Practice Paper on Percentages for CLAT 2020

  • 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?

  • There are 280 students in a school​​ out of which 65% of students are boys. Find the number of girls and boys in school.

  • 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.

  • 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.

  • If the price of rice decreased by 25% then Ram can buy 4 more kgs for Rs. 100. Find the new price of rice.

  • If 18% of one number is 27% of the other number. Difference between the two numbers is 1850. Find the numbers.

  • ​​ 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.

  • 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.

  • 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.

  • 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.

 

Answers and Solutions

  • Ans: 500

Solution:​​ 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

  • Ans: Boys- 182 and girls- 98

Solution:​​ Total no. of students = 280

As boys percentage is 65%,​​ 

  No. of boys = 280*65/100 = 182

  No of girls = 280-182 = 98

  • Ans: Rs. 8900

Solution:​​ 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

  • Ans: 6655

Solution:​​ 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

  • Ans: Rs. 6.25

Solution:​​ 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

  • Ans: 5550 and 3700

Solution:​​ 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

  • Ans: 10000

Solution:​​ 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

  • Ans: 66.66%

Solution:​​ 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%

  • Ans:​​ 15/7

Solution:​​ 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

  • Ans:​​ 30.33%

Solution:​​ 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%

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