Practice Questions: Percentage and its Applications | Quantitative Techniques for CLAT PDF Download

Q1: A student needs 40% marks to pass the exam, but he scored 178 marks and failed by 22 marks. What is the maximum marks of the exam?
Ans: 500
Let maximum marks be x, hence passing marks would be
40x/100 = 2x/5
He scored and 178 and needed 22 marks to pass the exam hence passing marks would be = 178+22 = 200
Equating above equations we get, 2x/5 = 200
x= 500

Q2: There are 280 students in a school out of which 65% of students are boys. Find the number of girls and boys in school.
Ans: Boys- 182 and girls- 98
Total no. of students = 280
As boys percentage is 65%,
No. of boys = 280*65/100 = 182
No of girls = 280-182 = 98

Q3: A man spends 25% of the money he had, then Rs. 175 of the remaining and then again spends 10% of the remainder. After all the spending he is left with Rs. 5850.
Ans: Rs. 8900
Let take the initial amount of money that man had with him to be x.
As his first spending was 25% of the initial amount,  then remaining money would be (1-25/100)x = 0.75x
Now the second spending was Rs. 175 of the remaining amount of 0.75x,
Therefore remainder after this spending would be 0.75x-175
His third spending was of 10% of the remainder, therefore remaining amount will be 90% of 0.75x-175 i.e., 0.9(0.75x-175)
As given in question the amount left with the man is Rs. 5850, thus we can equate the equation; 0.9(0.75x-175) = 5850
0.75x – 175 = 5850/0.9
0.75x – 175 = 6500
0.75x = 6675
x= 8900

Q4: If the present population of a town is 5500 and it is increasing every year at the rate of 10%. Find the population of the town after 3 years.
Ans: 6655
Present population is 5000
In the first year the population increase by 10%, therefore new population would become 5000(1+10/100) = 5000*11/10.
In the second year again the increase is of 10%, therefore again the new population would become 5000*11/10(1+10/100) = 5000*11*11/100
Similarly, for the third-year population would be
5000*11*11*11/1000 = 6655

Q5: If the price of rice decreased by 25% then Ram can buy 4 more kgs for Rs. 100. Find the new price of rice.
Ans: Rs. 6.25
Let price be P and Quantity be Q
Then Expenditure, E = P*Q = 100 ….(i)
As per Question, If price is reduced by 25% then consumption quantity increases by 4 kgs, therefore
E = (P(100-25)/100)*(Q+4)     
P*Q = (P(100-25)/100)*(Q+4)  (from eq (i))
Q = 0.75(Q+4)
0.25Q = 0.75*4
Q= 12 kg
P=100/12
As the new price reduced 25% of original price, we can take 75% of original price to be new one
New Price = (100/12)*75/100 = Rs. 6.25

Q6: If 18% of one number is 27% of the other number. Difference between the two numbers is 1850. Find the numbers.
Ans: 5550 and 3700
Let the two numbers be a and b
18% of a = 27%  of b
18a/100 = 27b/100
18a=27b ⇒ 2a=3b ⇒ a = 3b/2
Difference between the two numbers is 1850, therefore a-b = 1850
3b/2 – b  = 1850 ⇒ b/2 = 1850
b = 3700
a = 5550

Q7: During 2018, the population of the town increased by 5% and in 2019 the population decreased by 5%. At the end of 2019, the population of the town was 9975. Find the total population of the town at the beginning of 2018.
Ans: 10000
Let the population at the beginning of 2018 be x.
Increment of 5 % = x(1+5/100)
Decrement of 5% = x(1+5/100) ( 1-5/100)
Population at the end of 2019=  x(1+5/100) ( 1-5/100)
= 9975
x(1+5/100) ( 1-5/100) = 9975
x(105/100)(95/100) = 9975
x = 9975*10000/(95*105)
x = 10000

Q8: If Sneha scored 30% in her first science exam out of 150, then how much per cent she must score in her second exam out of 180 to score 50% overall marks.
Ans: 66.66%
In first paper he scored 30% out of 150 i.e., 30*150/100 = 45
Let p be the per cent of marks Sneha must score in her second paper.
Total marks= 150+180 = 330
She must score 50% of total marks to pass the exam,
i.e., 330*1/2 = 165
As he must score p% out of 180 in addition with 45 marks of the first exam, hence the equation becomes
45 + p*180/100 = 165
p18/10 = 120
p = 1200/18 = 66.66%

Q9: If the numerator of a fraction is increased by 5% and the denominator is decreased by 10%, then the fraction becomes 5/2. Find the original fraction.
Ans: 15/7
Let numerator by n and denominator be d.
Numerator increased by 5% =  n(1+5/100)
Denominator decreased by 10% = d(1-10/100)
New fraction = 5/2
[n(105/100)]/[d(9/10)]=5/2
105n/90d = 5/2
n/d = 450/210 = 15/7

Q10: Mohan’s monthly income is Rs. 16000 and his expenditure is Rs. 13500. In the upcoming month, his income increased by 15% while his expenses increased by 9%. Find the increase in his savings.
Ans: 30.33%
Mohan’s income increased by 15%, therefore new income would be =  16000 * (1+15/100)
= 16000*1.15 = 18400
His Expenses increased by 9%,
Therefore new expenses would be = 11500*(1+9/100)
=  11500*1.09 = 12535
His previous savings = 16000-11500 = 4500
His new savings = 18400-12535 = 5865
The percentage increase of his savings, here change in savings would go in the numerator and we will keep his previous saving in the denominator as it is the base for percentage change,
= (5865 – 4500)*100/4500
= 1365/45
= 30.33%.