PERCENTAGE

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
To express a/b as a percent: We have, a/b = (a/b x 100)
Thus, 1/4 = (1/4 x 100)% = 25%.

Percentage Increase/Decrease:
If the price of a commodity increases by R%, then the reduction in consumption so as not to
increase the expenditure is:
[R/ (100 + R) x 100]%
If the price of a commodity decreases by R%, then the increase in consumption so as not to
decrease the expenditure is:
[R/(100 – R) x 100]%

Results on Population:
Let the population of a town be P now and suppose it increases at the rate of R% per annum,
then:

Population after n years = 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 of 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) x 100]%.

If A is R% less than B, then B is more than A by [R/(100 – R) x 100]%.

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. 45%
B. 45 5/11%
C. 54 6/11%
D. 55%
Answer: Option B
Explanation:
Number of runs made by running = 110 – (3 x 4 + 8 x 6)
= 110 – (60)
= 50.
Required percentage = (50/110 x 100)% = 45 5/11%

Two students appeared at an examination. One of them secured 9 marks more than the
other and his marks was 56% of the sum of their marks. The marks obtained by them are:
A. 39, 30
B. 41, 32
C. 42, 33
D. 43, 34
Answer: Option C
Explanation:
Let their marks be (x + 9) and x.
Then, x + 9 = 56/100 (x + 9 + x)
25(x + 9) = 14(2x + 9)
3x = 99
x = 33
So, their marks are 42 and 33.

A fruit seller had some apples. He sells 40% apples and still has 420 apples. Originally, he had:
A. 588 apples
B. 600 apples
C. 672 apples
D. 700 apples
Answer: Option D
Suppose originally he had x apples.
Then, (100 – 40)% of x = 420.
60/100 x X = 420
x =(420 x 100/60) = 700.

What percentage of numbers from 1 to 70 have 1 or 9 in the unit’s digit?
A. 1
B. 14
C. 20
D. 21
Answer: Option C
Explanation:
Clearly, the numbers which have 1 or 9 in the unit’s digit, have squares that end in the digit 1.
Such numbers from 1 to 70 are 1, 9, 11, 19, 21, 29, 31, 39, 41, 49, 51, 59, 61, 69.
Number of such number =14
Required percentage = (14/70 x 100)% = 20%.

If A = x% of y and B = y% of x, then which of the following is true?
A. A is smaller than B.
B. A is greater 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. None of these
Answer: Option E
Explanation:
x% of y = (x/100 x y) = (y/100 X x) = y% of x
A = B.

If 20% of a = b, then b% of 20 is the same as:
A. 4% of a
B. 5% of a
C. 20% of a
D. None of these
Answer: Option A
Explanation:
20% of a = b=> 20 a = b.
b% of 20 = (b/100 x 20) = (20/100a x 1/100 x 20) = 4/100a = 4% of a.

In a certain school, 20% of students are below 8 years of age. The number of students above 8
years of age is of the number of students of 8 years of age which is 48. What is the total
number of students in the school?
A. 72
B. 80
C. 120
D. 150
E. 100
Answer: Option E
Explanation:
Let the number of students be x. Then,
Number of students above 8 years of age = (100 – 20)% of x = 80% of x.
80% of x = 48 + 2/3 of 48
80/100 x = 80
x = 100.

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 : 3
B. 1 : 1
C. 3 : 4
D. 4 : 3
Answer: Option D
Explanation:
5% of A + 4% of B = 2/3 (6% of A + 8% of B)
5/100 A + 4/100 B = 2/3 (6/100 A + 8/100 B)
1/20A + 1/25B = 1/25A + 4/75 B
(1/20 – 1/25) A =(4/75 – 1/25)B
1/100A = 1/75B
A/B = 100/75 = 4/34
Required ratio = 4 : 3

A student multiplied a number by 3/5 instead of 5/3.
What is the percentage error in the calculation?
A. 34%
B. 44%
C. 54%
D. 64%
Answer: Option D
Explanation:
Let the number be x.
Then, error = 5/3x – 3/5x = 16/15x.
Error% = (16/15x X 3/5x X 100)% = 64%.

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, the number of valid votes that the other
candidate got, was:
A. 2700
B. 2900
C. 3000
D. 3100
Answer: Option A
Explanation:
Number of valid votes = 80% of 7500 = 6000. Valid votes polled by other candidate = 45% of 6000
45/100 x 6000 = 2700.

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. 60%
C. 65%
D. 90%
Answer: Option A
Explanation:
Total number of votes polled = (1136 + 7636 + 11628) = 20400.
Required percentage = (11628/20400 x 100)% = 57%.

Two tailor’s X and Y are paid a total of Rs. 550 per week by their employer. If X is paid 120
percent of the sum paid to Y, how much is Y paid per week?
A. Rs. 200
B. Rs. 250
C. Rs. 300
D. None of these
Answer: Option B
Explanation:
Let the sum paid to Y per week be Rs. z.
Then, z + 120% of z = 550.
z + 120/100z = 550
11/5z = 550
z = (550 x 5/11) = 250.

Gauri went to the stationers and bought things worth Rs. 25, out of which 30 paise went on
sales tax on taxable purchases. If the tax rate was 6%, then what was the cost of the tax free
items?
A. Rs. 15
B. Rs. 15.70
C. Rs. 19.70
D. Rs. 20
Answer: Option C
Explanation:
Let the amount taxable purchases be Rs. x.
Then, 6% of x = (30/100 x 100/6) = 5.
Cost of tax free items = Rs. [25 – (5 + 0.30)] = Rs. 19.70

Rajeev buys good worth Rs. 6650. He gets a rebate of 6% on it. After getting the rebate, he
pays sales tax @ 10%. Find the amount he will have to pay for the goods.
A. Rs. 6876.10
B. Rs. 6999.20
C. Rs. 6654
D. Rs. 7000
Answer: Option A
Explanation:
Rebate = 6% of Rs. 6650 = Rs. (6/100 x 6650) = Rs. 399.
Sales tax = 10% of Rs. (6650 – 399) = Rs. (10/100 x 6251) = Rs. 625.10
100
Final amount = Rs. (6251 + 625.10) = Rs. 6876.10

The population of a town increased from 1,75,000 to 2,62,500 in a decade. The average
percent increase of population per year is:
A. 4.37%
B. 5%
C. 6%
D. 8.75%
Answer: Option B
Explanation:
Increase in 10 years = (262500 – 175000) = 87500.
Increase% = (87500/ 175000 x 100)% = 50%.
Required average = (50/10)% = 5%.

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