# RS Aggarwal Quantitative Aptitude PDF Free download: PERCENTAGE

Contents

**PERCENTAGE**

## IMPORTANT FACTS AND FORMULAE

** **

**Concept of Percentage :**By a certain percent ,we mean that many hundredths. Thus x percent means x hundredths, written as x%.

** **

**To express x% as a fraction : **We have , x% = x/100.

** **

Thus, 20% =20/100 =1/5; 48% =48/100 =12/25, etc.

**To express a/b as a percent :** We have, a/b =((a/b)*100)%.

** **

Thus, ¼ =[(1/4)*100] = 25%; 0.6 =6/10 =3/5 =[(3/5)*100]% =60%.

- If the price of a commodity increases by R%, then the reduction in consumption so asnot to increase the expenditure is

[R/(100+R))*100]%.

If the price of the commodity decreases by R%,then the increase in consumption so as to decrease the expenditure is

[(R/(100-R)*100]%.

**Results on Population :**Let the population of the town be P now and suppose it increases at the rate of

R% per annum, then :

- Population after nyeras = P [1+(R/100)]^n.
- Population n years ago = P /[1+(R/100)]^n.

**Results on Depreciation :**Let the present value of a machine be P. Suppose it depreciates at the rate

R% per annum. Then,

- Value of the machine after n years = P[1-(R/100)]
^{n}. - Value of the machine n years ago = P/[1-(R/100)]
^{n}.

- If A is R% more than B, then B is less than A by

[(R/(100+R))*100]%.

If A is R% less than B , then B is more than A by

[(R/(100-R))*100]%.

**SOLVED EXAMPLES**

** **

**Ex. 1. Express each of the following as a fraction :**

** **

# (i) 56% (ii) 4% (iii) 0.6% (iv) 0.008%

**sol. ** (i) 56% = 56/100= 14/25. (ii) 4% =4/100 =1/25.

(iii) 0.6 =6/1000 = 3/500. (iv) 0.008 = 8/100 = 1/1250.

**Ex. 2. Express each of the following as a Decimal :**

** **

**(i) 6% (ii)28% (iii) 0.2% (iv) 0.04%**

** **

**Sol. **(i) 6% = 6/100 =0.06. (ii) 28% = 28/100 =0.28.

(iii) 0.2% =0.2/100 = 0.002. (iv) 0.04%= 0.04/100 =0.004.

**Ex. 3. Express each of the following as rate percent :**

** **

**(i) 23/36 (ii) 6 ¾ (iii) 0.004**

** **

**Sol. ** (i) 23/36 = [(23/36)*100]% = [575/9]% = 63 8/9%.

(ii) 0.004 = [(4/1000)*100]% = 0.4%.

(iii) 6 ¾ =27/4 =[(27/4)*100]% = 675%.

** **

**Ex. 4. Evaluate :**

** **

**28% of 450+ 45% of 280****16 2/3% of 600 gm- 33 1/3% of 180 gm**

** **

**Sol. **(i) 28% of 450 + 45% of 280 =[(28/100)*450 + (45/100)*280] = (126+126) =252.

- 16 2/3% of 600 gm –33 1/3% of 180 gm = [ ((50/3)*(1/100)*600) – ((100/3)*(1/3)*280)]gm = (100-60) gm = 40gm.

** **

**Ex. 5. **

**(i) 2 is what percent of 50 ?**

**(ii) ½ is what percent of 1/3 ?**

**(iii)What percent of 8 is 64 ?**

**(iv)What percent of 2 metric tones is 40 quintals ?**

**(v)What percent of 6.5 litres is 130 ml?**

** **

**Sol.**

** **

(i) Required Percentage = [(2/50)*100]% = 4%.

(ii) Required Percentage = [ (1/2)*(3/1)*100]% = 150%.

(iii)Required Percentage = [(84/7)*100]% = 1200%.

- 1 metric tonne = 10 quintals.

Required percentage = [ (40/(2 * 10)) * 100]% = 200%.

- Required Percentage = [ (130/(6.5 * 1000)) * 100]% = 2%.

**Ex. 6. **

**Find the missing figures :**

** **

**(i) ?% of 25 = 20125 (ii) 9% of ? = 63 (iii) 0.25% of ? = 0.04**

** **

**Sol. **

** **

- Let x% of 25 = 2.125. Then , (x/100)*25 = 2.125

X = (2.125 * 4) = 8.5.

- Let 9% of x =6.3. Then , 9*x/100 = 6.3

X = [(6.3*100)/9] =70.

- Let 0.25% of x = 0.04. Then , 0.25*x/100 = 0.04

X= [(0.04*100)/0.25] = 16.

**Ex. 7.**

**Which is greatest in 16 ( 2/3) %, 2/5 and 0.17 ?**

** **

**Sol.** 16 (2/3)% =[ (50/3)* )1/100)] = 1/6 = 0.166, 2/15 = 0.133. Clearly, 0.17 is the greatest.

** **

**Ex. 8.**

**If the sales tax reduced from 3 1/2 % to 3 1/3%, then what difference does it make to a person who purchases an article with market price of Rs. 8400 ?**

** **

**Sol. ** Required difference = [3 ½ % of Rs.8400] – [3 1/3 % of Rs.8400]

= [(7/20-(10/3)]% of Rs.8400 =1/6 % of Rs.8400

= Rs. [(1/6)8(1/100)*8400] = Rs. 14.

**Ex. 9. An inspector rejects 0.08% of the meters as defective. How many will be examine to project ?**

** **

**Sol. ** Let the number of meters to be examined be x.

Then, 0.08% of x =2

[(8/100)*(1/100)*x] = 2

x = [(2*100*100)/8] = 2500.** **

** **

**Ex. 10. Sixty five percent of a number is 21 less than four fifth of that number. What is the number ?**

** **

**Sol. ** Let the number be x.

Then, 4*x/5 –(65% of x) = 21

4x/5 –65x/100 = 21

5 x = 2100

x = 140.

Ex.11. Difference of two numbers is 1660. If 7.5% of the number is 12.5% of the other number , find the number ?

**Sol. ** Let the numbers be x and y. Then , 7.5 % of x =12.5% of y

X = 125*y/75 = 5*y/3.

Now, x-y =1660

5*y/3 –y =1660

2*y/3= 1660

y =[ (1660*3)/2] =2490.

One number = 2490, Second number =5*y/3 =4150.

Ex. 12.

**In expressing a length 810472 km as nearly as possible with three significant digits , find the percentage error.**

**Sol. ** Error = (81.5 – 81.472)km = 0.028.

Required percentage = [(0.028/81.472)*100]% = 0.034%.

Ex. 13.

**In an election between two candidates, 75% of the voters cast thier thier votes, out of which 2% of the votes were declared invalid. A candidate got 9261 votes which were 75% of the total valid votes. Find the total number of votes enrolled in that election.**

**Sol.**

Let the number of votes enrolled be x. Then ,

Number of votes cast =75% of x. Valid votes = 98% of (75% of x).

75% of (98% of (75%of x)) =9261.

[(75/100)*(98/100)*(75/100)*x] =9261.

X = [(9261*100*100*100)/(75*98*75)] =16800.

**Ex.14. Shobha’s mathematics test had 75 problems i.e.10 arithmetic, 30 algebra and 35 geometry problems. Although she answered 70% of the arithmetic ,40% of the algebra, and 60% of the geometry problems correctly. she did not pass the test because she got less than 60% of the problems right. How many more questions she would have to answer correctly to earn 60% of the passing grade?**

** **

**Sol. **Number of questions attempted correctly=(70% of 10 + 40% of 30 + 60% 0f 35)

=7 + 12+21= 45

questions to be answered correctly for 60% grade=60% of 75 = 45

therefore required number of questions= (45-40) = 5.

**Ex.15. if 50% of (x-y) = 30% of (x+y) then what percent of x is y?**

** **

**Sol.**50% of (x-y)=30% of(x+y) ó (50/100)(x-y)=(30/100)(x+y)

ó5(x-y)=3(x+y) ó 2x=8y ó x=4y

therefore required percentage =((y/x) X 100)% = ((y/4y) X 100) =25%

** **

**Ex.16. Mr.Jones gave 40% of the money he had to his wife. he also gave 20% of the remaining amount to his 3 sons. half of the amount now left was spent on miscellaneous items and the remaining amount of Rs.12000 was deposited in the bank. how much money did Mr.jones have initially?**

**Sol. **Let the initial amount with Mr.jones be Rs.x then,

Money given to wife= Rs.(40/100)x=Rs.2x/5.Balance=Rs(x-(2x/5)=Rs.3x/5.

Money given to 3 sons= Rs(3X((20/200) X (3x/5)) = Rs.9x/5.

Balance = Rs.((3x/5) – (9x/25))=Rs.6x/25.

Amount deposited in bank= Rs(1/2 X 6x/25)=Rs.3x/25.

Therefore 3x/25=12000 ó x= ((12000 x 35)/3)=100000

So Mr.Jones initially had Rs.1,00,000 with him.

Short-cut Method : Let the initial amount with Mr.Jones be Rs.x

### Then,(1/2)[100-(3*20)]% of x=12000

ó (1/2)*(40/100)*(60/100)*x=12000

óx=((12000*25)/3)=100000

Ex 17 10% of the inhabitants of village having died of cholera.,a panic set in , during which 25% of the remaining inhabitants left the village. The population is then reduced to 4050. Find the number of original inhabitants.

**Sol:**

Let the total number of orginal inhabitants be x.

((75/100))*(90/100)*x)=4050 ó (27/40)*x=4050

óx=((4050*40)/27)=6000.

Ex.18 A salesman`s commission is 5% on all sales upto Rs.10,000 and 4% on all sales exceeding this.He remits Rs.31,100 to his parent company after deducing his commission . Find the total sales.

**Sol:**

Let his total sales be Rs.x.Now(Total sales) – (Commission )=Rs.31,100

x-[(5% of 10000 + 4% of (x-10000)]=31,100

x-[((5/100)*10000 + (4/100)*(x-10000)]=31,100

óx-500-((x-10000)/25)=31,100

óx-(x/25)=31200 ó 24x/25=31200óx=[(31200*25)/24)=32,500.

Total sales=Rs.32,500

** **

**Ex .19 Raman`s salary was decreased by 50% and subsequently increased by 50%.How much percent does he lose?**

**Sol:**

Let the original salary = Rs.100

New final salary=150% of (50% of Rs.100)=

Rs.((150/100)*(50/100)*100)=Rs.75.

Decrease = 25%

** **

**Ex.20 Paulson spends 75% of his income. His income is increased by 20% and he increased his expenditure by 10%.Find the percentage increase in his savings .**

**Sol:**

Let the original income=Rs.100 . Then , expenditure=Rs.75 and savings =Rs.25

New income =Rs.120 , New expenditure =

Rs.((110/100)*75)=Rs.165/2

New savings = Rs.(120-(165/2)) = Rs.75/2

Increase in savings = Rs.((75/2)-25)=Rs.25/2

Increase %= ((25/2)*(1/25)*100)% = 50%.

Ex21. The salary of a person was reduced by 10% .By what percent should his reduced salary be raised so as to bring it at par with his original salary ?

Sol:

Let the original salary be Rs.100 . New salary = Rs.90.

Increase on 90=10 , Increase on 100=((10/90)*100)%

= (100/9)%

Ex.22 When the price fo a product was decreased by 10% , the number sold increased by 30%. What was the effect on the total revenue ?

Sol:

Let the price of the product be Rs.100 and let original sale be 100 pieces.

Then , Total Revenue = Rs.(100*100)=Rs.10000.

New revenue = Rs.(90*130)=Rs.11700.

Increase in revenue = ((1700/10000)*100)%=17%.

Ex 23 . If the numerator of a fraction be increased by 15% and its denominator be diminished by 8% , the value of the fraction is 15/16. Find the original fraction.

Sol:

Let the original fraction be x/y.

Then (115%of x)/(92% of y)=15/16 => (115x/92y)=15/16

- ((15/16)*(92/115))=3/4

** **

Ex.24 In the new budget , the price of kerosene oil rose by 25%. By how much percent must a person reduce his consumption so that his expenditure on it does not increase ?

Sol:

Reduction in consumption = [((R/(100+R))*100]%

- [(25/125)*100]%=20%.

** **

**Ex.25 The population of a town is 1,76,400 . If it increases at the rate of 5% per annum , what will be its population 2 years hence ? What was it 2 years ago ?**

**Sol:**

Population after 2 years = 176400*[1+(5/100)]^2

=[176400*(21/20)*(21/40)]

= 194481.

Population 2 years ago = 176400/[1+(5/100)]^2

=[716400*(20/21)*(20/21)]= 160000.

Ex.26 The value of a machine depreiates at the rate of 10% per annum. If its present is Rs.1,62,000 what will be its worth after 2 years ? What was the value of the machine 2 years ago ?

**Sol.**

Value of the machine after 2 years

=Rs.[162000*(1-(10/100))^2] = Rs.[162000*(9/10)*(9/10)]

=Rs. 131220

Value of the machine 2 years ago

= Rs.[162000/(1-(10/100)^2)]=Rs.[162000*(10/9)*(10/9)]=Rs.200000

**Ex27. During one year, the population of town** **increased by 5% . If the total population is 9975 at the end of the second year , then what was the population size in the beginning of the first year ?**

Sol:

Population in the beginning of the first year

= 9975/[1+(5/100)]*[1-(5/100)] = [9975*(20/21)*(20/19)]=10000.

Ex.28 If A earns 99/3% more than B,how much percent does B earn less then A ?

**Sol:**

Required Percentage = [((100/3)*100)/[100+(100/3)]]%

=[(100/400)*100]%=25%

Ex. 29 If A`s salary is 20% less then B`s salary , by how much percent is B`s salary more than A`s ?

Sol:

Required percentage = [(20*100)/(100-20)]%=25%.

Ex30 .How many kg of pure salt must be added to 30kg of 2% solution of salt and water to increase it to 10% solution ?

**Sol:**

Amount of salt in 30kg solution = [(20/100)*30]kg=0.6kg

Let x kg of pure salt be added

Then , (0.6+x)/(30+x)=10/100ó60+100x=300+10x

ó90x=240 ó x=8/3.

Ex 31. Due to reduction of 25/4% in the price of sugar , a man is able to buy 1kg more for Rs.120. Find the original and reduced rate of sugar.

**Sol:**

Let the original rate be Rs.x per kg.

Reduced rate = Rs.[(100-(25/4))*(1/100)*x}]=Rs.15x/16per kg

120/(15x/16)-(120/x)=1 ó (128/x)-(120/x)=1

ó x=8.

So, the original rate = Rs.8 per kg

Reduce rate = Rs.[(15/16)*8]per kg = Rs.7.50 per kg

**Ex.32 In an examination , 35% of total students failed in Hindi , 45% failed in English and 20% in both . Find the percentage of those who passed in both subjects .**

Sol:

Let A and B be the sets of students who failed in Hindi and English respectively .

Then , n(A) = 35 , n(B)=45 , n(AÇB)=20.

So , n(AÈB)=n(A)+n(B)- n(AÇB)=35+45-20=60.

Percentage failed in Hindi and English or both=60%

Hence , percentage passed = (100-60)%=40%

**Ex33. In an examination , 80% of the students passed in English , 85% in Mathematics and 75% in both English and Mathematics. If 40 students failed in both the subjects , find the total number of students.**

**Sol:**

Let the total number of students be x .

Let A and B represent the sets of students who passed in English and Mathematics respectively .

Then , number of students passed in one or both the subjects

= n(AÈB)=n(A)+n(B)- n(AÇB)=80% of x + 85% of x –75% of x

=[(80/100)x+(85/100)x-(75/100)x]=(90/100)x=(9/10)x

Students who failed in both the subjects = [x-(9x/10)]=x/10.

So, x/10=40 of x=400 .

Hence ,total number of students = 400

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