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This mock test of Test: Percentages- 1 for Quant helps you for every Quant entrance exam.
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QUESTION: 1

Rajeev buys good worth Rs. 6650. He gets a rebate of 6% on it. After getting the rebate, he pays sales tax @ 10%. Find the amount he will have to pay for the goods.

Solution:

Rebate = 6 % of Rs. 6650 = Rs. (6 / 100 x 6650) = Rs. 399.

Sales tax = 10% of Rs.(6650 - 399) = Rs. (10 / 100 x 6251) = Rs.625.10

∴ Final amount = Rs. (6251 + 625.10) = Rs.6876.10

QUESTION: 2

30% of the men are more than 25 years old and 80% of the men are less than or equal to 50 years old. 20% of all men play football. If 20% of the men above the age of 50 play football, what percentage of the football players are less than or equal to 50 years?

Solution:

Let total number of men = 100

Then

80 men are less than or equal to 50 years old

(Since 80% of the men are less than or equal to 50 years old)

⇒ 20 men are above 50 years old (Since we assumed total number of men as 100)

20% of the men above the age of 50 play football

⇒ Number of men above the age of 50 who play football = 20 x 20 / 100 = 4

Number of men who play football = 20 (Since 20% of all men play football)

Percentage of men who play football above the age of 50 = 4/20 x 100 = 20%

=>Percentage of men who play football less than or equal to the age 50 = 100%−20%=80%

QUESTION: 3

Q As a percentage of p is equal to p as a percentage of [ p+ q] . Find q As a percentage of p

Solution:

Given, Q As a percentage of p is equal to p as percentage of ( p+ q)

i. e, ( q/ p) * 100 = [ p / (p+ q) ) ] * 100

Or, q/ p=p / ( p+ q) -----(1)

as q is some percentage of P, let's take , q = kp -- (2)

Putting (2) in (1) ;

kp / p= p /( p+ qp) or k = 1 /(1+k)-(3)

Solving the quadratic equation;

k^ 2 + k -1 = 0

We get, k= -1 +√5

Or k = -1 -√5

k = 1.24 /2 or -3.24/ 2 -- (4)

Ignoring the negative value, applying (1) in q As% of p = q/ p * 100

= k * 100

So, from equation - (4) ;

q As % of p = ( 1.24 / 2 ) * 100

= 62%

QUESTION: 4

A student multiplied a number by 3/5 instead of 5/3. What is the percentage error in the calculation?

Solution:

Let number is 100.

Actual Multiplication = (100 *5)/3 = 500/3.

Student multiplied = (100 *3)/5 = 300/5.

Error,

= (500/3) - (300/5)

= (2500 - 900)/15

= 1600/15

%Error = (1600 *100 *3)/(15*500) = 64%.

QUESTION: 5

If the price of petrol increases by 25% and Benson intends to spend only an additional 15% on petrol, by how much % will he reduce the quantity of petrol purchased?

Solution:

Assume that the initial price of 1 Litre petrol = Rs.100 ,Benson spends Rs.100 for petrol,

such that Benson buys 1 litre of petrol

After the increase by 25%, price of 1 Litre petrol = 100 x (100 + 25) / 100 = Rs. 125

Since Benson spends additional 15% on petrol,

amount spent by Benson =

100 x (100 + 15) / 100 = Rs.115

Hence Quantity of petrol that he can purchase = 115/125 Litre

Quantity of petrol reduced =

(1 - 115 / 125) Litre

Percentage Quantity of reduction =

(1 - 115 / 125) / 1 x 100

= 10 / 125 x 100 = 10 / 5 x 4 = 2 x 4 = 8%

QUESTION: 6

On my sister's 15th birthday, she was 159 cm in height, having grown 6% since the year before. How tall was she the previous year?

Solution:

Given that height on 15th birthday = 159 cm and growth = 6%

Let the previous year height = x

Then height on 15th birthday =

x x (100 + 6) / 100 = x + 106 / 100

⇒ 159 = x x 106 / 100

⇒ x = 159 x 100 / 106 = 1.5 x 100 = 150 cm

QUESTION: 7

P is able to do a piece of work in 15 days and Q can do the same work in 20 days. If they can work together for 4 days, what is the fraction of work left?

Solution:

1day's work of p = 1 / 15

1day's work of q = 1 / 20

1day's work of both = 7 / 60

4days work = 7 / 15

remaining work = 1 - 7 / 15 = 8 / 15

QUESTION: 8

In a certain school, 20% of students are below 8 years of age. The number of students above 8 years of age is 2/3 of the number of students of 8 years of age which is 48. What is the total number of students in the school?

Solution:

Let the number of students be *x*. Then,

Number of students above 8 years of age = (100 - 20)% of *x* = 80% of *x*.

∴ 80% of x = 48 + 2 / 3 of 48

⇒ 80 / 100 x = 80

⇒ x = 100.

QUESTION: 9

Arun got 30% of the maximum marks in an examination and failed by 10 marks. However, Sujith who took the same examination got 40% of the total marks and got 15 marks more than the passing marks. What were the passing marks in the examination?

Solution:

Let x is the maximum marks of the examination

Marks that Arun got = 30 % of x = 30x / 100

Given that Arun failed by 10 marks

⇒ Minimum Pass Mark =

30x / 100 + 10....(Equation 1)

Marks that Sujith got = 40 % of x = 40x / 100

Given that Sujith got 15 marks more than the passing marks

⇒ 40x / 100 = Minimum Pass Mark + 15

⇒ Minimum Pass Mark = 40x / 100 - 15...(Equation 2)

From equations 1 and 2, we have

30x / 100 + 10 = 40x / 100 - 15

⇒ 10x / 100 = 10 + 15 = 25

⇒ x / 10 = 25

⇒ Maximum marks of the examination = x = 250

Substituting the value of x in Equation 1, we have

Minimum Pass Mark = 30x / 100 + 10 = 30 x 250 / 100 + 10 = 75 + 10 = 85

QUESTION: 10

The population of a town increased from 1,75,000 to 2,62,500 in a decade. The average percent increase of population per year is:

Solution:

Increase in 10 years = (262500 - 175000) = 87500

Increase% = (87500 / 175000 x 100) % = 50%

∴ Required average = (50 / 10)% = 5%

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