MCQ Questions for Class 8 Maths Linear Equations In One Variable Quiz 9

The sum of three consecutive integers is equal to

$192$ . What is the product of these numbers?

0%
$216000$
0%
$7077888$
0%
$576$
0%
$110592$
0%
$262080$

Explanation

Let the three consecutive numbers be

$x,x+1,x+2$ .

To get the three consecutive numbers whose sum is

$192$ , we get the equation as,

$x+\left( x+1 \right) +\left( x+2 \right) =192$

$3x+3=192$

$3x=192-3$

$= 189$

$x=\dfrac { 189 }{ 3 }$

$x=63$
So the first number out of the three consecutive numbers is

$63$ ,
The three numbers are

$63,64$ and

$65$ .

The product of these numbers is,

$63\times 64\times 65=262080$
Hence the correct answer is option E.

Consider two consecutive positive integers. It is given that the result of adding the smaller integer and triple the larger integer is

$79$ . Find the

$2$ integers.

0%
18, 19
0%
19, 20
0%
20, 21
0%
26, 27
0%
39, 40

Explanation

Let the two consecutive positive integers are

$x,x+1$

The according to the question,

$\Rightarrow x+3(x+1)=79$

$\Rightarrow x+3x+3=79$

$\Rightarrow 4x=79-3$

$\Rightarrow 4x=76$

$\Rightarrow x=\dfrac{76}{4}=19$

Another integer

$=19+1=20$

So, the two integers are

$19,20$ .

When a number

$x$ is subtracted from

$36$ and then the difference is divided by

$x$ , the result is

$2$ . Find the value of

$x$ .

0%
$2$
0%
$4$
0%
$6$
0%
$12$

Explanation

Let the number be

$x$ .

According to the question:

$\dfrac { 36-x }{ x } =2\\ \Rightarrow 36-x=2x\\ \Rightarrow 36=3x$

$\Rightarrow x=12$

Hence option (D) is correct option

$13$ is added to one-half of a certain number, the result is

$37$ . The original number is equal to

0%
$24$
0%
$40$
0%
$48$
0%
$61$

Explanation

Let the original number is

$x$

Then according to the question, we have

$\dfrac{x}{2}+13=37$

$\Rightarrow \dfrac{x+26}{2}=37$

$\Rightarrow x+26=74$

$\Rightarrow x=74-26$

$\Rightarrow x=48$

Therefore, the original number is

$48$ .

$16+4x$ is

$10$ more than

$14$ , what is the value of

$8x$ ?

0%
$2$
0%
$6$
0%
$16$
0%
$80$

Explanation

According to question, we have

$16+4x=14+10$

$\Rightarrow 4x=8$

$\Rightarrow x=2$

So,

$8x=8\times2=16$
Therefore, option C is correct.

The monthly membership fee for an online television and movie service is

$9.80$

$. The cost of viewing television shows online is included in the membership fee, but there is an additional fee of$

$1.50$ to rent each movie online. For one month, Jill's membership and movie rental fees were

$12.80$ $ . How many movies did Jill rent online that month?

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

Explanation

Equation for monthly bill when

$m$ number of movies are rented per month

$= 9.80 + 1.5m$

Given,

$9.80 + 1.5m = 12.80$

$=> 1.5m = 3$

$=> m = 2$

Thus, Jill rented two movies that month

Last week Raul worked

$11$ more hours than Angelica. If they worked a combined total of

$59$ hours, how many hours did Angelica work last week?

0%
$24$
0%
$35$
0%
$40$
0%
$48$

Explanation

If Angelica worked

$a$ hours, then Raul worked

$a + 11$ hours.
Total hours worked

$= a + a + 11 = 59$

$\Rightarrow 2a = 48$

$\Rightarrow a = 24$ hours

$\dfrac{5}{x+3} = \dfrac{1}{x}+\dfrac{1}{2x}$ , calculate the value of

$x$ .

0%
$\dfrac{3}{14}$
0%
$\dfrac{1}{3}$
0%
$\dfrac{6}{13}$
0%
$\dfrac{3}{4}$
0%
$\dfrac{9}{7}$

Explanation

Given,

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

Taking RHS:

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

LCM of these is

$2x$

$\Rightarrow \dfrac { 2 }{ 2x } +\dfrac { 1 }{ 2x } \\ \Rightarrow \dfrac { 3 }{ 2x }$

Now taking LHS:

$\dfrac { 5 }{ x+3 }$

It is given, LHS

$=$ RHS

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

$\Rightarrow 5\times 2x=3\times (x+3)\\ \Rightarrow 10x=3x+9\\ \Rightarrow x=\dfrac {9}{7}$

A food truck sells salads for

$6.50$

$each and drinks for$

$2.00$ each. The food truck's revenue from selling a total of

$209$ salads and drinks in one day was

$836.50$ $. How many salads were sold that day?

0%
$77$
0%
$93$
0%
$99$
0%
$105$

Explanation

Let

$s$ be the number of salads sold.
Then number of drinks sold

$= 209 - s$

Since the earning was

$836.50$ , we have

$6.5s + 2(209-s) = 836.5$

$\Rightarrow 6.5s + 418 -2s = 836.5$

$\Rightarrow 4.5s = 418.5$

$\Rightarrow s = 93$

Thus,

$93$ salads were sold.

Solve the equation:

$\dfrac{12x+3}{2} = \dfrac{x+1}{4}$ .

0%
$-10$
0%
$15$
0%
$\dfrac {-5}{23}$
0%
$\dfrac {37}{12}$

Explanation

Given,

$\dfrac{12x+3}{2} = \dfrac{x+1}{4}$

$\Rightarrow 4(12x+3)=2(x+1)$ ...[Multiply both sides by

$4$ ]

$\Rightarrow 48x+12=2x+2$ ...[By Distribution Law]

Transposing

$x$ terms to one side, we get,

$\Rightarrow 48x-2x=2-12$

$\Rightarrow 46x=-10$

$\Rightarrow 23x=-5$

$\therefore x=-\dfrac{5}{23}$ .

Hence, option

$C$ is correct.

Solve the equation:

$-(x-5)+5(x-9)=2(x+8)-(2x+5)$ .

0%
$x=5$
0%
no solutions
0%
$x=\dfrac { 51 }{ 4 }$
0%
$x=-8$

Explanation

Given,

$-\left( x-5 \right) +5\left( x-9 \right) =2\left( x+8 \right) -\left( 2x+5 \right)$

$\implies$

$-x+5+5x-45=2x+16-2x-5$ ...[By Distribution Law]

$\implies$

$4x-40=11$ ....[Simplifying]

$\implies$

$4x=11+40$ ...[Adding

$40$ on both sides]

$\implies$

$4x=51$

$\implies$

$x=\cfrac { 51 }{ 4 }$ .

Hence, option

$C$ is correct.

$\dfrac{19}{5x+17} = \dfrac{19}{31}$ , then find

$x$ .

0%
$0.4$
0%
$1.4$
0%
$2.8$
0%
$3.4$

Explanation

Given

$\dfrac{19}{5x+17} = \dfrac{19}{31}$

$\Rightarrow 19\times 31=19(5x+17)\\ \Rightarrow 31=5x+17\\ \Rightarrow 5x=14\\ \Rightarrow x=\dfrac {14}{5}=2.8$

$\cfrac{7}{m-\sqrt{3}} = \cfrac{\sqrt{3}}{m} + \cfrac{4}{2m}$ , calculate the value of

$m$ .

0%

$m=\dfrac { -4\sqrt { 2 } -6 }{ 10-2\sqrt { 3 } }$
0%

$m=\dfrac { -4\sqrt { 3 } -6 }{ 10-2\sqrt { 3 } }$
0%
$m=\dfrac { -4\sqrt { 3 } +6 }{ 10-2\sqrt { 3 } }$
0%
$m=\dfrac { -4\sqrt { 3 } -6 }{ 10-2\sqrt { -3 } }$

Explanation

Given

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

$\Rightarrow 14m=(4+2\sqrt { 3 } )m-\sqrt { 3 } (4+2\sqrt { 3 } )$

$\Rightarrow (10-2\sqrt { 3 } )m=-4\sqrt { 3 } -6$

$\Rightarrow m=\dfrac { -4\sqrt { 3 } -6 }{ 10-2\sqrt { 3 } }$

Find the value of

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

0%
$0.71$
0%
$3.57$
0%
$5.8$
0%
$45$

Explanation

Given

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

Taking RHS:

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

LCM is

$5x$

$\Rightarrow \dfrac { 5 }{ 5x } +\dfrac { 4 }{ 5x } \\ \Rightarrow \dfrac { 9 }{ 5x }$

Now taking LHS:

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

LHS

$=$ RHS

$\dfrac { 2 }{ x+5 } =\dfrac { 9 }{ 5x }$

$\Rightarrow 5x\times 2=9(x+5)\\ \Rightarrow 10x=9x+45\\ \Rightarrow x=45$

Find the value of

$\dfrac {4}{y} + 4$ given that

$\dfrac {4}{y} + 4 = \dfrac {20}{y} + 20$

0%
$-1$
0%
$0$
0%
$1$
0%
$4$

Explanation

The value of

$\frac{20}{y}+20=5\left ( \frac{4}{y}+4 \right )$

Given

$\frac{4}{y}+4=5\left ( \frac{4}{y}+4 \right )$

Given one is possible only when the value of

$\frac{4}{y}+4=0$

Wayne would like to buy a school jacket priced at

$\$81$ , but the price of the jacket is

$\$59$ more than he has. In which of the following equations does

$x$ represent the number of dollars Wayne has?

0%
$x + 81 = 59$
0%
$x - 81 = 59$
0%
$x - 59 = - 81$
0%
$x - 81 = - 59$
0%
$x - 59 = 81$

Explanation

The price of jacket is

$$81$ .

Let the number of dollars Wayne had be

$x$ .

Given that

$81-x = 59$

Which implies

$x-81 = -59$

$\dfrac{3}{9}=\dfrac{3}{x+2}$ , what is the value of

$x$ ?

0%
$-\dfrac{5}{9}$
0%
$\dfrac{7}{3}$
0%
$3$
0%
$7$
0%
None of the above

Explanation

Given,

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

On cross multiplying, we get

$\Rightarrow 3(x+2)=9(3)$

$\Rightarrow (x+2)=(3)$

$\Rightarrow x=9-2$

$\Rightarrow x=7$

$\dfrac {2}{3x + 12} = \dfrac {2}{3}$ , then the value of

$x + 4$ is

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

Explanation

Given

$\dfrac { 2 }{ 3x+12 } =\dfrac { 2 }{ 3 }$

$\Rightarrow 3(2)=2(3x+12)$

$\Rightarrow 6=6x+24$

$\Rightarrow 6x=-18$

$\Rightarrow x=-3$

Therefore

$x+4=-3+4=1$

$\dfrac{18\times 72\times x}{48\times 315} = \sqrt{81}$ , find the value of

$x$ .

0%
$115$
0%
$150$
0%
$15$
0%
$105$

Explanation

$\dfrac{18\times 72\times x}{48\times 315}=\sqrt{81}$

$\implies \dfrac{9\times 72 \times x}{24\times 315}=9$

$\implies \dfrac{9\times 3\times x}{315}=9$

$\implies x=\dfrac{9\times 315}{9 \times 3}$

$\implies x=105$

$13$ is added to one-half of a certain number, the result is

$37$ . What is the original number?

0%
$24$
0%
$40$
0%
$48$
0%
$61$

Explanation

Let the original number be

$'x'$ .

Given that

$13$ is added to one half of

$'x'$ and the result is

$37$ .

In the Mathematical Equation form, we can write the above statement as

$\dfrac {x}{2}$

$+$

$13$

$=$

$37$

$\Rightarrow \dfrac {x}{2}$

$=$

$37$

$-$

$13$

$\Rightarrow \dfrac {x}{2}$

$=$

$24$

$\Rightarrow x$

$=$

$24$

$\times$

$2$

$\Rightarrow x$

$=$

$48$

Therefore, the original number is

$'48'$ .

$10 + x$ is

$5$ more than

$10$ , what is the value of

$2x$ ?

0%
$-5$
0%
$5$
0%
$10$
0%
$25$

Explanation

Given that

$10$

$+$

$x$ is

$5$ more than

$10$

Formally, we can express this

$10$

$+$

$x$

$=$

$10$

$+$

$5$

$\Rightarrow x$

$=$

$15$

$-$

$10$

$\Rightarrow x$

$=$

$5$

To find

$2x$ ,

$2x$

$=$

$2$

$\times$

$x$

$=$

$2$

$\times$

$5$

$=$

$10$

Therefore, the value of

$'2x'$ is

$'10'$ .

A triangle has a perimeter of

$13$ and one side of length

$3$ . If the lengths of the other two sides are equal, what is the length of each of them?

0%
$4$
0%
$5$
0%
$6$
0%
$7$
0%
$8$

Explanation

Let

$'x'$ be the length of one of the equal sides.

The perimeter of the triangle with sides a

$, b$ and

$c$ is

$P=a+b+c$ .

Substitute the values pf

$P=13$ and

$a=3, b=x , c=x$ in

$P=a+b+c$ as shown below:

$\Rightarrow 13=3+x+x$

$\Rightarrow 13=3+2x$

$\rightarrow 2x=13-3$

$\Rightarrow 2x=10$

$\Rightarrow x=5$

Therefore, the length of each of the side is

$5$ .

Hence, option B is correct.

A number is doubled and

$9$ is added. If the result is tripled it becomes

$75$ . What is that number?

0%
$3.5$
0%
$6$
0%
$8$
0%
$7$

Explanation

Let

$x =$ the number
So according to given conditions,

$3(2x+9)=75$

$\Rightarrow 6x+27=75$

$\Rightarrow 6x=75-27$

$\Rightarrow 6x=48$

$\Rightarrow x= \dfrac{48}{6}$

$\Rightarrow x= 8$

Hence. the number is

$8$ .

The acute angles of a right triangle have a ratio of

$2:3$ . What is the difference between the two angle measures?

0%
$42$ degrees
0%
$54$ degrees
0%
$18$ degrees
0%
$72$ degrees

Explanation

Use the fact that the sum of the angles in any triangle

$= 180^o$ degrees.

Let the measure of angle be

$2x$ and

$3x$

$2x + 3x + 90^o = 180^o$ (Combine like-terms).

$\Rightarrow 5x + 90^o = 180^o$ (Subtract

$90$ from both sides).

$\Rightarrow 5x = 90^o$ (Divide both sides by

$5$ ).

$\Rightarrow x = 18^o$

$\Rightarrow 2x = 36^o$

$\Rightarrow 3x = 54^o$

The measures of the acute angles in a right triangle whose ratio is

$2:3$ are:

$36$ degrees and

$54$ degrees.

Therefore, difference between the two angles

$=54^o-36^o=18^o$

Hence, option C is correct.

What is the solution of

$\displaystyle \frac{x-5}{2} - \frac{x-3}{5} = \frac{1}{2}$ ?

0%
$x = 5$
0%
$x = 7$
0%
$x = 8$
0%
$x = 9$

Explanation

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

$\dfrac{5x-25-2x+6}{5\times 2}=\dfrac{1}{2}$

$3x-19=5$

$3x=24$

$x=8$

Kiran scored

$80$ marks in his Science test which is

$\dfrac { 5 }{ 6 }$ of the total marks. What are the total marks for the test?

0%
$96$
0%
$100$
0%
$104$
0%
$108$

Explanation

Given : Kiran scored

$80$ marks in his science test which is

$\dfrac{5}{6}$ of total marks

Let the total marks be

$x$

Then

$\dfrac{5}{6}$ of total marks scored by Kiran in science

$=\dfrac{5}{6}\times x=\dfrac{5x}{6}$

But Kiran scored

$80$ marks

$\therefore \dfrac{5x}{6}=80$

Multiplying by

$6$ on both the sides, we get

$\dfrac{5x}{6}\times 6=80\times 6$

$5x=480$

Dividing by

$5$ on both sides, we get

$\dfrac{5x}{5}=\dfrac{480}{5}$

$\Rightarrow x=96$

Thus, the total marks is

$96$ .

The acute angles of a right triangle have a ratio of

$12 : 3.$ What is the difference between the two angle measures?

0%
$42^\circ$
0%
$54^\circ$
0%
$64^\circ$
0%
$72^\circ$

Explanation

Let the measure of acute angles be

$12x$ and

$3x$ respectively.

Now, we will apply the angle sum property of a triangle,

$12x+3x+90=180$

$\Rightarrow 15x+90^\circ=180^\circ$

$\Rightarrow 15x=90^\circ$

$\Rightarrow x=6^\circ$

So,

$\Rightarrow 12x=12\times6^\circ$

$=72^\circ$

$\Rightarrow 3x=3\times6^\circ$

$=18^\circ$

Therefore, the acute angles are

$72^\circ$ and

$18^\circ$ degrees.

The difference between these angles is

$72^\circ-18^\circ=54^\circ.$

The diagonal of a rectangle is thrice its smaller side. What is the ratio of its sides?

0%
$\sqrt {2} : 1$
0%
$2\sqrt {2} : 1$
0%
$3 : 2$
0%
$\sqrt {3} : 1$

Explanation

Let, The smaller side of rectangle be

$x$ m.

$\therefore$ the diagonal of rectangle is

$3x$ m.

$\therefore$ By Pythagorous theorem,

$(Diagonal)^2=(smaller \space side)^2+ (Greater \space side)^2$

$\implies (3x)^2= (x)^2+(Greater \space side)^2$

$\implies (Greater \space side)^2=9x^2-x^2=8x^2$

$\implies Greater side =\sqrt{8x^2}=2\sqrt2x$

$\therefore$ Required ratio=

$2\sqrt2x:x=2\sqrt2:1$

Two numbers are in the ratio

$3 : 8$ . If the sum of the numbers is

$165$ , then find the numbers.

0%
$45,120$
0%
$65,120$
0%
$35,130$
0%
None of these

Explanation

let the no. be

$3x$ and

$8x$

$3x+8x=165$

$\Rightarrow 11x=165$

$\Rightarrow x=15$

$\therefore$ No. are

$45$ and

$120$

A piece of string is

$40$ centimeters long. It is cut into three pieces. The longest piece is

$3$ times as long as the middle-sized and the shortest piece is

$23$ centimeters shorter than the longest piece. Find the length of the shortest piece(in cm).

0%
$27$
0%
$5$
0%
$4$
0%
$9$

Explanation

Let the length of largest piece be $3x$ .

Thus length of middle sized part will be $x$ .

According to the given condition, length of shortest part is $3x – 23$ .

Or, $3x + x + (3x – 23) = 40.$

$7x=63$

Or, $x = 9$

Or, the shortest piece $= 3(9) – 23 = 4$

CBSE Questions for Class 8 Maths Linear Equations In One Variable Quiz 9 - 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 9 - MCQExams.com

0:0:1

Practice Class 8 Maths Quiz Questions and Answers

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