MCQ Questions for Class 8 Maths Linear Equations In One Variable Quiz 5

The sum of two consecutive odd natural numbers is

$140$ . Find the bigger number.

0%
$85$
0%
$71$
0%
$91$
0%
$69$

Explanation

Let the integer be

$x$ .

The two consecutive odd number

$= x, x+2$

Sum

$= x+x+2 = 140$

$2x = 138\\ x = 69$

The length of larger number

$= x+2 = 71$ .

In a shooting competition a marksman receives

$50$ paise if he hits the mark and pays

$20$ paise if he misses it. He tried

$60$ shots and was paid Rs.

$1.30$ . How many times did he hit the mark?

0%
$10$
0%
$19$
0%
$14$
0%
$12$

Explanation

Let the Number of hits be

$x$

Number of misses will be

$60 - x$

Then according to the problem, we have

$0.50x -(60-x)0.20 = 1.30$

$\Rightarrow 0.50x -12.0 +0.2x = 1.30$

$\Rightarrow 0.7x = 13.30$

$\Rightarrow x = 19$

Number of times he hit the mark

$= 19$ .

Solve:

$8x+\displaystyle \frac{21}{4}= 3x+7$ , then

$x=$

0%
$\displaystyle \frac{2}{9}$
0%
$\displaystyle \frac{7}{20}$
0%
$\displaystyle \frac{6}{17}$
0%
$\displaystyle \frac{4}{21}$

Explanation

Given,

$8x+\cfrac { 21 }{ 4 } = 3x+7$

Multiply

$4$ on both the sides, we get

$32x +21 =12x +28$

Taking

$x$ to one side and constants to another side.

$\\ 20x = 7$

$\therefore x = \cfrac { 7 }{ 20 }$

Some part of a journey of

$555\ km$ was completed by a car with speed

$60\ km/h$ . Then the speed is increased by

$15\ km/h$ and the journey is completed. If it takes

$8$ hours to reach, find the time taken by

$60\ km/h$ speed.

0%
$11$
0%
$8$
0%
$13$
0%
$3$

Explanation

Let time taken to cover some distance with

$60$ km/hr be

$t$ .
Then,

$t (60) + (8-t)(75) = 555$

$\Rightarrow 75\times 8 - 15t = 555$

$\Rightarrow 600 - 555 = 15t$

$\Rightarrow t = 3$ hours
Distance

$=$

$3\times60= 180$ km

Therefore, the time taken by

$60$ km/hr speed is

$3$ hours.

$\displaystyle x=m+1$ and

$\displaystyle \frac{1}{3}(6x-3)-(8-3x)=11;$ find the value of

$m$ .

0%
$m=7$
0%
$m=6$
0%
$m=2$
0%
$m=3$

Explanation

Given,

$\dfrac {1}{3}(6x-3)-(8-3x)=11$

As L.C.M. is

$3$ , we have

$6x-3-24+9x=33$

$\Rightarrow 15x-27=33$

$\Rightarrow 15x=60$

$\Rightarrow x=4=3+1$

Thus comparing this with

$x=m+1$ , we get value of

$m$ as

$3$ .

The angles of a triangle are

$2\left ( x-7 \right )$ ,

$\displaystyle \frac{3}{2}\left ( x-1 \right )$ and

$3\left ( x+11 \right )$ . Find

$x$ and then show that the triangle is isosceles.

0%
$15$
0%
$18$
0%
$22$
0%
$25$

Explanation

Sum of the angles

$= 180$

$2(x-7) + \cfrac { 3 }{ 2 } (x-1) + 3(x+11) = 180\\ LCM = 2\\ 4(x-7) + 3(x-1) + 6(x+11) = 360\\ 4x - 28 + 3x - 3 + 6x +66 = 360\\ 13x +35 = 360\\ 13x = 325\\ x = 25$

Angle

$1 = 2(x-7) = 2(18) = 36$

Angle

$2 = \cfrac { 3 }{ 2 } (x-1) = \cfrac { 3 }{ 2 } (24) = 36$

Since two angles are equal, it is an isosceles triangle.

The measures of the angles of a triangle are

$2x,x+25$ and

$3x+5$ . Find

$x$ and then show that the triangle is isosceles.

0%
$25$
0%
$15$
0%
$28$
0%
$12$

Explanation

Measures of the angle

$= 2x, x+25, 3x+5$

Sum of the angles

$= 180$

$2x + x+25 +3x + 5 = 180\\ 6x + 30 = 180\\ 6x = 150\\ x = 25$

So, the angles of the triangle are

$2x = 50$

$x+25 = 50$

$3x +5 = 80$

Since two angles are equal, the triangle is isosceles.

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%
$39$ cm
0%
$45$ cm
0%
$80$ cm
0%
$40$ cm

Explanation

Let the Length of the line is

$x$ .

Therefore,

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

$\Rightarrow \dfrac {7x}{20}=28$

$\Rightarrow 7x=20\times 28$

$\Rightarrow x=80$

Therefore, the length of a line is

$80$ cm.

A boy has

$x$ coins of

$50$ paise each

$,\: 2x$ coins of

$25$ paise each,

$4x$ coins of

$10$ paise each and

$8x$ coins of

$5$ paise each. Find the number of

$5$ paise coins if the value of all the coins is Rs.

$9$ .

0%
$40$
0%
$20$
0%
$500$
0%
$300$

Explanation

According to the problem, we have

$x(0.50) + 2x (0.25) + 4x (0.10) + 8x (0.05) = 9.00$

$\Rightarrow\ 0.5x + 0.5x +0.4x +0.4x = 9.0$

$\Rightarrow\ 1.8x = 9$

$\Rightarrow x = 5$

Number of

$5$ paise coins

$= 8x = 40$ .

$(2ax + 1) (3x + 1) = 6a (x + 1)$ and

$x = 1$ , find the value of

$a$ .

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

Explanation

$(2ax + 1) (3x + 1) = 6a (x + 1)$

$\implies[2a(1)+1][3(1)+1]=6a[(1+1)]$

$\implies (2a+1)(3+1)=6a(2)$

$\implies (2a+1)(4)=12a$

$\implies 8a+4=12a$

$\implies 8a-12a=-4$

$\implies -4a=-4$

$\implies a=1$

Two numbers are in the ratio

$4 : 7$ . If thrice the larger be added to twice the smaller, the sum is

$59$ . Find the numbers.

0%
$8\displaystyle \frac{4}{29}$ and $14\displaystyle \frac{7}{29}$
0%
$8\displaystyle \frac{3}{34}$ and $14\displaystyle \frac{7}{5}$
0%
$8\displaystyle \frac{3}{34}$ and $14\displaystyle \frac{6}{25}$
0%
$8\displaystyle \frac{4}{29}$ and $14\displaystyle \frac{2}{19}$

Explanation

Let the numbers be

$4x ; 7x$

Given,

$3(7x) + 2(4x) = 59$

$=> 21x + 8x = 59$

$29x = 59$

$=> x = \cfrac {59}{29}$

Thus the numbers are

$4x = 4 \times \cfrac {59}{29} = \cfrac {236}{29} = 8 \cfrac {4}{29}$
and

$7x = 7 \times \cfrac {59}{29} = \cfrac {413}{29} = 14 \cfrac {7}{29}$

In an examination, a student scores

$4$ marks for every correct answer and loses

$1$ mark for every wrong answer. If he attempts in all

$60$ questions and secures

$130$ marks, the number of questions he attempts correctly, is ___

0%
$35$
0%
$38$
0%
$40$
0%
$42$

Explanation

Let the number of questions attempted correctly be

$x$ .

Then, number of incorrect ones

$= (60-x)$ .

$\therefore 4x-1\left( 60-x \right) =130$

$\Rightarrow 5x=190$

$\Rightarrow x=38$

So, number of questions student attempts correctly are

$38$ .

Amit is now

$6$ times as old as his son. Four years from now, the sum of their ages will be

$43$ years. Determine Amit's present age:

0%
$30$ years
0%
$32$ years
0%
$34$ years
0%
$28$ years

Explanation

Let present age of Amit's son be

$x$ yrs and age of Amit be

$6x$ yrs.
Four yrs from now,

$(6x+4)+(x+4)=43$ .....(Given condition)

$\therefore 7x=35$

$\therefore x=5$ ,
i.e., present age of Amit

$=6x=6\times 5=30$ yrs.
Hence, option

$A$ is the correct answer.

$\displaystyle \frac{1}{3}\left ( 7-4x \right )-\frac{1}{4}\left ( 11-5x \right )+1=x-\frac{1}{2}\left ( 3x-7 \right )$ , then

$x=$

0%
$1$
0%
$9$
0%
$6$
0%
$7$

Explanation

$\cfrac { 1 }{ 3 } (7-4x) -\cfrac { 1 }{ 4 } (11-5x) + 1 =x - \cfrac { 1 }{ 2 } (3x-7)\\ LCM = 12\\ 4(7-4x) - 3(11-5x) +12 = 12x - 6(3x-7)\\ 28 - 16x -33 +15x +12 = 12x -18x +42\\ 7 -x = 42 -6x\\ 5x = 35\\ x = 7$

The ten's digit of a two digit number exceeds its unit's digit by

$4$ . If the ten's digit and the unit's digit are in the ratio

$3 : 1$ , find the number.

0%
$62$
0%
$51$
0%
$65$
0%
$85$

Explanation

Let the units digit

$= x$

Let the tens digit

$= x+4$

Tens digit : One digit ::

$3:1$

$x+4 : x :: 3:1$

$\Rightarrow x+4 = 3x$

$\Rightarrow 2x = 4$

$\Rightarrow x = 2$

Tens digit

$= 6$

Therefore, the number is

$62$ .

Solve the following equation:

$\displaystyle \frac{4}{5}\, \left(x\, +\, \frac{5}{6} \right)\, +\, \frac{2}{3} \left(x\, -\, \frac{1}{4}\right)\, =\, {1}\frac{1}{9}$

0%
$\displaystyle \frac{5}{19}$
0%
$\displaystyle \frac{5}{16}$
0%
$\displaystyle \frac{5}{14}$
0%
$\displaystyle \frac{5}{12}$

Explanation

$\displaystyle \frac{4}{5}\left ( x+\cfrac{5}{6} \right )+\cfrac{2}{3}\left ( x-\cfrac{1}{4} \right )=1\cfrac{1}{9}$

$=>\cfrac { 4x }{ 5 } +\cfrac { 20 }{ 30 } +\cfrac { 2x }{ 3 } -\cfrac { 2 }{ 12 } =\cfrac { 10 }{ 9 }$

$=>\cfrac { 4x }{ 5 } +\cfrac { 2 }{ 3 } +\cfrac { 2x }{ 3 } -\cfrac { 1 }{ 6 } -\cfrac { 10 }{ 9 } =0$

$=>\cfrac { 72x+60+60x-15-100 }{ 90 } =0$

$=>72x+60x=-60+15+100$

$=>132x=55$

$=>x=\cfrac{55}{132}$

$=>x=\cfrac{5}{12}$

A farmer divides his herd of

$x$ cows among his

$4$ sons so, that first son gets-one-half of the herd, the second son get, one-fourth, the third son gets one-fifth, and the fourth son gets

$7$ cows, then the value of

$x$ is :

0%
$100$
0%
$140$
0%
$160$
0%
$180$

Explanation

Total number of cows

$= x$

According to the question, we have
A

$= \displaystyle{\frac{x}{2}}$ ,

$B = \displaystyle{\frac{x}{4}}$ , C

$= \displaystyle{\frac{x}{5}}$ ,

$D = 7$

$\therefore$

$\displaystyle{\frac{x}{2} + \frac{x}{4} + \frac{x}{5}} + 7 = x$

$\Rightarrow$

$\displaystyle{\frac{10x + 5x + 4x}{20}} = x - 7$

$\Rightarrow19x = 20x - 140$

$\Rightarrow 140 = x$
Hence, option 'B' is correct.

Out of six consecutive numbers the sum of first three is

$27$ . What is the sum of next three?

0%
$30$
0%
$40$
0%
$36$
0%
$45$

Explanation

Let the first three consecutive numbers be

$x, x+1$ and

$x + 2$ .
Then,

$x + x + 1 + x + 2 = 27$

$\Rightarrow 3x = 24$

$\displaystyle \Rightarrow x = 8$

$\displaystyle \therefore$ The first three numbers are

$8,9,10$ .

$\displaystyle \Rightarrow$ The next three numbers are

$11, 12, 13$ .
Hence, the required sum

$= 11 + 12 + 13 = 36$ .

I bought a number of books one day, four times of them on the next day, and eight times on the third day, In all I bought

$78$ books. How many books did I buy on the second day?

0%
$6$
0%
$8$
0%
$16$
0%
$24$
0%
$36$

Explanation

Let the number of books bought of the first day be

$a$
Thus, number of books on second day

$=4a$
and the number of books on third day

$=8a$
Now, given

$a+4a+8a=78$

$\Rightarrow 13a=78$

$\Rightarrow a=\dfrac{78}{13}$

$\Rightarrow a=6$
Thus, the number of books bought of second day

$=4\times 6=24$

The difference of the squares of two consecutive even natural numbers isTaking x as the smaller of the two numbers, form an equation in x, and hence find the larger of the two numbers.

0%
$18$
0%
$24$
0%
$23$
0%
$21$

Explanation

Given that,

The difference of squares of two consecutive even natural numbers is

$92$ .

The smaller number is

$x$ .

To find out,

$(i)$ The equation in

$x$ representing the given information.

$(ii)$ The larger of the two numbers.

$(i)$ If the smaller number is

$x$ , then the larger number will be

$x+2$

According to the question,

$(x+2)^2-x^2=92$

Hence,

$(x+2)^2-x^2=92$ is the equation that represents the given information.

$(ii)$ Solving

$(x+2)^2-x^2=92$

We know that,

$(a+b)^2=a^2+2ab+b^2$

Applying the same identity, we get:

$(x^2+4x+4)-x^2=92$

$\Rightarrow x^2+4x+4-x^2=92$

$\Rightarrow 4x+4=92$

Dividing both sides by

$4$ , we get:

$x+1=23$

$x=23-1$

$=22$

Hence,

$x=22$

$\therefore\ x+2=22+2$

$=24$

Hence, the larger of the two numbers is

$24$ .

The difference between two numbers is

$8$ and their ratio is

$1:5$ . Which of the following is the smaller number?

0%
$24$
0%
$8$
0%
$10$
0%
$2$

Explanation

Let the numbers be

$x$ and

$5x$ .

Given,

$5x-x=8$

$\Rightarrow 4x=8$

$\Rightarrow \displaystyle x=\frac{8}{4}=2$
Hence, smaller number is

$2$ .

The solution of the equation

$\displaystyle \frac{3a-2}{3}+\frac{2a+3}{2}=a+\frac{7}{6}$ is

0%
$a=2$
0%
$\displaystyle a=\frac{1}{3}$
0%
$a=-2$
0%
$\displaystyle a=\frac{1}{2}$

Explanation

Given,

$\displaystyle \frac{3a-2}{3}+\frac{2a+3}{2}=a+\frac{7}{6}$
L.C.M. of denominators is

$6$ .
Multiplying each term by

$6$ , we get

$\displaystyle 6\times \left ( \frac{3a-2}{3} \right )+6\times \left(\frac{2a+3}{2}\right)=6\times a+6\times \frac{7}{6}$

$\Rightarrow 2(3a-2)+3(2a+3)=6a+7$

$\Rightarrow 6a-4+6a+9=6a+7$

$\Rightarrow 12a+5=6a+7$

$\Rightarrow 6a=2$

$\Rightarrow \displaystyle a=\frac{2}{6}=\frac{1}{3}$
Hence, required solution is

$a=\cfrac{1}{3}$ .

The sum of one-half, one third and one-fourth of a number exceeds the number by

$12$ . What is the number?

0%
$72$
0%
$144$
0%
$98$
0%
$156$

Explanation

Let the number be

$x$ .

Then,

$\displaystyle \left ( \frac{x}{2}+\frac{x}{3}+\frac{x}{4} \right )-x=12$ ,

L.C.M of

$2, 3, 4=12$ ,

Thus multiplying above equation with

$12$ both sides,

$\Rightarrow 12 \times \dfrac {x}{2}+12 \times \dfrac {x}{3}+12 \times \dfrac {x}{4}-12 \times x-12 \times 12$

$\Rightarrow 6x+4x+3x-12x=144$

$\Rightarrow 13x-12x=144\Rightarrow x=144$ .

Hence, the required number is

$144.$

Find the value of

$y$ in the equation :

$\displaystyle \frac{(2-3y)+4y}{9y-(8y+7)}=\frac{4}{5}$

0%
$18$
0%
$9$
0%
$-38$
0%
$-32$

Explanation

Given,

$\displaystyle \frac{(2-3y)+4y}{9y-(8y+7)}=\frac{4}{5}$

$\Rightarrow \displaystyle \frac{2+y}{y-7}=\frac{4}{5}$

$\Rightarrow 5(2+y)=4(y-7)$ ....( cross multiplying )

$\Rightarrow 10+5y=4y-28$

$\Rightarrow y=-38$
Hence,. the solution is

$y=-38$

Solve for

$x$ :

$\displaystyle \frac{2x+3}{2}=5-\frac{2(x-4)}{3}$ .

0%
$\dfrac{5}{3}$
0%
$\dfrac{12}{29}$
0%
$\dfrac{37}{10}$
0%
None of these

Explanation

Given,

$\displaystyle \frac{2x+3}{2}=5-\frac{2(x-4)}{3}$

Multiplying all the terms by the LCM of

$2, 3$ , i.e.,

$6$ , we get

$\displaystyle 6\times \frac{(2x+3)}{2}=6\times 5-\frac{6\times 2(x-4)}{3}$

$\Rightarrow 3(2x+3)=30-4(x-4)$

$\Rightarrow \displaystyle 6x+9=30-4x+16$

$\Rightarrow 6x+9=-4x+46$

$\Rightarrow \displaystyle 6x+4x=46-9$

$\Rightarrow 10x=37$

$\Rightarrow x=\dfrac{37}{10}$

Hence, required solution is

$x=\cfrac{37}{10}$ .

X buys pens and pencils at Rs. 5 and Rs. 1 per piece respectively. For every two pens, he buys three pencils. He sold pens and pencils at 12

$\%$ and 10

$\%$ profit respectively. If his total sale is Rs. 725, then the number of pencils exceeds the number of pens by

0%
$25$
0%
$50$
0%
$75$
0%
$90$

Explanation

Let number of pens be

$x$

Hence, number of pencils will be

$\dfrac{3}{2}x$

Cost of one pen is

$Rs.\ 5$

Hence, total C.P. of pens

$= 5x$

Also, cost of one pencil is

$Rs.\ 1$

Hence, total C.P. of pencils

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

It is also given that, pens are sold at

$12\%$ profit,

Hence,

$S.P. = 112\% \ of\ 5x$

$= \dfrac{112}{100} \times 5x$

$=\dfrac{28x}{5}$

Also, pencils are sold at

$10\%$ profit,

Hence,

$S.P. = 110\% \ of\ \dfrac{3}{2}x$

$= \dfrac{330}{200}x$

$= \dfrac{33x}{20}$

Now, total sale is

$Rs.\ 725$ .

Hence,

$\dfrac{28x}{5} + \dfrac{33x}{20} = 725$

$112 x + 33x = 725 \times 20$

$145 x = 14500$

$x = 100$

Thus, number of pens

$= 100$

Number of pencils

$= \dfrac{3}{2} \times 100$

$= 150$

So, the number of pencils exceeds the number of pens by

$50$ .

Hence, option B is correct.

A candidate in an examination was asked to find

$\displaystyle \dfrac{5}{14}$ of a certain number. By mistake he found

$\displaystyle \dfrac{5}{4}$ of it. Thus his answer was

$25$ more than the correct answer. What was the number?

0%
$28$
0%
$56$
0%
$65$
0%
$20$

Explanation

Let the number be x
Then

$\displaystyle x\times \dfrac{5}{4}-x\times \dfrac{5}{14}=25$

$\displaystyle \Rightarrow \dfrac{35x-10x}{28}=25$

$\Rightarrow \dfrac{25x}{28}=25$

$\displaystyle \Rightarrow x=\dfrac{25\times 28}{25}=28$

Solve for

$x$ :

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

0%
$9$
0%
$14$
0%
$16$
0%
$11$

Explanation

Given,

$\displaystyle \frac{3x}{4}-\frac{(x-1)}{3}=\frac{(x-2)}{2}$
L.C.M. of denominator

$=12$

$\therefore$ Multiplying each term by

$12$ , we get

$\displaystyle 12\times \frac{3x}{4}-12\times \frac{(x-1)}{3}=12\times \frac{(x-2)}{2}$

$\Rightarrow 9x-4(x-1)=6(x-2)$

$\Rightarrow 9x-4x+4=6x-12$

$\Rightarrow 5x+4=6x-12$

$\Rightarrow x=16$
Hence, required value of

$x$ is

$16$ .

The solution of the the equation

$5(3x+5)=2(7x-4)$ is

0%
$x=17$
0%
$x=33$
0%
$x=-33$
0%
$x=-17$

Explanation

Given,

$5(3x+5)=2(7x-4)$

$\therefore 15x+25=14x-8$

Putting

$x$ terms on one side and constants on another side.

$\therefore 15x-14x=-8-25$

$\therefore x=-33$
Hence, option C is correct.

If two third, one half and one seventh of a number is added to itself the result is

$37$ . Find the number.

0%
$\displaystyle 14\frac{2}{97}$
0%
$\displaystyle 16\frac{2}{97}$
0%
$\displaystyle 18\frac{2}{97}$
0%
None

Explanation

Let,

The number be

$x$

Two-third of a number

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

One-half of a number

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

One-seventh of a number

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

A.T.Q,

$\dfrac{2}{3}x+\dfrac{1}{2}x+\dfrac{1}{7}x+x=37$

$\implies \dfrac{28x+21x+6x+42x}{42}=37$

$\implies \dfrac{97}{42}x=37$

$\implies x=\dfrac{37 \times 42}{97}$

$\implies x=\dfrac{1554}{97}=16\dfrac{2}{97}$

Answer : B

CBSE Questions for Class 8 Maths Linear Equations In One Variable Quiz 5 - 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 5 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

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