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

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  1. 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.

  2. 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?

  3. 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?