**Percentage**Percent means for every 100

So, when percent is calculated for any value, it means that we calculate the value for every 100 of the reference value.

percent is denoted by the symbol %. For example, x percent is denoted by x%

*x*%=*x/*100Example : 25%=25/100=1/4

- To express
*x/y*as a percent,we have*x/y*=(*x/y*×100)%Example : 1/4=(1/4×100)%=25%

- If the price of a commodity increases by R%, the reduction in consumptionso as not to increase the expenditure = [
*R/*(100+*R*)×100]% - If the price of a commodity decreases by R%, the increase in consumptionso as not to decrease the expenditure = [
*R/*(100−*R*)×100]% - If the population of a town = P and it increases at the rate of R% per annum, thenPopulation after n years =
*P(*(1+*R)/*100))n - If the population of a town = P and it increases at the rate of R% per annum, thenPopulation before n years =
*P(*(1+*R)/*100))n - If the present value of a machine = P and it depreciates at the rate of R% per annum,

ThenValue of the machine after n years = *P(*(1-*R)/*100))n

- If the present value of a machine = P and it depreciates at the rate of R% per annum,

ThenValue of the machine before n years = *P(*(1-*R)/*100))n

__Solved Examples__

__Level 1__

1. If A = x% of y and B = y% of x, then which of the following is true? | |

A. None of these | B. A is smaller than B. |

C. Relationship between A and B cannot be determined. | D. If x is smaller than y, then A is greater than B. |

E. A is greater than B. |

**Answer** : Option A

**Explanation** :

B = *yx/*100……………..(Equation 2)

From these equations, it is clear that A = B

2.If 20% of a = b, then b% of 20 is the same as: | |

A. None of these | B. 10% of a |

C. 4% of a | D. 20% of a |

** **

**Answer **:Option C

**Explanation** :

20% of a = b

=> b = 20a/100

b% of 20 = 20b/100=(20a/100) × 20/100

=(20×20×a)/(100×100)=4a/100 = 4% of a

3.Two numbers A and B are such that the sum of 5% of A and 4% of B is two-third of the sum of 6% of A and 8% of B. Find the ratio of A : B. | |

A. 2 : 1 | B. 1 : 2 |

C. 1 : 1 | D. 4 : 3 |

**Answer **:Option D

**Explanation** :

5% of A + 4% of B = 2/3(6% of A + 8% of B)

5A/100+4B/100=2/3(6A/100+8B/100)

⇒5A+4B=2/3(6A+8B)

⇒15A+12B=12A+16B

⇒3A=4B

⇒AB=43⇒A:B=4:3

4.The population of a town increased from 1,75,000 to 2,62,500 in a decade. What is the average percent increase of population per year? | |

A. 4% | B. 6% |

C. 5% | D. 50% |

**Answer** :Option C

**Explanation** :

Increase in the population in 10 years = 2,62,500 – 1,75,000 = 87500

% increase in the population in 10 years = (87500/175000)×100=8750/175=50%

Average % increase of population per year = 50%/10=5%

5.Three candidates contested an election and received 1136, 7636 and 11628 votes respectively. What percentage of the total votes did the winning candidate get? | |

A. 57% | B. 50% |

C. 52% | D. 60% |

**Answer** :Option A

**Explanation** :

Votes received by the winning candidate = 11628

Total votes = 1136 + 7636 + 11628 = 20400

Required percentage = (11628/20400)×100=11628/204=2907/51=969/17=57%

6.A fruit seller had some oranges. He sells 40% oranges and still has 420 oranges. How many oranges he had originally? | |

A. 420 | B. 700 |

C. 220 | D. 400 |

**Answer** :Option B

**Explanation** :

He sells 40% of oranges and still there are 420 oranges remaining

=> 60% of oranges = 420

⇒(60×Total Oranges)/100=420

⇒Total Oranges/100=7

⇒ Total Oranges = 7×100=700

7.A batsman scored 110 runs which included 3 boundaries and 8 sixes. What percent of his total score did he make by running between the wickets? | |

A. 499/11 % | B. 45 % |

C. 500/11 % | D. 489/11 % |

**Answer** :Option C

**Explanation** :

Total runs scored = 110

Total runs scored from boundaries and sixes = 3 x 4 + 8 x 6 = 60

Total runs scored by running between the wickets = 110 – 60 = 50

Required % = (50/110)×100=500/11%

8.What percentage of numbers from 1 to 70 have 1 or 9 in the unit’s digit? | |

A. 2023% | B. 20% |

C. 21% | D. 2223% |

**Answer** :Option B

**Explanation** :

Total numbers = 70

Total numbers in 1 to 70 which has 1 in the unit digit = 7

Total numbers in 1 to 70 which has 9 in the unit digit = 7

Total numbers in 1 to 70 which has 1 or 9 in the unit digit = 7 + 7 = 14

Required percentage = (14/70)×100=140/7=20%

__Level 2__

1.In an election between two candidates, one got 55% of the total valid votes, 20% of the votes were invalid. If the total number of votes was 7500, what was the number of valid votes that the other candidate got? | |

A. 2800 | B. 2700 |

C. 2100 | D. 2500 |

**Answer** :Option B

**Explanation** :

Total number of votes = 7500

Given that 20% of Percentage votes were invalid

=> Valid votes = 80%

Total valid votes = (7500×80)/100

1^{st} candidate got 55% of the total valid votes.

Hence the 2^{nd} candidate should have got 45% of the total valid votes

=> Valid votes that 2nd candidate got = (total valid votes ×45)/100

=7500×(80/100)×(45/100)=75×(4/5)×45=75×4×9=300×9=2700

2.In a competitive examination in State A, 6% candidates got selected from the total appeared candidates. State B had an equal number of candidates appeared and 7% candidates got selected with 80 more candidates got selected than A. What was the number of candidates appeared from each State? | |

A. 8200 | B. 7500 |

C. 7000 | D. 8000 |

**Answer** :Option D

**Explanation** :

State A and State B had an equal number of candidates appeared.

In state A, 6% candidates got selected from the total appeared candidates

In state B, 7% candidates got selected from the total appeared candidates

But in State B, 80 more candidates got selected than State A

From these, it is clear that 1% of the total appeared candidates in State B = 80

=> total appeared candidates in State B = 80 x 100 = 8000

=> total appeared candidates in State A = total appeared candidates in State B = 8000

3.In a certain school, 20% of students are below 8 years of age. The number of students above 8 years of age is 2/3 of the number of students of 8 years of age which is 48. What is the total number of students in the school? | |

A. 100 | B. 102 |

C. 110 | D. 90 |

** **

** **

**Answer** :Option A

**Explanation** :

Let the total number of students = x

Given that 20% of students are below 8 years of age

then The number of students above or equal to 8 years of age = 80% of x —–(Equation 1)

Given that number of students of 8 years of age = 48 —–(Equation 2)

Given that number of students above 8 years of age = 2/3 of number of students of 8 years of age

=>number of students above 8 years of age = (2/3)×48=32—–(Equation 3)

From Equation 1,Equation 2 and Equation 3,

80% of x = 48 + 32 = 80

⇒80x/100=80

⇒x100=1⇒x=100

4.In an examination, 5% of the applicants were found ineligible and 85% of the eligible candidates belonged to the general category. If 4275 eligible candidates belonged to other categories, then how many candidates applied for the examination? | |

A. 28000 | B. 30000 |

C. 32000 | D. 33000 |

**Answer** :Option B

**Explanation** :

Let the number of candidates applied for the examination = x

Given that 5% of the applicants were found ineligible.

It means that 95% of the applicants were eligible (∴ 100% – 5% = 95%)

Hence total eligible candidates = 95x/100

Given that 85% of the eligible candidates belonged to the general category

It means 15% of the eligible candidates belonged to other categories(∴ 100% – 85% = 15%)

Hence Total eligible candidates belonged to other categories

=(total eligible candidates×15)/100=(95x/100)×(15/100)

=(95x×15)/(100×100)

Given that Total eligible candidates belonged to other categories = 4275

⇒(95x×15)/(100×100)=4275

⇒(19x×15)/(100×100)=855

⇒(19x×3)/(100×100)=171

⇒(x×3)/(100×100)=9

⇒x/(100×100)=3

⇒x=3×100×100=30000

5.A student multiplied a number by 3/5 instead of 5/3.What is the percentage error in the calculation? | |

A. 64% | B. 32% |

C. 34% | D. 42% |

**Answer** :Option A

**Explanation** :

Let the number = 1

Then, ideally he should have multiplied 1 by 5/3.

Hence the correct result was 1 x (5/3) = (5/3)

By mistake, he multiplied 1 by 3/5.

Hence the result with the error = 1 x (3/5) = (3/5)

Error = 5/3−3/5=(25−9)/15=16/15

percentage error = (Error/True Value)×100={(16/15)/(5/3)}×100

=(16×3×100)/(15×5)=(16×100)/(5×5)=16×4=64%

6.The price of a car is Rs. 3,25,000. It was insured to 85% of its price. The car was damaged completely in an accident and the insurance company paid 90% of the insurance. What was the difference between the price of the car and the amount received ? | |

A. Rs. 76,375 | B. Rs. 34,000 |

C. Rs. 82,150 | D. Rs. 70,000 |

**Answer** :Option A

**Explanation** :

Price of the car = Rs.3,25,000

Car insured to 85% of its price

=>Insured price=(325000×85)/100

Insurance company paid 90% of the insurance

⇒Amount paid by Insurance company =(Insured price×90)/100

=325000×(85/100)×(90/100)=325×85×9=Rs.248625

Difference between the price of the car and the amount received

= Rs.325000 – Rs.248625 = Rs.76375

7.If the price of petrol increases by 25% and Benson intends to spend only an additional 15% on petrol, by how much % will he reduce the quantity of petrol purchased? | |

A. 8% | B. 7% |

C. 10% | D. 6% |

**Answer** :Option A

**Explanation** :

Assume that the initial price of 1 Litre petrol = Rs.100 ,Benson spends Rs.100 for petrol,

such that Benson buys 1 litre of petrol

After the increase by 25%, price of 1 Litre petrol = (100×(100+25))/100=Rs.125

Since Benson spends additional 15% on petrol,

amount spent by Benson = (100×(100+15))/100=Rs.115

Hence Quantity of petrol that he can purchase = 115/125 Litre

Quantity of petrol reduced = (1−115/125) Litre

Percentage Quantity of reduction = ((1−115/125))/1×100=(10/125)/×100=(10/5)×4=2×4=8%

8.30% of the men are more than 25 years old and 80% of the men are less than or equal to 50 years old. 20% of all men play football. If 20% of the men above the age of 50 play football, what percentage of the football players are less than or equal to 50 years? | |

A. 60% | B. 70% |

C. 80% | D. 90% |

**Answer** :Option C

**Explanation** :

Let total number of men = 100

Then

80 men are less than or equal to 50 years old

(Since 80% of the men are less than or equal to 50 years old)

=> 20 men are above 50 years old (Since we assumed total number of men as 100)

20% of the men above the age of 50 play football

⇒Number of men above the age of 50 who play football = (20×20)/100=4

Number of men who play football = 20 (Since 20% of all men play football)

Percentage of men who play football above the age of 50 = (4/20)×100=20%

=>Percentage of men who play football less than or equal to the age 50 = 100%−20%=80%

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