business mathematics

Percentage means "expressing one quantity as per cent".


1.To convert a common fraction into per cent, we multiply the fraction with 100. And put % symbol to the result.


2.To convert a percent into fraction we must remove % symbol., divide it with 100 and simplify the product.


3.If the population of a town(or the length of a tree) is P and annual increase is r% then:
*population after n years = P[1+(r/100)]^n
*population n years ago = P/(1+r/100)^n


4.If the population of a town(or the length of a tree) is P and annual decrease is r% then:
*population after n years = P[1-(r/100)]^n
*population n years ago = P/(1-r/100)^n

PERCENTAGES - PROFIT & LOSS practice problems:

For percentage concept click: http://business-maths.blogspot.com/2008/08/percentages.html
For percentage solved problems click: http://business-maths.blogspot.com/2008/11/problems-on-percentages.html
For profit and loss concept click here
For profit and loss solved problems click here

Practice problems:

1.In a criket tournament , a team wanted its over all success rate to be 75%.After
playing half the total number of matches,it recorded only 50 % success.What per-
centage of the remaining matches should it win to achieve an overall success rate
of 75%?
a)50% b)75% c)100% d)80%


2.A man sells shoes by first quoting the selling price at a certain mark up on the
cost price.He later offers a discount on this price such that the discount
percent is the same as the mark up percent.If his cost price per pair of shoes is
Rs.300,then he makes an overall___in the transaction
a)profit b)loss c)no profit no loss d)cannot be determined

3.The respective scores of two candidates in an examination are 50% and 80% of the
maximum possible.The second candidate passes by 60 marks where as the first one
fails by 120 marks.What is the pass mark?
a)360 b)420 c)480 d)300

4.Two dining sets are sold at the same selling price resulting in a 20% gain on the
first and and 1 30% gain on the second.By what is the cost price of the fist more
than that of the second?
a)8 1/2% b)8 1/3% c)17% d)25%

5.By giving a discount of 30%,a trader makes a profit of 20%.If the trader gives a
discount of Rs 400,he would make a 20% loss.Find the marked price of the article
a)Rs 600 b)Rs 750 c)Rs 900 d)none of these

6.A bag contains balls of three different colours-Red,Blue and Green.The number of
blue balls is 12 1/2% less than the number of red balls and 40% more than the
number of green balls.By what percent is the sum of the blue and green balls less
than the sum of the red and blue balls
a)20% b)25% c)37.5% d)6 1/4%

7.The price of sugar is increased by 50%.By what percentage should the consumption
of sugar be reduced so that the expenditure on sugar is increased by 20%?
a)20% b)10% c)33 1/3% d)25%

8.A trader purchased two air conditioners.He sold both of them-one at 20% profit and
the other at 20% loss.Find the overall profit or loss percentage in the transaction
a)4% loss
b)4% profit
c)Neither profit nor loss
d)cannot be determined

9.If a profit made by selling an article at Rs.3000 is twice the loss incurred when
the article was sold at Rs.900,then find the cost price of the article
a)Rs 1800 b)Rs 1500 c)Rs 1600 d)Rs 1200

10.A trader,in an attempt to increase his sales,announces 20% discount on an article
and hence reduces his profit per article from Rs 60 to Rs 40.Find the cost price
of the article.
a)Rs 40 b)Rs 60 c)Rs 100 d)cannot be determined

11.Which of the following choices is a better bargain for a customer?
a)A single discount of 40 %
b)A discount of 25% followed by another discount of 15 %
c)Three successive disconts of 20 %,10 % and 10 % respectively
d)All the above are equally good

12.3(0.05% of 1.25) is what percent of 7.5% of 25*10^-3?
a)3% b)33 1/3% c)100 % d)1%

13.A and B start a business by investing some capital each.At the end of the year,A
receives 20% of the total profit as commission while the rest of the profit is
shared in the ratio of their respective investments.A finally receives 50% more
than what B receives.If A's capital is Rs.10,576,then B's capital is
a)Rs14,101 b)Rs 8813 c)Rs 6610 d)none of these

14.A trader,by selling 30 suitcases,gains the selling price of 5 suitcases.Find the
profit percentage per suitcase sold
a)16 2/3% b)28 4/7 % c)20% d)25%

15.A trader makes 44% profit after allowing a discount of 20% on the marked price.What is the profit percent,if a discount of 40% is allowed?
a)22% b)8% c)4.2% d)27.7%

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 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 /
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
=75/2=37.5