digi-notes-percentage -maths-4-12-2015.pdf

8/20/2019 Digi-Notes-Percentage -Maths-4-12-2015.pdf

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The Digi-notes (Percentage) as presented here is prepared strictly according

to IBPS-PO, IBPS-Clerk, IBPS-SO and other PO/Clerk level competitive

exams. The main aim of the Digi-notes is to develop the notion of the

percentage so that candidate would be able to handle the quantitative

problems of the real world easily and conveniently and hence in their

competitive exams. Candidates are advised to handle the Digi-notes by pre-

assuming that they are going to learn the rule of the language of the

percentage so that you; yourself would be able to minimize the steps

required to conclude the answer of the question and hence develop theSHORT TRICKS.

All the best…

Preface


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1.

Preface …………………………………………..………….……… 2

2. Contents ………………………………………………….………… 3

3. Introduction …………………………………………….………… 4

4. What is Percentage? ……………………………….…………. 4

5.

Strategy……………………………………………………….…….. 5

6. The Percentage calculation Techniques ……….…..... 6

7. The Application of Unitary Method …………………… 11

8. The Application of Venn diagram ………………………. 14

9. Percentage Increase/Percentage Decrease.…..……. 19

10. The Technique of Net Effect of Successive

Percentages…………………………………..…………………… 24

11. The Percentage Change in the Product of Two or More

Quantities ………………………………………………………...……. 27

Contents


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Introduction:

Everything is RELATIVE in this world; nothing is ABSOLUTE. Therefore;

to analyse and solve the quantitative problems in day to day life;

percentage plays an important role as it provides a convenient method

to relate two or more entities in the real world. It is extensively used in

data analysis in any organization.

Percentage is the backbone of the entire Quantitative Aptitude. It is

spread everywhere in the entire Quantitative Aptitude. If anyone

analyses a paper of 50 questions; he/she would find that; nearly 80%

questions (I stated it in terms of percentage) involves the concept of


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percentage calculation extending from profit/loss/discount; simple

interest; compound interest; its application in Alligation; in time work

and so on to Data Interpretation. So candidates should be well

equipped with how to speak the language of percentage while dealingwith the question.

What is Percent?


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From the mathematically perception of the percent; we could conclude

that the mathematical meaning of the sign “%” is “”. When we

would see the sign “%” we would think as there is a multiplication by

“ ”. Strategy:

We would handle the questions of the percentage according to the

following strategy (Our main aim would be to develop those logical

steps which would make us psychologically strong enough to evolve

the so called SHORT TRICKS of a question):

1. The percentage calculation technique (Most candidates do not

know that they are weak in the percentage calculation technique

which is extensively used in Data Analysis; and they think that

Data Analysis is very tough).

2. The application of Unitary Method; when the base of the given

percentages are same.

3.

The Application of Venn diagram; in questions in which the data

overlaps.

4. The Percentage Increase and/or Percentage Decrease. When the

base of the given percentages are different; the application of

initial/Final value, again it is another form of Unitary Method.


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5. The Technique of Net Effect of Successive Percentages.

6. The percentage change in PRODUCT of two or more quantities.

The Percentage Calculation Technique:

`

Just Keep one thing in mind; while calculating the percentage value of

a given quantity (Q); try to extract the information that from which

quantity it is compared (C) and then put it in the denominator and then

multiply it by 100 and finally put “%” sign after that (just to nullify the

multiplication by 100, its equivalent “% ≡ ” should be multiplied.).To sum up:

From the above discussion we could conclude and easily digest the

following in our mind:


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Let us discuss it with the help of some examples:

Q.1. Find 15% of Rs.250?

Sol:

= . =..


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Q.2: Change % into equivalent fraction?

Q.3: Change

into equivalent percentage?

Q.4: 2 days is what percent of 2 weeks?

Q.5: The salary of a person is increased from Rs.4050 to

Rs.4500. Find the percentage increased in his salary?

Sol:

= ⁄ =

Sol:

= ×% = %

Sol:

= × × % = %

Sol: % ↑ = ×% = %


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Q.6: The salary of a person is decreased from Rs.4500 to

Rs.4050. Find the percentage decreased in his salary?

Q.7: The ratio of two numbers P and Q is 3:5. Find:

1. P is what percent of Q?

2.

Q is what percent of P?

3.

P is what percent of (P+Q)?

4.

Q is what percent of (P+Q)?

5.

P is what percent less than Q?

6. Q is what percent more than P?

Q8: P, Q and R are three numbers. The ratio of P and Q is 2:3

and that of Q and R 2:1. Find:

Sol:% ↓ = ×% =%

Sol 1: × % = % Sol 2: × % = % Sol 3: × % = % Sol 4: × % = % Sol 5:

× % = %

Sol 6: × % = %


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1.

(P+Q) is what percent of (Q+R)?

2. (P+R) is what percent less than (P+Q+R)?

Sol: Learn the rule for combining two different ratios having

one quantity common in the given ratios:

1. = = = × % = %

2. = =

= × % = %


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Q.9: The salary of Ram is 20% more than the salary of Mohan.

Find the salary of Mohan is what percent less than the salary of

Ram?

Q.10: The salary of Ram is 20% less than the salary of Mohan.

Find the salary of Mohan is what percent more than the salary

of Ram?

Q.11: A, B and C are three persons. The income of A is 40% more

than the income of B. While the income of B is 20% less than the

income of C. Find:

1. The income of C is what percent less than the income of A?

2. The income of C is what percent less than the income of A?

Short Trick: : = : = : = ×% = %

Short Trick:

: =

:

=

:

= ×% =%


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Q.12: 20% of the salary of Amit is equal to 30% of the salary of

Bunty. While 80% of the salary of Bunty is equal to 40% of the

salary of Chintu. Find the total salary of Amit and Chintu is what

percent less than the total salary of Amit, Bunty and Chintu?

Short Trick: = ×% = %

= ×%

=%


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Q.13: A positive numbers is by mistake multiplied by 5 instead

of being divided by 5.

1. Find the result obtained is what percent of the required

correct value?

2.

Find the result obtained is what percent more than the

required correct value (OR Find the percentage error in the

calculation)?

Q.14: A positive numbers is by mistake divided by 3/5 instead

of being multiplied by 5/3. Find the percentage error in the

calculation.

Short Trick:

=

= = =×= Sol 1:

= × % =%

Sol 2:

= ×% =%


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The Application of Unitary Method:

Recall your unitary method which we are used to apply to solve the

questions involving the quantities having either direct or inverse

relation among them. Here we have to apply unitary method involving

the quantities having direct relation. Let us discuss it with the help of

the following example:

The cost of 13 pen is 520. Find the cost of 15 pens.

We were used to approach the question as below in our childhood

classes:

Short Trick:

=

=× = = ⁄ = Sol:

= ⁄

×% = % ≡


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13 .520

1 .40

15 .600

Pen Rs

Pen Rs

Pen Rs

We are used to concise it in the following form:

.52015 .600

13

Rs Rs

i.e.

Given ValueCorresponding No of Unit Asked

Corresponding No of Unit

The same approach we will apply in the percentage questions in which

the base (from which comparison is made) of all the given percentagesare same. That is:

Let us solve some examples based on the discussion above:

Q 1: 59% of a number is 760 more than the 40% of the same number.

Find 25% of the number.

Q. 2: The sum of 35% and 24% of a number is 472. Find 3/4th of the

number.

x (Percentage value asked)

Short Trick: Sol: % ×%

=

% ×%

=


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Q. 3: The difference between a number and its 39% is 732; then find

the number.

Q. 4: Find the difference of 28.50% and 37.50% of 75.

Q. 5: A fruit seller had some apples. On a particular day he sold 36% of

the apples and he still remains with 448 apples. Find how many

apples had he sold on that day?

Short Trick: Sol:

=%

= % ×% = +% ×% =

Short Trick: Sol:

=

% × =

%×%

=

Short Trick: Sol:

= % × %

= % × ..% = .


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Q.6: In an exam; 35% of the appeared students failed while 1625

students passed. Find the difference between the number of the

passed and the number of the failed students in the exam.

Q.7: In a village; 45% of the total population are males while 30% of the

total population are females. If the number of the children in the village

is 1500; find the difference of the number of the males and the number

of the females in the village?

Short Trick: Sol:

=

% × %

= % ×% =

Short Trick: Sol: % = % % = % % = % ∴ % × %

=

% ×%=

Short Trick: Sol:

%=% %% =%

∴ % × % = % ×%=


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Q.8: Raju spends 15% of his monthly salary on house rent; 20% on food;

10% on travelling; 25% on children’s education and 5% on

miscellaneous. After all these expenditures he deposited Rs.6250 in his

savings bank account. Find how much amount did he spend on foodand house rent per month?

Q.9: In an election between two candidates; a candidate who got 84%

of the total voters on the voting list; won by a majority of 476 votes.

Find the total number of the voters on the voting list?

Q.10: In an exam, 6% candidates passed from class 11th. From class 12th,

an equal number of candidates appeared for the exam and 7%

Short Trick: Sol: % =% % % % % % =% ∴ % × %

= % × %=.

Short Trick: Sol: % =% % = % % = % ∴ % × %

= % ×%=


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candidates got passed with 80 more candidates got passed than class

11th. Find the number of candidates appeared from each class.

Q.11: In an examination, a candidate must get 80% marks to pass. If a

candidate, who gets 210 marks, fails by 50 marks, find the maximum

marks.

Q.12: A student has to secure 40% marks to pass. If he gets 90 marks

which is 40 marks more than the passing marks, find the maximum

marks set for the examination.

Q.13: In an examination it is required to get 290 of the aggregate

marks to pass. A student gets 209 marks and is failed by 12% marks.

What are the maximum aggregate marks a student can get?

Short Trick: Sol: = % × %

= −% ×%=

Short Trick:

Sol: = % ×% = +% ×%=

Short Trick:

Sol: = % ×% = −% ×%=


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Q.14: A candidate who gets 30% of the marks in a test fails by 50 marks.

Another candidate who gets 320 marks fails by 30 marks. Find the

maximum marks.

Q.15: A candidate scores 25% and fails by 30 marks, while another

candidate who scores 50% marks, gets 20 marks more than the

minimum required marks to pass the examination. Find the maximum

marks for the examination.

Q.16: In an exam, A scores 280 marks which are 20 more than passing

marks. B got 80% marks which are 60 more than the passing marks.

What is the passing percentage of the exam?

Short Trick:

Sol: = % ×% = −% ×%=

Short Trick:

Sol: .= % ×% = + −% ×%=

Short Trick:

Sol: .= % ×%

=−−%−% ×%=


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Q.17: A student got a total of 70% marks in an exam of 4 subjects. In

the three subjects he scores 81, 66 and 60 marks. If the maximum marks

of each subject are 100, then find the marks obtained in 4 th subject.

The Application of Venn diagram:

There are some questions involving the percentage in which a certain

keywords (Only; Either-or; Both; Either-or…not both etc.) are used

which tests our ability to perceive these keywords in practical

problems.

Short Trick:

Sol: % = % × = %−+ × =%

Short Trick: Sol: = = ×% × % =


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You are aware from your childhood classes about the Venn diagram;

and learned a formula related to it for two sets A and B having a

common value as described below:

∪ = ∩ First of all you forget about this formula. If you will try to apply it in thequestions you will involve yourself only in mathematical equations and

calculations. You would not be able to perceive the question logically

that what is going on psychologically behind the scene.

Let us equip ourselves with the psychological steps to conclude the

question easily and with the help of mental calculation:

First; let us learn how to read a Venn diagram according to the givenkeywords as described below:

Which Portion

Represent

Description Answer

A Entire Portion x + y

Only A ≡A butnot B

Remove B from A x


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A and B Always the

Intersection

(Common)

Portion

y

Either A or B (OR

simply A or B)

Only A + Only B +

Both

x + z +

y

A or B but not

both

Exclude Both

from A or B

x + z

You could perceive and digest it easily into your mind with the help of

the following diagram:

For the better perception; consider the following example:

Q.1: In an Office; 35 people like tea, 55 people like coffee while 15

people like both tea and coffee.

1. Find the number of the people who:

1) Like Tea

2) Like Coffee

3)

Like Only Tea4) Like Only Coffee

5) Like Both Tea and Coffee

6) Like Either Tea or Coffee

7) Like Either Tea or Coffee but not Both


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2. Find the Total number of the people in the office?

Solution: As there is a case of overlapping in the question: it would be

easy to analyse the situation with the help of the Venn diagram:

1. See the solution below extracted from the Venn diagram:

1) 20 + 15 = 35

2)

40 + 15 = 55

3) 20

4) 40

5) 15

6) 20 + 40 + 15 = 75

7) 20 + 40 = 60

2. Data Inadequate as there could be some people in the office who

might do not like both tea and coffee; about which no information

is given in the question.

Q.2: In an Office; 35 people like tea, 55 people like coffee while 15

people like both tea and coffee. 10 people like neither tea nor coffee

(i.e. do not like both)

1. Find the number of the people who:

1)

DO NOT LIKE Tea2) DO NOT LIKE Coffee

3) DO NOT LIKE Only Tea

4) DO NOT LIKE Only Coffee

5) DO NOT LIKE Both Tea and Coffee

6) DO NOT LIKE Either Tea or Coffee


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7) DO NOT LIKE Either Tea or Coffee but not Both

2. Find the Total number of the people in the office?

Sol: Now its turn to talk about the complement part about the Venn

diagram. This discussion would give us the idea about the pivot pointfor the entire Double Venn diagram questions. Let us discuss it (Try to

perceive the following diagram):

2. See the solution below extracted from the DO NOT LIKE Venn

diagram:

1) 40 + 10 = 50

2)

20 + 10 = 303) 40

4) 20

5) 10

6) 40 + 20 + 10 = 70

7) 40 + 20 = 60

3. Now this time we are given the information about how many

people are there in the office who DO NOT LIKE both tea andcoffee which is 10.

Now from the LIKE and DO NOT LIKE Venn diagram:

Total = [LIKE only T + LIKE only C + LIKE both] + DO NOT LIKE both

Total = Either-or LIKE + DO NOT LIKE both = (20 + 40 + 15) + 10 =

85


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Or we can also say that (Try to convince yourself by visualizing the

situation practically)

Total = Either-or DO NOT LIKE + LIKE both = (40 + 20 + 10) + 15 =

85

We could diagrammatically fit it in our mind as:

Conclusion:

Now the learning is complete; let us apply the learning in percentage

questions:

Q.3: In an exam; 50% of the appeared students failed in English, 45%failed in Hindi while 15% failed in both the subjects. Find the

percentage of the students who PASSED in:

1. Both English and Hindi.

2. Either English or Hindi.


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Sol: From the above Discussion (Attention: Pass is the complement of

Fail and vice-versa):

1. Pass in both = 100% - Failed in Either E or H = 100% - 80% = 20%

2. Pass in Either E or H = 100% - Failed in Both = 100% - 15% = 85%

Q.4: In a school; 55% of the students play Tennis, 50% play Hockeywhile 20% play both the games.If 90 students play neither Tennis nor

Hockey; find the number of the students who:

1. DO NOT PLAY Tennis

2. DO NOT PLAY Hockey

3. DO NOT PLAY Only Tennis

4. DO NOT PLAY Only Hockey

5.

DO NOT PLAY Both the games6. DO NOT PLAY Either Tennis or Hockey

7. DO NOT PLAY Either Tennis or Hockey but not both the games.

8. TOTAL number of the students in the hostel.

9. PLAY Tennis

10. PLAY Hockey

11. PLAY Only Tennis

12.

PLAY Only Hockey13. PLAY Both the games

14. PLAY Either Tennis or Hockey

15. PLAY Either Tennis or Hockey but not both the games.


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Sol: We could conclude the following from the discussion till now:

Now Consider the DO NOT PLAY Diagram and conclude the answers

using unitary Method:

1. = % × % = 2. = % × % = 3. = % × % = 4. = % × % = 5.

= % × % = 6. = % × % = 7. % × % = 8.

=

% ×%=

9. = % × % = 10. = % × % =

11. = % × % = 12. = % × % =


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13. = % × % = 14. = % ×

% =

15. =% × % =

Q5: In a factory, 72% people like tea while 44% people like coffee. If

20% people like neither tea nor coffee and144 people like both tea and

coffee; find the total number of the people in the office; if.

Solution:

= % ×%= Q.6: In a factory, 72% people like tea while 44% people like coffee. If

EACH of them like either tea or coffee and 64 people like both tea and

coffee; find the number of the people who like only coffee.

Sol: EACH of them like either T or C There are NO people in the office

who DO NOT LIKE both T and C


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= % × % = Percentage Increase and/or Percentage Decrease:

We will apply the concept of initial and final value of given fraction.

What we would mean by the INITIAL and FINAL vale of a fraction :

± Initial = D

Final = D ± N • + Increase • - Decrease


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For Example

: ≡ ≡ =

: ≡

≡ =

So for quick calculation you would have to learn the fractional valueof some percentages. Some are as below:


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100% = 1 50 % = 25 % =

1 % = 3 % = 62 % = 6

% = 16 % = 3 % = 66 % = 1 % = 9 % = 1 % = 2 % = 8 % =


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The rest is the story of unitary method as discussed below with the

help of atomic examples (Later we will discuss the combination of

these first four cases):

Q.1: Rs.640 is increased by %. Find the amount after increase.

Q.2: An amount is increased by % and becomes Rs.6250.Find the initial amount.

Q.3: Rs.275 is decreased by %. Find the amount after decrease.

Short Trick:

I = Rs.640 % ≡ + F = × =

Short Trick:

Sol: I % ≡ + F = Rs.6250 = × =

Short Trick:

Sol: I = Rs.275 % ≡ − F = × =


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Q.4: An amount is decreased by % and becomes Rs.250. Find theinitial amount.

Learning complete, now Let us apply it as a combination in differenttypes of questions:

Q.5: A man loses % of his money and after spending % ofthe remaining; he is left with Rs.1050. Find how much had he at first?

Q.6: A man had Rs.39200 in his locker. During first year; he deposited

25% of the amount in his locker; while during next year he deposited

%

of the increased amount in his locker. Find the total amount

deposited by him during a span of 2 years?

Short Trick:

Sol: I % ≡ − F = Rs.250 = × =

Short Trick:

Sol: Amount − Remainder − Rs.1050 ∴= . × × × =.


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Q.7: The present population of a town is 108000. During first year; the

population of the town increases by % while decreases by % during second year. During third year the population increasesby %. Find the population of the town after 3 years.

Q.8: 10% people of a village having died of cholera. Due to fear; 25% ofthe remaining people left the village; then the population of the town

reduces to 4050. Find the decrease in the number of the people in the

town?

Short Trick:

Sol: 39200

+ Increased Amount

+

Amount after 2 years

=

. × × × =. ∴

=..=.

Short Trick:

Sol:

108000+ ? − ? + After 3 years = × × × × × =


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1. Q.9: The value of a machine depreciates at the rate of 10% per

annum. If its present worth is Rs.3645000; find its worth.1. After

3 years.2.Before 3 years.

Q.10: The value of a land appreciates at the rate of % per annum.If its present worth is Rs.1984500; find its worth

Short Trick:

Sol: Initially−

Remaining People−

4050

= . × × × = ∴ = =

Short Trick:

Sol:1

Rs.3645000 − ? − ? − After 3years

= × × × × × =. Short Trick:

Sol:2

Before 3 years − ? − ? − Rs.3645000 = × × × × × =.


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1. After 2 years.

2. Before 2 years.

Q.11: Amit received a certain amount from his father. He

spends 20% of the amount on hostel expenses; 1/3 rd of the

remaining on stationary; % off the amount on food and25% of the remaining amount he spends on other expenses.

After all these expenditures he is left with Rs1200.Find the

total expenditure made by Amit.

Short Trick:

Sol:1 Rs.2381400+ ? + After 2 years = . × × × =.

Short Trick:

Sol:2 Before 2 years+ ? + Rs.2381400

= . × × ×

=.


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Q.12: Raju spends 40% of his monthly salary on house rent; 10%

of the remaining on travelling; % of the remaining. Afterall these expenditures he saves Rs.2250. Find how muchamount did he spends on food per month?

Q.13: Shalini spends

%

of his monthly income on house

rent; 30% on food and 20% of the remaining on other expenses.

After all these expenditures he deposited Rs.16000 in his

savings bank account. Find her monthly salary?

Short Trick:

Sol: Received Amount − R − R − R − Rs.1200 = . × × × × × × × =. ∴ =..=.

Short Trick:

Sol: No use of 40% and 10% to conclude the answer

Remaining Amount ≡ % ≡ − Saves = Rs.2250 After Spending on Travelling = . × × =.


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Net Effect of Successive Percentages:

Let us discuss it with the help of an example;

The salary of a person is first increased by 10% and thereafter it is

reduced by 5%. Find the net percentage change in his salary.By conventional method:

%

+

×

%−

× ×

%5.4

1005.104

100100

95

100

110

100%

S

S S

i

i f N

Short Trick:

Sol:

Salary % % R %≡ − Rs.16000 = % ≡ − ∴ = . × × × =.


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Assume that the net effect of 10% increase and 5% decrease is N%

Now look at the following scenario:

% %

+ × %− × × OR + ? %− × + Hence we can write:

%5.4100100

)5(10)5(10

100

)5(1010)5(100100

100

)5(1010010)5(100100100100

100

)5(100

100

10100

100

100`

100

95

100

110

100

100

xy y x N

N

N

S N

S

S N

S

Hence we can conclude that; if we are given two successive

percentages x% and y%, its net effect will be given by: %100

xy y x N


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By putting the values of x and y with sign i.e.; for percentage increase

put the values with positive sign and for the percentage decrease put

the values with negative sign.

Now we would be able to apply in the following questions withoutgoing into the details:

Q.1: A man increases the price of his article by 17%. After that

he allows the discount of 17% on the increased price to his

customers. Find his net profit or loss percentage in the deal.

Q.2: The population of a town decreases by 12% during the year

1988. If the percentage increase in the population of the town

in the beginning of 1990 is observed to be 1.2%; find at what %

it is increased during 1989?

Question 3: Find the net discount of two successive discounts

of 10% and 20%?

Short Trick:

Sol: % = × =.% ≡.%

Short Trick:

Sol: . = × → = %

Short Trick:

Sol: % = × =%≡%


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Percentage Change in the Product of Two or More Quantities:

Let us discuss it with the help of the following example:

Q: In the product, = × , a is increased by 10% while b isincreased by 20%. The new product so formed would be whatpercent more than the initial product?

Sol: By conventional Method:

% ∆ = −

×%

→ % ∆ = × × × × × ×% → % ∆ = × ×%

→ % ∆ = % OR (In Short Way):% ∆ = × =% Q: In the product; = × × ; a is increased by 10% whileb is increased by 20% and c is decreased by 10%. The new

product so formed would be what percent more than the initial

product?

Sol: %∆ = %; %; % %∆ = %; % % %∆ = % %


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%∆ = % % %∆=.% Hence we could conclude that we can also use the net effect methodfor finding the percentage change in the product of two or more

quantities.

We will extensively use it for finding the percentage change in AREA

(Product of Two Quantities) of 2D figures and the percentage change

in the VOLUME (Product of Three Quantities) of 3D figures. We will

we discuss the questions related to percentage change in AREA of 2D

figures (Volume questions are asked in SSC and equivalent exams; sowe will skip its discussion here)

Q.1: The length of a rectangle is increased by 10% while its breadth the

reduced by 5%. The area of the new rectangle would be what

percentage more than the area of the original area of the rectangle?

Sol:

% ∆ = ×

= . %

Q.2: The adjacent sides of a square are increased by 10% and 20%

respectively. The area of the rectangle so formed would be what

percent more than the area of the square?

Sol: % ∆ = × =% Q.3: Each side of a square is decreased by 30%. Find the percentage

change in the area of the square?

Sol: % ∆ = × =% Q.4: The length of a rectangle is increased by 10%. By what percentage

its breadth should be reduced in order to


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1. Increase the area by 4.5%.

2. Maintain the same area

Sol.1:

. % = ×

→ = % Sol.2: % = × → = % Q.5: The radius of a circle is increased by 2 cm; from 5 cm to 7 cm. Find

the percentage change in the area of the circle.

Sol: %∆= × = % ∴ % ∆ = × =% Q.6: The base and height of a triangle is increased by 10% and 20%;

then the area of the triangle increases to 529 meter square. Find the

original area of the triangle.

Sol: % ∆ = × =% ∴ = % × % = Q.7: The perimeter of a square is increased by 10%. Find the percentage

change in the area of the square.

Sol: %∆ ≡ % ∆ = % ∴ % ∆ = × =% Q.8: The circumference of a circle is decreased by 20%. Find the

percentage change in the area of the circle.


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Sol: %∆ ≡ % ∆= % ∴ % ∆ = ×

=%

Q.9: The perimeter of a rectangle is increased by 30%. Find the

percentage change in the area of the rectangle.

Sol: %∆( ) ≡ %∆ ∴

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digi-notes-percentage -maths-4-12-2015.pdf

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