Problems on Percentages
Posted by
Ravi Kumar at Monday, November 24, 2008
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Problems on Percentages
Problems on Percentages
Simple 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 Â¾
SOLUTION:
6 Â¾ =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. Â½ is what percent of 1/3?
SOLUTION:
=( Â½) / (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 Â½ % 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 Â½ % of 8400 â€“ 3 1/3 % of 8400
=(7/210/3)% 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 â€“ 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 xy=1660
5/3yy=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= 81.581.472=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=4540 = 5
17 . If 50% of (x â€“ y) = 30% of (x + y) then what percent
of x is y ?
SOLUTION:
50/100(xy) =30/100(x+y)
Â½ (xy)= 3/10(x+y)
5x5y=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 /
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) â€“ 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
==> 10000/79x100/x=10.5
==> 100007900=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 *(100x)/100 b=lb
160(100x)=100 *100
or 100x =10000/160
=125/2
so x = 100125/2
=75/2=37.5