APTITUDE- Problems on PERCENTAGES

PERCENTAGES

EXAMPLE PROBLEMS:

1 . Express the following as a fraction.
a) 56%
SOLUTION:
56/100=14/25
b) 4%
SOLUTION:
4/100=1/25
c) 0.6%
SOLUTION:
0.6/100=6/1000=3/500
d) 0.08%
SOLUTION:
0.08/100=8/10000=1/1250

2.Express the following as decimals
a) 6%
SOLUTION:
6% = 6/100=0.06
b) 0.04%
SOLUTION:
0.04% = 0.04/100=0.0004

3 . Express the following as rate percent.
i).23/36
SOLUTION:
= (23/36*100) %
= 63 8/9%
ii).6 ¾
SOLUTION:
6 ¾ =27/4
(27/4 *100) % =675 %

4.Evaluate the following:
28% of 450 + 45% of 280 ?
SOLUTION:
=(28/100) *450 + (45/100) *280
= 28 * 45 / 5
= 252

5.2 is what percent of 50?
SOLUTION:
Formula : (IS / OF ) *100 %
= 2/50 *100
= 4%

6.½ is what percent of 1/3?
SOLUTION:
=( ½) / (1/3) *100 %
= 3/2 *100 %
= 150 %

7.What percent of 2 Metric tonnes is 40 Quintals?
SOLUTION:
1 metric tonne =10 Quintals
So required percentage = (40/(2*10)) *100 %
= 200%

8.Find the missing figure .
i) ? % of 25 = 2.125
SOLUTION :
Let x% of 25 = 2.125. then
(x/100) *25 =2.125
x = 2.125 * 4
= 8.5
ii) 9% of ? =6.3
SOLUTION:
Let 9 % of x = 6.3.
Then 9/100 of x= 6.3
so x = 6.3 *100/7
= 70.

9.Which is the greatest in 16 2/3 %, 2/15,0.17?
SOLUTION:
16 2/3 % = 50/3 %
=50/3 * 1/100
=1/6
= 0.166
2 / 15 =0.133
So 0.17 is greatest number in the given series.

10.If the sales tax be reduced from 3 ½ % to 3 1/3 % ,then what difference does it make to a person who purchases an article with marked price of RS 8400?
SOLUTION:
Required difference = 3 ½ % of 8400 – 3 1/3 % of 8400
=(7/2-10/3)% of 8400
=1/6 % of 8400
= 1/6* 1/100* 8400
= Rs 14.
11. A rejects 0.08% of the meters as defective .How many will he examine to reject 2?
SOLUTION:
Let the number of meters to be examined be x.
Then 0.08% of x=2.
0.08/100*x= 2
x= 2 * 100/0.08
=2 * 100 * 100/8
= 2500
12.65 % of a number is 21 less than 4/5 of that number. What is the number?
SOLUTION: Let the number be x.
4/5 x- (65% of x) = 21
4/5x – 65/100 x=21
15x=2100
x=140

13. Difference of two numbers is 1660.If 7.5 % of one number is 12.5% of the other number. Find two numbers?
SOLUTION:
Let the two numbers be x and y.
7.5% of x=12.5% of y
So 75x=125 y
3x=5y
x=5/3y.
Now x-y=1660
5/3y-y=1660
2/3y=1660
y=2490
So x= 2490+1660
=4150.
So the numbers are 4150 , 1660.

14. In expressing a length 81.472 KM as nearly as possible with 3 significant digits ,Find the % error?
SOLUTION:
Error= 81.5-81.472=0.028
So the required percentage = 0.028/81.472*100%
= 0.034%

15. In an election between two persons ,75% of the voters cast their votes out of which 2% are invalid. A got 9261 which 75% of the total valid votes. Find total number of votes?
SOLUTION:
Let x be the total votes.
valid votes are 98% of 75% of x.
So 75%(98%(75% of x))) = 9261
==> 75/100 *98 /100 * 75 100 *x = 9261
x= 1029 * 4 *100 *4 / 9
= 16800
So total no of votes = 16800

16 . A’s maths test had 75 problems i.e 10 arithmetic, 30 algebra and 35 geometry problems. Although he answered 70% of arithmetic , 40% of algebra and 60 % of geometry problems correctly he didn’t pass the test because he got less than 60% of the problems right. How many more questions he would have needed to answer correctly to get a 60% passing grade.

SOLUTION:
70% of 10 =70/100 * 10
=7
40% of 30 = 40 / 100 * 30
= 12
60 % of 35 = 60 / 100 *35
= 21
So correctly attempted questions = 7 + 12 + 21
=40.
Questions to be answered correctly for 60% grade
=60% of 75
= 60/100 *75
=45.
So required questions=45-40 = 5

17 . If 50% of (x – y) = 30% of (x + y) then what percent of x is y ?
SOLUTION:
50/100(x-y) =30/100(x+y)
½ (x-y)= 3/10(x+y)
5x-5y=3x+3y
x=4y
So Required percentage =y/x*100 %
=y/4y *100 %
= 25%.

18 . If the price of tea is increased by 20% ,find how much percent must a householder reduce her consumption of tea so as not to increase the expenditure?
SOLUTION:
Reduction in consumption= R/(100+R) *100%
=20/120 *100
= 16 2/3 %

19.The population of a town is 176400 . If it increases at the rate of 5% per annum ,what will be the population 2 years hence? What was it 2 years ago?
SOLUTION:
Population After 2 years = 176400[1+5/100]2
=176400 * 21/20 *21/20
=194481
Population 2 years ago = 176400/(1+5/100)2
= 176400 * 20/21 *20/ 21
=160000

20.1 liter of water is add to 5 liters of a 20 % solution of alcohol in water . Find the strength of alcohol in new solution?
SOLUTION:
Alcohol in 5 liters = 20% of 5
=1 liter
Alcohol in 6 liters of new mixture = 1liter
So % of alcohol is =1/6 *100=16 2/3%

21. If A earns 33 1/3 more than B .Then B earns less than A by what percent?
SOLUTION:
33 1/3 =100 / 3
Required Percentage = (100/3)/(100 + (100/3)) *100 %
= 100/400 *100 = 25 %

22. A school has only three classes which contain 40,50,60 students respectively . The pass percent of these classes are 10, 20 and 10 respectively . Then find the pass percent in the school.
SOLUTION:
Number of passed candidates =
10/100*40+20/100 *50+10/100 * 60
=4+10+6
=20
Total students in school = 40+50+60 =150
So required percentage = 20/150 *100
= 40 /3
=13 1/3 %

23. There are 600 boys in a hostel . Each plays either hockey or football or both .If 75% play hockey and 45 % play football ,Find how many play both?
SOLUTION:
n(A)=75/100 *600
=450
n(B) = 45/100 *600
= 270
n(A^B)=n(A) + n(B) – n(AUB)
=450 + 270 -600
=120
So 120 boys play both the games.

24.A bag contains 600 coins of 25p denomination and 1200 coins of 50p denomination. If 12% of 25p coins and 24 % of 50p coins are removed, Find the percentage of money removed from the bag ?
SOLUTION:
Total money = (600 * 25/100 +1200 *50/100)
=Rs 750
25p coins removed = 12/100 *600
=72
50p coins removed = 24/100 *1200
=288
So money removed =72 *1/4 +288 *1/2
= Rs 162
So required percentage=162/750 *100
=21 .6%

25. P is six times as large as Q.Find the percent that Q is less than P?
SOLUTION:
Given that P= 6Q
So Q is less than P by 5Q.
Required percentage= 5Q/P*100 %
=5/6 * 100 %
=83 1/3%

26.For a sphere of radius 10 cm ,the numerical value of surface area is what percent of the numerical value of its volume?
SOLUTION:
Surface area = 4 *22/7 *r2
= 3/r(4/3 * 22/7 * r3)
=3/r * VOLUME
Where r = 10 cm
So we have S= 3/10 V
=3/10 *100 % of V
= 30 % of V
So surface area is 30 % of Volume.

27. A reduction of 21 % in the price of wheat enables a person to buy 10 .5 kg more for Rs 100.What is the reduced price per kg.
SOLUTION:
Let the original price = Rs x/kg
Reduced price =79/100x /kg
==> 100/(79x/100)-100/x =10.5
==> 10000/79x-100/x=10.5
==> 10000-7900=10.5 * 79 x
==> x= 2100/10.5 *79
So required price = Rs (79/100 *2100/10.5 *79) /kg
= Rs 2 per kg.

28.The length of a rectangle is increased by 60 % .By what percent would the width have to be decreased to maintain the same area?
SOLUTION:
Let the length =l,Breadth= b.
Let the required decrease in breadth be x %
then 160/100 l *(100-x)/100 b=lb
160(100-x)=100 *100
or 100-x =10000/160
=125/2
so x = 100-125/2

If you have questions, please ask below

5 Comments


  1. Naidu says:

    please keep model problem to download in ur website.
    thank u

  2. Victor Samuel M says:

    Pls give the important formulae of that chapter before solving the problems…so that the student can easily get into that concept on his own.

  3. sweta says:

    need problems that include discount in percentage e.g.:Employees of a discount appliance store receive an additional 20% off of the lowest price on an item. If an employee purchases a dishwasher during a 15% off sale, how much will he pay if the dishwasher originally cost $450?

  4. SUBHADEEP DAS says:

    a gnerally wears his father’s coat. unfortunately his cousin b poked him one day that he was wearing a coat of length more than his height by 15%. if the length of a’s father coat is 120cm then find the actual length of his coat?

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