MCQ Questions for Class 8 Maths Linear Equations In One Variable Quiz 11

A number when added to its half gives

$36$ . Find the number.

0%
$24$
0%
$28$
0%
$20$
0%
$18$

Explanation

Let the number be

$x$

$\Rightarrow x+\cfrac { x }{ 2 } =36\\ \Rightarrow \cfrac { 3x }{ 2 } =36\\ \Rightarrow x=24$

The difference between

$\left(\displaystyle\frac{3}{5}\right)^{th}$ of

$\left(\displaystyle\frac{2}{3}\right)^{rd}$ of a number and

$\left(\displaystyle\frac{2}{5}\right)^{th}$ of

$\left(\displaystyle\frac{1}{4}\right)^{th}$ of the same number is

$288$ . What is the number?

0%
$960$
0%
$850$
0%
$895$
0%
$995$

Explanation

$\Rightarrow$ Let the number be

$x$ .

According to the given condition,

$\dfrac{3}{5}\times \dfrac{2}{3}\times x-\dfrac{2}{5}\times \dfrac{1}{4}\times x=288$

$\Rightarrow$

$\dfrac{2}{5}x-\dfrac{2}{20}x=288$

$\Rightarrow$

$\dfrac{8x-2x}{20}=288$

$\Rightarrow$

$6x=5760$

$\therefore$

$x=\dfrac{5760}{6}=960$

$\therefore$ The required number is

$960$

There are four containers that are arranged in the ascending order of their heights. If the height of the smallest container given in the figure is expressed as

$\dfrac{7}{25} x = 10.5$ cm. Find the height of the largest container.

0%
$37.5$ cm
0%
$42.5$ cm
0%
$30.5$ cm
0%
$45.5$ cm

Explanation

$\Rightarrow$ We can see

$x$ is height of largest container.

$\Rightarrow$

$\dfrac{7}{25}x=10.5$

$\Rightarrow$

$7x=262.5$

$\therefore$

$x=\dfrac{262.5}{7}=37.5\,cm$

$\therefore$ Height of the largest container is

$37.5\,cm$ .

A sum of money is divided among Peter, Anita and Janet in the ratio

$13:12:7$ . Calculate how much Anita gets, if the amount Peter gets more than Janet is Rs.

$360$ ?

0%
Rs. $180$
0%
Rs. $640$
0%
Rs. $720$
0%
Rs. $480$

Two numbers are in the ratio

$\displaystyle1\frac{1}{2} : 2\frac{2}{3}$ . But if each number is increased by

$15$ . then the ratio becomes

$\displaystyle1\frac{2}{3} : 2 \frac{1}{2}$ . The numbers are:

0%
$27, 42$
0%
$36, 48$
0%
$27, 48$
0%
$27, 36$

Explanation

The ratio of two numbers

$1\dfrac{1}{2}:2\dfrac{2}{3}=\dfrac{3}{2}:\dfrac{8}{3}$

$=9:16$

Let the two numbers are

$9x$ and

$16x$ .

According to the given condition,

$\Rightarrow$

$\dfrac{9x+15}{16x+15}=1\dfrac{2}{3}:2\dfrac{1}{2}=\dfrac{5}{3}:\dfrac{5}{2}$

$\Rightarrow$

$\dfrac{9x+15}{16x+15}=\dfrac{2}{3}$

$\Rightarrow$

$3\times (9x+15)=2\times (16x+15)$

$\Rightarrow$

$27x+45=32x+30$

$\Rightarrow$

$5x=15$

$\therefore$

$x=3$

$\therefore$ The numbers are

$9x=9\times 3=27$ and

$16x=16\times 3=48$ .

If a boat goes

$7$ km upstream in

$42$ minutes and the speed of the stream is

$3$ kmph, then the speed of boat in still water is:

0%
$4.2$ km/hr
0%
$9$ km/hr
0%
$13$ km/hr
0%
$21$ km/hr

Explanation

As the boat goes 7 km upstream in 42 minutes,

Rate upstream

$= \cfrac{7}{42} \times 60 = 10 \; kmph$

Speed of stream

$= 3 \; kmph \; \left\{ Given \right\}$

Let speed of boat in still water be

$x \; km/hr$ .

As we know that,

speed upstream = speed of boat in still water - speed of stream

$\therefore$ speed upstream

$= \left( x - 3 \right) kmph$

$\Rightarrow \; \left( x - 3 \right) = 10$

$\Rightarrow \; x = 10 + 3 = 13 \; kmph$

Hence, speed of boat in still water is 13 kmph.

Hamid has three boxes of different fruits. Box

$A$ weights

$2\dfrac{1}{2}\ kg$ more than Box

$B$ and Box

$C$ weights

$10\dfrac{1}{4}\ kg$ more than Box

$B$ . The total weight of the boxes is

$48\dfrac{3}{4}$ . How many

$kgs$ does Box A weights?

0%
$15\dfrac {1}{2}\ kg$
0%
$14 \dfrac {1}{2}\ kg$
0%
$13\dfrac {1}{2}\ kg$
0%
$none\ of\ these$

Explanation

Let weight of

$B = x$

$\Rightarrow$ Weight of

$A =x+2\dfrac12$

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

$\Rightarrow$ Weight of

$C =x+10\dfrac 14$

$= x +{\dfrac{{41}}{4}}$

Total weight of bottles is

$48\dfrac 34=\dfrac {195}{4}$

$\Rightarrow x + x + x + {\dfrac{5}{2}} + {\dfrac{{41}}{4}} =\dfrac{195}{4}$

$\Rightarrow 3x+\dfrac{41+10}{4}=\dfrac{{195}}{4}$

$\Rightarrow 3x+\dfrac{51}{4}=\dfrac{{195}}{4}$

$\Rightarrow 3x=\dfrac{{195}}{4}- \dfrac{51}{4}$

$\Rightarrow 3x=\dfrac{{144}}{4}$

$\Rightarrow 3x=36$

$\Rightarrow x=12$

$\therefore$ Weight of

$B=12\ kg$

And weight of

$A = 12 + \dfrac{5}{2} = 14.5\ kg$

A crate of mangoes contains one bruised mango for every

$30$ mangoes in the crate. If

$3$ out of every

$4$ bruised mango are considered unsaleable, and there are

$12$ unsaleable mangoes in the crate, how many mangoes are there in the crate?

0%
$480$
0%
$500$
0%
$420$
0%
$520$

Explanation

Let

$x$ be the total number of mangoes in the crate, Bruised mongo

$=$

$\cfrac{1}{{30}}x$

unsaleable mangoes

$\cfrac{3}{4}\left( {\cfrac{1}{{30}}x} \right)$

$\Rightarrow$

$\cfrac{1}{{40}}x\, = 12$

$\Rightarrow$

$x = 12 \times 40$

$\Rightarrow$

$x= \,480$

$\dfrac{2x-3}{2}-\dfrac{(x+1)}{3}=\dfrac{3x-8}{4}$

0%
$x=4$
0%
$x=-2$
0%
$x=2$
0%
$x=-4$

Explanation

$\cfrac { 2x-3 }{ 2 } -\cfrac { x+1 }{ 3 } =\cfrac { 3x-8 }{ 4 }$

$\Longrightarrow (6x-9-2x-2)\times 4=(3x-8)\times 6$

$\Longrightarrow16x-44=18x-48$

$\Longrightarrow 2x=4$

$\Longrightarrow x=2$

$t = x+2$ , find the value of x .If

$2t-7 +\dfrac{3(t-1)}{2}=3$

0%
$\frac{23}{7}$
0%
$\frac{3}{7}$
0%
$\frac{9}{7}$
0%
$\frac{37}{7}$

Explanation

$(2t-7)+(\dfrac{3t-3}{2})=3\\(\dfrac{4t-14+3t-3}{2})=3\\7t-17=6\\\therefore 7t=6+17\\t=(\dfrac{23}{7})\\then x+2=(\dfrac{23}{7})\\\therefore x=(\dfrac{23}{7})-2\\=(\dfrac{23-14}{7})\\=(\dfrac{9}{7})$

Find the two consecutive odd numbers whose sum is 76.

0%
$77$ and $39$
0%
$37$ and $39$
0%
$37$ and $99$
0%
$56$ and $39$

Explanation

Let two consecutive odd numbers be

$x$ and

$x+2$

According to question

$x+x+2=76$

$2x+2=76$

$2x=74$

$x=37$

$x+2=37+2=39$

So, the numbers are

$37$ and

$39$

The sum of

$2$ consecutive numbers is

$13$ . Find the numbers.

0%
$6$ and $7$ .
0%
$5$ and $7$ .
0%
$6$ and $5$ .
0%
$7$ and $9$ .

Explanation

Let the two consecutive numbers be

$x$ and

$x+1$

According to the question

$x+x+1=13$

$2x+1=13$

$2x=13-1$

$2x=12$

$x=\dfrac{12}{2}$

$x=6$

$x+1=6+1=7$

Therefore the two consecutive numbers are

$6$ and

$7$ .

There is

$34000$ rupees in a beg in the denomination of

$Rs 50,Rs 100,Rs 500$ and

$Rs 1000$ currencynotes. If the ratio of the number of these notes is

$2:3:4:1$ then find the number of notes of

$500$ denomination.

0%
$20$
0%
$40$
0%
$30$
0%
$50$

Explanation

Let the number of notes be denoted by

$'x'$

We have given ratio =

$2:3:4:1$

$\therefore$ Acc to question:-

$34,000=50(2x)+100(3x)+500(4x)+1000(x)$

$=100x+300x+2000x+1000x$

$34,000=3400 x$

$\dfrac{34,000}{3400}=x$

$x=10$ notes

$\therefore$ No. of

$500$ denominator notes

$=4\times 10$

$\Rightarrow 40$

option

$B$ is correct.

A man travelled two-fifth of his journey by train, one-third by bus, one-fourth by car and the remaining

$3\ km$ on foot. What is the length of his total journey?

0%
$180$
0%
$150$
0%
$120$
0%
$140$

Explanation

Let the length of his total journey be

$x$

So,

$\dfrac{2}{5}x+\dfrac{1}{3}x+\dfrac{1}{4}x+3=x$

solving

$x\left(\dfrac{2}{5}+\dfrac{1}{3}+\dfrac{1}{4}\right)-x=-3$

$x\left(\dfrac{2}{5}+\dfrac{1}{3}+\dfrac{1}{4}-1\right)=-3$

$x\left(\dfrac{2\times 12+20+1\times 15-60}{60}\right)=-3$

$x\left(\dfrac{24+20+15-60}{60}\right)=-3$

$x\left(\dfrac{-1}{60}\right)=-3$

$\dfrac{x}{60}=3$

$\therefore x=180\ km$

The sum of three consecutive even numbers is

$276$ . Find the numbers.

0%
$96,92,94$
0%
$91,92,94$
0%
$90,94,98$
0%
$90,92,94$

Explanation

Solution:- (D)

$90, 92, 94$

Let the numbers be

$2a, 2a + 2, 2a + 4$ .

Now according to the question-

$2a + \left( 2a + 2 \right) + \left(2a + 4 \right) = 276$

$2a + 2a + 2 + 2a + 4 = 276$

$6a + 6 = 276$

$a + 1 = \cfrac{276}{6}$

$a = 46 - 1 = 45$

Therefore, the numbers are-

$2a = 2 \times 45 = 90$

$2a + 2 = 2 \times 45 + 2 = 92$

$2a + 4 = 2 \times 45 + 4 = 94$

Hence the numbers are

$90, 92$ and

$94$ .

If twelve less than

$3$ times a number is

$27$ , then the number is ___

0%
$29$
0%
$39$
0%
$65$
0%
$13$

Explanation

Let

$x$ be a number.

$\Rightarrow\,3x-12=27$

$\Rightarrow\,3x=27+12$

$\Rightarrow\,3x=39$

$\Rightarrow\,x=\dfrac{39}{3}=13$

$\therefore\,x=13$

If one fourth, half and one third of the number are added to the number itself, then result is equal to

$25$ . Find the number.

0%
$10$
0%
$11$
0%
$12$
0%
$13$

$\dfrac{1}{8}+\dfrac{1}{9}=\dfrac{x}{10}$ , then

$x$ is equal to

0%
$100$
0%
$1.36$
0%
$170$
0%
$2.36$

Explanation

$\dfrac{1}{8} + \dfrac{1}{9} = \dfrac{x}{10}$

$\implies 10\left(\dfrac{1}{8} + \dfrac{1}{9} \right) = x$

$\implies 10 \left(\dfrac{9 + 8}{72} \right) = x$

$\implies \dfrac{170}{72} = x$

$\implies x = 2.36$

If the number exceeds

$\dfrac{2}{3}$ of itself by

$7$ ,then

0%
$21$
0%
$44$
0%
$2$
0%
$7$

Explanation

Let the no. be

$x$

$2x/3=x-7\\\implies x=21$

Find the solution to the equation

$\dfrac{4x}{5}-\dfrac{7}{4}=\dfrac{x}{5}+\dfrac{x}{4}$ .

0%
$5$
0%
$6$
0%
$7$
0%
$0$

Explanation

Given,

$\dfrac{4x}{5}-\dfrac{7}{4}=\dfrac{x}{5}+\dfrac{x}{4}$ .

$\Rightarrow \left(\dfrac{4x}{5}-\dfrac{7}{4}\right)= \left(\dfrac{x}{5}+\dfrac{x}{4}\right)$

$\Rightarrow\left(\dfrac{(4\times4x)-(7\times5)}{20}\right)=\left(\dfrac{5x+4x}{20}\right)$ ... [Cross-multiplying the denominators]

$\Rightarrow\left(\dfrac{16x-35}{20}\right)=\left(\dfrac{9x}{20}\right)$

$\Rightarrow \,16x-35=9x$ .

Transposing

$x$ terms to one side, we get,

$\Rightarrow \,16x-9x=35$

$\Rightarrow \,7x=35$

$\Rightarrow\,x=\dfrac{35}{7}=5$

$\therefore\,x=5$ .

Hence, option

$A$ is correct.

Rashmi has a sum of

$Rs. x$ . She spend

$Rs. 800$ on grocery,

$Rs. 600$ on cloths and

$Rs. 500$ on education and received as

$Rs. 200$ as a gift. How much money (in

$Rs$ ) is left with her?

0%
$x - 1700$
0%
$x - 1900$
0%
$x + 200$
0%
$x - 2100$

Explanation

Total money

$=Rs. x$
Money spent

$=Rs. 800$ on grocery
Money spent

$=Rs. 600$ on cloths
Money spent

$=Rs. 500$ on education
Money left with Rashmi

$=x-(Rs. 800+ Rs. 600+ Rs. 500)$

$=x-Rs. 1900$
She received a gift of

$=Rs. 200$

$\therefore$ Money left

$=x-1900+200$

$=x-1700$

Hence correct option is

$A$ .

Which of the following is a linear equation in one variable?

0%
$2x+1=y-3$
0%
$2t-1=3t+5$
0%
$2x-1=x^{2}$
0%
$x^{2}-x+1=0$

Explanation

$\textbf{Step - 1: Checking all options}$

$A) 2x+1=y-3$

$\text{We can see that the given equation has two variables x and y in and }$

$\text{hence is not a linear equation in}$

$\text{one variable}$

$B) 2t-1=3t+5$

$\text{We can clearly see that the given equation has only one variable t and}$

$\text{hence is a linear equation in one }$

$\text{variable.}$

$C) 2x-1=x^2$

$\text{We can see that the given equation is not linear as its variable has}$

$\text{a power greater than one}$

$D) x^2-x+1=0$

$\text{We can see that the given equation is not linear as its variable has}$

$\text{a power greater than one}$

$\textbf{Thus, the equation }\mathbf{ 2t-1=3t+5}\textbf{ is a linear equation in one variable.}$

The sum of two numbers isif one number is double the other find the numbers.

0%
$24,48$
0%
$25,50$
0%
$23,46$
0%
None f these

Explanation

Given that,

The sum of two numbers is $72$ .

One of the numbers is double that of the other.

To find out,

The numbers.

Let one of the numbers be $x$ .

Hence, the other number will be $2x$ .

The sum of the two numbers is $72$ .

Hence, $x+2x=72$

$3x=72\\$

$x=\dfrac{72}{3}$

$=24$

Hence, $2x=2\times 24$

$=48$

Hence, the numbers are $24$ and $48$ .

The sum of three consecutive integers isWhat is the sum of the squares of the first number and the last number?

0%
$1201$
0%
$576$
0%
$1236$
0%
$1154$

Explanation

Given that, the sum of three consecutive integers is

$72$

To find out: The sum of squares of the first and last number.

Let the three consecutive numbers be

$x, \ x+1$ , and

$x+2$

$\therefore \ x +\left (x+1\right) + \left(x+2\right) = 72$

$\Rightarrow 3x + 3 = 72$

$\Rightarrow 3x = 69$

$\Rightarrow x =\dfrac{69}{3}= 23$

Thus the numbers are

$23, 24, 25$ .

$\therefore \$ The sum of squares of first and last numbers

$= 23^2 + 25^2= 529 +625 = 1154$ .

Hence, the sum of squares of the first and last number is

$1154$ .

On increasing the salary of a man by

$25\%$ , it becomes

$Rs. 20000$ . What was his original salary?

0%
$Rs. 15000$
0%
$Rs. 16000$
0%
$Rs. 18000$
0%
$Rs. 25000$

Explanation

Let original salary of him

$=x$

The value increased

$=25\%$ of

$Rs. x$

$=\left(\dfrac {25}{100} \right )\times x$

$=\left (\dfrac {25x}{100} \right)$

$=\left(\dfrac x4\right )$

Salary after increment

$=(x+\left (\dfrac x4\right))$

$=\dfrac {(4x+x)}4$

$=\left(\dfrac {5x}4\right )$

His increased salary

$=Rs. 20000$

$\left (\dfrac {5x}4 \right)=20000$

$x=\dfrac {(20000\times 4)}5$

$x=4000\times 4$

$x=Rs. 16000$

$(2n + 5) = 3(3n - 10)$ , then

$n = ?$

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

Explanation

Given,

$(2n + 5) = 3(3n - 10)$ .

$\Rightarrow$

$2n + 5 = 9n - 30$ .

Transposing

$2n$ to RHS and it becomes

$-2n$ and

$-30$ to LHS it becomes

$30$

$\Rightarrow$

$30 + 5 = 9n - 2n$

$\Rightarrow$

$35 = 7n$ .

Multiplying both sides by

$\dfrac{1 }{7}.$

$\Rightarrow$

$35 \times \dfrac{1 }{ 7} = 7n \times \dfrac{1 }{ 7}$

$\Rightarrow$

$5 = n$ .

Hence, option

$A$ is correct.

State weather the statement is true(T) or false (F):
Two different equations can never have the same answer.

0%
True
0%
False

Explanation

False,

Two different equations may have the same answer.

eg.

$2x+1=2$ and

$2x-5=-4$ are the two linear equations whose solutions are

$\cfrac12$ .

The sum of three consecutive multiples of

$7$ is

$357$ . Find the smallest multiple.

0%
$112$
0%
$126$
0%
$119$
0%
$116$

Explanation

Let us take three consecutive multiples of

$7$ as,

$7x,(7x+7),(7x+14)$ [

$x$ is a natural number]

$7x+(7x+7)+(7x+14)=357$

$\Rightarrow 21x+21=357$

$\Rightarrow 21(x+1)=357$

$\Rightarrow \frac{21(x+1)}{21}=\frac{357}{21}$ [Diving both sides by

$21$ ]

$\Rightarrow x+1=17$

$x=17-1$

$x=16$
Hence, the smallest multiple of

$7$ is

$7*16=112$

The digit in the ten's place of a two-digit number is

$3$ more than the digit in the units place. Let the digit at the unit's place be

$b$ . Then the number is

0%
$11b+30$
0%
$10b+30$
0%
$11b+3$
0%
$10b+3$

Explanation

We take unit place to be

$b$ ,

So, the digit at ten's place =

$(3+b)$ ,

Hence, the number =

$10(3+b)+b = 30+10b+b = 11b+30$ .

Linear equation in one variable has

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

Explanation

$\textbf{Step - 1: Defining}$

$\text{A linear equation is a relation between variables of degree 1. }$

$\text{A linear equation in one variable is a relation of a variable of power 1}$

$\textbf{Hence a linear equation in one variable has only one variable with power 1}$

CBSE Questions for Class 8 Maths Linear Equations In One Variable Quiz 11 - 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 11 - MCQExams.com

0:0:2

Practice Class 8 Maths Quiz Questions and Answers

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