Concept of Arithmetic_PERCENTAGE

8/9/2019 Concept of Arithmetic_PERCENTAGE

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180 Concep t o f A r i t h me t i c

Chapter-10

Percentage

Introduction

Per cent is a kind of fraction of which denominator

is 100. The word per cent is an abbreviation of

the Latin phrase percentum which means per

hundred or hundredths. Thus the term per cent

means per hundred or for every hundred. For

example,

( i ) When we say that a man gives 30 per cent

of his income as income tax, this meansthat he gives Rs 30 out of every hundred

rupees of his income.

( ii) A trader makes a profit of 15 per cent means

he gains Rs 15 on every hundred rupees of

his investment.

( i i i ) A boy scored 70 per cent marks in his final

examination means that he obtained 70

marks out of every hundred marks.

The term per cent wi ll be writ ten as the

symbol (%).

Per cent and Vulgar Fraction

Since per cent is a form of fraction, hence per

cent can be expressed as fraction and vice versa.

This can be illustrated through examples. See the

examples given below:Ex. 1: Exp r ess t h e gi v en f r ac t i ons as pe r c en t .

(i)2

1 (i i )

4

3 ( i i i )

5

1 1 ( iv)

3

2

Soln: To convert a fraction into a per cent we

express the given fraction with

denominator as 100 or multiply the

fraction by 100 and put the per cent

sign (%).

(i)100

50

1002

1001

2

1

= 50%

or, 1002

1 = 50%

(ii)100

75

1004

1003

4

3

= 55%

or, 1004

3 = 55%

(iii)100

220

1005

10011

5

11

= 220%

or, 1005

11 = 220%

(iv) %3

266

1003

266

1003

200

1003

1002

3

2

or, %3

266

3

200100

3

2

Ex. 2: E x p r e s s t h e f o l l o w i n g p e r c en t a s a

f r a c t i o n .

(i ) 25 % (i i ) %3

18 (i i i ) 14 0%

Soln: To convert a per cent into a fraction, we

divide it by 100 and remove the per cent

sign %. Thus,

(i) 25% =4

1

100

25

(ii) 12

1

1003

25

%3

25

%3

1

8

(iii) 140% =5

7

100

140

Per cent and Decimal Fraction

Ex. 3: Conver t t h e ever y dec ima l i n t o per c en t .

(i ) 0 .35 (i i ) 0 .12 5 (i i i ) 2 .25

Soln: In order to convert a given decimal into a

per cent, express it as a fraction with

denominator as 100 or move the decimal

point on the right side by two digits and

put the per cent sign %. Thus,

(i) 0.35 =

100

35

100

10035.0

= 35%

(ii) 0.125 =100

5.12

100

100125.0

= 12.5%

(iii) 2.25 =100

225

100

10025.2

= 225%


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181Percentage

Ex. 4: Ex p r e ss t h e ev e r y g i v e n pe r c en t a s a

d ec i m a l f r a c t i o n .

(i ) 38 % (i i ) 16 .5% (i i i ) 32 0.5 %

Soln: To convert a given per cent in decimal

form, we express it as a fraction withdenominator as 100 and then the fraction

is written in decimal form. Thus,

(i) 38% =100

38 = 0.38

(ii) 16.5% =100

5.16 = 0.165

(iii) 320.5% =100

5.320 = 3.205

Per cent and Ratio

Ex. 5: C on v er t t h e f o l l o w i n g r a t i o s i n t o p e r

c en t .

(i ) 6 : 5 (i i ) 3 : 12 (i i i ) 9 : 20Soln: In order to convert a given ratio into a per

cent, we first convert the given ratio into

fraction and then multiply the fraction by

100. Thus,

(i) 6 : 5 = 1005

6

5

6 = 120%

(ii) 3 : 12 = 10012

3

12

3 = 25%

(iii) 9 : 20 = 10020

9

20

9 = 45%

Ex. 6: E x p r e s s t h e f o l l o w i n g p e r c en t a s

r a t i o s .

(i ) 52 % (i i ) 27 .5% (i i i ) 8%Soln: To convert a given per cent into a ratio,

we first convert the per cent into a fraction

and then express it as a ratio. Thus,

(i) 52% =25

13

100

52 = 13 : 25

(ii) 27.5% =40

11

1000

275

100

5.27 = 11 : 40

(iii) 8% =5

2

100

8 = 2 : 5

Finding per cent of a Quantity

Many a time, we need to find a given per cent of a

given quantity. For example, we might be asked tofind the number of girl students in a school, given

that there are 28% girl students in the school and

total number of students in the school is 1000. We

shall explain the process of finding such a

percentage through some examples.

Ex . 7 : F i nd 2 0 % o f Rs 15 0 .

Soln: We have, 20% of Rs 150

=100

20of Rs 150 = Rs 150

100

20 = Rs 30

Thus, 20% of Rs 150 is Rs 30.

Ex. 8: F i n d t h e n um b er o f g i r l s t u d e n t s i n a

s ch o o l , i f t h e r e a r e 2 8% gi r l s t u d e n t s

i n t h e sc h o ol a n d t h e t o t a l n umb e r o f

s t u d en t s i n t h e s ch o o l i s 1 0 0 0 .

Soln: Required number of girl students

= 28% of 1000 = 1000100

28 = 280

Ex. 9: In an orchard, %3

21 6 of the trees are

apple trees. If the total number of trees

in the orchard is 240, find the number

of other types of trees in the orchard.

Soln: Total number of trees = 240 number of apple trees

= %3

216 of 240 = %

3

50of 240

= 240100

1

3

50 = 40

Therefore the number of other trees

= 240 40 = 200

Thus, number of trees of other types in

the orchard is 200.

Ex. 10: F i n d 1 0 % mo r e t h a n R s 9 0 .


=100

10of Rs 90 = Rs 90

100

10 = Rs 9

10% more than Rs 90= Rs 90 + Rs 9 = Rs 99

Ex. 11: F i n d 2 0% l es s t h a n R s 7 0 .


=100

20of Rs 70 = Rs 70

100

20 = Rs 14

20% less than Rs 70= Rs 70 Rs 14 = Rs 56

Expressing One Quantity as a per cent ofAnother Quantity

We come across many situations where we have to

express a quantity as a per cent of another quantity.

For example,

We may be asked to find the percentage of marksobtained by Peter, if he obtains 285 marks out of a

maximum of 500 marks. This is equivalent to

finding what per cent one number is of the other.

We shall explain the method of finding what per

cent one number is of another through some

examples.


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Ex. 12: Wha t p er c en t i s n umb er 8 o f n um b er

2 5 ?

Soln: Recall that by per cent, we mean per 100.

We, therefore, proceed as follows:

Out of 25 ie per 25, the number is 8.

per 100, the number is 10025

8 = 32

Thus, the required per cent is 32. In other

words, 8 is 32% of 25.

Note: To find what per cent the fir st

number is of second, we divide the first

number by the second number and

multiply the result by 100.

Alternative Method:

Let the number 8 be x% of number 25.

x% of 25 = 8

or, 25100

x

= 8

or, x=25

1008= 32

Thus, the required per cent is 32. In other

words, 8 is 32% of 25.

Ex. 13: Pet e r o b t a i n e d 3 8 5 m a r k s o u t o f a

m a x i m u m o f 5 0 0 m a r k s . F i n d t h e

pe r c en t a ge of ma r k s obt a i ned by Pe t er .

Soln: Required percentage of marks

= 100500

285 = 57

Thus, Peter obtained 57% marks.

Ex. 14: Wha t p er c en t o f 2 5 k g i s 3 .5 k g?

Soln: Note that we wish to find here what per

cent 3.5 is of 25.

Required percentage = 10025

5.3

= 100250

35 = 14

Thus, 3.5 kg is 14% of 25 kg.

Ex. 15: A b a s k e t c o n t a i n s 3 0 0 m a n g o es . 7 5

ma ngoes wer e d i s t r i b u t ed am ong some

s t u d e n t s . F i n d t h e p er c en t a g e of

m a n g oe s l e f t i n t h e b a s k et .

Soln: Original number of mangoes = 300

Mangoes distributed = 75

Mangoes left in the basket= 300 75 = 225

Percentage of mangoes left

= 100300

225 = 75

Thus, 75% mangoes are left in the basket.

Ex. 16: J a m s h e d o bt a i n ed 5 5 3 m a r k s ou t o f

7 00 an d Geet a ob t a i n e d 486 m a r k s ou t

o f 6 0 0 i n M a t h em a t i c s . W h o s e

p e r f o r m a n c e i s b et t e r ?

Soln: To compare the performance, we shall

convert the marks into per cent.

Thus, we have,

Marks obtained by Jamshed

= %100700

553 = 79%

Marks obtained by Geeta

= %100600

486 = 81%

Now, 81 > 71

Therefore, Geetas performance is better.

Ex. 17: The e xc i s e du t y on a c er t a i n i t em ha s

been r educed t o Rs 3480 f r om Rs 5220 .

F i n d t h e p er c en t a g e r e d u c t i o n i n t h e

ex c i s e du t y f o r t h a t i t em .

Soln: Amount of reduction in excise duty

= Rs 5220 Rs 3480 = Rs 1740

Percentage of reduction is calculated on

initial excise duty ie Rs 5220. Required per cent reduction

= %1005220

1740 = %

3

100= %

3

133

Questions Based on Population Growth

If the original population of a town is P, and the

annual increase is r%, then, the population in n

years will be obtained by the following formulae:

( i ) Required population =

nr

P

1001

Since, population after one year becomes

100

rPP =

1001

rP

That is, the population P at the beginning

of the year is multiplied by

1001

r in the

course of the years.

Now, the population at the beginning of

the second year is

1001

rP

Population after 2 years = Population atthe beginning of second year + increase in

it

=100100

1100

1 rr

Pr

P

=

1001

1001

rrP =

2

1001

rP

: : :

: : :

The population in n year =

nr

P

1001


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183Percentage

( ii) If the annual decrease be r%, then the

population in n years =

nr

P

1001

( i i i ) If the annual increase be r%, then thepopulation n years ago

n

n

r

PrP

1001

1001

( iv) If the annual increase be r per thousand,

then the population in n years

=

nr

P

10001

Ex. 18: T h e p o p u l a t i o n of a t ow n 3 y ea r s a g o

w a s 4 0 9 6 0 . I f t h e p r e sen t p o p u l a t i o n

o f t h e t o w n i s 4 9 1 3 0 , t h en f i n d t h e

annua l p e r c en t i n c r ea s e i n popu l a t i o n .

Soln: See the formula (iii) mentioned above,

Population n years ago

nr

PP

1001

100

RateAnnual1

)(PopulationPresentTime

or, 40960 = 3

1001

49130

r

or,

3

1001

r=

40960

49130=

4096

4913=

3

16

17

or,100

1 r =1617 =

1611

or,100

r=

16

1

r=16

100=

4

25= %

4

16

Ex. 19: T h e p op u l a t i o n o f a t ow n i s 26 2 4 0 0 0 .

I f t h e p o pu l a t i o n o f t h e t o w n i s

i n c r e a s i n g a t t h e r a t e o f 2 5 p e r

t h o u s a n d p er a n n um , t h e n f i n d

(i ) w h a t w a s t h e pop u l a t i on o f t h e

t own 1 y ea r ago?

(i i ) w h a t w i l l b e t h e po p ul a t i on o f t h e

t ow n a f t e r 3 y e a r s ?

Soln: (i) From the formulae [(iii ) and iv)]mentioned above,

Population n years ago = nr

P

10001

required population

40

41

2624000

1000

251

26240001

=41

402624000 = 2560000

(ii) From the formulae [(i) and (iv)]

mentioned above,

Population after n years

=

nr

P

10001

required population

=

3

1000

2512624000

=

3

40412624000

=404040

4141412624000

= 28225761

Depreciation

The value of a machine or of any other arti cle

subject to wear and tear decreases with time.

Relative decrease in the value of a machine is called

its depreciation . Depreciation per unit time is

called the rate of depreciation. Thus, if V is the

value of a machine at a certain time and R% per

annum is the rate of depreciation then the value

of machine after n years =

nRV

1001 .

Ex. 20: T h e v a l u e of a r e s i d e n t i a l f l a t

c o n s t r u c t e d a t a c o s t o f Rs 100 000 i s

d e p r e ci a t i n g a t t h e r a t e o f 1 0 % p e r

a n n um . Wh a t w i l l b e i t s v a l u e 3 y ea r s

a f t e r c o n st r u c t i o n ?

Soln: We have,

V = Initial value = Rs 100000

R = Rate of depreciation = 10% per annum

Value after 3 years =

3

1001

RV

= Rs

3

100

101100000

= Rs

3

10

11100000


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= Rs

3

10

9100000

= Rs

109

109

109100000

= Rs 72900

Hence, value of the flat after 3 years

= Rs 72900.

Ex. 21: T h e p r e sen t p r i c e o f a s c o ot e r i s R s

72 90 . I f i t s v a l u e decr eas es ever y y ea r

b y 1 0% , t h e n f i n d i t s v a l u e b ef o r e 3

y e a r s .

Soln: Let the value of the scooter be Rs P before

three years. Then, its present value is Rs

3

100

101

P

But the present value is given as Rs 7290.

Therefore,

7290 =

3

100

101

P

or, 7290 =

3

10

11

P

or, 7290 = P 3

3

10

9

or, P = 3

3

9

107290 = 10000.

Hence, the value before 3 years was Rs

10000.

Rate of Increase or Decrease ChangesEvery Year

I. If Pbe the population of a city or a town at

the beginning of a certain year and the rate

of increase or decrease is R1% for the first

n1 years, r

2% for the next n

2 years and so

on and Rk% for the last n

k years, then the

population at the end of (n1+ n

2+ ..... + n

k)

years is given by

Pn= ...

1001

1001

2121

nnRR

P

knkR

.....

1001 ;

where n= n1+ n

2+ ..... + n

kand (+) sign is

used for increase and () sign is used for

decrease.

II. If V0 is the value of an article at certain

time and the rate of appreciation or

depreciation is R1% for first n

1 years, R

2%

for next n2years and so on and R

k% for the

last nk years, then the value at the end of

n1 + n

2 + ..... + n

k years is given by

Vn= ...1001

1001

21

210

nn

RRV

knkR

1001..... ;

where n= n1+ n

2 + ..... + n

kand () sign is

used for depreciation and (+) sign is used

for appreciation.

Ex. 22: T h e p o p u l a t i o n o f a t ow n w a s 1 6 0 0 0 0

t h r ee yea r s ago . I f i t h ad i n c r eas ed by

3 % , 2 . 5 % a n d 5 % i n t h e l a s t t h r e e

y ea r s , f i n d t h e p r es en t p o p u l a t i o n o f

t h e t o w n .

Soln: Let P be the present population of the

town. Then,

P = 160000

1005.21

10031

100

51

or, P = 160000 20

21

40

41

100

103

or, P = 2 103 41 21 = 177366.

Hence, present population of the town

= 177366.

Ex. 23: 1 0 0 0 0 w o r k er s w e r e em p l o y e d t o

c on st r u c t a r i v er b r i d g e i n f o u r y ea r s .

A t t h e en d o f f i r s t y e a r , 1 0% wo r k e r s

w e r e r et r en c h e d . At t h e e n d o f t h e

second yea r , 5% o f t he wo r k e r s a t t h a t

t i m e w e r e r e t r e n c h e d . H o w ev er t o

c omp l e t e t he p r o j ect i n t im e , t h e num ber

o f w o r k e r s w a s i n c r e a s ed b y 1 0% a t

t h e e n d of t h e t h i r d y ea r . H o w m a n y

w o r k e r s w er e w o r k i n g d u r i n g t h e

f o u r t h y ea r ?

Soln: We have,

Initial number of workers = 10000

Reduction of workers at the end of first

year = 10%

Reduction of workers at the end of second

year = 5%

Increase of workers at the end of third

year = 10%

Number of workers working during the

fourth year

=

100

101

100

51

100

10110000

=10

11

20

19

10

910000 = 9405

Hence, the number of workers working

during the fourth year was 9405.


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185Percentage

Ex. 25: I f 2 3 % of a n u m b er i s 4 6 , f i n d t h e

n um b e r .

Soln: Let the number be x.23% of x = 46

or, x100

23= 46

or, x=23

10046= 200

Thus, the number is 200.

Ex. 26: I n an e x am i n a t i o n , Nee t a s ec u r e d 372

ma r k s . I f s h e se cu r e d 63% ma r k s , f i n d

t h e m a x i m u m m a r k s .

Soln: Let the maximum marks be x.

Neetas marks = 62% of x

Neeta secured 372 marks

62% of x = 372

or, x100

62 = 372

or, x=62

100372= 600

Hence, maximum marks are 600.

Ex. 27: Wha t i s t h e r a t i o 5 : 4 e qu a l t o w h e n

expess ed as a pe r c en t ?

Soln: Fractional equivalent of 5 : 4 is equal to4

5.

Percentage equivalent of4

5 is 100

4

5

= 125%

Hence, the ratio 5 : 4 is equal to 125%.Ex. 28: T h e s t r e n gt h o f s t u d e n t s i n a s ch o o l

i n c r e a ses f r o m 9 0 0 t o 9 3 6 . F i n d t h e

pe r c en t a g e i n c r ea s e i n t h e s t r e n g t h o f

t h e s t u d e n t s .

Soln: Increase in strength of students

= 936 900 = 36

We have to find what per cent is 36 of

900?

required per cent increase

= 100900

36 = 4%

Hence, the strength of students increases

by 4%.

Ex. 29: Raju and Nita get 294 and 372 marks

respectively in an examination. If Nita

got 62% marks, then find the maximum

marks and the per cent marks obtained

by Raju.

Soln: Let the maximum marks be x.

Marks obtained by Nita = 62% of x

Nita gets 372 marks

62% of x = 372

or, x10062 = 372

x=62

100372= 600

Per cent marks obtained by Raju

= 100600

294 = 49

Hence, maximum marks is 600 and Raju

got 49% of the maximum marks.

Ex. 30: A r e du c t i o n o f 1 0% i n t h e p r i c e o f t e a

en a b l es a t r a d er t o o bt a i n 2 5 k g mo r e

i n Rs 22500 . Wha t i s t h e r e du ced p r i c e

o f t e a p er k g . A l s o f i n d t h e o r i g i n a l

p r i c e o f t h e t e a pe r k g .

Soln: Reduction in the price of tea = 10%

10% of Rs 22500 =100

1022500= Rs 2250

Now, in Rs 2250 a trader can obtain 25

kg of tea.

Reduced price of 25 kg of tea = Rs 2250

Some More Solved Examples

Ex. 24: A n ew c a r c o s t s Rs 3 6 0 0 0 0 . I t s p r i c e

dep r e c i a t e s a t t h e r a t e o f 1 0% a y ea r

d u r i n g t h e f i r s t t w o y ea r s a n d a t t h e

r a t e o f 2 0% a y ea r t h e r e a f t e r . W h a t

w i l l b e t h e p r i c e o f t h e c a r a f t e r 3 y e a r s ?

Soln: We have,

Cost of the car = Rs 360000

Rate of depreciation in first two years

= 10% per annum.

Rate of depreciation in the third year

= 20%

Price of the car after 3 years

= Rs

100

101

100

101360000

100

201

= Rs

5

11

10

11

10

11360000

= Rs

5

4

10

9

10

9360000 = Rs 233280.

Hence, the price of the car after 3 years

= Rs 233280.


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Reduced price of 1 kg of tea

=25

2250= Rs 90

Let the original price of 1 kg of tea be Rsx.

Now, reduction in original price

= 10% of x= Rs100

10x

Reduced price of tea per kg

= Rs

100

10xx

Now, according to the question,

100

10xx = 90

or, 90x = 100 90

x= 100

Hence, original price of tea per kg= Rs 100

Ex. 31: A r educ t i on o f2

11 2 pe r c en t i n t h e p r i c e

o f m a n g o e s e n a b l e s a p u r c h a s e r t o

ob t a i n 4 m o r e f o r a r u pee. Wha t i s t h e

r e du c ed p r i c e? How m any c ou l d he get

f o r 5 0 paise before the reduction in

price?

Soln: Owing to the reduction in price the

purchaser saves2

112 per cent or

8

1 of

Re 1. With this sum he gets 4 mangoes at

the reduced price.

The reduced price of a mango

= Re 48

1 = Re

32

1

Again

2

112100 per cent or,

8

11

or,8

7 of the original price of a mango

= Re32

1

The original price of a mango

=7

8 Re

32

1= Re

28

1

He could get

28

1

2

1 or 14 mangoes

for half a rupee ie, 50 paise, before the

reduction in price.

Ex. 32: The p r i c e o f s uga r goes up by 2 0%. By

h ow m u c h p er c e nt m u s t a h o u s e w i f e

r e d u c e h e r c on s u m p t i o n s o t h a t t h e

e x pend i t u r e does no t i n c r e a s e?

Soln: Let the consumption of sugar originallybe 100 kg and its price be Rs 100. Then,

New price of 100 kg sugar = Rs 120.

[ Price increases by 20%]

Now, Rs 120 can fetch 100 kg sugar.

Rs 100 can fetch =

100

120

100

=3

250kg sugar

Reduction in consumption

=

3

250100 % =

3

50% =

3

216 %.

Alternative Method:

Let the price and consumption each be100 units.

Then, his earlier expenditure was

= Rs (100 100)

Now, the new price = 120 units

To maintain the expenditure, suppose he

reduces his consupmtion by x%, then his

total expenditure

= Rs [120 (100 - x)]

From the question, we have,

100 100 = 120 (100 - x)

or, 120x = 120 100 - 100 100

or, x=120

)100120(100 =

3

216 %.

Ex. 33: Th e p r i c e of e d i b l e o i l i n c r e a s e s f r om

R s 1 3 t o R s 1 5 p e r k g . B y h ow m u c h pe r c en t m us t a h ou s ew i f e r e du c e he r

c o n sum p t i o n s o t h a t t h e ex p e n d i t u r e

d o e sn o t i n c r e a s e?

Soln: Let the consumption of edible oil originally

be 100 kg.

Then, her earlier expenditure

= 100 13 = Rs 1300

Now, due to increase in price,

Rs 1500 can fetch 100 kg edible oil

Rs 1300 can fetch 13001500

100

=3

260kg


=3

260100 =

3

40kg

per cent reduction =3

40% =

3

113 %


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187Percentage

Ex. 34: I n an e l e ct i o n bet w een t wo c and i d a t e s

A a n d B , A g o t 6 0% of t h e t o t a l v a l i d

v o t e s. 1 5 % o f t h e t o t a l v o t e s w e r e

de cl a r ed i n v a l i d . I f t h e t o t a l n umbe r o f

v ot e s i s 5 0 0 0 0 0 , f i n d t h e n um b e r of v a l i d v o t e s p o l l e d i n f a v o u r o f t h e

c a n d i d a t e B .

Soln: Total number of invalid votes

= 15% of 500000

=100

15 500000 = 75000

Total number of valid votes polled

= 500000 75000 = 425000

Percentage of valid votes polled in favour

of the candidate A = 60%

Percentage of valid votes polled in favourof the candidate B = 40%

= 40% of 425000

= 100

40

425000 = 170000

the number of valid votes polled in favourof the candidate B is 170000.

Ex. 35: T h e t a x o n a c ommod i t y i s d i m i n i s h ed

by 15%, and i t s consum p t i on i n c r eas es

b y 10% .

(i ) F i n d t h e d ec r ea se per cen t i n t h e

r e v enue de r i v ed f r om i t .

(i i ) W i t h w h a t i n c r ea s e p er c en t i n i t s

c on s um p t i o n w o u l d t h e r e v en u e

r em a i n t h e s am e?

Soln: (i )The new tax is 85%, or20

17of the original

tax. The new consumption is 100

110

or 10

11

of the original consumption.

the new revenue =20

17 of

10

11 of the

original revenue

=200

187 of the original revenue

= %2

193 of the original revenue.

the required decrease

= 100% %2

193

= %2

16 = 6.5%

Alternative Method:

Let the original tax be Rs 100 and the

consumption be 100 units.

Original revenue = Rs (100 100) = Rs 10000


New tax = Rs (100 15) = Rs 85 and

the new consumption = Rs (100 + 10)

= Rs 110

New Revenue = Rs (85 110) = Rs 9350 Decrease in Revenue= Rs (10000 9350 =) Rs 650

per cent decrease = 10010000

650 = 6.5%

(ii) Since the new tax is20

17 of the original

tax, the revenue would remain the same

if the new consumption becomes17

20 of

the original consumption.

the required increase in consumption

= 117

20

= 17

3

= 17

11

17 %.

Alternative Method:

Let the original tax be Rs 100 and the

consumption be 100 units.

Original revenue = Rs (100 100) = Rs 10000

New tax = Rs (100 15) = Rs 85

Let the new consumption be x units.


10000 = 85x

[ Revenue remains the same]

or, x=85

10000

or, Increase in consumption

= 85

150010085

10000

= %17

1117

17

300

Ex. 36: Mohan s i n c ome i s 10% l es s t h a n t h a t

o f S oh a n . T h e n w h a t p e r c en t i s

S o h a n s i n c om e m o r e t h a n M o h a n s

i n c om e ?

Soln: Let Sohans income be Rs 100

Income of Mohan = Rs (100 10) = Rs 90

Sohans income is Rs 10 more than that

of Mohan.

If Mohans income is Rs 90, Sohans

income is Rs 10 more.

If Mohans income is Rs 100, Sohans

income is Rs

100

90

10

9

111 more.

Hence Sohans income is9

111 % more than

that Mohans income.


9/26


Ex. 37: R a k e sh s i n c om e i s 2 5 % m o r e t h a n

t h a t o f Rohan . Wha t pe r c en t i s Rohan s

i n c om e l e ss t h a n R a k e sh s i n c om e?

Soln: Let Rohans income be Rs 100. Then,

Rakeshs income = Rs 125.If Rakeshs income is Rs 125, Rohans

income = Rs 100

If Rakeshs income is Re 1, Rohans

income = Re125

100

If Rakeshs income is Rs 100, Rohans

income = Rs

100

125

100 = Rs 80.

Hence, Rohans income is 20% less than

that of Rakesh.

Ex. 38: Ra n i s w e i g h t i s 2 5 % t h a t o f M e en a s

a n d 4 0 % t h a t o f T a r a s . W h a t

pe r c en t age o f Ta r a s we i gh t i s Meena s

w e i g h t ? Soln: Let Meenas weight be x kg and Taras

weight be y kg.

Then, Ranis weight = 25% of Meenas

weight = x100

25 .... (i)

Also, Ranis weight = 40% of Taras weight

= y100

40 ... (ii)

From (i) and (ii), we get

yx 100

40

100

25

or, 25x = 40y

[Multiplying both sides by 100]or, 5x= 8y

[Dividing both sides by 5]

or, x= y5

8 .... (iii)

We have to find Meenas weight as the

percentage of Taras weight ie

1601005

81005

8

100 y

y

y

x

[Using (iii)]

Hence, Meenas weight is 160% of Taras

weight.

Ex. 39: I n a s c hoo l , t h e ages o f 20% st u den t s

a r e l e ss t h a n 5 y e a r s . 6 4 gi r l s h a v e m o r e t h a n 5 y e a r s o f a g e a n d t h i s

n um b er i s3

2o f numbe r o f b o y s m o r e

t h a n 5 y e a r s o f a g e. Fi n d t h e t o t a l

n um ber o f s t u d en t s i n t h e s choo l ?

Soln: Let the total number of students be x

Number of students less than 5 years of

age = 20% of x

= x100

20

= 5

x

Number of students above 5 years of

age =

5

xx =

5

4x

Now according to the question,

Number of girls above 5 years of age = 64

and

Number of boys above 5 years of age

=2

364= 96

Total number of students above 5 years of

age = 96 + 64 = 160

or, 5

4x

= 160

x=4

5160= 200

Total number of students in the school= 200

Ex. 40: I f 2yx a n dz

1x , t h e n f i n d t h e per

c en t c ha nge i n t h e va l ue o f x wh en t he

v a l u e o f y i n c r e a ses b y 2 0% a n d t h a t

o f z dec r eas es by 25 %.

Soln: 2yx andz

x1

x =z

yk

2; where k= constant.

On increasing the value of y by 20%, the

increased value of y

=100

120y=

5

6y

On decreasing the value of z by 25%, the

decreased value of z

=100

75z=

4

3z

New value of x =

4

3

5

62

z

y

k

=z

yk

3

4

25

36 2 =

z

yk

2

25

48

Change in the value of x

=z

yk

z

yk

22

25

48 =

z

yk

2523

2


10/26

189Percentage

per cent change in the value of x

=

z

kyz

ky2

2 100

25

23 = 92%

Ex. 41: A me t r e i s t a k en t o equa l 3 9 . 3 7 01 i n c h .

I f t h e v a l u e i s app r o x im a t e d t o2 0

73 9

i n c h , f i n d t h e r e l a t i v e e r r o r .

Soln: According to the question,

1 metre = 39.3701 inch

In approximation, 1 metre

=20

739 inch = 39.35 inch

Hence, error = 39.3701 = 39.35

= 0.0201

Error on 39.35 = 0.0201

Error on 1 =35.39

0201.0

Error on 100 = 10035.39

0201.0 = 0.05%.

Ex. 42: I n a n e x a m i n a t i o n 4 2 % ca n d i d a t e s

f a i l e d i n H i n d i a n d 5 2 % f a i l e d i n

E n g l i s h , 1 7 % f a i l e d i n b o t h t h e

s u b j ec t s . I f 6 9 c a n d i d a t e s p a s se d i n

bo t h t h e sub j ec t s , f i n d t h e t o t a l nu mbe r

o f c a n d i d a t e s a p p e a r ed i n t h e

e x a m i n a t i o n .

Soln: Let the number of candidates appeared be

100.

Number of candidates who failed inHindi = 42

and the number of candidates who failedin English = 52

and the number of candidates who failed

in both the subjects = 17

Number of candidates who failed eitherin Hindi only or in English only and in

both subjects

= (42 17) + (52 17) + 17 = 77

Hence the number of candidates who

passed in both the subjects

= 100 - 77 = 23

If 23 candidates passed, number of

candidates appeared = 100

If 69 candidates passed, number ofcandidates appeared

= 23

69100 = 300.

Ex. 43: An a l l o y c on t a i ns 36% z i nc , 40% copper

a n d t h e r es t i s n i c k el . F i n d i n g r am s

t h e quan t i t y o f ea c h o f t h e c on t e n t s i n

a s amp l e o f 1 k g a l l o y .

Soln: We have, Zinc in the alloy = 36%

Copper in the alloy = 40%

Nickel in the alloy= [100 (36 + 40)]% = 24%

Now, quantity of zinc in 1 kg of alloy

= 36% of 1 kg

= 36% of 1000 grams

=

1000

100

36 = 360 grams

quantity of copper in the alloy

= 40% of 1 kg

= 40% of 1000 grams

=

1000

100

40 = 400 grams

and, quantity of nickel in the alloy

= 24% of 1 kg

= 24% of 1000 grams

=

1000

100

24 = 240 grams

Ex. 44: A num ber i s i nc r eas ed by 10% and t heni t i s d e cr ea s e d b y 1 0% . F i n d t h e n e t

i n c r eas e o r d ecr eas e pe r c en t .

Soln: Let the number be 100.

Increase in the number = 10%

= 10% of 100 = 10

Increased number = 100 + 10 = 110.This number is decreased by 10%.

Therefore, decrease in the number

= 10% of 110 =

110

100

10 = 11

New number = 110 11 = 99Thus, net decrease = 100 99 = 1

Hence, net percentage decrease

= %100100

1

= 1%

Ex. 45: I f t h e p r i c e i s i n c r e a s ed b y 1 0% a n d

t h e sa l e i s decr eas ed by 5 %, t h en wh a t

w i l l be t h e ef f ec t on i nc ome?

Soln: Let the price be Rs 100 per goods and the

sale is also of 100 goods.

So, the money obtained after selling all

the 100 goods = Rs (100 100 =) 10,000.

Now, the increased price is Rs 110 per

goods and the decreased sale is 95 goods.

So, the money obtained after selling all

the 95 goods

= Rs (110 95) = Rs 10,450.

increase in income= 10,450 10,000 = Rs 450

% increase =10000

100450 = 4.5%

Ex. 46: T h e co st o f m a n u f a c t u r i n g t h e ca r i s

m a d e u p o f t h r e e i t e m sc o st o f

m a t e r i a l s , l a b o u r a n d o v er h e a d s. I n

1 9 8 2 , t h e c o st o f t h e s e i t em s w a s i n

t h e r a t i o o f 5 : 4 : 3 . I n 198 3 , t he cos t


11/26


o f m a t e r i a l r o s e b y 1 6% , t h e co st o f

l abou r i nc r eased by 10%, bu t ov er h eads

we r e r educed by 8%. Fi n d t h e i n c r eas e

pe r c en t i n t h e p r i c e o f c a r .

Soln: Suppose cost of materials, labour,overheads were Rs 5x, Rs 4x and Rs 3x in

1982 respectively.

Total cost of manufacturing in 1982

= Rs (5x+ 4x+ 3x) = Rs 12x

Increase in cost of material in 1983 = 16%

Cost of material in 1983

= Rs

100

1655 xx = Rs x

5

29

Similarly, cost of labour in 1983

= Rs

100

1044 xx = Rs x

5

22

and cost of overheads in 1983

= Rs

100

833 xx = Rs x

25

69

Total cost of manufacturing in 1983

= Rs

xxx

25

69

5

22

5

29= Rs x

25

324

Thus, increase in price in 1983 over 1982

= Rs

xx 12

25

324= Rs x

25

24

or, Percentage increase in price

= 100

12

25

24

x

x

= 8%

Ex. 47: I n a n ex a m i n a t i o n , a c a n d i d a t e

o b t a i n e d 3 2% ma r k s a n d f a i l e d by 1 6

ma r k s . Ano t h er c and i d a t e s ec u r ed 36 %

m a r k s a n d o bt a i n e d 1 0 m a r k s m o r e

t h a n t h e m i n i m u m m a r k s t o p a s s .

D et e r m i n e t h e n u m b e r o f m i n i m u m

m a r k s t o p a s s t h e ex am i n a t i o n ?

Soln: The unsuccessful candidate obtains 32%

marks, he fails by 16 marks.

The successful candidate obtains 36%

marks.

He gets 10 marks more than pass marks.

Difference in percentage

= 36 32 = 4

Difference in marks = 16 + 10 = 26

Now 4% of maximum marks = 26

100% of maximum marks = 1004

26 = 650

Marks obtained by successful candidate

=100

36650 = 234

Since he gets 10 marks more than pass

marks.

pass marks = 234 - 10 = 224.Alternative Method:

Let the maximum marks be x.Now, according to the question,

32% of x+ 16 = 36% of x 10

or, 36% of x 32% of x = 16 + 10

or, 4% of x = 26

or, x=4

10026= 650

Hence, minimum marks to pass the exam

= 32% of x + 16

= 32% of 650 + 16

= 16650100

32

= 208 + 16 = 224

Ex. 48: I n an e x am i n a t i o n , A ob t a i n s 10% l e ss

t h a n t h e m i n i m u m n u m b er o f m a r k s

r e qu i r e d f o r p a s s i n g , B ob t a i n s9

11 1 %

l e s s t h a n A , and C1 7

34 1 % l e ss t h a n

t h e numbe r o f ma r k s ob t a i n e d b y A and

B t o ge t h e r . Does C pa s s o r f a i l ?

S o l n : Suppose maximum marks = 100 and pass

percentage = 40

According to the question,

A secures 10% less than pass marks.

As marks =100

)10100(40

=1009040 = 36

Also, B obtains9

111 % marks less than A

obtains.

Bs marks =100

9

100100

36

=9100

80036

= 32 marks

Total marks obtained by A and B

= 36 + 32 = 68.

Now, C obtains17341 %, marks less than

the marks obtained A and B together


12/26

191Percentage

Cs marks =100

17

341100

68

=17100

100068

= 40 marks

Hence, C passes.

Ex. 49: I n a n ex am i n a t i o n m a x i m u m m a r k s i s

1 0 0 0 . I n t h i s ex am i n a t i o n , A o bt a i n s

2 0% l e ss t h a n B , B ob t a i n s 1 0% mo r e

t h a n C , C ob t a i n s 5% l e s s t h a n D an d

D ob t a i ns 20% l ess t han E . I f A ob t a i ns

4 1 8 m a r k s , t h e n f i n d t h e p er c en t o f

t h e t o t a l m a r k s E o bt a i n e d .

Soln: Let E obtain 100 marks.

Marks obtained by D= 100 (20% of 100)

= 80 marks

Marks obtained by C

= 80 (5% of 80)

= 80

100

580 = 76 marks

Marks obtained by B= 76 + (10% of 76)

= 76 + 7.6 = 83.6 marks

Marks obtained by A= 83.6 (20% of 83.6 )

= 83.6 16.72 = 66.88 marks

If A obtains 66.88 marks, then marks

obtained by E = 100

If A obtains 418 marks, then marks

obtained by E = 41888.66

625 = 625

per cent marks obtained by E

= 1001000

625 = 62.5%

Ex. 50: I n an e l ec t i on be tw een tw o cand i da t es ,

a c and i d a t e who get s 40% o f t h e t o t a l

v o t e s p o l l e d i s d e f e a t e d b y 1 5 , 0 0 0

v o t e s. F i n d t h e numbe r o f v o t es po l l ed

t o w i n n i n g c a n d i d a t e.

Soln: Let the number of votes be 100.

Votes cast in favour of defeated candidate

= 40

Votes cast in favour of winningcandidate = 100 40 = 60

Difference of votes = 60 40 = 20

Actual difference of votes = 15,000

If the difference of votes is 20, votesreceived by winning candidate = 60

If the difference of votes is 1, votes

received by winning candidate

=20

60= 3.

If the difference of votes is 15,000, votes

received by winning candidate

= 3 15,000 = 45,000.

Alternative Method:

Let the number of votes be x.Votes cast in favour of defeated candidate

=100

40x

Votes cast in favour of winning candidate

=100

60x

Difference of votes =100

40

100

60 xx =

100

20x


100

20x = 15000

or, x = 15000 5

required answer =100

60515000

= 45000

Ex. 51: I n an e l e ct i o n bet w een t wo c and i d a t e s

A and B , A go t 65% o f t h e t o t a l v o t e s

c as t and w on t h e el ect i on by 2748 vo t es.

F i n d t h e t o t a l n um b er o f v ot e s c a st i f

n o v o t e i s de cl a r e d i n v a l i d .

Soln: Let the total number of votes be x.

Th en ,

Number of votes polled in favour of A

=100

65x

Number of votes polled in favour of B

=100

35x

difference of votes =100

35

100

65 xx =

100

30x


100

30x = 2748

x=30

1002748= 9160

Ex. 52: After spending 85% of his income and

giving 10% of the remainder in charity,

a man has Rs 607.50 left with him.Find his income.

So l n : Let his income be Rs 100.

Then, expenditure = Rs 85

Remainder = Rs (100 - 85) = Rs 15.


13/26


Now, 10% of Rs 15 = Rs

15

100

10

= Rs 1.50

It is given that the man gives 10% of theremainder in charity. Therefore,

contribution to charity = Rs 1.50

Balance left with the man

= Rs (15 1.50) = Rs 13.50

Now,

If the amount left with the man is Rs

13.50, his income = Rs 100.

If the amount left with the man is Rs607.50, his income

= Rs

50.607

50.13

100= Rs 4500.

Hence, his income = Rs 4500.

Ex. 53: T h r e e p e r s on s A , B a n d C w h o s e

s a l a r i e s t o g et h er am oun t t o Rs 14400 ,

s p en d 8 0 , 8 5 a n d 7 5 p e r c en t o f t h e i r s a l a r i e s r e sp e ct i v el y . I f t h e i r s a v i n g s

a r e a s 8 : 9 : 2 0 , f i n d t h ei r r e spe ct i v e

s a l a r i e s .

Soln: A saves (100 - 80) or 20% of his salary, B

saves (100 - 85) or 15% of his salary,

C saves (100 - 75) or 25% of his salary,

100

20of As salary :

100

15of Bs salary :

100

25 of Cs salary

= 8 : 9 : 20 ... (1)

From (1),

5

1of As salary :

20

3of Bs salary

= 8 : 9.

5

91of As salary =

20

83of Bs salary

As salary : Bs salary =920

583

=3

2= 2 : 3.

Again from (1),

100

15of Bs salary :

100

25of Cs salary

= 9 : 20.

100

2015 of Bs salary =100

925 of Cs

salary.

Bs salary : Cs salary

=1002015

100925

=

4

3= 3 : 4.

the salaries of A, B and C are as2 : 3: 4.

Dividing Rs 14400 in the ratio of

2 : 3 : 4. we get,

As salary =9

2 of Rs 14400 = Rs 3200

Bs salary =9

3 of Rs 14400 = Rs 4800

Cs salary =9

4 of Rs 14400 = Rs 6400

Ex. 54: A ma n spends 75% of h i s i nc ome , wh en

i n c om e i s i n c r e a s ed b y 2 0 % , h e

i n c r e a s es h i s ex p e n d i t u r e b y 1 0% . B y

h o w m u c h p e r ce n t i s h i s s a v i n g s

i n c r e a s e d ?

Soln: Let mans income be Rs x.

Mans expenditure = Rs 100

75

x = Rs 4

3x

His savings = Rs

4

3xx = Rs

4

x

After income is increased by 20%, new

income

= Rs

100

201x

= Rs

100

120x= Rs

5

6x

New expenditure

= Rs

100

101

4

3x

= Rs

10

11

4

3x = Rs

40

33x

New savings = Rs

40

33

5

6 xx

= Rs x

40

3348

= Rs8

3x

Percentage increase in savings

=%100

4

14

1

8

3

x

xx

= %1002

23

x

xx = 50%


14/26

193Percentage

Practice Exercise

pass. Find the total marks and the minimumper cent marks to pass the examination.

15. Two candidates A an d B cont ested an

election. At the elction 10% of the people on

the voting list did not vote. A defeated B by

308 votes and by counting it is found that A

had been supported by 47% of the whole

number on the voters list. Find the number

of votes obtained by A and B.

16. In an election 10% of the people in the voters

list did not participate. 60 votes were declared

invalid. There are only two candidates A and

B. A defeated B by 308 votes. It was found

that 47% of the people listed in the voters

list voted for A. Find the total number of votes

polled.

17. In an examination in which maximum marksare 2500. In this examination Ram got 50%

more marks than Shyam, Shyam got %3

216

less marks than Hari and Hari got %3

133

more marks than Krishna. If Ram got 1500

marks, then what percentage of marks was

obtained by Krishna?

18. In an examination in which maximum marks

are 500, A got 10% less than B, B got 25%

more than C, C got 20% less than D. If A got

360 marks, what percentage of marks was

obtained by D?

19. In an examination, pass marks are 36% ofmaximum marks. If an examinee gets 17

marks and fails by 10 marks in the

examination, what are the maximum marks?

20. If the import duty on motor cars be reduced

by 40 per cent of its present amount, by how

much per cent must the import of cars be

increased in order that,

(i) the revenue may be unaltered;

(ii) the revenue may be increased by 10 per

cent?

21. A shoe has six pairs of eyelets. The two eyelets

which form a pair are8

3cm apart, when the

shoe is laced while each pair of eyelets is

half a cm from the next pair. A lace measuring10 cm is threaded through the bottom holes

and carried over in the form of a letter X to

the next pair of holes and so on to the top.

Find what percentage of the whole length of

the laces left over at the top pair of holes to

be tied into a bow.

1. 55% of the population of a town are males. Ifthe total population of the town is 64100,

find the population of females in the town.

2. Find the per cent of pure gold in 22 carat

gold, if 24 carat gold is hundred per cent pure

gold.

3. In a fabric, cotton and synthetic fibres are in

the ratio of 2 : 3. What is the percentage of

cotton fibre in the fabric?

4. 2% of the employees in a factory are females

and the number of male employees is 264.

Find the total number of employees. Also, find

the number of female employees.

5. A man spends 92% of his monthly income. If

he saves Rs 220 per month, what is his

monthly income?

6. In an examination 94% of the candidatespassed and 114 failed. How many candidates

appeared?

7. If the price of an article is raised by 10%,

find by how much per cent must a consumer

reduce his consumption of that article so as

not to increase his expenditure on that article.

8. Due to a fall of 10% in the rate of sugar, 500

gm more sugar can be purchased for Rs 140.

Find thd original rate and the reduced rate of

the sugar.

9. Expenditure of Ashok is 20% less than that

of Ajeet. What per cent is Ajeets expenditure

more than Ashoks expenditure?

10. A man loses 20% of his money. After spending

25% of the remainder, he has Rs 480.00 left.

How much money did he originally have?11. There is an er ror in the measurement of

length and width of the floor of a room.

Measurement of length is 5% more than the

real length and the measurement of width is

3% less than the real width. Find the per

cent error in the area of the floor.

12. In an examination 49% students failed in

English and 36% students failed in Hindi. If

15% failed in both the subjects, find the

percentage of students who passed in both

the subjects. If total number of students who

passed the examination is 450, then find the

number of students who appeared in the

examination.

13. In an examination, 60% passed in English,

52% in Mathematics, while 32% failed inboth. If 176 students passed in both the

subjects, find the number of candidates who

sat for the examination.

14. In an examination a candidate obtained 30%

marks and failed by 16 marks. Another

candidate secured 45% marks and obtained

15 marks more than the minimum marks to


15/26


22. In an election 20% of the total voters did not

take part. There are three candidates A, B

and C in the election. A got 40% of the total

votes polled and B got 23% of the total number

of votes. A won the election by 7000 votesfrom his nearest competitor. Find the total

number of voters in the election. How many

of them take part in the election and who

was the nearest competitor of A? And also

find the number of votes obtained by each

candidate A, B and C separately.

23. A motorist reduces his distance covered

annually (in km) by x% when the price of

petrol increased by y%. Find the increase per

cent in his annual petrol bill.

24. Tax on commodity is decreased by 10% and

thereby its consumption increases by 8%. Find

the increase or decrease in the revenue

obtained from the commodity.

25. The price of sugar has been increased by 32%

but a family reduced its consumption so thatthe expenditure on it was only 10% more than

before. If the earlier monthly consumption of

sugar of the family was 10 kg, what is its

monthly consumption now?

26. Ram ordered for 6 black toys and some

additional brown toys. The prices of black

toy is2

12 times that of a brown toy. While

preparing the bill, the clerk interchanged the

number of black toys which increased the bill

by 45%. Find the number of brown toys.

27. Out of five questions in a paper,20

1th of

students answered all questions and201 th

none. Of the rest4

1th answered only four

and5

1answered only one question. If

2

124 %

of the total number of students answered only

three questions and 200 answered only two,

what was the total number of students?

28. A man spends 80% of his income. With the

increase in the cost of living his expenditure

increases by 37.5% and his income increases

by

3

216 %. Find his present savings.

29. A salesmans commission is 5% on all sales

upto Rs 10000 and 4% on all sales exceeding

that. He remits Rs 31100 to his parent

company after deducting his commission.

Find his total sales.

30. The number of males per hundred females in

a country was 120. At the next census, the

total population increased by 5% and the

female population increased by 10%. Find the

increase per cent in the male population.

31. The ratio of the number of boys to that of girls

in a school is 3 : 2. If 20% boys and 25%girls are scholarship holders, find the

percentage of the school students who are

not scholarship holders.

32. Two numbers are respectively 20 per cent and

50 per cent more than a third number. What

percentage is the first of the second?

33. Two numbers are 30% and 37% less than a

third number respectively. The second

number is what per cent less than the first

number?

34. The cost of manufacturing a car is made up of

three items : cost of raw material, labour and

overheads. In a year the cost of these items

were in the ratio 4 : 3 : 2. Next year the cost

of raw material rose by 10%, labour cost

increased by 8% but the overheads reducedby 5%. Find the percentage increase in the

price of the car.

35. The annual increase in the population of a

town is 5%. If the present population of the

town is 231525, what was it 3 years ago?

36. The number of inhabitants in a town increases

at a certain rate per cent. The number at

present is 375000 and the number 3 years

ago was 352947. Find the rate per cent?

37. The value of a machine is Rs 2,00,000. At

the end of every year its value reduces at the

rate of 2% of that at the beginning of the year.

Find its value at the end of third year.

38. A building worth Rs 1331000 is constructed

on a land worth Rs 729000. If the land

appreciates at 10% per annum and thebuilding depreciates at 10% per annum, find

the time after which the values of the land

and the building will be the same.

39. 24000 blood donors were registered with a

charitable hospital. The number of donors

increased at the rate of 5% every six month.

Find the time period at the end of which the

total number of blood donors becomes 27783.

40. The population of a vi llage is 20000. If the

annual birth rate is 4% and the annual death

rate 2%, calculate the population after two

years.

41. The populat ion of a town 2 years ago was

62500. Due to migration of cities, it decreases

every year at the rate of 4% per annum. Find

its present population.42. Total population of a country is 294 106 out

of which 150 million are males and the rest

females. Out of every 1000 males, 98 can read

and write but only 5.3% of the total population

can do so. Find what percentage of women

in the whole women population of the country

can read and write.


16/26

195Percentage

1. We have, population of the town = 64100

55% of the population = 64100100

55 = 35255.

It is given that 55% of the population are

males. Therefore, number of males = 35255.

Hence, number of females in the town

= Total Population No. of males

= 64100 35255 = 288452. In 22 carat gold, pure gold is 22 parts of 24

parts.

per cent of pure gold in 22 carat gold

= %10024

22

= %

3

291

= %10024

22

= %

3

291

3. It is given that the cotton and synthetic fibres

are in the ratio 2 : 3. So, let cotton and

synthetic fibres be 2x and 3x respectively.

Total quantity of fibre = 2x + 3x = 5x

Thus, in 5x fibres, cotton fibres = 2x

Percentage of cotton fibres

= %1005

2

x

x = 40%

4. Let the total number of employees be 100.

Then, the number of female employees = 12

Number of male employees= (100 12) = 88

Answers and explanations

Now, if the number of male employees is 88,

total number of employees = 100

If the number of male employees is 264, total

number of employees = 30026488

100

Hence, the total number of employees = 300

Number of female employees

= (300 264) = 36.Alternative Method:

Let the total number of employees be x.

It is given that the number of female

employees = 12%

Therefore, number of male employees

= (100 12)% = 88%

It is also given that the total number of male

employees is 264.

88% of x = 264

or, 264100

88 x

or, x=88

100264

or, x = 300.Hence, the total number of employees = 300

Number of female employees= 300 264 = 36.

5. Let the total income be Rs x.

We have, expenditure = 92%

Savings = (100 92)% = 8%

43. The population of a town increases by 12%

during the first year and decreases by 10%

during the second year. If the present

population of a town is 50400, what was it 2

years ago?44. A litre of water is evaporated from 6 litres of

sugar solution containing 4% of sugar and

the rest water. Find the percentage of sugar

in the remaining solution.

45. The daily wages of a worker are increased by

10% but the number of hours worked by him

per week is decreased by 10%. If originally

he was getting Rs 2000 per week, what will

he get per week now?

46. The population of a town consists of 40%

men, 35% women and 25% children. If there

are 28000 more women than children, find

the number of men, women and children

separately.

47. The present population of a certain city is

30,00,000. If the annual birth and death ratesare 3.6% and 1.6% respectively, find the

population of city after 3 years.

48. The price of an article is reduced by 10%. To

restore it to its former value, by how much

per cent should the new price be increased?

49. If the income of a person decreases by 50%,

find the percentage increase of his incomeso that there is no loss or gain.

50. In an examination out of 600000 candidates

5% remain absent, 30% of the appeared

candidates failed. The ratio of candidates who

passed in first division, in second division

and in third division is 1 : 2 : 3. Find the

number of successful candidates in different

divisions.

51. There is a census after an interval of 10 years

starting from 1961. There are increase in

population at the rate of 10%, 15% and 40%

of a certain city in 1971, 1981 and 1991. If

the total population in 1991 be 2567950, find

the population of the city in 1961.

52. After 30 kg of water had been evaporated from

a solution of salt and water, which had 15%salt, the remaining solution had 20% salt.

Find the weight of the original solution.


17/26


It is given that the man saves Rs 220. This

means that 8% of the total income is Rs 220.

ie 220100

8 x x=

8

100220

x = 2750Hence, the mans monthly income = Rs 2750.

Alternative Method:

Let his monthly income be Rs 100.

Then, his expenditure = Rs 92.

His savings = Rs (100 - 92) = Rs 8.Now, if the savings is Rs 8, then income

= Rs 100.

If the savings is Rs 220, then income

= Rs

220

8

100 = Rs 2750.

Hence, the mans monthly income = Rs 2750.

6. Suppose 100 candidates appeared in the

examination.

Then, number of passed candidates = 94Number of failed candidates = (100 - 94) = 6.

Thus , if th e number of fai lu res is 6,

candidates appeared = 100

If the number of failures is 114, candidates

appeared =

114

6

100 = 1900

Hence, 1900 candidates appeared in the

examination.

Alternative Method:

We have, pass percentage = 94%

percentage of failures = 6%Let the number of candidates appeared be x.

It is given that 114 candidates failed in the

examination and the percentage of failures is

6% .Therefore, 6% of x = 114

or, 114100

6 x

or, x= 19006

100114

Hence, 1900 candidates appeared in the

examination.

7. Let the consumption of an article originally

be 100 kg and its price be Rs 100. Then,

New price of 100 kg of the article = Rs 110

[ Price increases by 10%]

Now, Rs 110 can fetch 100 kg of the article

Rs 100 can fetch

=

100110

100=

11

1000kg of the article


= %11

1000100

= %

11

100= %

11

19

8. Reduction in rate of sugar = 10%

500 gm more sugar can be purchased for

Rs 140.

1 kg more sugar can be purchased for

= (140 2 =) Rs 280.

10% of 280 = 280100

10 = Rs 28

Now, from Rs 28 one can purchase 1 kg of

sugar.

reduced rate of 1 kg of sugar = Rs 28.Suppose original price of 1 kg of sugar = Rs x

Reduction in price = 10% of x =100

10x

Reduced price of the sugar per kg =

100

x10x

= Rs100

90x


100

90x= 28

or, x=90

10028=

9

280

ie price of sugar per kg = Rs9

280= Rs

9

131

9. Let the expenditure of Ajeet be Rs 100.

Ashoks expenditure = Rs (100 20)= Rs 80

Ajeets expenditure is Rs 20 more than thatof Ashok.

When Ashoks expenditure is Rs 80, then

Ajeets expenditure is Rs 20 more. When Ashoks expenditure is Rs 100, then

Ajeets expenditure is80

10020 = 25% more.

10. Suppose, he originally had Rs 100.

Amount lost = 20% of Rs 100 = Rs 20.

Remainder = Rs (100 - 20) = Rs 80.

Expenditure = 25% of the remainder

= 25% of Rs 80

= Rs

80

100

25= Rs 20.

Remainder = Rs (80 - 20) = Rs 60.

If remainder is Rs 60, he originally had Rs

100.

If remainder is Re 1, he originally had Rs

60

100.

If remainder is Rs 480, he originally had

Rs

480

60

100 = Rs 800.

Hence, the man had Rs 800.


18/26

197Percentage

Alternative Method:

Suppose he originally had Rs x.

Amount lost = 20% of Rs x= x100

20= Rs

5

x

Remainder = Rs

5

xx = Rs

5

4x

Expenditure = 25% of the remainder

= 25% of Rs5

4x

=5

4

100

25 x = Rs

5

x

Remainder = =

55

4 xx = Rs

5

3x


5

3x= Rs 480

or, x=3

5480= Rs 800

Hence, the man had Rs 800.

11. Let the real length and width be 100 metres

and 100 metres respectively.

real area of the floor = 100 100 = 10000 sq m

Now, measurement of length = (100 + 5 =) 105m

and the measurement of width = (100 3 =)

97 m

measurement of area= (105 97 =) 10185 sq m

Error in area = (10185 10000 =) 185 sq m

In area of 10000 sq m error in area= 185 sq m

In area of 100 sq m error in area =10000

100185

= 1.85%

12. Suppose, number of students who appeared

in the examination = 100

Number of students failed in English = 49Number of students failed in Hindi = 36

Number of students who failed in both the

subjects = 15

Number of students who failed in English

Only

= 49 15 = 34

Number of students who failed in Hindi only

= 36 15 = 21Total number of failed students

= 34 + 21 + 15 = 70

Total number of students who passed inboth the subjects = (100 70) = 30

When 30 students are passed, then total

number of students = 100

When 450 students are passed, the total

number of students =30

450100

= 1500 students

13. Let the number of candidates = 100According to the question,

Number of candidates who, passed in English

= 60

number of candidates who failed in English= 100 60 = 40

Again number of candidates, who passed in

Maths = 52

Number of candidates, who failed in Maths= 100 52 = 48

Again number of candidates, who failed in

both = 32

Number of candidates, who failed inEnglish only = 40 32 = 8

Number of candidates, who failed in Maths

only = 48 32 = 16

Hence, the number of candidates, who failedin both subjects = 8 + 16 + 32 = 56

Number of candidates, who passed in bothsubjects

= 100 - 56 = 44

If 44 candidates passed in both subjects, total

candidates = 100

If 1 candidate passed in both subjects, total

candidates =44

100

If 176 candidates passed in both subjects,

total candidates = 17644

100 = 400.

14. Suppose total marks is x and the minimum

marks to pass the examination is y. marks obtained by the first candidate

= 30% of x= x100

30=

10

3x

Since, he failed by 16 marks

10

3x= y 60

or, 3x 10y = 600 ....(i)

Now, marks obtained by the second candidate

= 45% of x=100

45x =

20

9x

Since second candidate obtained 15 marks

more than the minimum marks to pass.

20

9x= y+ 15

or, 9x 20y = 300 ....(ii)

Now, multiplying equation (i) by 2 and

subtracting it from equation (ii), we have

3x= 1500

x= 500 Total marks = 500


19/26


Now, putting the value of x in equation (i), we

have,

1500 10y = 600

10y = 2100

y=10

2100= 210

Minimum marks to pass = 210

In 500 total marks one should obtain 210

minimum marks to pass the examination.

In 100 total marks one should obtain

100

500

21042 minimum marks to pass the

examination.

Hence the minimum per cent marks to pass

the examination is 42%.

Alternative Method:

Let the total marks be 100.

Marks obtained by first candidate = 30 and

the marks obtained by the second candidate= 45

Difference in their obtained marks

= 45 30 = 15

But the real difference of his obtained marks

is 60 + 15 = 75

When difference is 15, then the total marks

is 100.

When difference is 75, the total marks is

75

15

100 500.

Marks obtained by first candidate = 30% of

500

= 150

Minimum marks = 150 + 60 = 210

Required per cent = 100500

210 = 42%

15. Let total number of voters be 100.

the number of people who participated inthe election = 100 10 = 90

Votes obtained by A = 47

Votes obtained by B = 90 47 = 43

Difference between the obtained votes of A

and B = 47 43 = 4

When A obtains 4 more votes than B, then

the total number of votes = 100

When A obtains 308 more votes than B,then the total number of votes

=

4

308100 = 7700

Votes obtained by A = 47% of 7700

=100

477700 = 3619

Votes obtained by B = 43% of 7700

=100

437700 = 3311

16. Let the total number of people in the voters

list be x.

10% people did not participate in voting.

Number of people participated in the voting.

= x 10% of x =10

xx =

10

9x

Now, according to the question, 60 votes were

declared invalid.

Total number of valid votes polled = 6010

9

x

Now, according to the question

Total number of votes obtained by A

= 47% of x= x100

47

Now, number of votes obtained by B

= xx

100

4760

10

9 = 60

100

43x

Again according to the question,

60x

100

43

100

x47= 308

or,100

4x= 308 60 = 248

x=4

100248= 6200

total number of votes polled = 90% of x= 90% of 6200

= 6200100

90 = 5580

17. Let Krishna get 100 marks

Marks obtained by Hari

= 100 +3

133 % of 100 = 100 +

3

133

= 100 +3

100=

3

400marks

Marks obtained by Shyam

=3

216

3

400 % of

3

400

=1003

50

3

400

3

400

=

9

200

3

400 =

9

1000marks

Marks obtained by Ram

=9

1000+ 50% of

9

1000

=100

50

9

1000

9

1000 =

3

500marks


20/26

199Percentage

When marks obtained by Ram is3

500, then

marks obtained by Krishna = 100

When marks obtained by Ram is 1500, then

marks obtained by Krishna =500

15003100

= 900

per cent marks obtained by Krishna

1002500

900 = 36%

18. Suppose marks obtained by D = 100

Marks obtained by C = 100 (20% of 100)

= 100

100

10020= 80

Marks obtained by B = 80 + (25% of 80)

= 80 +

100

25

80 = 100

Marks obtained by A = 100 (10% of 100)

= 100

100

10010 = 90

When marks obtained by A is 90, then

marks obtained by D = 100

When marks obtained by A is 60, then

marks obtained by D = 36090

100 = 400

per cent marks obtained by D = 100500

400

= 80%

19. Let the maximum marks be x.

An examinee gets 17 marks and fails by 10marks.

pass marks = 17 + 10 = 27Now, according to the question,

36% of maximum marks = Pass marks

or, x100

36= 27

or, x=36

10027= 75

Here maximum marks = 75

20. (i) The new duty is 60 per cent or5

3 of the

former duty.

Therefore, in order that the revenue may

remain unaltered the number of cars imported

must be3

5 of the present number. Therefore

the import must increase by

1

3

5or by

3

266

per cent.

(ii) The new revenue is to be100

110or

10

11of the

former revenue.

Therefore the number of cars imported must

be

5

3

10

11 or

6

11 of the present number.

Therefore the import must increase by

1

6

11

or6

5or by

3

183 per cent.

Alternative Method:

Let x be the number of cars imported

originally, and let Rs y be the duty on each

car. The present revenue derived from the

import is Rs xy.

The new duty on each car

=10060

of Rs y= Rs y53

.

(i)The new rev enue is the same as befo re ,

namely, Rs xy;

the number of cars to be imported

=100

3

2166

100

3

2166

300

500

3

5

5

3 xxx

y

xy of the

present number.

The import must increase by3

266 per cent.

(ii) The new revenue = Rs xy100

110

;

the number of cars to be imported

=100

3

1183

100

3

1183

300

550

5

3

100

110 xxyxy of

the present number

The import must increase by3

183 per cent.

21. According to the question,

Distance between corresponding eyelets =8

3m

Distance between two pairs =2

1cm

Now when the lace goes from one eyelet on

one side to another eyelet of the next pair to

form X, the distance is equal to the

hypotenuse of a right-angled triangle with

sides equal to8

3cm and

2

1cm.


21/26


Hence, length of the lace forming one

X =

22

2

1

8

32

=

4

1

64

92

=64

252 =

8

52 =

8

10and there are 5 such Xs.

Lace used by these =8

105 =

8

50 cm

Now lace used for joining two corresponding

eyelets at the bottom =8

3cm

Total length of the lace used

=8

3

8

50 =

8

53cm

Lace left out = 8

53

10

= 8

27

cm

Hence, percentage of lace left over

=8

27010

8

27100

108

27

= 33.75%

22. Suppose total number of voters = 100

Number of voters who polled the vote

= 100 20 = 80

Votes obtained by A = 40% of the votes polled

= 40% of 80 = 80100

40 = 32

Votes obtained by B

= 23% of the total number of votes

= 23% of 100 = 23Votes obtained by C = 80 (32 + 23) = 25

Clearly, after A, C obtains the maximum

number of votes. Hence C is the nearest

competitor of A.

Difference of votes obtaind by A and C

= 32 25 = 7

When A obtains 7 votes more than C, then

total votes = 100

When A obtains 7000 votes more than C,

then total votes = 70007

100 = 100000 votes

Hence total number of votes = 100000

Number of voters who took part in the election

= 100000 20% of 100000

= 80000Votes obtained by A = 40% of the votes polled

= 40% of 80000 = 32000

Votes obtained by B

= 23% of total number of votes

= 23% of 100000 = 23000

Votes obtained by C = 80000 (32000 + 23000)

= 25000

23. Solve as Q.No. 20. (Now see the solution

given below.)

Let the annual distance covered by the motorist

be 100 km and the price of petrol be Rs 100

per km. Annual bill of petrol = Rs (100 100)Now, according to the question, distance is

reduced by x% and the price of petrol is

increased by y%.

New annual bill of petrol= Rs (100 x)(100 + y)

Percentage increase in annual petrol bill

= 100100100

)100100()y100)(x100(

= 100100100

100100

100

y100

100

x100

= 1001

100

y1

100

x1

= 1001100100

xy

100

y

100

x1

=

100

xyxy %

24. Solve as Q.No. 23. Now, see the solution

given below.

Let the total tax on the commodity be Rs 100

per unit and total consumption be 100 units.

total revenue collection = Tax per unit number of units consumed

= (100 100 =) Rs 10000

If the tax is decreased by 10%, new tax rate

will be Rs (100 10) = Rs 90 per unit

If consumption be increased by 8%, new

consumption will be (100 + 8 =) 108 units

New revenue collection = Rs 90 108

= Rs 9720

Decrease in revenue collection

= Rs (10000 9720) = Rs 280

Percentage decrease in revenue collection

= 10010000

280 = 2.8%

25. See the solution given in Q.No. 7. It will

become easier to slove this question.

Suppose price of sugar is Rs 100 per kg.

Earlier monthly expenditure of the family

= (100 10) = Rs. 1000

Suppose, monthly consumption of sugar of

the family now is Rs x per kg.

Increased price of sugar = 100 + (32% of 100)

= Rs 132

Monthly expenditure of the family now

= Rs 132x

Increase in monthly expenditure

= Rs (132x 1000)


22/26

201Percentage

Percentage increase in monthly expenditure

= 1001000

1000132

x


1001000

1000132

x= 10

or, 132x = 1100

x=132

1100=

3

25=

3

18 kg

26. Let the number of brown toys be x and the

cost of a brown toy be Re 1

Cost of a black toy =2

51 = Rs

2

5

Total cost of black toys =2

56 = Rs 15

Total cost of brown toys = x 1 = Rs x

Total cost of both the toys = Rs (15 + x)

Considering the mistake by clerk, cost of black

toys = Rs2

5x

And cost of brown toys = 6 1 = Rs 6

Total combined cost (faulty) = Rs

6

2

x5


6

2

x5

100

145)x15(

or, 62

5

20

29

20

2915

xx

or, xx20

9

2

56

4

87

or, x20

1

4

63

or, x =15

27. Let the total number of students be x

Number of students who answered all the

questions =20

x

Number of students who answered no

question at all =20

x

Remaining number of students who answered

one, two, three or four questions

=10

9

20

18

2020

xxxxx


4

1 of

40

9

10

9 xx students answered only four

questions

5

1 of

50

9

10

9 xx students answered only one

question

2

124 % of x =200

49x students answered only

three questions

Total number of students who answered only

two questions

=200

49

50

9

40

9

10

9 xxxx

=200

493645180 xxxx

=200

130180 xx=

200

50x=

4

x

As the number of students who answered only

two questions is 200.

4x = 200

or x = 800

The total number of students is 800.

28. Let his income be Rs 100.

Expenses = 80% of 100 = Rs 80

savings = (100 80 =) Rs 20Increase in expenditure

= %2

137 of 80 = %

2

75 of 80

= 80200

75 = Rs 30

Increased expenditure = 80 + 30 = Rs 110

Increased income = 3

216100

= Rs3

2116 = Rs

3

350

Savings = 1103

2116 = Rs

3

26 = Rs

3

20

% savings = 100350

3

3

20 =

7

40=

7

55 %

29. Let total sales of the salesman be Rs x.

Total commission up to sale of Rs 10000

=100

510000 = 500.

Total commis sion for sa les exce eding Rs10000

=25

1000

100

4)10000(

xx


3110025

10000500

xx


23/26


or, 25x (12500 + x 10000) = 25 31100

or, 24x 2500 = 25 31100

or, x =24

3120025 = 32500.

30. Let the present female population be 100

Male population at present will be 120(given)

Total population at present= 100 + 120 = 220

After census the new total population

= 220 + 5% of 220

=100

105220 = 231

After census, female population

= 100 + 10% of 100 = 110

After census, population of male

= 231 110 = 121

Increase in male population = 121 120 = 1

Percentage increase in male population

= 100120

1 =

6

5%

31. Suppose the number of boys and girls in the

school are 3x and 2x respectively.

Number of scholarship holder students= 20% of 3x + 25% of 2x = 1.1x

Hence, the required percentage

=

10023

1.123

xx

xxx

= 1005

9.3

x

x = 78%

32. Let the third number be x

20% of x =5100

20 xx

first number =5

6

5

5

5

xxxxx

Now, 50% of x =2100

50 xx

second number =2

3

2

xxx

Let the first number be y% of the second

number then, y% of5

6

2

3 xx

or,5

61002

3 xyx

or, %8053

21006

x

xy

Hence, first number is 80 per cent of the

second number.

33. Let the third number be x.

Then, the first number = x 30% of x

= xx100

30 =

10

7x

Second number = x 37% of x

= xx100

37 =

100

63x

Difference between the numbers

=100

63

10

7 xx =

100

6370 xx=

100

7x

Required percentage = 100

10

7100

7

x

x

= 10%

34. Let the cost of raw material, labour and

overheads be Rs 4x, Rs 3x and Rs 2x

respectively.Increased price of raw material

=

100

101004x =

100

1104 x = Rs

5

22x

Increased cost of labour

=

100

81003x =

100

1083 x = Rs

25

81x

Reduced overheads

=

100

51002x =

100

952 x = Rs

10

19x

New total cost of manufacturing

= Rs

10

19

25

81

5

22 xxx

= Rs

50

95162220 xxx= Rs

50

477x

Initial cost of manufacturing

= Rs (4x +3x+2x) = Rs 9x

increase in cost = Rs

x

x9

50

477

= Rs50

27x

required per cent increase

=x

x

950

10027

= 6%

35. By the formula, if the annual increase be r%,

then the population n years ago

=

nr

P

1001 = n

r

P

1001


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203Percentage

Here, P = 231525, r = 5% and n = 3 years

Required population =3

100

51

231525

=212121

202020231525

= 200000

36. If the original population of a town is P, and

the annual inrease is r%, then the population

in n years will be obtained by the formula

given below:

Required population =

nr

P

1001

Here, required population = 375000,

P= 352947 and n= 3 years

375000 =3

1001352947

r

or,352947

375000=

3

1001

r

or,117649

125000=

3

1001

r

or,

3

49

50

=

3

1001

r

or,49

50=

1001

r

or, 149

50 =100

r

or,49

1=

100

r

or,10049

1 r

r=49

100=

49

22

required rate =49

22 %

37. Value of machine at the end of third year

=3

100

21200000

=100100100

989898200000

= Rs 188238.408

38. Let t be the time in which the value of the

land and the building be the same.

Value of land when it appreciates at 10% per

annum in time t

=

t

100

101729000 =

t

10

11729000

Value of building when it depreciates at 10%

per annum in time t

=

t

100

1011331000 =

t

10

91331000

The value of land and building wil l be the

same in t years, ie

tt

10

91331000

10

11729000

or, 729

1331

729000

1331000

9

10

10

11

t

=

3

9

11

999

111111

or,

3

9

11

9

11

t

or,t= 3 years

Hence, the value of land and building will be

the same in 3 years.

39. We have,

P = Initial number of donors = 24000

A = Final number of donors = 27783

R = Rate of increase = 5% every six month

= 10% per annum

Let the total time be n years. Then,

A=

nR

P

2

1001

or, 27783 =

n2

200

10124000

or, 27783 =

n2

20

2124000

or,24000

27783=

n2

20

21

or,

3

2021

=

n2

2021

or, 2n = 3

or,n=2

3

Hence, required time period =2

3 years.


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40. We have, annual birth rate = 4% and annual

death rate = 2%

Annual growth = (4 2)% = 2%Thus, we have

P = Initial population = 20000;R = Rate of growth = 2% per annum;

n= 2 years

Population after 2 years =

nR

P

1001

=

2

100

2120000

=

2

50

1120000

=

2

50

5120000

=

50

51

50

5120000 = 20808

Hence, population of the town after 2 years

= 20808.

41. We have,

Population two years ago = 62500Rate of decrease of population = 4% per annum.

Present population =

2

100

4162500

=

2

25

1162500

=

2

25

2462500

=25

24

25

2462500 = 57600.

Hence, present population = 57600.

42. Total populat ion = 294000000

Number of males = 150000000

Number of females = 294000000 150000000

= 144000000

Number of persons who can read and write

=100

3.5294000000= 15582000

Number of literate males =1000

98150000000

= 14700000

Number of literate females= 15582000 14700000 = 882000

Percentage of literate females

=144000000

100882000 = %

80

49 = 0.6125%

43. Let the population of the town 2 years ago be

P. Since the population increases in the firstyear and decr eases in th e second year,

therefore,

Population after 2 years

=

1001

1001 21

rrP ;

where r1 = 12% and r

2 = 10%

Now according to the question,

1001

1001 21

rrP = 50400

or, 50400 =

100

101

100

121P

or, 50400 =10

9

25

28P

or,P=928

102550400

= 50000

44. 100 litres of solution contains 4 parts of

sugar.

sugar in 6 litres of solution = 6100

4

= 0.24 parts

Now, 1 litre water is evaporated,

Now, 5 litres of solution contains 0.24 parts

of sugar.

100 litres of solution contains = 1005

24.0

= 4.8%

45. Suppose the daily wages and the number of

hours worked by the worker per week are Rs

x and y hours respectively. Then xy = 2000

Also,100

)10100(

100

)10100( yx

= Rs xy

100

99

will be the new wages per week.

Hence, the new wages per week

= 2000100

99 = Rs 1980

46. In the population of the town,Women = 35%

Children = 25% and

Men = 40%

Now, according to the question, the number

of women is 28000 more than that of

children.

(35% 25%) of total population = 28000or, 10% of total population = 28000

or,

100

10 of total population = 28000

Total population =10

10028000 = 280000

Number of men= 40% of the total population

=100

28000040 = 112000

Number of women

= 35% of the total population

=100

28000035 = 98000


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205Percentage

Number of children

= 25% of the total population

=100

28000025 = 70000

47. Birth rate = 3.6% and death rate = 1.6%

Resultant growth = 3.6% 1.6% = 2%

Population of the town after 3 years

=

3

100

213000000

=

50

51

50

51

50

513000000 = 3183624

48. Let the original price of an article be Rs x.

If the price is reduced by 10%, then new price

= Rs x

Concept of Arithmetic_PERCENTAGE

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