a way that A gets double of what B gets and B gets double of what C gets.

0%
Rs. $1720$ , Rs. $860$ , Rs. $430$
0%
Rs. $1640$ , Rs. $920$ , Rs. $450$
0%
Rs. $1690$ , Rs. $825$ , Rs. $495$
0%
Rs. $1800$ , Rs. $400$ , Rs. $810$

Explanation

Let the amount of money C gets be Rs. x

$\therefore$ Amount of money B gets

$=$ Rs.

$2x$
Amount of money A gets

$=Rs. 4x$
According to question,

$4x+2x+x=3010$

$\Rightarrow 7x=3010\Rightarrow x=\displaystyle\frac{3010}{7}=430$

$\therefore$ Amount of money A gets

$=Rs. (4\times 430)$

$=Rs. 1720$
Amount of money B gets

$=Rs. (2\times 430)$

$=Rs. 860$
And amount of money C get

$=Rs. 430$ .

Which of the following statement do not hold in solving the equation

$15 + 3x = 3$ ?

0%
$3x = 3 -15$
0%
$15 - 3 = -3x$
0%
$15+\dfrac {3x}{3}=3$
0%
$\dfrac {15}{3}$ + $\dfrac {3x}{3}$ = $\dfrac {3}{3}$

Explanation

The given equation is,

$\Rightarrow$

$3x=-12$ , so

$x=-4$

$\Rightarrow$

$(A)$

$3x=3-15$ , here we get

$x=-4$

$\Rightarrow$

$(B)$

$15-3=-3x$ , here we get

$x=-4$

$\Rightarrow$

$(C)$

$15+\dfrac{3x}{3}=3$ , here we get

$x=-12$

$\Rightarrow$

$(D)$

$\dfrac{15}{3}+\dfrac{3x}{3}=\dfrac{3}{3}$ , here we get

$x=-4$ .

$\therefore$ Option

$(C)$ do not hold in solving given equation.

Solve for x :

$\dfrac {6x-2}{9}+\dfrac {3x+5}{18} = \dfrac {1}{3}.$

0%
$\dfrac {1}{3}$
0%
$\dfrac {2}{3}$
0%
$\dfrac {3}{5}$
0%
$\dfrac {8}{3}$

Explanation

$\dfrac{6x-2}{9}+\dfrac{3x+5}{18}=\dfrac{1}{3}$

$\Rightarrow$

$\dfrac{12x-4+3x+5}{18}=\dfrac{1}{3}$

$\Rightarrow$

$\dfrac{15x+1}{18}=\dfrac{1}{3}$

$\Rightarrow$

$15x+1=\dfrac{1}{3}\times 18$

$\Rightarrow$

$15x+1=6$

$\Rightarrow$

$15x=5$

$\Rightarrow$

$x=\dfrac{5}{15}$

$\therefore$

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

A container is

$\dfrac{1}{8}$ full of water. After

$10$ cups of water are added, the container is

$\dfrac{3}{4}$ full. What is the volume of the container, in cups?

0%
$13\dfrac{1}{3}$
0%
$13\dfrac{1}{2}$
0%
$15$
0%
$16$
0%
$40$

Explanation

Let the container full of water be

$\dfrac {1}{8}x$ .

After

$10$ cups added, the container is

$\dfrac {3}{4}$ full.

$\dfrac {x}{8}+10=\dfrac {3x}{4}$

$\Rightarrow \dfrac {6x}{8}-\dfrac {x}{8}=10$

$\Rightarrow 5x=80$

$\Rightarrow x=16$

Therefore, volume of container in cups is

$16$ .

A boy was asked to multiply a given number by

$\dfrac{8}{17}.$ Instead, he divided the given number by

$\dfrac{8}{17}$ and got the result

$225$ more than what he should have got if he had multiplied the number by

$\dfrac{8}{17}.$ The given number was

0%
$8$
0%
$17$
0%
$64$
0%
$136$

Explanation

Let the number be

$x$

Result when calculated wrongly

$=\dfrac{17x}{8} \quad \quad ...(I)$

Result when calculated correctly

$=\dfrac{8x}{17}\quad \quad ...(II)$

$(I)-(II)=225$

$\Rightarrow \dfrac{17x}{8}-\dfrac{8x}{17}=225$

$\Rightarrow \dfrac{289-64}{136}x=225$

$\Rightarrow \dfrac{225}{136}x=225$

$\Rightarrow x=136$

$\dfrac {9}{5}$ of a number is 45, what is

$\dfrac {1}{5}$ of the same number?

0%
$5$
0%
$25$
0%
$30$
0%
$81$

Explanation

Let the number be

$x$ .

$\Rightarrow$ Then, according to the given condition,

$\Rightarrow$

$\dfrac{9}{5}\,\,of\,\,x=45$

$\Rightarrow$

$\dfrac{9}{5}\times x=45$

$\Rightarrow$

$x=45\times \dfrac{5}{9}$

$\Rightarrow$

$x=25$

$\therefore$ Number is

$25$ .

$\therefore$

$\dfrac{1}{5}$ of the number =

$\dfrac{1}{5}\times 25=5$

$\dfrac {2}{5}(5x+1)+\dfrac {3}{5} = 1$ , then what is the value of 5 x?

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

Explanation

$\dfrac{2}{5}(5x+1)+\dfrac{3}{5}=1$

$\Rightarrow$

$2x+\dfrac{2}{5}+\dfrac{3}{5}=1$

$\Rightarrow$

$2x+\dfrac{5}{5}=1$

$\Rightarrow$

$2x+1=1$

$\Rightarrow$

$2x=0$

$\therefore$

$x=0$

$\therefore$

$x=0$ then,

$5x=0$

Arpan cut a cake in two equal parts and one of these two pieces is again cut into equal parts. Each small part is of

$20$ grams. If there are seven parts of the whole cake, what was the weight of the original cake?

0%
$140$ grams
0%
$280$ grams
0%
$240$ grams
0%
$300$ grams

Explanation

Let the weight of the cake is

$x$ grams.

If he cuts it into two equal parts, then the weight of one part will be

$\dfrac{x}{2}$ .

Again he cuts one part into equal parts.

Since total seven parts are there, so he has cut one piece into

$6$ equal parts.

The weight of the small part is

$\dfrac{x}{2\times 6}$ and it is given that the weight of small part is

$20$ grams.

$\Rightarrow 20= \dfrac{x}{12}$

$\Rightarrow x=240$

Weight of cake is

$240$ grams.

The denominator of a rational number is greater than its numerator byIf 3 is subtracted from the numerator and 2 is added to the denominator, the new number becomes

$\displaystyle \frac{1}{5}$ . Then the original fraction was _________.

0%
$\displaystyle \frac{7}{11}$
0%
$\displaystyle \frac{3}{5}$
0%
$\displaystyle \frac{5}{8}$
0%
$\displaystyle \frac{4}{7}$

Explanation

Let, the numerator of rational fraction be

$x$

$\therefore$ the denomiantor

$=x+3$

A.T.Q,

$\dfrac{x-3}{x+3+2}=\dfrac{1}{5}$

$\implies \dfrac{x-3}{x+5}=\dfrac{1}{5}$

$\implies 5x-15=x+5$

$\implies 5x-x=5+15$

$\implies 4x=20$

$\implies x=\dfrac{20}{4}=5$

$\therefore$ The original rational fraction is

$\dfrac{5}{5+3}=\dfrac{5}{8}$

The sum of 4 consecutive numbers isThen the last number is

0%
22
0%
27
0%
36
0%
25

Explanation

solution:

let the first number be

$x$ then all 4 consecutive numbers will be

$x , x+1,x+2,x+3$

sum will be

$x + x+1 + x+2 + x+3 = 4x +6$

$4x +6 = 94$

$4x = 88$

$x = 22$

then , the last number is

$x+3 = 25$

hence the correct opt: D

The two digit number having the unit digit

$x + 5$ and ten's digit

$x - 5$ is _______.

0%
$11x + 55$
0%
$11x - 45$
0%
$9x - 55$
0%
$2x + 10$

Explanation

In a two digit number,
The unit digit

$= x + 5$
Ten's digit

$= x - 5$

$\therefore$ The number

$= 1(x + 5) + 10 (x - 5)$

$= x + 5 + 10x - 50$

$= 11x - 45$ .

$A$ has certain amount in his account. He gives half of this to his eldest son and one third of the remaining to his youngest son. The amount left with him now is

0%
$\cfrac { 1 }{ 3 }$ of the original
0%
$\cfrac { 2 }{ 5 }$ of the original
0%
$\cfrac { 3 }{ 4 }$ of the original
0%
$\cfrac { 1 }{ 6 }$ of the original

Explanation

Let

$A$ have

$x$ amount.

Hence, according to the problem,

$x-\cfrac { x }{ 2 } -\left(\cfrac { 1 }{ 3 }\right)\cfrac { 1 }{ 2 } x$

$\Rightarrow \cfrac { 1 }{ 2 } x-\cfrac16 x$

$\Rightarrow \cfrac { 2 }{ 6 } x$

$\Rightarrow \cfrac { 1 }{ 3 } x$

It is the one-third of the original amount

$x$ .

Hence, the amount left with him now is

$\cfrac { 1 }{ 3 }$ of the original.

CBSE Questions for Class 8 Maths Linear Equations In One Variable Quiz 10 - 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 10 - MCQExams.com

0:0:1

Practice Class 8 Maths Quiz Questions and Answers

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