MCQ Questions for Class 8 Maths Linear Equations In One Variable Quiz 4

What should be subtracted from

(\frac{3}{4} - \frac{2}{3})

$\displaystyle\left(\displaystyle\frac{3}{4}-\frac{2}{3}\right)$ to get

\frac{- 1}{6}

$\displaystyle\frac{-1}{6}$ ?

0%
$\displaystyle -\frac{6}{13}$
0%
$\displaystyle\frac{1}{4}$
0%
$\displaystyle\frac{2}{7}$
0%
$\displaystyle -\frac{1}{8}$

Explanation

From the given question, we can write that

\frac{3}{4} - \frac{2}{3} - a = - \frac{1}{6}

$\dfrac{3}{4} - \dfrac{2}{3} -a = -\dfrac{1}{6}$

We have to find

a

$a$

a = \frac{3}{4} - \frac{2}{3} + \frac{1}{6}

$a=\dfrac{3}{4}-\dfrac{2}{3}+ \dfrac{1}{6}$

a = \frac{3 \times 3 - 2 \times 4 + 1 \times 2}{12}

$a=\dfrac{3 \times 3 - 2 \times 4 + 1 \times 2}{12}$

a = \frac{9 - 8 + 2}{12}

$a=\dfrac{9-8+2}{12}$

a = \frac{3}{12}

$a=\dfrac{3}{12}$

a = \frac{1}{4}

$a=\dfrac{1}{4}$

Solve for

x

$x$ :

\frac{9 x}{7 - 6 x} = 15

$\dfrac {9x}{7-6x}=15$

0%
$\dfrac{99}{105}$
0%
$-\dfrac{105}{99}$
0%
$\dfrac{105}{99}$
0%
$-\dfrac{99}{105}$

Explanation

\frac{9 x}{7 - 6 x} = 15

$\dfrac{9x}{7 - 6x} = 15$

9 x = 15 (7 - 6 x) = 105 - 90 x

$9x = 15(7 - 6x) = 105 - 90x$

9 x + 90 x = 105

$9x + 90x = 105$

99 x = 105

$99x = 105$

x = \frac{105}{99}

$x = \dfrac{105}{99}$

The sum of three consecutive multiples of

11

$11$ is

363

$363$ . Find these multiples.

0%
$110,121,132$
0%
$99,77,187$
0%
$110,99,154$
0%
none

Explanation

Let the three consecutive multiples of

11

$11$ are

11 x,

$11x,$

11 x + 11

$11x+11$ and

11 x + 22

$11x+22$

Their sum is

363

$363$

11 x + 11 x + 11 + 11 x + 22 = 363

$11x+11x+11+11x+22=363$

33 x + 33 = 363

$33x+33=363$

33 x = 363 - 33

$33x=363-33$

33 x = 330

$33x=330$

x = \frac{330}{33}

$x=\dfrac{330}{33}$

x = 10

$x=10$

First multiple

= 11 x = 11 \times 10 = 110

$=11x=11\times 10=110$

Second multiple

= 11 x + 11 = 110 + 11 = 121

$=11x+11=110+11=121$

Third multiple

= 11 x + 22 = 110 + 22 = 132

$=11x+22=110+22=132$

One-third of one half of one-fifth of a number is

15

$15$ then the number is

0%
$50$
0%
$150$
0%
$15$
0%
$450$

Explanation

Let the no. is

x

$x$

One-third of one half of one fifth of

x

$x$ is

15

$15$

That is,

\frac{1}{3}

$\dfrac{1}{3}$ of

\frac{1}{2}

$\dfrac{1}{2}$ of

\frac{1}{5}

$\dfrac{1}{5}$ of

x = 15

$x=15$

\frac{1}{3} \times \frac{1}{2} \times \frac{1}{5} \times x = 15

$\dfrac{1}{3}\times \dfrac{1}{2}\times \dfrac{1}{5}\times x=15$

\frac{1}{30} \times x = 15

$\dfrac{1}{30}\times x=15$

x = 15 \times 30 = 450

$x=15\times 30=450$

\frac{5.6 (x + 3)}{0.7} = 24,

$\cfrac { 5.6(x+3) }{ 0.7 } =24,$ then the value of

x

$x$ is

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

Explanation

$5.6\dfrac{(x+3)}{0.7}=24$

$5.6x+16.8=16.8$

$x=0$

\sqrt{1 + \frac{x}{289}} = 1 \frac{1}{17}

$\sqrt { 1+\dfrac { x }{ 289 } } =1\dfrac { 1 }{ 17 }$ then

x =

$x=$

0%
$1$
0%
$13$
0%
$35$
0%
$15$

Explanation

We have,

$\sqrt{1+\dfrac{x}{289}}=1\dfrac{1}{17}$

$\sqrt{\dfrac{289+x}{289}}=\dfrac{18}{17}$

On squaring both sides, we get

${{\left( \sqrt{\dfrac{289+x}{289}} \right)}^{2}}={{\left( \dfrac{18}{17} \right)}^{2}}$

$\dfrac{289+x}{289}=\dfrac{324}{289}$

$289+x=324$

$x=324-289$

$x=35$

Find three consecutive numbers whose sum is

$108$ .

0%
$35, 46, 37$
0%
$5, 36, 37$
0%
$35, 36, 37$
0%
None of these

Explanation

Let, the three consecutive numbers be

$x-1, x, x+1.$

According to question,

$x-1 + x + x+1 = 108$

$\Rightarrow 3x = 108$

$\Rightarrow x = \dfrac{108}{3}$

$\Rightarrow x =36$

Hence, the required numbers are,

$35, 36, 37.$

Solve for

$x$ :-

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

0%
$x=\dfrac{5}{14}$
0%
$x=\dfrac{3}{14}$
0%
$x=\dfrac{33}{14}$
0%
None of these

Explanation

$\dfrac{{2x - 1}}{2}\,\,\, - \dfrac{{x + 3}}{3}\,\, = \dfrac{{x - 2}}{5}\Rightarrow \dfrac{6x-3-2x-6}{6}=\dfrac{x-2}{5}\Rightarrow 20x-45=6x-12\Rightarrow 14x=33\Rightarrow x=\dfrac{33}{14}$

$4$ more than twice a number is

$6$ less than that number then what is the number?

0%
$2$
0%
$-10$
0%
$10$
0%
$16$

Explanation

Lets assume the number is

$x$ .

According to the question statement, we get

$4+2x=x-6$

$2x-x = -6-4$ .

$x = -10$ .

If you subtract

$\cfrac{1}{2}$ from a number and multiply the result by

$\cfrac{1}{2}$ , you get

$\cfrac{1}{8}$ . The number obtained is

0%
$\dfrac 39$
0%
$\dfrac 34$
0%
$\dfrac {3}{13}$
0%
$\dfrac 32$

Explanation

Let the number be

$x$

We need to subtract

$\dfrac {1}{2}$ from

$x$ and multiply the resultant by

$\dfrac {1}{2}$ ,

So, we get

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

The resultant is equal to

$\dfrac {1}{8}$

$\left (x-\dfrac {1}{2}\right)\dfrac {1}{2}=\dfrac {1}{8}$

$(2x-1)\cfrac { 1 }{ 4 } = \cfrac { 1 }{ 8 }$

$4x - 2 = 1$

$4x = 3$

$x = \cfrac { 3 }{ 4 }$

Half of a herd of deer are grazing in the field and three fourths of the remaining are playing nearby. The rest 9 are drinking water from the pond. Find the number of deer in the herd.

0%
65
0%
36
0%
72
0%
74

Explanation

Let the total no of deer be

$x$

So, we have

$x - \left(\dfrac { 1 }{ 2 } x +\dfrac { 3 }{ 4 } \times \dfrac { 1 }{ 2 } x\right) =9\\ x-\left(\dfrac { 1 }{ 2 } x +\dfrac { 3 }{ 8 } x\right) =9\\ x-\dfrac { 7 }{ 8 } x =9\\ \dfrac { 1 }{ 8 } x =9;$

Thus,

$x =72$ deer in the herd.

It costs Rs.

$10$ /kilometer to fly and Rs.

$2$ /kilometer to drive. If one travels

$200$ km covering

$x$ km of the distance by flying and the rest by driving, then the cost of the trip is

0%
Rs. $2,000$
0%
Rs. $24,000$
0%
Rs. $(8x + 400)$
0%
Rs. $(12x + 400)$

Explanation

Let distance travelled by flying is

$x$ km.
Hence, the distance travelled by driving

$= (200-x)$ km.
The cost of flying

$=$ Rs.

$10$ per km
Hence, the cost of flying

$x$ km

$= 10 \times x =10x$
The cost of driving

$=$ Rs.

$2$ per km
The cost of driving

$(200-x)$ km

$= 2 \times (200-x) = 400 -2x$
Hence, the total cost

$=10x + 400-2x =$ Rs.

$8x+400$

Baichungs father is

$26$ years younger than Baichungs grandfather and

$29$ years older than Baichung. The sum of the ages of all the three is

$135$ years. What is the age of each one of them?

0%
Baichungs age: 14 years; Baichungs fathers age: $28$ years; Baichungs grandfathers age $= 42$ years
0%
Baichungs age: $17$ years; Baichungs fathers age: $46$ years; Baichungs grandfathers age $= 72$ years
0%
Baichungs age: $16$ years; Baichungs fathers age: $38$ years; Baichungs grandfathers age $= 52$ years
0%
Baichungs age: $25$ years; Baichungs fathers age: $58$ years; Baichungs grandfathers age $= 82$ years

Explanation

Let the father's age be

$x$ .

Grandfather's age

$= x+26$

Baichung age

$= x -29$

Sum of ages

$= 135$

$x+26 + x + x -29 = 135$

$3x = 135 +3 = 138$

$x = 46$

So, grandfather's age

$= 46 + 26 = 72$ yrs

Father's age

$= 46$ yrs

Baichungs age

$= 46 - 29 = 17$ yrs

The present ages of a father and his son are in the ratio

$7 : 3$ and the ratio of their ages will be

$2 : 1$ after

$10$ years. Then, the present age of father (in years) is

0%
$42$
0%
$56$
0%
$70$
0%
$77$

Explanation

Let the present ages of father and son be

$7x$ and

$3x$ .

After

$10$ years, their ages will be

$7x+10$ and

$3x+10$ .

According to the question, we have

$\dfrac {7x+10}{3x+10}=\dfrac {2}{1}$

$1(7x+10)=2(3x+10)$

$7x+10=6x+20$

$7x-6x=20-10$

$x=10$

Then present age of father is

$7x$ , i.e.

$7\times 10=70$ years.

The ages of Rahul and Haroon are in the ratio

$5:7$ . Four years later the sum of their ages will be

$56$ years. What is the present age of rahul?

0%
$14$
0%
$18$
0%
$20$
0%
$11$

Explanation

Let the ages of Rahul and Haroon be

$5x$ and

$7x$ .

Four years later, their ages will be

$5x+4$ and

$7x+4$ .

Given, the sum of their ages is

$56$ .

$\therefore 5x+4+7x+4 = 56$

$12x = 56 -8$

$12 x = 48$

$\therefore x = 4$

Therefore, age of Rahul will be

$5x=5\times 4=20$ and that of Haroon will be

$7x=7\times 4=28$ .

Thus, Rahul is

$20$ years and Haroon is

$28$ years old.

Lakshmi is a cashier in a bank. She has currency notes of denominations Rs.

$100$ , Rs.

$50$ and Rs.

$10$ , respectively. The ratio of the number of these notes is

$2:3:5$ . The total cash with Lakshmi is Rs.

$4,00,000$ . How many notes of each denomination does she have?

0%
Rs. $100$ $\rightarrow 2000$ notes; Rs. $50 \rightarrow 3000$ notes; Rs. $10 \rightarrow 5000$ notes
0%
Rs. $100 \rightarrow 1500$ notes; Rs. $50 \rightarrow 2000$ notes; Rs. $10\rightarrow 3000$ notes
0%
Rs. $100 \rightarrow 1000$ notes; Rs. $50 \rightarrow 2500$ notes; Rs. $10 \rightarrow 4000$ notes
0%
Rs. $100 \rightarrow 3000$ notes; Rs. $50 \rightarrow 2000$ notes; Rs. $10 \rightarrow 4000$ notes

Explanation

Let the number of

$100, 50$ and

$10$ Rs notes be

$2x, 3x, 5x$ .

Total cash

$=$ Rs.

$4,00,000$

$\therefore 2x(100) + 3x(50) + 5x(10) =4,00,000$

$200x +150x +50x = 4,00,000$

$400x = 4,00,000$

$\therefore x = 1,000$

Noumber of Rs.

$100$ notes

$= 2 \times 1000=2000$

Number of Rs.

$50$ notes

$=3 \times 1000= 3000$

Number of Rs.

$10$ notes

$=5\times 1000= 5000$

One of the two digits of a two digit number is three times the other digit. If you interchange the digits of this two-digit number and add the resulting number to the original number, you get

$88$ . What is the original number?

0%
$26$
0%
$12$
0%
$43$
0%
$16$

Explanation

Let one of the digits be

$x$ .

Then the other digit is

$3x$ .

Original number

$= 10x +3x = 13x$

Interchanged number

$= 10(3x) +x = 31x$

$\Rightarrow 13x+31x = 88$

$\Rightarrow 44x = 88$

$\Rightarrow x = 2$

The other digit

$= 3x= 3\times 2 = 6$

Therefore the original number is

$26$ .

The sum of three non-zero prime numbers is

$100$ . One of them exceeds the other by

$36$ . Then the largest number is

0%
$73$
0%
$91$
0%
$67$
0%
$57$

Explanation

The sum of three prime numbers can only be even if one of the prime numbers is even, the only prime number that is even is

$2$ .
So, the sum of the other two prime numbers must be

$98$ .

$x+x+36=98$

$\Rightarrow 2x=62$

$\Rightarrow x=31$
Numbers are

$2, 31$ and

$67$ .
The highest among them is

$67$ .

Three consecutive integers are such that when they are taken in increasing order and multiplied by

$2, 3$ and

$4$ respectively, they add up to

$74$ . Find these numbers.

0%
$4, 5, 6$
0%
$5, 6, 7$
0%
$8, 9, 10$
0%
$7, 8, 9$

Explanation

Let the consecutive integers be

$x, x+1, x+2$

Given that they are taken in increasing order and multiplied by

$2,3,4$ .

$\therefore (x)2 + 3(x+1) + 4(x+2) = 74$

$2x+3x+3 + 4x + 8 = 74$

$9x = 74 -11 = 63$

$\therefore x = 7$

Then, the numbers would be

$x=7$ ,

$x+1=7+1=8$ ,

$x+2=7+2=9$ .

So, the integers are

$7, 8, 9$ .

The sum of three consecutive odd numbers is

$57$ . Find the numbers.

0%
$17, 19$ and $21$
0%
$19, 15$ and $25$
0%
$13, 17$ and $26$
0%
$21, 13$ and $29$

Explanation

Let the

$3$ consecutive odd numbers be

$x, x+2, x+4$

$x+x+2 +x+4 = 57\\ 3x = 51\\ x = 17$

So the numbers are

$17, 19$ and

$21.$

Three consecutive whole numbers are such that if they be divided by

$5, 3$ and

$4$ respectively; the sum of their quotients is

$40$ . Find the numbers.

0%
$20, 35$ and $22$
0%
$50, 51$ and $52$
0%
$40, 11$ and $25$
0%
$5, 5$ and $8$

Explanation

Let the three consecutive whole numbers be

$x,x+1,x+2$
Now, according to question

$\cfrac x5+\cfrac{x+1}3+\cfrac{x+2}4=40$

$\Rightarrow 12x+20x+20+15x+30=2400$

$\Rightarrow 47x=2350$

$\Rightarrow x=50$
Hence, the three consecutive whole numbers are

$50,51$ and

$52$ .

A person travelled

$\displaystyle \frac{5}{8}$ th of the distance by train

$\displaystyle \frac{1}{4}$ th by bus and the remaining 15 km by boat .The total distance travelled by him was

0%
$90$ km
0%
$120$ km
0%
$150$ km
0%
$180$ km

Explanation

Let total distance be x km

$\displaystyle x-\left ( \frac{5x}{8}+\frac{x}{4} \right )=15$

$\displaystyle x-\frac{7x}{8}=15$
x = 120 km

A student was asked to divide a number by

$6$ and add

$12$ to the quotient. He, however, first added

$12$ to the number and then divided it by

$6$ , getting

$112$ as the answer. The correct answer should have been.

0%
$122$
0%
$118$
0%
$114$
0%
$124$

Explanation

Let the number be

$x$ .

Then as per the given condition, the student was asked to find out the result of

$\dfrac {x}{6}+12$ ....(1)

But he write it as

$\dfrac {x+12}{6}$ and got result as

$112$ .

Therefore,

$\dfrac {x+12}{6}=112$

$\Rightarrow x+12=112\times 6$

$\Rightarrow x=660$

So, the number is

$660$

So, the correct result would be

$\dfrac {x}{6}+12=\dfrac {660}{6}+12=122$ .

Hence, option A is correct.

The sum of the digits of a two digit number is

$6$ and its ten's digit is twice its unit digit. Find the number.

0%
$58$
0%
$20$
0%
$42$
0%
$76$

Explanation

Let the unit's digit be

$x$ , then the ten's digit =

$2x$

Given sum of digits

$=6$

$\implies x+2x = 6\\\;\;\;\;\;\;\;\;\; 3x = 6\\ \;\;\;\;\;\;\;\;\; x=2$

Ten's digit

$= 2x = 4$

Required number =

$10(2x)+1(x)=10(4)+1(2)=42$

Separate

$178$ into two parts so that the first part is

$8$ less than twice the second part.

0%
$126$ and $42$
0%
$176$ and $102$
0%
$126$ and $82$
0%
$116$ and $62$

Explanation

Let the second part be

$x$

First part is

$2x-8$ .

Therefore from question, we have

Sum of two parts is

$178$

Therefore,

$x+2x-8=178$

$\Rightarrow 3x=178+8$

$\Rightarrow 3x=186$

$\Rightarrow x=62$

First part is

$2x-8=2(62)-8=124-8=116$

Therefore, the first and second parts are

$116$ and

$62$ .

The ten's digit of a two digit number is three times the unit digit. The sum of the number and the unit digit is

$32$ . Find the number.

0%
$31$
0%
$12$
0%
$19$
0%
$39$

Explanation

Let the units digit be

$x$ .

The tens digit

$= 3x$ .

Original number

$= 10(3x) +x = 31x$

Sum of number and unit digit

$= 32$

$31x + x = 32\\ \Rightarrow 32x = 32\\ \Rightarrow x = 1$

Original number is

$31(x) = 31$ .

Twice a number is decreased by

$15$ , equals

$25$ . Find the number.

0%
$20$
0%
$25$
0%
$30$
0%
$35$

Explanation

Let the number be

$x$

According to given problem, we have

$2x-15=25$

$2x=25+15$

$2x=40$

$\therefore x=20$

$28$ is

$12$ less than

$4$ times a number. Find the number.

0%
$10$
0%
$12$
0%
$14$
0%
$8$

Explanation

Let the number be

$x$

From question, we know

$28$ is

$12$ less than

$4$ times a number.

Therefore,

$28=4x-12$

$\therefore 4x=28+12$

$\therefore x=\dfrac {40}{4}=10$

Therefore, the number is

$10$ .

AB is a segment. The point P is on the perpendicular bisector of segment AB such that length of AP exceeds length of AB by

$7$ cm. If the perimeter of

$\Delta ABP$ is

$38$ cm. Find the sides of

$\Delta ABP$ .

0%
$5$ cm, $12$ cm and $19$ cm
0%
$8$ cm, $15$ cm and $15$ cm
0%
$2$ cm, $17$ cm and $11$ cm
0%
$6$ cm, $14$ cm and $18$ cm

Explanation

Let the length of the

$AB = x$

$\therefore AP = x+7 = BP$

$\because p$ is on perpendicular bisector.

The perimeter of the

$\Delta ABP = AB+AP+BP = x+x+7+x+7 = 38 cm$

On solving, it gives

$x = 8cm$

$\therefore$ three sides are

$8, 8+7, 8+7 = 8cm, 15cm, 15 cm$

If Megha's age is increased by three times her age, the result is 60 years. Find her age.

0%
$11$ years
0%
$20$ years
0%
$15$ years
0%
$19$ years

Explanation

Let Megha's age be

$x$ years

Given that if Megha's age is increased by three times, the result is

$60$ years.

$\therefore x + 3x = 60$

$\Rightarrow 4x = 60$

$\therefore \ x = 15$

Hence, Megha's age is

$15$ years.

CBSE Questions for Class 8 Maths Linear Equations In One Variable Quiz 4 - 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 4 - MCQExams.com

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Practice Class 8 Maths Quiz Questions and Answers

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