Percentage and its applications /COMMERCIAL MATHEMATICS

Percentage and ItsApplications
Notes
MODULE - 2
Commercial
Mathematics
Mathematics Secondary Course 203
8
PERCENTAGE AND ITS APPLICATIONS
You must have seen advertisements in newspapers, television and hoardings etc of the
followingtype:
“Sale, up to 60% off”.
“Voters turnout in the poll was over 70%”.
“Rameshgot93%aggregateinclassXIIexamination”.
“Banks have lowered the rate of interest on fixed deposits from 8.5% to 7%”.
In all the above statements, the important word is ‘percent’. The word ‘percent’ has been
derived from the Latin word ‘percentum’ meaning per hundred or out of hundred.
In this lesson, we shall study percent as a fraction or a decimal and shall also study its
applications in solving problems of profit and loss, discount, simple interest, compound
interest, rate of growth and depreciation etc.
OBJECTIVES
Afterstudyingthislesson,youwillbeableto
• illustrate the concept of percentage;
• calculate specified percent of a given number or a quantity;
• solve problems based on percentage;
• solve problems based on profit and loss;
• calculate the discount and the selling price of an article, given marked price of
the article and the rate of discount;
• solve inverse problems pertaining to discount;
• calculate simple interest and the amount, when a given sum of money is invested
for a specified time period on a given rate of interest;

Notes
MODULE - 2
Commercial
Mathematics
Mathematics Secondary Course204
• illustrate the concept of compound interest vis-a-vis simple interest;
• calculate compound interest, the amount and the difference between compound
and simple interest on a given sum of money at a given rate and for a given time
period; and
• solve real life problems pertaining to rate of growth and decay, using the formula
of compound interest, given a uniform or variable rate.
EXPECTED BACKGROUND KNOWLEDGE
• Fourfundamentaloperationsonwholenumbers,fractionsanddecimals.
• Comparisonoftwofractions.
8.1 PERCENT
Recall that a fraction
4
3
means 3 out of 4 equal parts.
13
7
means 7 out of 13 equal parts
and
100
23
means 23 out of 100 equal parts.
A fraction whose denominator is 100 is read as percent, for example
100
23
is read as
twenty three percent.
The symbol ‘%’ is used for the term percent.
A ratio whose second term is 100 is also called a percent,
So, 33 : 100 is equivalent to 33%.
Recallthatwhilecomparingtwofractions,
5
3
and
2
1
,wefirstconvertthemtoequivalent
fractionswithcommondenominator(L.C.M.ofthedenominators).
thus
10
6
2
2
5
3
5
3
=×= , and
10
5
5
5
2
1
2
1
=×=
Now, because
2
1
5
3
10
5
10
6
>∴>

Notes
MODULE - 2
Commercial
Mathematics
We could have changed these fractions with common denominator 100 as
%60or
100
60
20
20
5
3
5
3
=×=
%50or
100
50
50
50
2
1
2
1
=×=
and so,
2
1
5
3
> as 60% is greater than 50%.
8.2 CONVERSION OF A FRACTION INTO PERCENT AND
VICE VERSA
Intheabovesection,wehavelearntthat,toconvertafractionintopercent,wechangethe
fraction into an equivalent fraction with denominator 100 and then attach the symbol %
withthechangednumeratorofthefraction.Forexample,
%75
100
1
75
100
75
25
25
4
3
4
3
=×==×= and
%16
100
1
16
100
16
4
4
25
4
25
4
=×==×=
Note: To write a fraction as percent, we may multiply the fraction by 100, simplify
it and attach % symbol. For example,
%16%100
25
4
25
4
=×=
Conversely,
To write a percent as a fraction, we drop the % sign, multiply the number by
100
1
(or divide the number by 100) and simplify it. For example,
100
3
%3,
100
17
100
1
17%17,
100
47
100
1
47%47 ==×==×=
100
%,
10
21
100
210
%210,
20
9
100
45
100
1
45%45
x
x =====×= .

Notes
MODULE - 2
Commercial
Mathematics
8.3 CONVERSION OF DECIMAL INTO A PERCENT AND
VICE VERSA
Letusconsiderthefollowingexamples:
%35
100
1
35
100
35
35.0 =×==
%470
100
1
470
100
470
10
47
7.4 =×===
%9.45
100
1
10
459
1000
459
459.0 =×==
%63.0
100
1
100
63
10000
63
0063.0 =×==
Thus, to write a decimal as a percent, we move the decimal point two places to the
right and put the % sign
Conversely,
To write a percent as a decimal, we drop the %sign and insert ormove the decimal
point two places to the left. For example,
43% = 0.43 75% = 0.75 12% = 0.12
9% = 0.09 115% = 1.15 327% = 3.27
0.75% = 0.0075 4.5% = 0.045 0.2% = 0.002
Let us take a few more examples:
Example 8.1: Shweta obtained 18 marks in a test of 25 marks. What was her percentage
ofmarks?
Solution: Total marks = 25
Marks obtained = 18
∴ Fraction of marks obtained =
25
18
∴ Marks obtained in percent =
25
18
×
100
72
4
4
= = 72%
Alternatively:
Marks obtained in percent =
25
18
× 100% = 72%

Notes
MODULE - 2
Commercial
Mathematics
Example 8.2: One-fourth of the total number of shoes in a shop were on discount sale.
What percent of the shoes were there on normal price?
Solution: Fraction of the total number of shoes on sale =
4
1
∴ Fraction of the total number of shoes on normal price =
4
3
4
1
1 =−
= %75
100
75
25
25
4
3
==× or %75%100
4
3
=×
Example 8.3: Out of 40 students in a class, 32 opted to go for a picnic. What percent of
students opted for picnic?
Solution: Total number of students in a class = 40
Number of students, who opted for picnic = 32
∴ Number of students, in percent, who opted for picnic
= %80%100
40
32
=×
Example 8.4: In the wordARITHMETIC, what percent of the letters are I’s?
Solution: Total number of letters = 10
Number of I’s = 2
∴ Percent of I’s = %20%100
10
2
=×
Example 8.5: Amixture of 80 litres, of acid and water, contains 20 litres of acid. What
percentofwaterisinthemixture?
Solution: Totalvolumeofthemixture=80litres
Volume of acid = 20 litres
∴Volume of water = 60 litres
∴ Percentage of water in the mixture = %75%100
80
60
=×
CHECK YOUR PROGRESS 8.1
1. Converteachofthefollowingfractionsintoapercent:
(a)
25
12
(b)
20
9
(c)
12
5
(d)
15
6
(e)
625
125

Notes
MODULE - 2
Commercial
Mathematics
(f)
10
3
(g)
300
108
(h)
150
189
(i)
25
72
(j)
1250
1231
2. Writeeachofthefollowingpercentsasafraction:
(a) 53% (b) 85% (c) 16
8
7
% (d) 3.425% (e) 6.25%
(f) 70% (g) 15
4
3
% (h) 0.0025% (i) 47.35% (j) 0.525%
3. Writeeachofthefollowingdecimalsasapercent:
(a) 0.97 (b) 0.735 (c) 0.03 (d) 2.07 (e) 0.8
(f) 1.75 (g) 0.0250 (h) 3.2575 (i) 0.152 (j) 3.0015
4. Writeeachofthefollowingpercentsasadecimal:
(a) 72% (b) 41% (c) 4% (d) 125% (e) 9%
(f) 410% (g) 350% (h) 102.5% (i) 0.025% (j) 10.25%
5. Gurpreetgothalftheanswerscorrect,inanexamination.Whatpercentofheranswers
were correct?
6. Prakhar obtained 18 marks in a test of total 20 marks. What was his percentage of
marks?
7. Harish saves ` 900 out of a total monthly salary of ` 14400. Find his percentage of
saving.
8. A candidate got 47500 votes in an election and was defeated by his opponent by a
margin of 5000 votes. If there were only two candidates and no votes were declared
invalid,findthepercentageofvotesobtainedbythewinningcandidate.
9. In the word PERCENTAGE, what percent of the letters are E’s?
10. In a class of 40 students, 10 secured first division, 15 secured second division and 13
justqualified.Whatpercentofstudentsfailed.
8.4 CALCULATION OF PERCENT OF A QUANTITY OR A
NUMBER
To determine a specified percent of a number or quantity, we first change the percent to a
fractionoradecimalandthenmultiplyitwiththenumberorthequantity.Forexample:
25% of 90 =
100
25
× 90 = 22.50
or 25% of 90 = 0.25 × 90 = 22.50
60% of Rs. 120 = 0.60 × Rs. 120 = Rs. 72.00
120% of 80 kg = 1.20 × 80 kg = 96 kg

Notes
MODULE - 2
Commercial
Mathematics
Letustakesomeexamplesfromdailylife:
Example 8.6: In an examination, Neetu scored 62% marks. If the total marks in the
examination are 600, then what are the marks obtained by Neetu?
Solution: Here we have to find 62% of 600
∴ 62% of 600 marks = 0.62 × 600 marks = 372 marks
∴ Marks obtained by Neetu = 372
Example 8.7: Naresh earns ` 30800 per month. He keeps 50% for household expenses,
15%forhispersonalexpenses,20%forexpenditureonhischildrenandtheresthesaves.
What amount does he save per month?
Solution: Expenditure on Household = 50%
Expenditure on self = 15%
Expenditureonchildren=20%
Total expenditure = (50 + 15 + 20)% = 85%
∴ Savings (100 – 85)% = 15%
∴ 15% of ` 30800 = ` (0.15 × 30800)
= ` 4620
Example 8.8: What percent of 360 is 144?
Solution: Let x% of 360 = 144
144360
100
=×∴
x
Or %40100
360
144
=×=x
Alternatively, 144 out of 360 is equal to the fraction
360
144
∴ Percent = %40%100
360
144
=×
Example 8.9: If 120 is reduced to 96, what is the reduction percent?
Solution: Here, reduction = 120 – 96 = 24
∴ Reduction percent = %20%100
120
24
=×

Notes
MODULE - 2
Commercial
Mathematics
Example 8.10: The cost of an article has increased from `450 to `495.Bywhatpercent
did the cost increased?
Solution: The increase in Cost Price = ` (495 – 450)
= ` 45
Increase percent = %10100
450
45
=×
Example 8.11: 60%ofthestudentsinaschoolaregirls.Ifthetotalnumberofgirlsinthe
schoolis690,findthetotalnumberofstudentsintheschool.Also,findthenumberofboys
intheschool.
Solution: Let the total number of students in the school be x
Then, 60% of x = 690
1150
60
100690
or690
100
60
=
×
==×∴ xx
∴ Total number of students in the school = 1150
∴ Hence number of boys = 1150 – 690 = 460
Example 8.12: A’s income is 25% more than that of B. B’s income is 8% more than that
of C. IfA’s income is ` 20250, then find the income of C.
Solution: Let income of C be ` x
Income of B = x + 8% of x
= x
x
x ×=+
100
108
100
8
IncomeofA =
100
108
of25%
100
108 xx
+
=
100
125
100
108
×
x
20250
100
125
100
108
=××∴ x
or 15000
125
100
108
100
20250 =××=x
∴ Income of C is ` 15000.

Notes
MODULE - 2
Commercial
Mathematics
Example 8.13: Areduction of 10% in the price of tea enables a dealer to purchase 25 kg
moreteafor`22500.Whatisthereducedpriceperkgoftea?Also,findtheoriginalprice
per kg.
Solution: 10% of ` 22500 = =×22500
100
10
` 2250
∴ Reduced price of 25 kg tea = ` 2250
∴ Reduced price per kg = `
25
2250
= ` 90 per kg.
Since, the reduction was 10% so the original price = ` 100 per kg.
Example 8.14: Astudent got 45% marks in the first paper and 70% in the second paper.
How much percent should he get in the third paper so as to get 60% as overall score?
Solution: Let each paper be of 100 marks.
∴ Marks obtained in first paper = 45% of 100 = 45
Marks obtained in second paper = 70% of 100 = 70
Total marks (in three papers) he wants to obtain = 60% of 300
= 180300
100
60
=×
∴ Marks to be obtained in third paper = 180 – (45 + 70)
= 180 – 115 = 65
Example 8.15: Find the sum which when increased by 15% becomes ` 19320.
Solution: Let the sum be ` x
∴ x + 15% of x = 19320
19320
100
115
or19320
100
15
==+
xx
x
16800
115
10019320
=
×
=∴x
Hence, the required sum = ` 16800.
1. Find: (i) 16% of 1250 (ii) 47% of 1200
2. A family spends 35% of its monthly budget of ` 7500 on food. How much does the
familyspendonfood?

Notes
MODULE - 2
Commercial
Mathematics
3. In a garden, there are 500 plants of which 35% are trees, 20% are shrubs, 25% are
herbs and the rest are creepers. Find out the number of each type of plants.
4. 60 is reduced to 45. What percent is the reduction?
5. If 80 is increased to 125, what is the increase percent?
6. Ramanhastoscoreaminimum40%marksforpassingtheexamination.Hegets178
marksandfailsby22marks.Findthemaximummarks.
7. It takes me 45 minutes to go to school and I spend 80% of the time travelling by bus.
How long does the bus journey last?
8. In an election, between 2 candidates 25% voters did not cast their votes.A candidate
scored 40% of the votes polled and was defeated by 900 votes. Find the total number
of voters.
9. A rise of 25% in the price of sugar compels a person to buy 1.5 kg of sugar less for
` 240. Find the increased price as well as the original price per kg of sugar.
10. A number is first increased by 20% and then decreased by 20%. What is the net
increase or decrease percent?
11. ‘A’scored 12 marks, while B scored 10 marks, in the first terminal examination. If in
the second terminal examination (with same total number of marks) ‘A’scored 14
marks and ‘B’ scored 12 marks, which student showed more improvement?
12. 30,000 students appeared in a contest. Of them 40% were girls and the remaining
boys. If 10% boys and 12% girls won the contest with prizes, find the percentage of
students who won prizes.
13. Sunil earns 10% more than Shailesh and Shailesh earns 20% more than Swami. If
Swami earns `3200 less than Sunil, find the earnings of each.
8.5 APPLICATION OF PERCENTAGE
In our day to day life, we come across a number of situations wherein we use the concept
of percent. In the following section, we discuss the application of percentage in different
fields,likeproblemsinprofitandloss,discount,simpleinterest,compoundinterest,rateof
growthanddepreciation.
8.5.1 Profit and Loss
Let us recall the terms and formulae related to profit and loss.
Cost Price (C.P.): The Price at which an article is purchased, is called its cost price.
Selling Price (S.P.): The Price at which an article is sold, is called its selling price.
Profit (Gain): When S.P. > C.P., then there is profit, and
Profit = S.P. – C.P.

Notes
MODULE - 2
Commercial
Mathematics
Loss: When C.P. > S.P., then there is loss, and
Loss = C.P. – S.P.
Formulae Profit % = %100
C.P.
Loss
Loss%%,100
C.P.
Profit
⎟
⎠
⎞
⎜
⎝
⎛
×=⎟
⎠
⎞
⎜
⎝
⎛
×
S.P. =
( ) ( ) ( )( )
100
Loss%100C.P.
100
Profit%100C.P. −
=
+×
C.P. =
( )
( )
( )Loss%100
100S.P.
Profit%100
100S.P.
−
×
=
+
×
Note: Gain % or loss % is always calculated on C.P.
Letustakesomeexamplestoillustratetheapplicationsoftheseformulaeinsolvingproblems
related to profit and loss:
Example 8.16: Ashopkeeper buys an article for Rs. 360 and sells it for Rs. 270. Find his
gain or loss percent.
Solution: Here C.P. = Rs. 360, and S.P. = Rs. 270
Since C.P. > S.P., ∴ there is a loss.
Loss = C.P. – S.P. = Rs (360 – 270) = Rs. 90
Loss % %100
C.P.
Loss
⎟
⎠
⎞
⎜
⎝
⎛
×=
= %25100
360
90
=×
Example 8.17: Sudha purchased a house for ` 4,52,000 and spent ` 28,000 on its
repairs. She had to sell it for ` 4,92,000. Find her gain or loss percent.
Solution: Here C.P. = Cost price + Overhead charges
= ` (452000 + 28000) = ` 4,80,000
S.P. = ` 4,92,000
Since, S.P. > C.P., ∴ Gain = ` (492000 – 480000) = ` 12000
Gain % = %5.2%
2
5
480000
10012000
==
×
Example8.18:Bysellingabookfor`258,apublishergains20%.Forhowmuchshould
he sell it to gain 30%?

Notes
MODULE - 2
Commercial
Mathematics
Solution: S.P. = Rs. 258
Profit = 20%
C.P. =
Profit%100
100S.P.
+
×
= `
120
100258×
= ` 215
Now, if Profit = 30% and C.P. = Rs. 215, then,
S.P. =
( )
100
Profit%100C.P. +×
= `
100
301215×
= ` 279.50
Example 8.19: A man bought oranges at 25 for ` 100 and sold them at 20 for ` 100.
Find his gain or loss percent.
Solution: C.P. of 25 oranges = ` 100
∴ C.P. of 1 orange = `
25
100
= ` 4
and S.P. of 1 orange = `
20
100
= ` 5
∴ Profit on 1 orange = ` (5 – 4) = ` 1
Profit % = %25100
4
1
=×
Example 8.20: Aman sold two horses for ` 29700 each. On one he lost 10% while he
gained 10% on the other. Find his total gain or loss percent in the transaction.
Solution: S.P. of first horse = ` 29700
Loss = 10%
∴ C.P. = `
90
10029700×
= ` 33,000
S.P. of 2nd horse = ` 29700,
Profit = 10%
C.P. = `
110
10029700×
= ` 27,000
Total CP = ` (33000 + 27000) = ` 60,000
Total SP = ` (2 × 29700) = ` 59400
Net Loss = ` (60000 – 59400) = ` 600

Notes
MODULE - 2
Commercial
Mathematics
Loss % = 100
60000
600
× = 1%
Example8.21:Thecostpriceof15articlesisequaltothesellingpriceof12articles.Find
thegainpercent.
Solution: Let the C.P. of 15 articles be ` 15
then S.P. of 12 articles = ` 15
S.P. of 15 articles = ` 15
12
15
× = `
4
75
Gain = ` ⎟
⎠
⎞
⎜
⎝
⎛
−15
4
75
= `
4
15
Gain % = %25100
15
4/15
=×
Example 8.22: Awatch was sold at a profit of 12%. Had it been sold for ` 33 more, the
profit would have been 14%. Find the cost price of the watch.
Solution: Let the cost price of the watch be ` x
100
112
100
112
S.P.
xx
=
×
=∴
If the watch is sold for Rs. 33 more then S.P. = ⎟
⎠
⎞
⎜
⎝
⎛
+ 33
100
112x
New profit = 14%
114
10033
100
112
C.P.
×⎟
⎠
⎞
⎜
⎝
⎛
+
==∴
x
x
or 114x = 112 x + 3300 or 2x = 3300
x = 1650 ∴ C.P. = ` 1650
1. A shopkeeper bought an almirah from a wholesale dealer for ` 4500 and sold it for
` 6000. Find his profit or loss percent.

Notes
MODULE - 2
Commercial
Mathematics
2. Aretailer buys a cooler for ` 3800 but had to spend ` 200 on its transport and repair.
If he sells the cooler for ` 4400, determine, his profit percent.
3. A vendor buys lemons at the rate of 5 for `7 and sells them at `1.75 per lemon. Find
hisgainpercent.
4. A man purchased a certain number of oranges at the rate of 2 for `5 and sold them at
the rate of 3 for ` 8. In the process, he gained ` 20. Find the number of oranges he
bought.
5. By selling a bi-cycle for ` 2024, the shopkeeper loses 12%. If he wishes to make a
gain of 12% what should be the selling price of the bi-cycle?
6. By selling 45 oranges for `160, a woman loses 20%. How many oranges should she
sell for ` 112 to gain 20% on the whole?
7. A dealer sold two machines at ` 2400 each. On selling one machine, he gained 20%
and on selling the other, he lost 20%. Find the dealer’s net gain or loss percent.
8. Harish bought a table for ` 960 and sold it to Raman at a profit of 5%. Raman sold it
to Mukul at a profit of 10%. Find the money paid by Mukul for the table.
9. Amanbuysbananasat6for `5andanequalnumberat `15perdozen.Hemixesthe
twolotsandsellsthemat`14perdozen.Findhisgainorlosspercent,inthetransaction.
10. Ifthesellingpriceof20articlesisequaltothecostpriceof23articles,findthelossor
gainpercent.
8.5.2 Discount
You must have seen advertisements of the following types, especially during the festival
season.
SALE DIWALI BONANZA
discount upto 50% 20% discountonallitems.
Adiscountisareductioninthemarked(orlist)priceofanarticle.“20%discount”means
a reduction of 20% in the marked price of an article. For example, if the marked price of
an article is ` 100, it is sold for ` 80, i.e. `20 less than the marked price. Let us define the
terms,weshalluse:
MarkedPrice(orListprice): Themarkedprice(M.P.)ofanarticleisthepriceatwhich
thearticleislistedforsale.Sincethispriceiswritten(marked)onthearticle,soitiscalled
the marked price.
Discount: The discount is the reduction from the marked price of the article.
Net selling price (S.P.): In case of discount selling, the price of the article obtained by
subtractingdiscountfromthemarkedpriceiscalledtheNetSellingpriceorSellingprice
(S.P.).Letustakethefollowingexamples,toillustrate:
}

Notes
MODULE - 2
Commercial
Mathematics
Example 8.23: Acoat is marked at ` 2400. Find its selling price if a discount of 12% is
offered.
Solution: Here, Marked Price (M.P.) of the coat = ` 2400
Discount = 12%
Net selling price (S.P.) = M.P. – Discount
= ` 2400 – 12% of ` 2400
= ` 2400 – ` ⎟
⎠
⎞
⎜
⎝
⎛
×2400
100
12
= ` (2400 – 288)
= ` 2112
Thus, the net selling price of coat is ` 2112.
Example 8.24: A machine listed at ` 8400 is available for ` 6300. Find the rate of
discountoffered.
Solution: Here, Marked Price (M.P.) = ` 8400
Net selling price (S.P.) = ` 6300
Discountoffered = ` (8400 – 6300)
= ` 2100
Discount% = %100
8400
2100
× = 25%
Note: Discount is always calculated on Marked Price.
Example 8.25: Awholesaler’s list price of a fan is ` 1250 and is available to a retailer at
a discount of 20%. For how much should the retailer sell it, to earn a profit of 15%.
Solution: M.P. = ` 1250
Discount = 20% of ` 1250
= ` 1250
100
20
× = ` 250
∴ Cost Price of the retailer = ` (1250 – 250)
= ` 1000
Profit = 15%
∴ S.P. =
( )
100
Profit%100C.P. +
= `
100
1151000×
= ` 1150

Notes
MODULE - 2
Commercial
Mathematics
Example8.26: Ashopkeepermarkshisgoods25%morethantheircostpriceandallows
a discount of 10%. Find his gain or loss percent.
Solution: Let the C.P. of an article = ` 100
∴ Marked Price (M.P.) = ` 100 + 25% of ` 100
= ` 125
Discount offered = 10%
∴Net selling Price = ` 125 – 10% of ` 125
= ` 125 – ` ⎟
⎠
⎞
⎜
⎝
⎛
×125
100
10
= ` (125 – 12.50) = ` 112.50
∴ Gain = ` (112.50 – 100) = ` 12.50
Gain % = %5.12100
100
50.12
=×
Example 8.27: An article listed at `5400 is offered at a discount of 15%. Due to festival
season, the shopkeeper allows a further discount of 5%. Find the selling price of the
article.
Solution: M.P. = ` 5400, Discount = 15%
∴ SP = ` 5400 – 15% of ` 5400
= ` 5400 – `
100
15
× 5400
= ` (5400 – 810) = ` 4590
Festival discount = 5%
∴ Net selling Price = ` 4590 – 5% of ` 4590
= ` 4590 – `
100
5
× 4590
= ` (4590 – 229.50)
= ` 4360.50
∴ Net selling price of article = ` 4360.50.
Example 8.28: Aretailer buys books from a wholesaler at the rate of `300 per book and
markedthemat `400each.Heallowssomediscountandgetsaprofitof30%onthecost
price. What percent discount does he allow to his customers?

Notes
MODULE - 2
Commercial
Mathematics
Solution: C.P. of one book = ` 300
M.P. = ` 400
Profit = 30%
∴ S.P. =
( )
100
Profit%100C.P. +
= `
100
130300×
= ` 390
∴ Discount offered = ` (400 – 390) = ` 10
Discount % = %5.2100
400
10
=×
1. A shirt with marked price ` 375/- is sold at a discount of 15%. Find its net selling
price.
2. A pair of socks marked at ` 60 is being offered for ` 48. Find the discount percent
beingoffered.
3. Awashingmachineissoldatadiscountof10%onitsmarkedprice.Afurtherdiscount
of5%isofferedforcashpayment.Findthesellingpriceofthewashingmachineifits
marked price is ` 18000.
4. A man pays ` 2100 for a machine listed at ` 2800. Find the rate of discount offered.
5. Thelistpriceofatablefanis`840anditisavailabletoaretaileratadiscountof25%.
For how much should the retailer sell it to earn a profit of 15%.
6. A shopkeeper marks his goods 50% more than their cost price and allows a discount
of 40%, find his gain or loss percent.
7. A dealer buys a table listed at ` 2500 and gets a discount of 28%. He spends ` 100
on transportation and sells it at a profit of 15%. Find the selling price of the table.
8. A retailer buys shirts from a manufacturer at the rate of ` 175 per shirt and marked
them at ` 250 each. He allows some discount and earns a profit of 28% on the cost
price. What percent discount does he allow to his customers?
8.5.3 Simple Interest
When a person has to borrow some money as a loan from his friends. relatives, bank etc.
hepromisestoreturnitafteraspecifiedtimeperiodalongwithsomeextramoneyforusing
themoneyofthelender.
The money borrowed is called the Principal, usually denoted by P, and the extra money
paid is called the Interest, usually denoted by I.

Notes
MODULE - 2
Commercial
Mathematics
The total money paid back, that is, the sum of Principal and the Interest is called the
Amount, and is usually denoted byA.
Thus, A = P + I
The interest is mostly expresed as a rate percent per year (per annum).
Interestdependson,howmuchmoney(P)hasbeenborrowedandtheduratonoftime(T)
forwhichitisused.Interestiscalculatedaccordingtoamutuallyagreedratepercent,per
annum (R). [i.e. R = r % =
100
r
]
Thus, Interest = (Principal) × (Rate % per annum) × time
∴ I = P × R × T
Interestcalculatedasabove,iscalledsimpleinterest.Letustakesomeexamples,involving
simpleinterest.
Example 8.29: Find the simple interest in each of the following cases
P R T
(a) ` 8000 5% 2 yrs
(b) ` 20,000 15% 1
2
1
yrs
Solution: (a) I = P. R. T.
= ` ⎥⎦
⎤
⎢⎣
⎡
×× 2
100
5
8000 = ` 800
(b) I = ` ⎥⎦
⎤
⎢⎣
⎡
××
2
3
100
15
20000 = ` 4500
Example 8.30: Find at what rate of simple interest per annum will ` 5000 amount to
` 6050 in 3 years.
Solution: Here A = ` 6050, P = ` 5000, T = 3 yrs
∴ I = ` (6050 – 5000) = ` 1050
I = P × R × T or
TP
100I
r
TP
I
r%
×
×
=∴
×
=
7
35000
1001050
r =
×
×
= ∴ R = 7%

Notes
MODULE - 2
Commercial
Mathematics
Example 8.31: Asum amounts to ` 4875 at 12
2
1
% simple interest per annum after 4
years. Find the sum.
Solution: Here A = ` 4875, R = 12
2
1
% =
2
25
%, T = 4 yrs
I = P × R × T
I = ` ⎟
⎠
⎞
⎜
⎝
⎛
×× 4
200
25
P = `
2
P
∴ A = ` ⎟
⎠
⎞
⎜
⎝
⎛
+
2
P
P = `
2
3P
Thus,
2
3P
= ` 4875 or 3P = ` 9750 or P = ` 3250
Example 8.32: In how many years will a sum of ` 2000 yield an interest (Simple) of
` 560 at the rate of 14% per annum?
Solution: Here P = ` 2000, I = ` 560 R = 14%
I = P × R × T or T××=
100
14
2000560
years2
142000
100560
T =
×
×
=∴
Thus, in 2 years, a sum of ` 2000 will yield an interest of ` 560 at 14% per annum.
Example 8.33: Acertain sum of money at simple interest amounts to ` 1300 in 4 years
and to ` 1525 in 7 years. Find the sum and rate percent.
Solution: Here 1300 = P
100
4RP
+
××
...(i)
and 1525 = P
100
7RP
+
××
...(ii)
Subtracting(i)from(ii) 75
100
RP
or
100
3RP
225 =
×××
=
Puttingin(i)weget
1300 = 75 × 4 + P or P = ` (1300 – 300) = ` 1000

Notes
MODULE - 2
Commercial
Mathematics
Again, we have 7.5%
1000
10075
P
10075
Ror75
100
RP
=
×
=
×
==
×
∴ Principal = ` 1000 and rate = 7.5%
Alternatively:
Amount after 4 years = ` 1300
Amount after 7 years = ` 1525
∴ Interest for 3 years = ` [1525 – 1300] = ` 225
∴ Interest for 1 year = `
3
225
= ` 75
∴ 1300 = P + Interest for 4 yrs = P + 4 × 75 or P = ` (1300 – 300) = ` 1000
R = %5.7
11000
10075
=
×
×
Example8.34:Acertainsumofmoneydoublesitselfin10years.Inhowmanyyearswill
it become 2
2
1
times at the same rate of simple interest.
Solution: Let P = ` 100, T = 10 yrs, A = ` 200, ∴ I = ` 100
10%Ror
100
10R100
100 =
××
=∴
Now P = ` 100, R = 10% and A = ` 250 ∴ I = ` 150
yrs15TorT
100
10
100150 =××=∴
Thus, in 15 yrs, the sum will become 2
2
1
times.
Example 8.35: Out of ` 70,000 to invest for one year, a man invests ` 30,000 at 4% and
` 20,000 at 3% per annum simple interest. At what rate percent, should he lend the
remaining money, so that he gets 5% interest on the total amount he has?
Solution: Interest on total amount at 5% for one year
= ` 1
100
5
70,000 ×× = ` 3500
Interest on ` 30,000 at 4% for 1 year = ` 1
100
4
30000 ××
= ` 1200

Notes
MODULE - 2
Commercial
Mathematics
Interest on ` 20,000 at 3% for 1 year = ` 1
100
3
20000 ××
= ` 600
∴ Interest on remaining ` 20,000 for 1 yr = ` [3500 – 1200 – 600]
= ` 1700
8.5%
20000
1001700
Ror1
100
R
200001700 =
×
=××=∴
∴The remaining amount should be invested at 8.5% per annum.
1. Ramaborrowed`14000fromherfriendat8%perannumsimpleinterest.Shereturned
the money after 2 years. How much did she pay back altogether?
2. Rameshdeposited `15600 in a financial company, which pays simple interest at 8%
per annum. Find the interest he will receive at the end of 3 years.
3. Naveen lent ` 25000 to his two friends. He gave ` 10,000 at 10% per annum to one
ofhisfriendandtheremainingtootherat12%perannum.Howmuchinterestdidhe
receive after 2 years.
4. Shalini deposited ` 29000 in a finance company for 3 years and received ` 38570 in
all. What was the rate of simple interest per annum?
5. Inhowmuchtimewillsimpleinterestonasumofmoneybe
5
2
thofthesum,attherate
of 10% per annum.
6. Atwhatrateofinterestwillsimpleinterestbehalftheprincipalin5years.
7. A sum of money amounts to ` 1265 in 3 years and to ` 1430 in 6 years, at simple
interest. Find the sum and the rate percent.
8. Out of ` 75000 to invest for one year, a man invested ` 30000 at 5% per annum and
`24000at4%perannum.Atwhatpercentperannum,shouldheinvesttheremaining
money to get 6% interest on the whole money.
9. A certain sum of money doubles itself in 8 years. In how much time will it become 4
times of itself at the same rate of interest?
10. In which case, is the interest earned more:
(a) ` 5000 deposited for 5 years at 4% per annum, or
(b) ` 4000 deposited for 6 years at 5% per annum?

Notes
MODULE - 2
Commercial
Mathematics
8.5.4 Compound Interest
Intheprevioussection,youhavestudiedaboutsimpleinterest.Whentheinterestiscalculated
on the Principal for the entire period of loan, the interest is called simple interest and is
givenby
I = P × R × T
Butifthisinterestisdue(notpaid)afterthedecidedtimeperiod,thenitbecomesapartof
the principal and so is added to the principal for the next time preiod, and the interest is
calculated for the next time period on this new principal. Interest calculated, this way is
calledcompoundinterest.
Thetimeperiodafterwhichtheinterestisaddedtotheprincipalforthenexttimeperiodis
called the Conversion Period.
Theconversionperiodmaybeoneyear,sixmonthsorthreemonthsandtheinterestissaid
tocompounded,annually,semi-annuallyorquarterly,respectively.Letustakeanexample:
Example8.36:FindthecompoundinterestonasumofRs.2000,fortwoyearswhenthe
interestiscompoundedannuallyat10%perannum.
Solution: Here P = ` 2000 and R = 10%
∴ Interest for the first conversion time period (i.e. first year)
= ` 2000 ×
100
10
× 1 = ` 200
∴ Principal for the second year (or 2nd conversion period)
= ` (2000 + 200) = ` 2200
∴ Interest for the 2nd time period = ` 2200 ×
100
10
× 1 = ` 220
∴ Amount payable at the end of two years = ` (2200 + 220)
= ` 2420
∴ Total interest paid at the end of two years = ` (2420 – 2000)
= ` 420
or [` (200 + 220) = ` 420]
∴ Compound interest = ` 420
Thus,forcalculatingthecompoundinterest,theinterestdueaftereverycoversionperiod
is added to the principal and then interest is calculated for the next period.
8.5.4.1 Formula for Compound Interest
Let a sum P be borrowed for n years at the rate of r% per annum, then
Interest for the first year = P ×
100
r
× 1 =
100
Pr

Notes
MODULE - 2
Commercial
Mathematics
Amountafteroneyear = Principal for 2nd year =
100
Pr
P +
= ⎟
⎠
⎞
⎜
⎝
⎛
+
100
r
1P
Interest for 2nd year = ⎟
⎠
⎞
⎜
⎝
⎛
+
100
r
1P ×
100
r
× 1 = ⎟
⎠
⎞
⎜
⎝
⎛
+
100
r
1.
100
Pr
Amount after 2 years = ⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
+=⎟
⎠
⎞
⎜
⎝
⎛
++⎟
⎠
⎞
⎜
⎝
⎛
+
100
r
1
100
r
1P
100
r
1
100
Pr
100
r
1P
=
2
100
r
1P ⎟
⎠
⎞
⎜
⎝
⎛
+
Similarly, amount after 3 years =
3
100
r
1P ⎟
⎠
⎞
⎜
⎝
⎛
+ and so on.
Amount after n years =
n
100
r
1P ⎟
⎠
⎞
⎜
⎝
⎛
+
Thus, if Arepresents the amount and R represents r% or
100
r
, then
A = P(1 + R)n
=
n
100
r
1P ⎟
⎠
⎞
⎜
⎝
⎛
+
andcompoundinterest = A – P = P (1 + R)n
– P
= P[(1 + R)n
–1] or
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
−⎟
⎠
⎞
⎜
⎝
⎛
+ 1
100
r
1P
n
Note:Simpleinterestandcompoundinterestareequalforfirstyear(firstconversionperiod)
Example 8.37: Calculate the compound interest on ` 20,000 for 3 years at 5% per
annum,whentheinterestiscompoundedannually.
Solution: Here P = ` 20,000, R = 5% and n =3
∴ CI = P[(1 + R)n
–1]
= ` 20000
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
−⎟
⎠
⎞
⎜
⎝
⎛
+ 1
100
5
1
3
= `
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
−⎟
⎠
⎞
⎜
⎝
⎛
1
20
21
3
= ` ⎥⎦
⎤
⎢⎣
⎡ −
×
8000
80009261
20000
= ` 3152.50

Notes
MODULE - 2
Commercial
Mathematics
Example 8.38: Calculate the compound interest on `20,000 for 1
2
1
years at the rate of
10%perannum,whentheinterestiscompoundedsemi-annually.
Solution: Here P = ` 20,000, R = 10% per annum
= 5% per half year
and n = 1
2
1
yrs = 3 half years
∴ CI = P[(1 + R)n
–1] = `
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
−⎟
⎠
⎞
⎜
⎝
⎛
+ 1
100
5
120,000
3
= ` ⎥⎦
⎤
⎢⎣
⎡
−× 1
8000
9261
20,000 = ` 3152.50
Example 8.39: Calculate the compound interest on ` 20,000 for 9 months at the rate of
4% per annum, when the interest is compounded quarterly.
Solution: Here P = ` 20,000, R = 4% per annum
= 1% per quarter of year
and n = 3/4 yrs = 3 quarters
∴ CI = P[(1 + R)n
–1] = `
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
−⎟
⎠
⎞
⎜
⎝
⎛
+ 1
100
1
120,000
3
= ` 20,000 ×
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
−⎟
⎠
⎞
⎜
⎝
⎛
1
100
101
3
= `
100100100
3030120000
××
×
= ` 606.02
Example 8.40: calculate the amount and compound interest on ` 12000 for 1
2
1
years at
the rate of 10% per annum compounded annually.
Solution: Here P = ` 12000, R = 10% and n = 1
2
1
years
Sinceinterestiscompounded,annually,so,amountattheendof1yearisgivenby

Notes
MODULE - 2
Commercial
Mathematics
1
100
R
1PA ⎟
⎠
⎞
⎜
⎝
⎛
+= = `
1
100
10
112000 ⎟
⎠
⎞
⎜
⎝
⎛
+×
= `
10
11
12000× = ` 13200
∴ Principal for next 6 months = ` 13200
and Rate R = %5%
2
10
=
∴ A = `
1
100
5
113200 ⎟
⎠
⎞
⎜
⎝
⎛
+ = `
20
21
13200×
= ` 13860
∴Amount after 1
2
1
years = ` 13860
Compound interest = ` [13860 – 12000]
= ` 1860
Note: We can calculate the amount for 1
2
1
yrs as
A = `
11
100
5
1
100
10
112000 ⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
+
Example 8.41: At what rate percent per annum, will a sum of ` 15,625 become
` 17576 in three years, when the interest is compounded annually?
Solution: Here A = ` 17576, P = ` 15,625 and n = 3
Let R = r% per annum
3
100
r
11562517576 ⎟
⎠
⎞
⎜
⎝
⎛
+=∴
33
25
26
15625
17576
100
r
1 ⎟
⎠
⎞
⎜
⎝
⎛
==⎟
⎠
⎞
⎜
⎝
⎛
+∴
25
1
1
25
26
100
r
or
25
26
100
r
1 =−==⎟
⎠
⎞
⎜
⎝
⎛
+∴

Notes
MODULE - 2
Commercial
Mathematics
or 4
25
100
r ==
∴ Rate = 4% per annum.
Example 8.42: In how much time will a sum of ` 8000 amount to ` 9261 at 10% per
annum,compoundedsemi-annually?
Solution: Here A= ` 9261, P = ` 8000 and n = x semi yrs
R = 10% per annum = 5% semi annually
x
100
5
180009261 ⎟
⎠
⎞
⎜
⎝
⎛
+=∴
or 3x
20
21
20
21
or
20
21
8000
9261
x3x
=∴⎟
⎠
⎞
⎜
⎝
⎛
=⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
=
∴ Time = 3 half years = 1
2
1
years
Example 8.43: Find the difference between simple interest and compound interest for
1
2
1
years at4%perannum,forasumof`24000,whentheinterestiscompoundedsemi-
annually..
Solution: Here P = ` 24000, R = 4% per annum
T =
2
3
years R = 2% per semi-annually
n = 1
2
1
years =
2
3
years = 3 semi years
Simple Interest = P × R × T = `
2
3
100
4
24000 ××
= ` 1440.
For compound interest,A=
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
+
n
100
R
1P
A = `
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
+
3
100
2
124000

Notes
MODULE - 2
Commercial
Mathematics
A = `
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
3
50
51
24000 = ` ⎥⎦
⎤
⎢⎣
⎡
××
50
51
50
51
50
51
24000
= `
125
51515124 ×××
= ` 25468.99 or ` 25469
∴ CI = ` [25469 – 24000] = ` 1469
Difference = CI – SI = ` [1469 – 1440]
= ` 29
Example 8.44: Asum of money is invested at compound interest for 1
2
1
year at 4%
compounded annually. If the interests were compounded semi-annually, it would have
fetched ` 20.40 more than in the previous case. Find the sum.
Solution: Let the sum be ` x.
Here R=4%annually,or2%semi-annually
T = 1
2
1
yrs or 3 semi years
In first case
A = `
11
100
2
1
100
4
1 ⎥⎦
⎤
⎢⎣
⎡
+⎥⎦
⎤
⎢⎣
⎡
+x
= ` ⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
50
51
25
26
x = `
1250
1326 x
In 2nd case
A = `
3
100
2
1 ⎟
⎠
⎞
⎜
⎝
⎛
+x = `
3
50
51
⎟
⎠
⎞
⎜
⎝
⎛
x
= `
125000
132651
∴ Difference = ` ⎥⎦
⎤
⎢⎣
⎡
− xx
1250
1326
125000
132651
= `
125000
51x

Notes
MODULE - 2
Commercial
Mathematics
or
100
2040
125000
51x
=∴ x = `
51
125000
100
2040
× = ` 5000
∴ Sum = ` 50,000
1. Calculatethecompoundintereston`15625for3yearsat4%perannum,compounded
annually.
2. Calculate the compound interest on ` 15625 for 1
2
1
years at 8% per annum,
compoundedsemi-annually.
3. Calculate the compound interest on ` 16000 for 9 months at 20% per annum,
compoundedquarterly
4. Findthesumofmoneywhichwillamountto`27783in3yearsat5%perannum,the
interestbeingcompoundedannually.
5. Findthedifferencebetweensimpleinterestandcompoundinterestfor3yearsat10%
per annum, when the interest is compounded annually on `30,000.
6. The difference between simple interest and compound interest for a certain sum of
moneyat8%perannumfor 1
2
1
years, when the interest is compounded half-yearly
is ` 228. Find the sum.
7. A sum of money is invested at compound interest for 9 months at 20% per annum,
whentheinterestiscompoundedhalfyearly.Iftheinterestwerecompoundedquarterly,
it would have fetched ` 210 more than in the previous case. Find the sum.
8. Asumof`15625amountsto`17576at8%perannum,compoundedsemi-annually.
Findthetime.
9. Find the rate at which ` 4000 will give ` 630.50 as compound interest in 9 months,
interestbeingcompoundedquarterly.
10. A sum of money becomes ` 17640 in two years and ` 18522 in 3 years at the same
rateofinterest,compoundedannually.Findthesumandtherateofinterestperannum.
8.5.5 Rate of Growth and Depreciation
In our daily life, we come across the terms like growth of population, plants, viruses etc
and depreciation in the value of articles like machinery, crops, motor cycles etc.
The problems of growth and depreciation can be solved using the formula of compound
interestderivedintheprevioussection.

Notes
MODULE - 2
Commercial
Mathematics
If Vo
is the value of an article in the beginning and Vn
is its value after ‘n’conversion
periods and the rate of growth/depreciation for the period be denoted by r%, then we can
write
n
on
100
r
1VV ⎟
⎠
⎞
⎜
⎝
⎛
+= in case of growth, and
n
on
100
r
–1VV ⎟
⎠
⎞
⎜
⎝
⎛
= in case of depreciation.
If the rate of growth/depreciation varies for each conversion period, then
....
100
r
1
100
r
1
100
r
1VV 321
on ⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
+= for growth, and
....
100
r
–1
100
r
–1
100
r
1VV 321
on ⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
−= for depreciation.
Let us take some examples to illustrate the above concepts.
Example 8.45: The population of a city is 9765625. What will be its population after 3
years, if the rate of growth of population is 4% per year?
Solution: Here Vo
= 9765625, r = 4% and n = 3
3
3
100
4
19765625V ⎥⎦
⎤
⎢⎣
⎡
+=∴
3
25
26
9765625 ⎟
⎠
⎞
⎜
⎝
⎛
×=
= 10985000.
Hence, the population of that city after 3 years will be = 10985000.
Example8.46:Thecostofacarwas`3,50,000inJanuary2005.Iftherateofdepreciation
is 15% for the first year and 10% for the subsequent years, find its value after 3 years.
Solution: Here Vo
= ` 3,50,000
r1
= 15%, r2
= 10% and r3
= 10%
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
−=∴
100
r
–1
100
r
–1
100
r
1VV 321
o3

Notes
MODULE - 2
Commercial
Mathematics
= ` ⎟
⎠
⎞
⎜
⎝
⎛
−⎟
⎠
⎞
⎜
⎝
⎛
−⎟
⎠
⎞
⎜
⎝
⎛
−
100
10
1
100
10
1
100
15
1350000
= `
10
9
10
9
20
17
350000 ××× = ` 2,40,975/-
∴ The value of car after 3 years = ` 240975.
Example 8.47: Aplantgainsitsheightattherateof2%permonthofwhatwasitsheight
inthebeginningofthemonth.Ifitsheightwas1.2minthebeginningofJanuary2008,find
its height in the beginning ofApril 2008, correct upto 3 places of decimal.
Solution: Here Vo
= 1.2 m, r = 2%, n = 3
n
o3
100
r
1VV ⎟
⎠
⎞
⎜
⎝
⎛
+=∴
m1.2734
50
51
1.2
100
2
11.2
33
=⎟
⎠
⎞
⎜
⎝
⎛
=⎟
⎠
⎞
⎜
⎝
⎛
+=
= 1.273 m
Hence,heightofplantinthebeginningofApril=1.273m.
Example 8.48: The virus of a culture decreases at the rate of 5% per hour due to a
medicine. If the virus count in the culture at 11.00AM was 2.3 × 107
, find the virus count
at 1.00 PM on the same day.
Solution: Vo
= 2.3 × 107
, r = 5%, n =2
( )27
2
7
2 95.0103.2
100
5
1103.2 ××=⎟
⎠
⎞
⎜
⎝
⎛
−×=V
= 2.076 × 107
Hence, the virus count at 1.00 PM is 2.076 × 107
.
1. The population of a town is 281250. What will be its population after 3 years, if the
rate of growth of population is 4% per year?
2. The cost of a car was ` 4,36,000 in January 2005. Its value depreciates at the rate of
15% in the first year and then at the rate of 10% in the subsequent years. Find the
value of the car in January 2008.

Notes
MODULE - 2
Commercial
Mathematics
3. The cost of machinery is ` 360000 today. In the first year the value depreciates by
12% and subsequently, the value depreciates by 8% each year. By how much, the
value of machinery has depreciated at the end of 3 years?
4. The application of manure increases the output of a crop by 10% in the first year, 5%
in the second year and 4% in the third year. If the production of crop in the year 2005
was 3.5 tons per hectare, find the production of crop per hectare in 2008.
5. The virus of a culture decreases at the rate of 4% per hour due to a medicine. If the
virus count in the culture at 9.00AM was 3.5 × 108
, find the virus count at 11.00AM
on the same day.
6. Three years back, the population of a village was 50000.After that, in the first year,
the rate of growth of population was 5%. In the second year, due to some epidemic,
the population decreased by 10% and in the third year, the population growth rate
was noticed as 4%. Find the population of the town now.
LET US SUM UP
• Percent means ‘per hundred’.
• Percents can be written as fractions as well as decimals and vice-versa.
• To write a percent as a fraction, we drop the % sign and divide the number by 100.
• Towriteafractionasapercent,wemultiplythefractionby100,simplifyitandsuffix
the%sign.
• Todeterminethespecificpercentofanumberorquantity,wechangethepercenttoa
fractionoradecimalandthenmultiply.
• When the selling price is more than the cost price of the goods, there is a profit (or
gain).
• When the selling price is less than the cost price of the goods, there is a loss.
Profit (Gain) = S.P. – C.P. ; Loss = C.P. – S.P.
100
C.P.
Gain
Gain% ×= ; 100
C.P.
Loss
Loss% ×=
C.P.
100
Gain%100
S.P. ×
+
= ; C.P.
100
Loss%–100
S.P. ×=
• The simple interest (I.) on a principal (P) at the rate of R% for a time T years, is
calculated,usingtheformula
I. = P × R × T

Notes
MODULE - 2
Commercial
Mathematics
• Discount is a reduction in the list price of goods.
• Discount is always calculated on the marked price of the goods
• (Markedprice–discount),givestheprice,whichacustomerhastopaywhilebuying
anarticle.
• Two or more successive discounts are said to form a discount series.
• A discount series can be reduced to a single discount.
• Sales tax is charged on the sale price of goods.
• An instalment plan enables a person to buy costlier goods.
• In the case of compound interest
Amount (A) = P (1 + R)n
, where P is the Principal, R = rate% and n = time.
• Compountinterestisgreaterthansimpleinterest,exceptforthefirstconversionperiod.
• IfVo
is the value of an article in the beginning andVn
is its value after ‘n’conversion
periods and ‘r’be the rate of growth/depreciation per period, then
n
on
100
r
1VV ⎟
⎠
⎞
⎜
⎝
⎛
+= in case of growth, and
n
on
100
r
–1VV ⎟
⎠
⎞
⎜
⎝
⎛
= in case of depreciation.
• If the rate of growth/depreciation varies for each conversion period, then
....
100
r
1
100
r
1
100
r
1VV 321
on ⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
+= for growth, and
....
100
r
–1
100
r
–1
100
r
1VV 321
on ⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
−= for depreciation.
TERMINAL EXERCISE
1. Write each of the following as a percent
(a)
20
9
(b)
10
7
(c) 0.34 (d) 0.06
2. Writeeachofthefollowingasadecimal:
(a) 36% (b) 410% (c) 2% (d) 0.35%

Notes
MODULE - 2
Commercial
Mathematics
3. Writeeachofthefollowingasfraction:
(a) 0.12% (b) 2.5% (c) 25.5% (d) 255%
4. Findeachofthefollowing:
(a) 23% of 500 (b) 2.5% of 800 (c) 0.4% of 1000 (d)115%of400
5. What percent of 700 is 294?
6. By what percent is 60 more than 45?
7. What number increased by 10% of itself is 352?
8. Find the number whose 15% is 270.
9. What number decreased by 7% of itself is 16.74?
10. If three fourth of the students of a class wear glasses, what percent of the students of
the class do not wear glasses?
11. There are 20 eggs in a fridge and 6 of them are brown. What percent of eggs are not
brown?
12. 44%ofthestudentsofaclassaregirls.Ifthenumberofgirlsis6lessthanthenumber
of boys, how many students are there in the class?
13. During an election, 70% of the population voted. If 70,000 people cast their votes,
what is the population of the town?
14. A man donated 5% of his monthly income to a charity and deposited 12% of the rest
inaBank.IfhehasRs.11704leftwithhim,whatishismonthlyincome?
15. Ratan stores has a sale of ` 12000 on Saturday, while Seema stores had a sale of
` 15000 on that day. Next day, they had respective sales of ` 15000 and ` 17500.
WhichstoreshowedmoreimprovementinSales?
16. A candidate has to secure 45% marks in aggregate of three papers of 100 marks each
to get through. He got 35% marks in the first paper and 50% marks in the second
paper.Atleasthowmanymarksshouldhegetinthirdpapertopasstheexamination?
17. The price of sugar rises by 25%. By how much percent should a householder reduce
his consumption of sugar, so as not to increase his expenditure on sugar?
18. By selling 90 ball pens for `160,apersonloses20%.Howmanyballpensshouldhe
sell for Rs. 96, so as to have a gain of 20%?
19. A vendor bought bananas at 6 for 5 rupees and sold them at 4 for 3 rupees. Find his
gain or loss percent.
20. Amanboughttwoconsignmentsofeggs,firstat`18perdozenandanequalnumber
at ` 20 per dozen. He sold the mixed egges at ` 23.75 per dozen. Find his gain
percent.

Notes
MODULE - 2
Commercial
Mathematics
21. A man sells an article at a gain of 10%. If he had bought it for 10% less and sold it for
` 10 more, he would have gained 25%. Find the cost price of the article.
22. A pair of socks is marked at ` 80 and is being offered at ` 64. Find the discount
percentbeingoffered.
23. A dealer buys a table listed at ` 1800 and gets a discount of 25%. He spends ` 150
on transportation and sells it at a profit of 10%. Find the selling price of the table.
24. AT.V. set was purchased by paying ` 18750. If the discount offered by the dealer
was 25%, what was the marked price of the TV set?
25. A certain sum of money was deposited for 5 years. Simple interest at the rate of 12%
waspaid.Calculatethesumdepositedifthesimpleinterestreceivedbythedepositor
is ` 1200.
26. Simple intrest on a sum of money is
3
1
rd of the sum itself and the number of years is
thrice the rate percent. Find the rate of interest.
27. Inwhattimewill `2700yieldthesameinterestat4%perannumas `2250in4years
at 3% per annum?
28. Thedifferencebetweensimpleinterestonasumofmoneyfor3yearsandfor2years
at 10% per annum is ` 300. Find the sum.
29. Find the sum which when invested at 4% per annum for 3 years will becomes
` 70304, when the interest is compounded annually.
30. thedifferencebetweencompoundinterestandsimpleinterestat10%perannumin2
years (compounded annually) is `50. Find the sum.
31. A sum of money becomes ` 18522 in three years and ` 19448.10 in 4 years at the
same rate of interest, compounded annually. Find the sum and the rate of interest per
annum.
32. Findthesumofmoneywhichwillamountto`26460insixmonthsat20%perannum,
whentheinterestiscompoundedquarterly.
33. At what rate percent per annum will a sum of ` 12000 amount to ` 15972 in three
years,whentheinterestiscompoundedannually?
34. The price of a scooter depreciates at the rate of 20% in the first year, 15% in the
second year and 10% afterwards, what will be the value of a scooter now costing
` 25000, after 3 years.
35. The population of a village was 20,000, two years ago. It increased by 10% during
first year but decreased by 10% in the second year. Find the population at the end of
2 years.

Notes
MODULE - 2
Commercial
Mathematics
ANSWERS TO CHECK YOUR PROGRESS
8.1
1. (a) 48% (b) 45% (c) 41
3
2
% (d) 40% (e) 20%
(f) 30% (g) 36% (h) 126% (i) 288% (j) 98.48%
2. (a)
100
53
(b)
20
17
(c)
160
27
(d)
4000
137
(e)
16
1
(f)
10
7
(g)
400
63
(h)
40000
1
(i)
2000
947
(j)
4000
21
3. (a) 97% (b) 73.5% (c) 3% (d) 207% (e) 80%
(f) 175% (g) 2.5% (h) 325.75% (i) 15.2% (j) 300.15%
4. (a) 0.72 (b) 0.41 (c) 0.04 (d) 1.25 (e) 0.09
(f) 4.1 (g) 3.5 (h) 1.025 (i) 0.00025 (j) 0.1025
5. 50% 6. 90% 7. 6.25% 8. 47.5% 9. 30%
10. 5%
8.2
1. (a) 200 (b) 564
2. Rs. 2625 3. 175, 100, 125, 100 4. 25%
5. 56.25% 6. 500 7. 36minutes
8. 6000 9. Rs. 40, Rs. 32 10. 4% decrease
11. B 12. 10.8%
13. Rs. 13200, Rs. 12000, Rs. 10000
8.3
1. 33
3
1
%profit 2. 10% 3. 25% 4. 120
5. Rs. 2576 6. 21 7. 4% loss 8. Rs. 1108.80
9. 12% gain 10. 15% gain
8.4
1. Rs. 318.75 2. 20% 3. Rs. 15390 4. 25%
5. Rs. 724.50 6. 10% loss 7. Rs. 2185 8. 10.4%
8.5
1. Rs. 16240 2. Rs. 3744 3. Rs. 5600 4. 11%

Notes
MODULE - 2
Commercial
Mathematics
5. 4 years 6. 10% 7. Rs. 1100, 5% 8.
7
5
9 %
9. 24 years 10. b
8.6
1. Rs. 1951 2. Rs. 1951 3. Rs. 2522 4. Rs. 24000
5. Rs. 630 6. Rs. 46875 7. Rs. 80000 8. 1
2
1
years
9. 20% 10. Rs. 1600, 5%
8.7
1. 316368 2. Rs. 300186 3. Rs. 291456
4. 4.2042 tons/hectare 5. 3.2256×108
6. 49140
ANSWERS TO TERMINAL EXERCISE
1. (a) 45% (b) 70% (c) 34% (d) 6%
2. (a) 0.36 (b) 4.10 (c) 0.02 (d) 0.0035
3. (a)
2500
3
(b)
40
1
(c)
200
51
(d)
20
51
4. (a) 115 (b) 20 (c) 4 (d) 460
5. 42% 6. 25% 7. 320 8. 1800
9. 18 10. 25% 11. 70% 12. 50
13. 1 Lakh 14. Rs. 14000 15. Ratan Stores 16. 50
17. 20% 18. 36 19. 60% gain 20. 25%
21. Rs. 400 22. 20% 23. Rs. 1650 24. Rs. 25000
25. Rs. 2000 26.
3
1
3 % 27.
2
1
2 years 28. Rs. 3000
29. Rs. 62500 30. Rs. 5000 31. Rs. 16000, 5% 32. Rs. 24000
33. 10% 34. Rs. 13500 35. 19800

Percentage and its applications /COMMERCIAL MATHEMATICS

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Percentage and its applications /COMMERCIAL MATHEMATICS