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

In an examination, a student scores 4 marks for every correct answer and loses 1 mark for every wrong answer. If he attempts all 75 questions and secures 125 marks, the number of questions he attempted correctly, is

0%
35
0%
40
0%
42
0%
46

Explanation

If the number of correct answer be x, then
the number of incorrect answer is

$\displaystyle (75- x)$ .

$\displaystyle \therefore \quad 4x-\left( 75-x \right) =125\quad or\quad x=40$

$\displaystyle \frac { 4x }{ 9 } -8=8-\frac { 5x }{ 9 }$ , find

$x$ .

0%
$-16$
0%
$16$
0%
$9$
0%
$\dfrac{3}{16}$

Explanation

Given,

$\displaystyle \frac { 4x }{ 9 } -8=8-\frac { 5x }{ 9 }$ .

Transposing

$x$ terms to one side, we get,

$\Longrightarrow\dfrac { 4x }{ 9 } +\dfrac { 5x }{ 9 } =8+8$

$\Longrightarrow\displaystyle \frac { 9x }{ 9 } =16$

$\Longrightarrow x=16$ .

Hence, option

$B$ is correct.

$\displaystyle 5x-35=55+18x$ , then

$x$ .

0%
is a fraction
0%
is an integer
0%
is a rational number
0%
cannot be solved

Explanation

Given,

$5x-35=55+18x$ .

Transposing

$x$ terms to one side, we get,

$\Longrightarrow 5x-18x=55+35$

$\Longrightarrow -13x=90$

$\Longrightarrow x=-\dfrac{90}{13}$ , which is a fraction of the form

$\dfrac{p}{q}$ ;

$q \ne 0$ and

$p$ &

$q$ are co-primes.

Hence, options

$A$ and

$C$ are correct.

A man has Rs.

$x$ with him. He gave

$\dfrac{1}{4}$ th to his wife,

$\dfrac{1}{3}$ rd to his son and Rs.

$1000$ to his daughter. Find

$x$ .

0%
$2400$
0%
$2200$
0%
$2000$
0%
None of the above.

Explanation

The total amount of money with the man = Rs.

$x$ .

$\displaystyle \frac { 1 }{ 4 } x+\frac { 1 }{ 3 } x+1000=x$

$\Longrightarrow\displaystyle x-\frac { 1 }{ 4 } x-\frac { 1 }{ 3 } x=1000$

$\Longrightarrow\displaystyle \frac { 12x-3x-4x }{ 12 } =1000$

$\Longrightarrow\displaystyle 5x=1000\times 12$

$x$ = Rs.

$2400$

The sum of

$5$ ,

$9$ and a number is

$50$ . Find the number.

0%
$35$
0%
$36$
0%
$37$
0%
$38$

Explanation

Let the number be

$x$
According to the question,

$\displaystyle 5+9+x=50$

$\therefore \displaystyle x=50-14$

$\therefore \displaystyle x=36$

The sum of two numbers is

$30$ . One of the number exceeds the other by

$10$ . Find the number.

0%
$10, -20$
0%
$-10, 20$
0%
$10, 20$
0%
$-10, -20$

Explanation

Let the required number be

$x$ .
Then, the other number will be

$x+10$
According to the question, sum of the two numbers is

$30$ .

Therefore,

$\displaystyle x+x+10=30$

$\therefore \displaystyle 2x=20$

$\therefore \displaystyle x=10$
The numbers are

$10$ and

$10+10=20$ .

Solve

$8(3+2x)=13x$ and find the value of

$x$

0%
$-8$
0%
$8$
0%
$\dfrac{1}{8}$
0%
$\dfrac{-1}{8}$

Explanation

Given,

$\displaystyle 8(3+2x)=13x$

$\Rightarrow 24+16x=13x$ ...[By Distribution Law]

Transposing

$x$ terms to one side, we get,

$\Rightarrow\displaystyle 16x-13x = -24$

$\Rightarrow\displaystyle 3x=-24$

$\Rightarrow x=\dfrac{-24}{3}$

$\Rightarrow x=-8$

The ratio of adults to girls to boys on a class field trip was

$1:4:5$ . If the trip included

$6$ more boys than girls, how many adults were with the group?

0%
$3$
0%
$4$
0%
$6$
0%
$8$

Explanation

Given, the ratio of adults to girls to boys on a class field trip is

$1:4:5$
Let

$x$ be the common variable
Let number of adults be

$x$ , number of girls be

$4x$ and number of boys be

$5x$ .
Since the trip included

$6$ more boys than girls,

$5x-4x=6$

$x=6$
Therefore

$6$ adults were with the group.

The four consecutive numbers add up to

$74$ . What are these integers?

0%
$18, 19, 20, 21$
0%
$17, 18, 19, 20$
0%
$20, 18, 18, 19$
0%
$16, 17, 18, 19$

Explanation

Let the first integer

$= p$

$\displaystyle \therefore$ the second integer

$= p+1$
and the third consecutive integer

$= p+2$ and fourth

$= p+3$

$\displaystyle \because$ Sum of four consecutive numbers =

$74$

$\displaystyle p+(p+1)+(p+2)+(p+3)=74$

$\Longrightarrow \displaystyle 4p+6=74$

$\Longrightarrow 4p=68$

$\Longrightarrow \displaystyle p=\frac { 68 }{ 4 }$

$\therefore p=17$
1 st integer

$= 17$
2nd integer

$= 18$
3rd integer

$= 19$
4 th integer

$= 20$

So, option B is correct.

The sum of three consecutive even integers is

$72$ . Find the smallest number

0%
$24$
0%
$22$
0%
$26$
0%
None

Explanation

Let the three consecutive numbers be

$x, x+2$ and

$x+4$ .
Sum =

$\displaystyle x+x+2+x+4$

$\therefore 3x + 6 = 72$

$\Longrightarrow 3x=66$

$\Longrightarrow x=22$

The sum of

$6$ consecutive numbers is

$105$ . Find the smallest number.

0%
$13$
0%
$15$
0%
$14$
0%
$12$

Explanation

Let the 6 consecutive numbers be

$x, x+1, x+2, x+3, x+4$ and

$x+5$

$\Longrightarrow \displaystyle x+x+1+x+2+x+3+x+4+x+5=105$

$\Longrightarrow \displaystyle 6x+15=105$

$\therefore 6x=90$

$\therefore x=15$

The sum of three consecutive multiples of

$5$ is

$45$ . Which is the smallest of the three multiples?

0%
$10$
0%
$5$
0%
$15$
0%
$20$

Explanation

Let

$5x$ be the smallest multiple of

$5$ .
Then, the three consecutive multiples shall be

$5x$ ,

$5x+5$ and

$5x+10$
According to the question,

$\displaystyle 5x+5x+5+5x+10=45$

$\therefore \displaystyle 5x+5x+5x=45-15$

$\therefore \displaystyle 15x=30$

$\therefore \displaystyle x=2$

$\displaystyle \therefore$ Smallest multiple

$= 5x = 5\times 2 = 10$

The length of a rectangle is

$1\frac { 3 }{ 4 }$ of its breadth. If perimeter is

$66$ . find the length of rectangle.

0%
$12$
0%
$21$
0%
$20$
0%
$11$

Explanation

Let breadth of rectangle be

$x$
Length =

$\displaystyle 1\frac { 3 }{ 4 } \times x=\frac { 7 }{ 4 } x$ ....given

According to the question,

$\displaystyle 2\left( \frac { 7 }{ 4 } x+x \right) =66$

$\Longrightarrow 2\left( \dfrac { 7x+4x }{ 4 } \right) =66$

$\Longrightarrow\displaystyle 11x=66\times 2$

$\Longrightarrow\displaystyle x=12$

Length =

$\dfrac { 7 }{ 4 } \times 12=21$

Reduce the following linear equation:

$6t - 1 = t - 11$

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

Explanation

$6t - 1 = t - 11$

$6t - t = -11 + 1$

$5t = -10$

$t = -2$

Solve the linear equation:

$5x - 12 = 2x + 18$

0%
$x=8$
0%
$x=9$
0%
$x=10$
0%
$x=11$

Explanation

$5x - 12 = 2x + 18$

$5x - 2x = 18 + 12$

$3x = 30$

$x = 10$

Reduce the linear equation:

$x + 3-\dfrac{2x}{3}+\dfrac{x}{6}=0$

0%
$x=12$
0%
$x=10$
0%
$x=8$
0%
$x=-6$

Explanation

Given,

$x + 3-\dfrac{2x}{3}+\dfrac{x}{6}=0$
L.C.M of the denominator

$3$ and

$6$ is

$6$ .
Multiplying both the sides by

$6$ , we get

$6x + 18 - 4x + x = 0$

$7x - 4x + 18 = 0$

$3x = -18$

$x = -6$

Reduce the linear equation:

$\dfrac{x}{2}+\dfrac{2x}{4}= 10$

0%
$x=6$
0%
$x=7$
0%
$x=8$
0%
$x=10$

Explanation

Given,

$\dfrac{x}{2}+\dfrac{2x}{4}= 10$
L.C.M of the denominator

$2$ and

$4$ is

$4$ .
Multiplying both the sides by

$4$ , we get

$2x + 2x = 40$

$4x = 40$

$x = 10$

The sum of one fifth, one third and one ninth of a number is

$29$ . Find the number.

0%
$45$
0%
$29$
0%
$54$
0%
None of the above.

Explanation

Let the required number be

$x$

According to the question,

$\displaystyle \frac { x }{ 5 } +\frac { x }{ 3 } +\frac { x}{ 9 } =29$

L.C.M of

$5,3,9$ is

$45.$

Then the equation becomes :

$\displaystyle \frac { x \times 9 }{ 5 \times 9 } +\frac { x \times 15 }{ 3 \times 15 } +\frac { x \times 5}{ 9 \times 5 } =29$

$\Longrightarrow\displaystyle \frac { 9x+15x+5x }{ 45 } =29$

$\Longrightarrow\displaystyle 29x=29\times 45$

$\Longrightarrow\displaystyle x=\frac { 29\times 45 }{ 29 }$

$\therefore x=45$

Solve linear equation:

$m - \dfrac {m - 1}{2} = 1 - \dfrac {m - 2}{3}$ .

0%
$m = \dfrac {4}{5}$
0%
$m = \dfrac {3}{5}$
0%
$m = \dfrac {8}{5}$
0%
$m = \dfrac {7}{5}$

Solve for

$x$ :

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

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

Explanation

Given,

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

$\Rightarrow x+10x-5=4x+4-2$ ...[By Distribution Law]

$\Rightarrow 11x-5=4x+2$ .

Transposing

$x$ terms to one side, we get,

$\Rightarrow 11x-4x=2+5$

$\Rightarrow 7x=7$

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

Hence, option

$C$ is correct.

Ram is

$5$ times as old as Shyam. If their difference of age is

$8$ years, how old is Ram?

0%
$8$ years
0%
$10$ years
0%
$12$ years
0%
$5$ years
0%
None of these

Explanation

Let the age of Shyam be

$x$ so, age of Ram is

$5x$ .

As per the problem, the difference of their ages is

$8$ years.

$5x-x = 8$

$\Rightarrow 4x = 8$

$\Rightarrow x = 2$

So, Rams age

$= 5x$

$= 5\times 2$

$= 10$ years.

There are

$42$ integers in a group. If there are

$5$ times as many odd integers as there are even integers, how many numbers are even integers?

0%
$7$
0%
6
0%
5
0%
9

Explanation

Let numbers of even integers be

$x$

Then numbers of odd integers are

$5x$ .

$\therefore x+5x=42$

$6x=42$

$\Rightarrow x=\dfrac{42}{6}$

$=7$

Solve for

$t$ :

$\displaystyle\frac{t}{2}+4=\displaystyle\frac{3}{4}t-5$ .

0%
$4$
0%
$9$
0%
$18$
0%
$36$

Explanation

Given,

$\dfrac{1}{2}t+4=\dfrac{3}{4}t-5$ .

Transposing

$t$ terms to one side, we get,

$\Rightarrow \dfrac{1}{2}t-\dfrac{3}{4}t=-5-4$

$\Rightarrow \dfrac{2t-3t}{4}=-9$ …[Cross-multiplying the denominators on the LHS]

$\Rightarrow \dfrac{-t}{4}=-9$

$\Rightarrow -t=-36$ ...[Multiplying

$4$ on both sides]

$\Rightarrow t=36$ .

Hence, option

$D$ is correct.

The sum of four consecutive odd positive integers is

$160$ . Find the largest among them.

0%
37
0%
39
0%
41
0%
43
0%
45

Explanation

Let the four consecutive odd positive integers are

$x,x+2,x+4,x+6$

$\therefore x+x+2+x+4+x+6=160$

$4x+12=160$

$4x=160-12$

$4x=148$

$x=\dfrac{148}{4}$

$=37$

$\therefore$ Largest odd integer is

$37+6=43$ .

If the number of boys in a class is thrice the number of girls and the total number of boys is 84, calculate the number of girls in the class.

0%
$28$
0%
$20$
0%
$30$
0%
$252$

Explanation

Let the number of girls are

$x$

Then number of boys

$=3x$

$\Rightarrow 3x=84$

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

Then number of girls

$=28$

The sum of five positive, consecutive integers is

$115$ . Find the value of the smallest integer among these.

0%
$21$
0%
$22$
0%
$23$
0%
$24$
0%
$25$

Explanation

Let the five consecutive integers are

$x,x+1,x+2,x+3,x+4$

$\therefore x+x+1+x+2+x+3+x+4=115$

$\Rightarrow 5x+10=115$

$\Rightarrow 5x=115-10$

$\Rightarrow 5x=105$

$\Rightarrow x=\dfrac{105}{5}=21$

The smallest integer

$=21$

$\dfrac{2}{3}(5x+7)=8x$ ,then

$x$ is equal to:

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

Explanation

Given,

$\dfrac{2}{3}(5x+7)=8x$ .

$\Rightarrow \dfrac{10x}{3}+\dfrac{14}{3}=8x$ ...[By Distribution Law]

$\Rightarrow \dfrac{10x+14}{3}=8x$ …[Cross-multiplying the denominators on the LHS]

$\Rightarrow 10x+14=24x$ ...[Multiplying

$3$ on both sides].

Transposing

$x$ terms to one side, we get,

$\Rightarrow 10x-24x=-14$

$\Rightarrow -14x=-14$

$\Rightarrow x=\dfrac{-14}{-14}=1$ .

Hence, option

$A$ is correct.

A club has

$27$ employees. If there are seven more women than men in the club, calculate the number of men in the club.

0%
$10$
0%
$8$
0%
$6$
0%
$9$

Explanation

Let the number of men in the club

$=x$

Then number of women

$=$

$x+7$

Then according to the question,

$\Rightarrow x+x+7=27$

$\Rightarrow 2x=27-7$

$\Rightarrow 2x=20$

$\Rightarrow x=\dfrac{20}{2}=10$

Then number of men

$=10$

$280$ meals are to be prepared by three chefs. Every chef has its own speed but the combined output of all three is modelled by the equation

$8x+4x+2x=280$ . If

$x$ is a positive integer, which of the following could

$8x$ represent in the equation?

0%
The total meal output by the slowest chef, who made $40$ meals
0%
The total meal output by the fastest chef, who made $160$ meals
0%
The total meal output by the fastest chef, who made $80$ meals
0%
The different between the output between the slowest and fastest chef, which would be $120$ meals

Explanation

In given equation

$8x+4x+2x=280$

Then

$14x=280$

$\therefore x=20$

Then

$x$ is

$20$ meals

Then mean of

$8x$ in equation is

$8x=8\times 20=160$ meals

Then the meal output by the fastest chef, who made

$160$ meals.

The value of

$d$ which satisfies the expression

$\cfrac{4(d+3)-9}{8}=\cfrac{10-(2-d)}{6}$ is:

0%
$\cfrac{23}{16}$
0%
$\cfrac{23}{8}$
0%
$\cfrac{25}{8}$
0%
$\cfrac{25}{4}$

Explanation

Given,

$\dfrac {4(d+3)-9}{8}=\dfrac {10- (2-d)}{6}$ .

Cross multiplying, we get,

$24(d+3)-54=80-8(2-d)$

$\Rightarrow 24d+72-54=80-16+8d$ ...[By Distribution Law]

$\Rightarrow 24d+18=64+8d$ .

Transposing

$d$ terms to one side, we get,

$\Rightarrow 24d-8d=64-18$

$\Rightarrow 16d=46$

$\Rightarrow d=\dfrac{23}{8}$ .

Hence, option

$B$ is correct.

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

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Practice Class 8 Maths Quiz Questions and Answers

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