SSC Percentage.pdf

180 Concept of Arithmetic

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Chapter-10

Percentage

IntroductionPer cent is a kind of fraction of which denominatoris 100. The word ‘per cent’ is an abbreviation ofthe Latin phrase ‘percentum’ which means ‘perhundred’ or ‘hundredths’. Thus the term ‘per cent’means ‘per hundred’ or ‘for every hundred’. Forexample,

( i ) When we say that a man gives 30 per centof his income as income tax, this meansthat he gives Rs 30 out of every hundredrupees of his income.

( i i) A trader makes a profit of 15 per cent meanshe gains Rs 15 on every hundred rupees ofhis investment.

( i ii ) A boy scored 70 per cent marks in his finalexamination means that he obtained 70marks out of every hundred marks.

The term ‘per cent ’ will be written as thesymbol (%).

Per cent and Vulgar FractionSince per cent is a form of fraction, hence percent can be expressed as fraction and vice versa.This can be illustrated through examples. See theexamples given below:Ex. 1: Express the given fractions as per cent.

(i) 21

(ii) 43

(iii) 511

(iv) 32

Soln: To convert a fraction into a per cent weexpress the given fraction withdenominator as 100 or multiply thefraction by 100 and put the per centsign (%).

(i) 10050

10021001

21

= 50%

or, 10021 = 50%

(ii) 10075

10041003

43

= 55%

or, 10043 = 55%

(iii) 100220

100510011

511

= 220%

or, 1005

11 = 220%

(iv) %3266

1003266

1003

200

10031002

32

or, %3266

3200100

32

Ex. 2: Express the following per cent as afraction.

(i) 25% (ii) %318 (iii) 140%

Soln: To convert a per cent into a fraction, wedivide it by 100 and remove the per centsign %. Thus,

(i) 25% = 41

10025

(ii) 121

100325%

325%

318

(iii) 140% = 57

100140

Per cent and Decimal FractionEx. 3: Convert the every decimal into per cent.

(i) 0.35 (ii) 0.125 (iii) 2.25Soln: In order to convert a given decimal into a

per cent, express it as a fraction withdenominator as 100 or move the decimalpoint on the right side by two digits andput the per cent sign %. Thus,

(i) 0.35 = 10035

10010035.0

= 35%

(ii) 0.125 = 1005.12

100100125.0

= 12.5%

(iii) 2.25 = 100225

10010025.2

= 225%

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Ex. 4: Express the every given per cent as adecimal fraction.(i) 38% (ii) 16.5% (iii) 320.5%

Soln: To convert a given per cent in decimalform, we express it as a fraction withdenominator as 100 and then the fractionis written in decimal form. Thus,

(i) 38% = 10038

= 0.38

(ii) 16.5% = 1005.16

= 0.165

(iii) 320.5% = 1005.320

= 3.205

Per cent and RatioEx. 5: Convert the following ratios into per

cent.(i) 6 : 5 (ii) 3 : 12 (iii) 9 : 20

Soln: In order to convert a given ratio into a percent, we first convert the given ratio intofraction and then multiply the fraction by100. Thus,

(i) 6 : 5 = 10056

56

= 120%

(ii) 3 : 12 = 100123

123

= 25%

(iii) 9 : 20 = 100209

209

= 45%

Ex. 6: Express the following per cent asratios.(i) 52% (ii) 27.5% (iii) 8%

Soln: To convert a given per cent into a ratio,we first convert the per cent into a fractionand then express it as a ratio. Thus,

(i) 52% = 2513

10052

= 13 : 25

(ii) 27.5% = 4011

1000275

1005.27

= 11 : 40

(iii) 8% = 52

1008

= 2 : 5

Finding per cent of a QuantityMany a time, we need to find a given per cent of agiven quantity. For example, we might be asked tofind the number of girl students in a school, giventhat there are 28% girl students in the school andtotal number of students in the school is 1000. Weshall explain the process of f inding such apercentage through some examples.

Ex. 7: Find 20% of Rs 150.Soln: We have, 20% of Rs 150

= 10020

of Rs 150 = Rs 15010020

= Rs 30

Thus, 20% of Rs 150 is Rs 30.Ex. 8: Find the number of girl students in a

school, if there are 28% girl studentsin the school and the total number ofstudents in the school is 1000.

Soln: Required number of girl students

= 28% of 1000 = 100010028

= 280

Ex. 9: In an orchard, %3216 of the trees are

apple trees. If the total number of treesin the orchard is 240, find the numberof other types of trees in the orchard.

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

= %3216 of 240 = %

350

of 240

= 240100

13

50 = 40

Therefore the number of other trees= 240 – 40 = 200

Thus, number of trees of other types inthe orchard is 200.

Ex. 10: Find 10% more than Rs 90.Soln: We have, 10% of Rs 90

= 10010

of Rs 90 = Rs 9010010

= Rs 9

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

Ex. 11: Find 20% less than Rs 70.Soln: We have, 20% of Rs 70

= 10020

of Rs 70 = Rs 7010020

= Rs 14

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

Expressing One Quantity as a per cent ofAnother QuantityWe come across many situations where we have toexpress 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 amaximum of 500 marks. This is equivalent tofinding what per cent one number is of the other.We shall explain the method of finding what percent one number is of another through someexamples.


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Ex. 12: What per cent is number 8 of number25?

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 100258

= 32

Thus, the required per cent is 32. In otherwords, 8 is 32% of 25.

Note: To find what per cent the firstnumber is of second, we divide the firstnumber by the second number andmultiply 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 = 251008

= 32

Thus, the required per cent is 32. In otherwords, 8 is 32% of 25.

Ex. 13: Peter obtained 385 marks out of amaximum of 500 marks. Find thepercentage of marks obtained by Peter.

Soln: Required percentage of marks

= 100500285

= 57

Thus, Peter obtained 57% marks.Ex. 14: What per cent of 25 kg is 3.5 kg?Soln: Note that we wish to find here what per

cent 3.5 is of 25.

Required percentage = 10025

5.3

= 10025035

= 14

Thus, 3.5 kg is 14% of 25 kg.Ex. 15: A basket contains 300 mangoes. 75

mangoes were distributed among somestudents. Find the percentage ofmangoes left in the basket.

Soln: Original number of mangoes = 300Mangoes distributed = 75 Mangoes left in the basket

= 300 – 75 = 225 Percentage of mangoes left

= 100300225

= 75

Thus, 75% mangoes are left in the basket.Ex. 16: Jamshed obtained 553 marks out of

700 and Geeta obtained 486 marks outof 600 in Mathematics. Whoseperformance is better?

Soln: To compare the performance, we shallconvert the marks into per cent.Thus, we have,Marks obtained by Jamshed

= %100700553

= 79%

Marks obtained by Geeta

= %100600486

= 81%

Now, 81 > 71Therefore, Geeta’s performance is better.

Ex. 17: The excise duty on a certain item hasbeen reduced to Rs 3480 from Rs 5220.Find the percentage reduction in theexcise duty for that item.

Soln: Amount of reduction in excise duty= Rs 5220 – Rs 3480 = Rs 1740

Percentage of reduction is calculated oninitial excise duty ie Rs 5220. Required per cent reduction

= %10052201740

= %3

100= %

3133

Questions Based on Population GrowthIf the original population of a town is P, and theannual increase is r%, then, the population in nyears will be obtained by the following formulae:

( i ) Required population = nrP

1001

Since, population after one year becomes

100rPP =

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 init

= 1001001

1001 rrPrP

=

1001

1001 rrP =

2

1001

rP

: : : : : :

The population in n year = nrP

1001

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

population in n years = nrP

1001

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

n

n

rPrP

1001

1001

( iv) If the annual increase be r per thousand,then the population in n years

= nrP

10001

Ex. 18: The population of a town 3 years agowas 40960. If the present populationof the town is 49130, then find theannual per cent increase in population.

Soln: See the formula (iii) mentioned above,Population n years ago

nrPP

1001

100RateAnnual1

)(PopulationPresentTime

or, 40960 = 3

1001

49130

r

or, 3

1001

r = 40960

49130 = 4096

4913 =

3

1617

or, 1001 r = 16

17 = 16

11

or, 100r

= 161

r = 16100

= 425

= %416

Ex. 19: The population of a town is 2624000.If the population of the town isincreasing at the rate of 25 perthousand per annum, then find

(i) what was the population of thetown 1 year ago?

(ii) what will be the population of thetown after 3 years?

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

Population n years ago = nrP

10001

required population

4041

2624000

1000251

26240001

= 41402624000

= 2560000

(ii) From the formulae [(i ) and (iv)]mentioned above,Population after n years

= nrP

10001

required population

= 3

10002512624000

= 3

40412624000

= 4040404141412624000

= 28225761

DepreciationThe value of a machine or of any other articlesubject to wear and tear decreases with time.Relative decrease in the value of a machine is calledits depreciation. Depreciation per unit time iscalled the rate of depreciation. Thus, if V is thevalue of a machine at a certain time and R% perannum is the rate of depreciation then the value

of machine after n years = nRV

1001 .

Ex. 20: The value of a residential flatconstructed at a cost of Rs 100000 isdepreciating at the rate of 10% perannum. What will be its value 3 yearsafter construction?

Soln: We have,V = Initial value = Rs 100000R = Rate of depreciation = 10% per annum

Value after 3 years = 3

1001

RV

= Rs

3

10010

1100000

= Rs

3

101

1100000


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= Rs

3

109

100000

= Rs

109

109

109100000

= Rs 72900Hence, value of the flat after 3 years

= Rs 72900.Ex. 21: The present price of a scooter is Rs

7290. If its value decreases every yearby 10%, then find its value before 3years.

Soln: Let the value of the scooter be Rs P beforethree years. Then, its present value is Rs

3

100101

P

But the present value is given as Rs 7290.Therefore,

7290 = 3

100101

P

or, 7290 = 3

1011

P

or, 7290 = P × 3

3

109

or, P = 3

3

9107290

= 10000.

Hence, the value before 3 years was Rs10000.

Rate of Increase or Decrease ChangesEvery Year

I. If P be the population of a city or a town atthe beginning of a certain year and the rateof increase or decrease is R1% for the firstn1 years, r2% for the next n2 years and soon and Rk% for the last nk years, then thepopulation at the end of (n1 + n2 + ..... + nk)years is given by

Pn= ...100

1100

121

21

nn RRP

knkR.....

1001 ;

where n = n1 + n2 + ..... + nk and (+) sign isused for increase and (–) sign is used fordecrease.

II. If V0 is the value of an article at certaintime and the rate of appreciation ordepreciation is R1% for first n1 years, R2%

for next n2 years and so on and Rk% for thelast nk years, then the value at the end ofn1 + n2 + ..... + nk years is given by

Vn= ...100

1100

121

210

nn RRV

knkR

1001..... ;

where n = n1 + n2 + ..... + nk and (–) sign isused for depreciation and (+) sign is usedfor appreciation.

Ex. 22: The population of a town was 160000three years ago. If it had increased by3%, 2.5% and 5% in the last threeyears, find the present population ofthe town.

Soln: Let P be the present population of thetown. Then,

P = 160000 ×

1005.21

10031

10051

or, P = 160000 × 2021

4041

100103

or, P = 2 × 103 × 41 × 21 = 177366.Hence, present population of the town= 177366.

Ex. 23: 10000 workers were employed toconstruct a river bridge in four years.At the end of first year, 10% workerswere retrenched. At the end of thesecond year, 5% of the workers at thattime were retrenched. However tocomplete the project in time, the numberof workers was increased by 10% atthe end of the third year. How manyworkers were working during thefourth year?

Soln: We have,Initial number of workers = 10000Reduction of workers at the end of firstyear = 10%Reduction of workers at the end of secondyear = 5%

Increase of workers at the end of thirdyear = 10%Number of workers working during thefourth year

=

100101

10051

10010110000

= 1011

2019

10910000 = 9405

Hence, the number of workers workingduring the fourth year was 9405.

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Ex. 25: If 23% of a number is 46, find thenumber.

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

or, x10023

= 46

or, x = 2310046

= 200

Thus, the number is 200.Ex. 26: In an examination, Neeta secured 372

marks. If she secured 63% marks, findthe maximum marks.

Soln: Let the maximum marks be x.Neeta’s marks = 62% of xNeeta secured 372 marks 62% of x = 372

or, x10062

= 372

or, x = 62100372

= 600

Hence, maximum marks are 600.Ex. 27: What is the ratio 5 : 4 equal to when

expessed as a per cent?

Soln: Fractional equivalent of 5 : 4 is equal to 45

.

Percentage equivalent of 45

is 10045

= 125%Hence, the ratio 5 : 4 is equal to 125%.

Ex. 28: The strength of students in a schoolincreases from 900 to 936. Find thepercentage increase in the strength ofthe students.

Soln: Increase in strength of students= 936 – 900 = 36

We have to find what per cent is 36 of900? required per cent increase

= 10090036

= 4%

Hence, the strength of students increasesby 4%.

Ex. 29: Raju and Nita get 294 and 372 marksrespectively in an examination. If Nitagot 62% marks, then find the maximummarks and the per cent marks obtainedby Raju.

Soln: Let the maximum marks be x.Marks obtained by Nita = 62% of xNita gets 372 marks 62% of x = 372

or, x10062

= 372

x = 62100372

= 600

Per cent marks obtained by Raju

= 100600294

= 49

Hence, maximum marks is 600 and Rajugot 49% of the maximum marks.

Ex. 30: A reduction of 10% in the price of teaenables a trader to obtain 25 kg morein Rs 22500. What is the reduced priceof tea per kg. Also find the originalprice of the tea per kg.

Soln: Reduction in the price of tea = 10%

10% of Rs 22500 = 1001022500

= Rs 2250

Now, in Rs 2250 a trader can obtain 25kg of tea. Reduced price of 25 kg of tea = Rs 2250

Some More Solved Examples

Ex. 24: A new car costs Rs 360000. Its pricedepreciates at the rate of 10% a yearduring the first two years and at therate of 20% a year thereafter. Whatwill be the price of the car after 3years?

Soln: We have,Cost of the car = Rs 360000Rate 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

10010

110010

1360000

10020

1

= Rs

511

1011

1011360000

= Rs

54

109

109360000 = Rs 233280.

Hence, the price of the car after 3 years= Rs 233280.


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Reduced price of 1 kg of tea

= 252250

= Rs 90

Let the original price of 1 kg of tea be Rsx.Now, reduction in original price

= 10% of x = Rs 10010x

Reduced price of tea per kg

= Rs

10010xx

Now, according to the question,

10010xx = 90

or, 90x = 100 × 90 x = 100Hence, original price of tea per kg

= Rs 100

Ex. 31: A reduction of 2112 per cent in the price

of mangoes enables a purchaser toobtain 4 more for a rupee. What is thereduced price? How many could he getfor 50 paise before the reduction inprice?

Soln: Owing to the reduction in price the

purchaser saves 2112 per cent or 8

1 of

Re 1. With this sum he gets 4 mangoes atthe reduced price. The reduced price of a mango

= Re 481 = Re 32

1

Again

2112100 per cent or,

811

or, 87

of the original price of a mango

= Re 321

The original price of a mango

= 78

× Re 321

= Re 281

He could get

281

21

or 14 mangoes

for half a rupee ie, 50 paise, before thereduction in price.

Ex. 32: The price of sugar goes up by 20%. Byhow much per cent must a house wifereduce her consumption so that theexpenditure does not increase?

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 =

100120100

= 3250

kg sugar

Reduction in consumption

=

3250100 % = 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 unitsTo maintain the expenditure, suppose hereduces his consupmtion by x%, then histotal 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

= 3216 %.

Ex. 33: The price of edible oil increases fromRs 13 to Rs 15 per kg. By how muchper cent must a housewife reduce herconsumption so that the expendituredoesnot increase?

Soln: Let the consumption of edible oil originallybe 100 kg.Then, her earlier expenditure

= 100 × 13 = Rs 1300Now, due to increase in price,Rs 1500 can fetch 100 kg edible oil

Rs 1300 can fetch 13001500100

= 3260

kg


= 3260100 = 3

40 kg

per cent reduction = 340

% = 3113 %

187Percentage

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Ex. 34: In an election between two candidatesA and B, A got 60% of the total validvotes. 15% of the total votes weredeclared invalid. If the total number ofvotes is 500000, find the number ofvalid votes polled in favour of thecandidate B.

Soln: Total number of invalid votes= 15% of 500000

= 10015

× 500000 = 75000

Total number of valid votes polled= 500000 – 75000 = 425000

Percentage of valid votes polled in favourof the candidate A = 60% Percentage of valid votes polled in favourof the candidate B = 40%

= 40% of 425000

= 10040

× 425000 = 170000

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

Ex. 35: The tax on a commodity is diminishedby 15%, and its consumption increasesby 10%.

(i) Find the decrease per cent in therevenue derived from it.

(ii) With what increase per cent in itsconsumption would the revenueremain the same?

Soln: (i) The new tax is 85%, or 2017

of the original

tax. The new consumption is 100110

or 1011

of the original consumption.

the new revenue = 2017

of 1011

of the

original revenue

= 200187

of the original revenue

= %2193 of the original revenue.

the required decrease

= 100% – %2193

= %216 = 6.5%

Alternative Method:Let the original tax be Rs 100 and theconsumption be 100 units. Original revenue = Rs (100 × 100)

= Rs 10000

Now, according to the question,New tax = Rs (100 – 15) = Rs 85 andthe 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 is 2017

of the original

tax, the revenue would remain the same

if the new consumption becomes 1720

of

the original consumption. the required increase in consumption

= 11720

= 173

= 171117 %.

Alternative Method:Let the original tax be Rs 100 and theconsumption be 100 units. Original revenue = Rs (100 × 100)

= Rs 10000New tax = Rs (100 – 15) = Rs 85Let the new consumption be x units.Now, according to the question,

10000 = 85x[ Revenue remains the same]

or, x = 8510000

or, Increase in consumption

= 851500100

8510000

= %171117

17300

Ex. 36: Mohan’s income is 10% less than thatof Sohan. Then what per cent isSohan’s income more than Mohan’sincome?

Soln: Let Sohan’s income be Rs 100Income of Mohan = Rs (100 – 10) = Rs 90

Sohan’s income is Rs 10 more than thatof Mohan. If Mohan’s income is Rs 90, Sohan’sincome is Rs 10 more. If Mohan’s income is Rs 100, Sohan’s

income is Rs

1009010

9111 more.

Hence Sohan’s income is 9111 % more than

that Mohan’s income.


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Ex. 37: Rakesh’s income is 25% more thanthat of Rohan. What per cent is Rohan’sincome less than Rakesh’s income?

Soln: Let Rohan’s income be Rs 100. Then,Rakesh’s income = Rs 125.If Rakesh’s income is Rs 125, Rohan’sincome = Rs 100If Rakesh’s income is Re 1, Rohan’s

income = Re 125100

If Rakesh’s income is Rs 100, Rohan’s

income = Rs

100125100

= Rs 80.

Hence, Rohan’s income is 20% less thanthat of Rakesh.

Ex. 38: Rani’s weight is 25% that of Meena’sand 40% that of Tara’s. Whatpercentage of Tara’s weight is Meena’sweight?

Soln: Let Meena’s weight be x kg and Tara’sweight be y kg.Then, Rani’s weight = 25% of Meena’s

weight = x10025

.... (i)

Also, Rani’s weight = 40% of Tara’s weight

= y10040

... (ii)

From (i) and (ii), we get

yx 10040

10025

or, 25x = 40y[Multiplying both sides by 100]

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

or, x = y58

.... (iii)

We have to find Meena’s weight as thepercentage of Tara’s weight ie

160100581005

8

100 y

y

yx

[Using (iii)]Hence, Meena’s weight is 160% of Tara’sweight.

Ex. 39: In a school, the ages of 20% studentsare less than 5 years. 64 girls havemore than 5 years of age and this

number is 32

of number of boys more

than 5 years of age. Find the totalnumber of students in the school?

Soln: Let the total number of students be xNumber of students less than 5 years ofage = 20% of x

= x10020

= 5x

Number of students above 5 years of

age =

5xx = 5

4x

Now according to the question,Number of girls above 5 years of age = 64andNumber of boys above 5 years of age

= 2364

= 96

Total number of students above 5 years ofage = 96 + 64 = 160

or, 54x

= 160

x = 45160

= 200

Total number of students in the school= 200

Ex. 40: If 2yx and z1x , then find the per

cent change in the value of x when thevalue of y increases by 20% and thatof z decreases by 25%.

Soln: 2yx and zx 1

x = z

yk2

; where k = constant.

On increasing the value of y by 20%, theincreased value of y

= 100120y

= 56y

On decreasing the value of z by 25%, thedecreased value of z

= 10075z

= 43z

New value of x = 4

356 2

z

y

k

= z

yk34

2536 2

= z

yk2

2548

Change in the value of x

= z

ykz

yk22

2548

= z

yk25

232

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per cent change in the value of x

=

zkyz

ky2

2 10025

23 = 92%

Ex. 41: A metre is taken to equal 39.3701 inch.

If the value is approximated to 20739

inch, find the relative error.Soln: According to the question,

1 metre = 39.3701 inchIn approximation, 1 metre

= 20739 inch = 39.35 inch

Hence, error = 39.3701 = 39.35 = 0.0201

Error on 39.35 = 0.0201

Error on 1 = 35.390201.0

Error on 100 = 10035.39

0201.0 = 0.05%.

Ex. 42: In an examination 42% candidatesfailed in Hindi and 52% failed inEnglish, 17% failed in both thesubjects. If 69 candidates passed inboth the subjects, find the total numberof candidates appeared in theexamination.

Soln: Let the number of candidates appeared be100. Number of candidates who failed inHindi = 42and the number of candidates who failedin English = 52and the number of candidates who failedin both the subjects = 17 Number of candidates who failed eitherin Hindi only or in English only and inboth subjects

= (42 – 17) + (52 – 17) + 17 = 77Hence the number of candidates whopassed in both the subjects

= 100 - 77 = 23If 23 candidates passed, number ofcandidates appeared = 100 If 69 candidates passed, number ofcandidates appeared

= 2369100

= 300.

Ex. 43: An alloy contains 36% zinc, 40% copperand the rest is nickel. Find in gramsthe quantity of each of the contents ina sample of 1 kg alloy.

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

=

100010036

= 360 grams

quantity of copper in the alloy= 40% of 1 kg= 40% of 1000 grams

=

100010040

= 400 grams

and, quantity of nickel in the alloy= 24% of 1 kg= 24% of 1000 grams

=

100010024

= 240 grams

Ex. 44: A number is increased by 10% and thenit is decreased by 10%. Find the netincrease or decrease per cent.

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 =

11010010

= 11

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

= %100100

1

= 1%

Ex. 45: If the price is increased by 10% andthe sale is decreased by 5%, then whatwill be the effect on income?

Soln: Let the price be Rs 100 per goods and thesale is also of 100 goods.So, the money obtained after selling allthe 100 goods = Rs (100 × 100 =) 10,000.Now, the increased price is Rs 110 pergoods and the decreased sale is 95 goods.So, the money obtained after selling allthe 95 goods

= Rs (110 × 95) = Rs 10,450. increase in income

= 10,450 – 10,000 = Rs 450

% increase = 10000100450

= 4.5%

Ex. 46: The cost of manufacturing the car ismade up of three items—cost ofmaterials, labour and overheads. In1982, the cost of these items was inthe ratio of 5 : 4 : 3. In 1983, the cost


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of material rose by 16%, the cost oflabour increased by 10%, but overheadswere reduced by 8%. Find the increaseper cent in the price of car.

Soln: Suppose cost of materials, labour,overheads were Rs 5x, Rs 4x and Rs 3x in1982 respectively.Total cost of manufacturing in 1982

= Rs (5x + 4x + 3x) = Rs 12xIncrease in cost of material in 1983 = 16% Cost of material in 1983

= Rs

1001655 xx = Rs x

529

Similarly, cost of labour in 1983

= Rs

1001044 xx = Rs x

522

and cost of overheads in 1983

= Rs

100833 xx = Rs x

2569

Total cost of manufacturing in 1983

= Rs

xxx

2569

522

529

= Rs x25324

Thus, increase in price in 1983 over 1982

= Rs

xx 12

25324

= Rs x2524

or, Percentage increase in price

= 100122524

x

x = 8%

Ex. 47: In an examination, a candidateobtained 32% marks and failed by 16marks. Another candidate secured 36%marks and obtained 10 marks morethan the minimum marks to pass.Determine the number of minimummarks to pass the examination?

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 = 4Difference in marks = 16 + 10 = 26Now 4% of maximum marks = 26

100% of maximum marks = 100426

= 650

Marks obtained by successful candidate

= 10036650 = 234

Since he gets 10 marks more than passmarks. 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 – 10or, 36% of x – 32% of x = 16 + 10or, 4% of x = 26

or, x = 410026

= 650

Hence, minimum marks to pass the exam= 32% of x + 16= 32% of 650 + 16

= 1665010032

= 208 + 16 = 224Ex. 48: In an examination, A obtains 10% less

than the minimum number of marks

required for passing, B obtains 9111 %

less than A, and C 17341 % less than

the number of marks obtained by A andB together. Does C pass or fail?

Soln: Suppose maximum marks = 100 and passpercentage = 40According to the question,A secures 10% less than pass marks.

A’s marks = 100)10100(40

= 1009040 = 36

Also, B obtains 9111 % marks less than A

obtains.

B’s marks = 100

9100100

36

= 910080036

= 32 marks

Total marks obtained by A and B= 36 + 32 = 68.

Now, C obtains 17341 %, marks less than

the marks obtained A and B together

191Percentage

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C’s marks = 100

17341100

68

= 17100100068

= 40 marks

Hence, C passes.Ex. 49: In an examination maximum marks is

1000. In this examination, A obtains20% less than B, B obtains 10% morethan C, C obtains 5% less than D andD obtains 20% less than E. If A obtains418 marks, then find the per cent ofthe total marks E obtained.

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 –

100580 = 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 marksobtained by E = 100 If A obtains 418 marks, then marks

obtained by E = 41888.66

625 = 625

per cent marks obtained by E

= 1001000625

= 62.5%

Ex. 50: In an election between two candidates,a candidate who gets 40% of the totalvotes polled is defeated by 15,000votes. Find the number of votes polledto winning candidate.

Soln: Let the number of votes be 100.Votes cast in favour of defeated candidate= 40 Votes cast in favour of winningcandidate = 100 – 40 = 60Difference of votes = 60 – 40 = 20Actual difference of votes = 15,000If the difference of votes is 20, votesreceived by winning candidate = 60If the dif ference of votes is 1, votesreceived by winning candidate

= 2060

= 3.

If the difference of votes is 15,000, votesreceived by winning candidate

= 3 × 15,000 = 45,000.Alternative Method:Let the number of votes be x.Votes cast in favour of defeated candidate

= 10040x

Votes cast in favour of winning candidate

= 10060x

Difference of votes = 10040

10060 xx

= 10020x

According to the question,

10020x

= 15000

or, x = 15000 × 5

required answer = 10060515000

= 45000Ex. 51: In an election between two candidates

A and B, A got 65% of the total votescast and won the election by 2748 votes.Find the total number of votes cast ifno vote is declared invalid.

Soln: Let the total number of votes be x.Then,Number of votes polled in favour of A

= 10065x

Number of votes polled in favour of B

= 10035x

difference of votes = 10035

10065 xx

= 10030x


10030x

= 2748

x = 301002748

= 9160

Ex. 52: After spending 85% of his income andgiving 10% of the remainder in charity,a man has Rs 607.50 left with him.Find his income.

Soln: Let his income be Rs 100.Then, expenditure = Rs 85 Remainder = Rs (100 - 85) = Rs 15.


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Now, 10% of Rs 15 = Rs

1510010

= Rs 1.50It is given that the man gives 10% of theremainder in charity. Therefore,contribution to charity = Rs 1.50Balance left with the man

= Rs (15 – 1.50) = Rs 13.50Now,If the amount left with the man is Rs13.50, his income = Rs 100. If the amount left with the man is Rs607.50, his income

= Rs

50.607

50.13100

= Rs 4500.

Hence, his income = Rs 4500.Ex. 53: Three persons A, B and C whose

salaries together amount to Rs 14400,spend 80, 85 and 75 per cent of theirsalaries respectively. If their savingsare as 8 : 9 : 20, find their respectivesalaries.

Soln: A saves (100 - 80) or 20% of his salary, Bsaves (100 - 85) or 15% of his salary,C saves (100 - 75) or 25% of his salary,

10020

of A’s salary : 10015

of B’s salary :

10025

of C’s salary

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

From (1), 51

of A’s salary : 203

of B’s salary

= 8 : 9.

591

of A’s salary = 2083

of B’s salary

A’s salary : B’s salary = 920583

= 32

= 2 : 3.

Again from (1),

10015

of B’s salary : 10025

of C’s salary

= 9 : 20.

1002015

of B’s salary = 100925

of C’s

salary. B’s salary : C’s salary

= 1002015100925

= 43

= 3 : 4.

the salaries of A, B and C are as2 : 3 : 4.Dividing Rs 14400 in the ratio of2 : 3 : 4. we get,

A’s salary = 92

of Rs 14400 = Rs 3200

B’s salary = 93

of Rs 14400 = Rs 4800

C’s salary = 94

of Rs 14400 = Rs 6400

Ex. 54: A man spends 75% of his income, whenincome is increased by 20%, heincreases his expenditure by 10%. Byhow much per cent is his savingsincreased?

Soln: Let man’s income be Rs x.

Man’s expenditure = Rs 10075

x = Rs 43x

His savings = Rs

43xx = Rs 4

x

After income is increased by 20%, newincome

= Rs

10020

1x

= Rs

100120x

= Rs 56x

New expenditure

= Rs

100101

43 x

= Rs

1011

43 x = Rs 40

33x

New savings = Rs

4033

56 xx

= Rs x

403348

= Rs 83x

Percentage increase in savings

= %100

41

41

83

x

xx

= %1002

23

xxx

= 50%

193Percentage

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Practice Exercise

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

15. Two candidates A and B contested anelection. At the elction 10% of the people onthe voting list did not vote. A defeated B by308 votes and by counting it is found that Ahad been supported by 47% of the wholenumber on the voter’s list. Find the numberof votes obtained by A and B.

16. In an election 10% of the people in the voters’list did not participate. 60 votes were declaredinvalid. There are only two candidates A andB. A defeated B by 308 votes. It was foundthat 47% of the people listed in the voters’list voted for A. Find the total number of votespolled.

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

more marks than Shyam, Shyam got %3216

less marks than Hari and Hari got %3133

more marks than Krishna. If Ram got 1500marks, then what percentage of marks wasobtained by Krishna?

18. In an examination in which maximum marksare 500, A got 10% less than B, B got 25%more than C, C got 20% less than D. If A got360 marks, what percentage of marks wasobtained by D?

19. In an examination, pass marks are 36% ofmaximum marks. If an examinee gets 17marks and fails by 10 marks in theexamination, what are the maximum marks?

20. If the import duty on motor cars be reducedby 40 per cent of its present amount, by howmuch per cent must the import of cars beincreased 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 are 83

cm apart, when the

shoe is laced while each pair of eyelets ishalf a cm from the next pair. A lace measuring10 cm is threaded through the bottom holesand carried over in the form of a letter X tothe next pair of holes and so on to the top.Find what percentage of the whole length ofthe laces left over at the top pair of holes tobe 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 caratgold, if 24 carat gold is hundred per cent puregold.

3. In a fabric, cotton and synthetic fibres are inthe ratio of 2 : 3. What is the percentage ofcotton fibre in the fabric?

4. 2% of the employees in a factory are femalesand the number of male employees is 264.Find the total number of employees. Also, findthe number of female employees.

5. A man spends 92% of his monthly income. Ifhe saves Rs 220 per month, what is hismonthly income?

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

7. If the price of an article is raised by 10%,find by how much per cent must a consumerreduce his consumption of that article so asnot to increase his expenditure on that article.

8. Due to a fall of 10% in the rate of sugar, 500gm more sugar can be purchased for Rs 140.Find thd original rate and the reduced rate ofthe sugar.

9. Expenditure of Ashok is 20% less than thatof Ajeet. What per cent is Ajeet’s expendituremore than Ashok’s expenditure?

10. A man loses 20% of his money. After spending25% of the remainder, he has Rs 480.00 left.How much money did he originally have?

11. There is an error in the measurement oflength and width of the f loor of a room.Measurement of length is 5% more than thereal length and the measurement of width is3% less than the real width. Find the percent error in the area of the floor.

12. In an examination 49% students failed inEnglish and 36% students failed in Hindi. If15% failed in both the subjects, find thepercentage of students who passed in boththe subjects. If total number of students whopassed the examination is 450, then find thenumber of students who appeared in theexamination.

13. In an examination, 60% passed in English,52% in Mathematics, while 32% failed inboth. If 176 students passed in both thesubjects, find the number of candidates whosat for the examination.

14. In an examination a candidate obtained 30%marks and failed by 16 marks. Anothercandidate secured 45% marks and obtained15 marks more than the minimum marks to


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22. In an election 20% of the total voters did nottake part. There are three candidates A, Band C in the election. A got 40% of the totalvotes polled and B got 23% of the total numberof votes. A won the election by 7000 votesfrom his nearest competitor. Find the totalnumber of voters in the election. How manyof them take part in the election and whowas the nearest competitor of A? And alsofind the number of votes obtained by eachcandidate A, B and C separately.

23. A motorist reduces his distance coveredannually (in km) by x% when the price ofpetrol increased by y%. Find the increase percent in his annual petrol bill.

24. Tax on commodity is decreased by 10% andthereby its consumption increases by 8%. Findthe increase or decrease in the revenueobtained 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 thanbefore. If the earlier monthly consumption ofsugar of the family was 10 kg, what is itsmonthly consumption now?

26. Ram ordered for 6 black toys and someadditional brown toys. The prices of black

toy is 212 times that of a brown toy. While

preparing the bill, the clerk interchanged thenumber of black toys which increased the billby 45%. Find the number of brown toys.

27. Out of five questions in a paper, 201

th of

students answered all questions and 201

th

none. Of the rest 41

th answered only four

and 51

answered only one question. If 2124 %

of the total number of students answered onlythree questions and 200 answered only two,what was the total number of students?

28. A man spends 80% of his income. With theincrease in the cost of living his expenditureincreases by 37.5% and his income increases

by 3216 %. Find his present savings.

29. A salesman’s commission is 5% on all salesupto Rs 10000 and 4% on all sales exceedingthat. He remits Rs 31100 to his parentcompany after deducting his commission.Find his total sales.

30. The number of males per hundred females ina country was 120. At the next census, the

total population increased by 5% and thefemale population increased by 10%. Find theincrease per cent in the male population.

31. The ratio of the number of boys to that of girlsin a school is 3 : 2. If 20% boys and 25%girls are scholarship holders, find thepercentage of the school students who arenot scholarship holders.

32. Two numbers are respectively 20 per cent and50 per cent more than a third number. Whatpercentage is the first of the second?

33. Two numbers are 30% and 37% less than athird number respectively. The secondnumber is what per cent less than the firstnumber?

34. The cost of manufacturing a car is made up ofthree items : cost of raw material, labour andoverheads. In a year the cost of these itemswere in the ratio 4 : 3 : 2. Next year the costof raw material rose by 10%, labour costincreased by 8% but the overheads reducedby 5%. Find the percentage increase in theprice of the car.

35. The annual increase in the population of atown is 5%. If the present population of thetown is 231525, what was it 3 years ago?

36. The number of inhabitants in a town increasesat a certain rate per cent. The number atpresent is 375000 and the number 3 yearsago was 352947. Find the rate per cent?

37. The value of a machine is Rs 2,00,000. Atthe end of every year its value reduces at therate 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 constructedon a land worth Rs 729000. If the landappreciates at 10% per annum and thebuilding depreciates at 10% per annum, findthe time after which the values of the landand the building will be the same.

39. 24000 blood donors were registered with acharitable hospital. The number of donorsincreased at the rate of 5% every six month.Find the time period at the end of which thetotal number of blood donors becomes 27783.

40. The population of a village is 20000. If theannual birth rate is 4% and the annual deathrate 2%, calculate the population after twoyears.

41. The population of a town 2 years ago was62500. Due to migration of cities, it decreasesevery year at the rate of 4% per annum. Findits present population.

42. Total population of a country is 294 × 106 outof which 150 million are males and the restfemales. Out of every 1000 males, 98 can readand write but only 5.3% of the total populationcan do so. Find what percentage of womenin the whole women population of the countrycan read and write.

195Percentage

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1. We have, population of the town = 64100

55% of the population = 6410010055

= 35255.

It is given that 55% of the population aremales. Therefore, number of males = 35255.Hence, number of females in the town

= Total Population – No. of males= 64100 – 35255 = 28845

2 . In 22 carat gold, pure gold is 22 parts of 24parts. per cent of pure gold in 22 carat gold

= %1002422

= %

3291

= %1002422

= %

3291

3. It is given that the cotton and synthetic fibresare in the ratio 2 : 3. So, let cotton andsynthetic fibres be 2x and 3x respectively.Total quantity of fibre = 2x + 3x = 5xThus, in 5x fibres, cotton fibres = 2x Percentage of cotton fibres

= %10052

xx

= 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 explanationsNow, if the number of male employees is 88,total number of employees = 100If 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 femaleemployees = 12%Therefore, number of male employees

= (100 – 12)% = 88%It is also given that the total number of maleemployees is 264. 88% of x = 264

or, 26410088

x

or, x = 88100264

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 presentpopulation of a town is 50400, what was it 2years ago?

44. A litre of water is evaporated from 6 litres ofsugar solution containing 4% of sugar andthe rest water. Find the percentage of sugarin the remaining solution.

45. The daily wages of a worker are increased by10% but the number of hours worked by himper week is decreased by 10%. If originallyhe was getting Rs 2000 per week, what willhe get per week now?

46. The population of a town consists of 40%men, 35% women and 25% children. If thereare 28000 more women than children, findthe number of men, women and childrenseparately.

47. The present population of a certain city is30,00,000. If the annual birth and death ratesare 3.6% and 1.6% respectively, find thepopulation of city after 3 years.

48. The price of an article is reduced by 10%. Torestore it to its former value, by how muchper 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 candidates5% remain absent, 30% of the appearedcandidates failed. The ratio of candidates whopassed in first division, in second divisionand in third division is 1 : 2 : 3. Find thenumber of successful candidates in differentdivisions.

51. There is a census after an interval of 10 yearsstarting from 1961. There are increase inpopulation at the rate of 10%, 15% and 40%of a certain city in 1971, 1981 and 1991. Ifthe total population in 1991 be 2567950, findthe population of the city in 1961.

52. After 30 kg of water had been evaporated froma solution of salt and water, which had 15%salt, the remaining solution had 20% salt.Find the weight of the original solution.


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It is given that the man saves Rs 220. Thismeans that 8% of the total income is Rs 220.

ie 220100

8 x x = 8

100220

x = 2750Hence, the man’s 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

8100

= Rs 2750.

Hence, the man’s 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 the number of failures is 6,candidates appeared = 100If the number of failures is 114, candidates

appeared =

114

6100

= 1900

Hence, 1900 candidates appeared in theexamination.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 theexamination and the percentage of failures is6% .Therefore, 6% of x = 114

or, 114100

6 x

or, x = 19006100114

Hence, 1900 candidates appeared in theexamination.

7. Let the consumption of an article originallybe 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

=

100110100

= 111000

kg of the article


= %11

1000100

= %

11100

= %1119

8. Reduction in rate of sugar = 10% 500 gm more sugar can be purchased forRs 140. 1 kg more sugar can be purchased for

= (140 × 2 =) Rs 280.

10% of 280 = 28010010

= Rs 28

Now, from Rs 28 one can purchase 1 kg ofsugar. 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 = 10010x

Reduced price of the sugar per kg =

100x10x

= Rs 10090x


10090x

= 28

or, x = 9010028

= 9280

ie price of sugar per kg = Rs 9280

= Rs 9131

9. Let the expenditure of Ajeet be Rs 100. Ashok’s expenditure = Rs (100 – 20)

= Rs 80 Ajeet’s expenditure is Rs 20 more than thatof Ashok. When Ashok’s expenditure is Rs 80, thenAjeet’s expenditure is Rs 20 more. When Ashok’s expenditure is Rs 100, then

Ajeet’s expenditure is 8010020

= 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

8010025

= Rs 20.

Remainder = Rs (80 - 20) = Rs 60.If remainder is Rs 60, he originally had Rs100.If remainder is Re 1, he originally had Rs

60100

.

If remainder is Rs 480, he originally had

Rs

480

60100

= Rs 800.

Hence, the man had Rs 800.

197Percentage

KKUNDAN

Alternative Method:Suppose he originally had Rs x.

Amount lost = 20% of Rs x = x10020

= Rs 5x

Remainder = Rs

5xx = Rs 5

4x

Expenditure = 25% of the remainder

= 25% of Rs 54x

= 54

10025 x

= Rs 5x

Remainder = =

554 xx

= Rs 53x


53x

= Rs 480

or, x = 35480

= 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 mNow, measurement of length = (100 + 5 =) 105 mand 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 = 10000100185

= 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 = 36Number of students who failed in both thesubjects = 15Number of students who failed in EnglishOnly

= 49 – 15 = 34Number 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 totalnumber of students = 100 When 450 students are passed, the total

number of students = 30450100

= 1500 students13. Let the number of candidates = 100

According to the question,Number of candidates who, passed in English

= 60 number of candidates who failed in English

= 100 – 60 = 40Again number of candidates, who passed inMaths = 52 Number of candidates, who failed in Maths

= 100 – 52 = 48Again number of candidates, who failed inboth = 32 Number of candidates, who failed inEnglish only = 40 – 32 = 8Number of candidates, who failed in Mathsonly = 48 – 32 = 16Hence, the number of candidates, who failedin both subjects = 8 + 16 + 32 = 56 Number of candidates, who passed in bothsubjects

= 100 - 56 = 44If 44 candidates passed in both subjects, totalcandidates = 100If 1 candidate passed in both subjects, total

candidates = 44100

If 176 candidates passed in both subjects,

total candidates = 17644

100 = 400.

14. Suppose total marks is x and the minimummarks to pass the examination is y. marks obtained by the first candidate

= 30% of x = x10030

= 103x

Since, he failed by 16 marks

103x

= y – 60

or, 3x – 10y = –600 ....(i)Now, marks obtained by the second candidate

= 45% of x = 10045

x = 209x

Since second candidate obtained 15 marksmore than the minimum marks to pass.

209x

= y + 15

or, 9x – 20y = 300 ....(ii)Now, multiplying equation (i) by 2 andsubtracting it from equation (ii), we have 3x = 1500 x = 500 Total marks = 500


KKUNDAN

Now, putting the value of x in equation (i), wehave, 1500 – 10y = –600 10y = 2100

y = 102100

= 210

Minimum marks to pass = 210 In 500 total marks one should obtain 210minimum marks to pass the examination. In 100 total marks one should obtain

100500210

42 minimum marks to pass the

examination.Hence the minimum per cent marks to passthe examination is 42%.Alternative Method:Let the total marks be 100. Marks obtained by first candidate = 30 andthe marks obtained by the second candidate= 45Difference in their obtained marks

= 45 – 30 = 15But the real difference of his obtained marksis 60 + 15 = 75 When difference is 15, then the total marksis 100. When difference is 75, the total marks is

75

15100

500.

Marks obtained by first candidate = 30% of500

= 150 Minimum marks = 150 + 60 = 210

Required per cent = 100500210

= 42%

15. Let total number of voters be 100. the number of people who participated inthe election = 100 – 10 = 90Votes obtained by A = 47Votes obtained by B = 90 – 47 = 43Difference between the obtained votes of Aand B = 47 – 43 = 4 When A obtains 4 more votes than B, thenthe total number of votes = 100 When A obtains 308 more votes than B,then the total number of votes

= 4308100

= 7700

Votes obtained by A = 47% of 7700

= 100477700 = 3619

Votes obtained by B = 43% of 7700

= 100437700 = 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 = 10xx = 10

9x

Now, according to the question, 60 votes weredeclared invalid.

Total number of valid votes polled = 60109

x

Now, according to the questionTotal number of votes obtained by A

= 47% of x = x10047

Now, number of votes obtained by B

= xx1004760

109

= 6010043

x

Again according to the question,

60x10043

100x47

= 308

or, 1004x

= 308 – 60 = 248

x = 4100248

= 6200

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

= 620010090

= 5580

17. Let Krishna get 100 marks Marks obtained by Hari

= 100 + 3133 % of 100 = 100 + 3

133

= 100 + 3100

= 3400

marks

Marks obtained by Shyam

= 3216

3400

% of 3400

= 100350

3400

3400

= 9200

3400

= 91000

marks

Marks obtained by Ram

= 91000

+ 50% of 91000

= 10050

91000

91000

= 3500

marks

199Percentage

KKUNDAN

When marks obtained by Ram is 3500

, then

marks obtained by Krishna = 100 When marks obtained by Ram is 1500, then

marks obtained by Krishna = 50015003100

= 900 per cent marks obtained by Krishna

1002500900

= 36%

18. Suppose marks obtained by D = 100 Marks obtained by C = 100 – (20% of 100)

= 100 –

10010020

= 80

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

= 80 +

1002580 = 100

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

= 100 –

10010010

= 90

When marks obtained by A is 90, thenmarks 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 = 100500400

= 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, x10036

= 27

or, x = 3610027

= 75

Here maximum marks = 75

20. (i) The new duty is 60 per cent or 53

of the

former duty.Therefore, in order that the revenue mayremain unaltered the number of cars imported

must be 35

of the present number. Therefore

the import must increase by

135

or by 3266

per cent.

(ii) The new revenue is to be 100110

or 1011

of the

former revenue.Therefore the number of cars imported must

be

53

1011

or 611

of the present number.

Therefore the import must increase by

1

611

or 65

or by 3183 per cent.

Alternative Method:Let x be the number of cars importedoriginally, and let Rs y be the duty on eachcar. The present revenue derived from theimport is Rs xy.The new duty on each car

= 10060

of Rs y = Rs y53

.

(i) The new revenue is the same as before,namely, Rs xy; the number of cars to be imported

= 100

32166

10032166

300500

35

53

xxxy

xy of the

present number.

The import must increase by 3266 per cent.

(ii) The new revenue = Rs xy100110

;

the number of cars to be imported

= 100

31183

10031183

300550

53

100110

xxyxy of

the present number

The import must increase by 3183 per cent.

21. According to the question,

Distance between corresponding eyelets = 83

m

Distance between two pairs = 21

cm

Now when the lace goes from one eyelet onone side to another eyelet of the next pair toform X, the distance is equal to thehypotenuse of a right-angled triangle with

sides equal to 83

cm and 21

cm.


KKUNDAN

Hence, length of the lace forming one

X = 22

21

832

= 41

649

2

= 6425

2 = 852 = 8

10 and there are 5 such X’s.

Lace used by these = 8105 = 8

50 cm

Now lace used for joining two corresponding

eyelets at the bottom = 83

cm

Total length of the lace used

= 83

850

= 853

cm

Lace left out = 85310 = 8

27cm

Hence, percentage of lace left over

= 8

27010827100

108

27

= 33.75%

22. Suppose total number of voters = 100Number of voters who polled the vote

= 100 – 20 = 80Votes obtained by A = 40% of the votes polled

= 40% of 80 = 8010040

= 32

Votes obtained by B= 23% of the total number of votes= 23% of 100 = 23

Votes obtained by C = 80 – (32 + 23) = 25Clearly, after A, C obtains the maximumnumber of votes. Hence C is the nearestcompetitor of A.Difference of votes obtaind by A and C

= 32 – 25 = 7 When A obtains 7 votes more than C, thentotal votes = 100 When A obtains 7000 votes more than C,

then total votes = 70007

100 = 100000 votes

Hence total number of votes = 100000Number of voters who took part in the election

= 100000 – 20% of 100000= 80000

Votes 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 solutiongiven below.)Let the annual distance covered by the motoristbe 100 km and the price of petrol be Rs 100per km. Annual bill of petrol = Rs (100 × 100)Now, according to the question, distance isreduced by x% and the price of petrol isincreased by y%. New annual bill of petrol

= Rs (100 – x)(100 + y) Percentage increase in annual petrol bill

= 100100100

)100100()y100)(x100(

= 100100100100100

100y100

100x100

= 1001100

y1

100x

1

= 1001100100

xy100

y100

x1

=

100xyxy %

24. Solve as Q.No. 23. Now, see the solutiongiven below.Let the total tax on the commodity be Rs 100per unit and total consumption be 100 units. total revenue collection = Tax per unit ×number of units consumed

= (100 × 100 =) Rs 10000If the tax is decreased by 10%, new tax ratewill be Rs (100 – 10) = Rs 90 per unitIf consumption be increased by 8% , newconsumption will be (100 + 8 =) 108 unitsNew revenue collection = Rs 90 × 108

= Rs 9720Decrease in revenue collection

= Rs (10000 – 9720) = Rs 280Percentage decrease in revenue collection

= 10010000

280 = 2.8%

25. See the solution given in Q.No. 7. It willbecome easier to slove this question.Suppose price of sugar is Rs 100 per kg.Earlier monthly expenditure of the family

= (100 × 10) = Rs. 1000Suppose, monthly consumption of sugar ofthe family now is Rs x per kg.Increased price of sugar = 100 + (32% of 100)

= Rs 132Monthly expenditure of the family now

= Rs 132xIncrease in monthly expenditure

= Rs (132x – 1000)

201Percentage

KKUNDAN

Percentage increase in monthly expenditure

= 1001000

1000132

x


1001000

1000132

x = 10

or, 132x = 1100

x = 1321100

= 325

= 318 kg

26. Let the number of brown toys be x and thecost of a brown toy be Re 1

Cost of a black toy = 251 = Rs 2

5

Total cost of black toys = 256 = Rs 15

Total cost of brown toys = x × 1 = Rs xTotal cost of both the toys = Rs (15 + x)Considering the mistake by clerk, cost of black

toys = Rs 25

x

And cost of brown toys = 6 × 1 = Rs 6

Total combined cost (faulty) = Rs

6

2x5


6

2x5

100145)x15(

or, 62

520

29202915

xx

or, xx209

256

487

or, x201

463

or, x =1527. Let the total number of students be x

Number of students who answered all the

questions = 20x

Number of students who answered no

question at all = 20x

Remaining number of students who answeredone, two, three or four questions

= 109

2018

2020xxxxx


41

of 409

109 xx

students answered only four

questions

51

of 509

109 xx

students answered only one

question

2124 % of x = 200

49x students answered only

three questionsTotal number of students who answered onlytwo questions

= 20049

509

409

109 xxxx

= 200493645180 xxxx

= 200130180 xx

= 20050x

= 4x

As the number of students who answered onlytwo questions is 200.

4x

= 200

or x = 800The 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

= %2137 of 80 = %

275

of 80

= 8020075

= Rs 30

Increased expenditure = 80 + 30 = Rs 110

Increased income = 3216100

= Rs 32116 = Rs 3

350

Savings = 11032116 = Rs 3

26 = Rs 320

% savings = 1003503

320

= 740

= 755 %

29. Let total sales of the salesman be Rs x.Total commission up to sale of Rs 10000

= 100510000 = 500.

Total commission for sales exceeding Rs10000

= 251000

1004)10000(

xx


311002510000500

xx


KKUNDAN

or, 25x – (12500 + x – 10000) = 25 × 31100or, 24x – 2500 = 25 × 31100

or, x = 243120025

= 32500.

30. Let the present female population be 100 Male population at present will be 120(given) Total population at present

= 100 + 120 = 220After census the new total population

= 220 + 5% of 220

= 100105220 = 231

After census, female population= 100 + 10% of 100 = 110

After census, population of male= 231 – 110 = 121

Increase in male population = 121 – 120 = 1Percentage increase in male population

= 100120

1 = 6

5%

31. Suppose the number of boys and girls in theschool are 3x and 2x respectively. Number of scholarship holder students

= 20% of 3x + 25% of 2x = 1.1xHence, the required percentage

= 100

231.123

xx

xxx

= 10059.3

xx

= 78%

32. Let the third number be x

20% of x = 510020 xx

first number = 56

55

5xxxxx

Now, 50% of x = 210050 xx

second number = 23

2xxx

Let the first number be y% of the second

number then, y% of 56

23 xx

or, 56

10023 xyx

or, %8053

21006

x

xy

Hence, first number is 80 per cent of thesecond number.

33. Let the third number be x.Then, the first number = x – 30% of x

= xx10030

= 107x

Second number = x – 37% of x

= xx10037

= 10063x

Difference between the numbers

= 10063

107 xx

= 1006370 xx

= 1007x

Required percentage = 100

107

1007

x

x

= 10%

34. Let the cost of raw material, labour andoverheads be Rs 4x , Rs 3x and Rs 2xrespectively.Increased price of raw material

=

10010100

4x = 1001104 x = Rs 5

22x

Increased cost of labour

=

1008100

3x = 1001083 x = Rs 25

81x

Reduced overheads

=

10051002x = 100

952 x = Rs 1019x

New total cost of manufacturing

= Rs

1019

2581

522 xxx

= Rs

50

95162220 xxx = Rs 50

477x

Initial cost of manufacturing= Rs (4x +3x +2x) = Rs 9x

increase in cost = Rs

xx 950

477

= Rs 5027x

required per cent increase

= xx

95010027

= 6%

35. By the formula, if the annual increase be r%,then the population n years ago

= nrP

1001 = nr

P

1001

203Percentage

KKUNDAN

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

Required population = 3

10051

231525

= 212121202020231525

= 200000

36. If the original population of a town is P, andthe annual inrease is r%, then the populationin n years will be obtained by the formulagiven below:

Required population = nrP

1001

Here, required population = 375000, P = 352947 and n = 3 years

375000 = 3

1001352947

r

or, 352947375000

= 3

1001

r

or, 117649125000

= 3

1001

r

or, 3

4950

=

3

1001

r

or, 4950

= 1001 r

or, 14950

= 100r

or, 491

= 100r

or, 100491 r

r = 49100

= 4922

required rate = 4922 %

37. Value of machine at the end of third year

= 3

10021200000

= 100100100989898200000

= Rs 188238.408

38. Let t be the time in which the value of theland and the building be the same.Value of land when it appreciates at 10% perannum in time t

= t

100101729000 =

t

1011729000

Value of building when it depreciates at 10%per annum in time ‘t’

= t

1001011331000 =

t

1091331000

The value of land and building will be thesame in t years, ie

tt

1091331000

1011729000

or, 7291331

7290001331000

910

1011

t

= 3

911

999111111

or, 3

911

911

t

or, t = 3 yearsHence, the value of land and building will bethe same in 3 years.

39. We have,P = Initial number of donors = 24000A = Final number of donors = 27783R = Rate of increase = 5% every six month

= 10% per annumLet the total time be n years. Then,

A = nRP

2

1001

or, 27783 = n2

20010124000

or, 27783 = n2

202124000

or, 2400027783

= n2

2021

or, 3

2021

=

n2

2021

or, 2n = 3

or, n = 23

Hence, required time period = 23

years.


KKUNDAN

40. We have, annual birth rate = 4% and annualdeath rate = 2% Annual growth = (4 – 2)% = 2%Thus, we haveP = Initial population = 20000;R = Rate of growth = 2% per annum;n = 2 years

Population after 2 years = nRP

1001

= 2

1002120000

=

2

501120000

= 2

505120000

= 50

51505120000 = 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

1004162500

= 2

251162500

=

2

252462500

= 2524

252462500 = 57600.

Hence, present population = 57600.42. Total population = 294000000

Number of males = 150000000Number of females = 294000000 – 150000000

= 144000000Number of persons who can read and write

= 1003.5294000000

= 15582000

Number of literate males = 100098150000000

= 14700000 Number of literate females

= 15582000 – 14700000 = 882000 Percentage of literate females

= 144000000100882000

= %8049

= 0.6125%

43. Let the population of the town 2 years ago beP. Since the population increases in the firstyear and decreases in the second year,therefore,Population after 2 years

=

1001

1001 21 rrP ;

where r1 = 12% and r2 = 10%

Now according to the question,

1001

1001 21 rrP = 50400

or, 50400 =

10010

110012

1P

or, 50400 = 109

2528

P

or, P = 928102550400

= 50000

44. 100 litres of solution contains 4 parts ofsugar.

sugar in 6 litres of solution = 6100

4

= 0.24 partsNow, 1 litre water is evaporated,Now, 5 litres of solution contains 0.24 partsof sugar.

100 litres of solution contains = 100524.0

= 4.8%45. Suppose the daily wages and the number of

hours worked by the worker per week are Rsx and y hours respectively. Then xy = 2000

Also, 100)10100(

100)10100( yx

= Rs xy10099

will be the new wages per week.Hence, the new wages per week

= 200010099

= Rs 1980

46. In the population of the town,Women = 35%Children = 25% andMen = 40%

Now, according to the question, the numberof women is 28000 more than that ofchildren. (35% – 25%) of total population = 28000or, 10% of total population = 28000

or,

10010

of total population = 28000

Total population = 1010028000

= 280000

Number of men= 40% of the total population

= 10028000040

= 112000

Number of women= 35% of the total population

= 10028000035

= 98000

205Percentage

KKUNDAN

Number of children= 25% of the total population

= 10028000025

= 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

100213000000

=

5051

5051

50513000000 = 3183624

48. Let the original price of an article be Rs x.If the price is reduced by 10%, then new price

= Rs x

100

10100 = Rs x

109

To restore it to its former price, let the newprice be increased by y%.According to the question,

100

100109 yx = x

or, (100 + y) = 91000

or, y = %9111

9100100

91000

49. Let the income of the person be Rs x. Income after 50% decrease

= 210050

10050100 xxx

He has to increase his income by Rs

22xxx

, if he wants to get his previous

back ie to make neither profit nor loss.

% increase in income =

100

2

2x

x

= 100%

50. Total number of examinees = 600000If 5% remains absent, number of candidatespresent or appeared

=

1005100600000

= 6000 × 95 = 570000

Number of candidates failed in theexamination

= 10030570000 = 171000

Number of successful candidates whopassed in the exam in various divisions

= 570000 – 171000 = 399000But the ratio of candidates who passed infirst, second and third division = 1 : 2 : 3Number of candidates passed in first division

= 399000321

1

= 66500

Number of candidates passed in seconddivision

= 399000321

2

= 133000

Number of candidates passed in third division

= 399000321

3

= 199500

51. Let the total population of the city in 1961 bex. Then, according to the question,

2567950 =

10040

110015

110010

1x

= 57

2023

1011

x

x = 7231110002567950

= 1450000

52. Let the weight of the original solution be xkg.

Weight of salt = 10015x

kg

Weight of water = 10085x

kg

After evaporating 30 kg of water, percentageof salt = 20Now, according to the question,

10030

10015

x

x

= 20

or, 3015x

x = 20

or, 15x = 20x – 600or, 5x = 600

x = 5600

= 120 kg

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