# Percentage

### Important Formulas

1. 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 = 1 . 100 5

 To express a as a percent: We have, a = a x 100 %. b b b

 Thus, 1 = 1 x 100 % = 25%. 4 4

2. 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 x 100 % (100 + R)

If the price of a commodity decreases by R%, then the increase in consumption so as not to decrease the expenditure is: R x 100 % (100 – R)

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

 1. Population after n years = P 1 + R n 100

2. Population n years ago = P 1 + R n 100

4. Results on Depreciation:Let the present value of a machine be P. Suppose it depreciates at the rate of R% per annum. Then:

 1. Value of the machine after n years = P 1 – R n 100

2. Value of the machine n years ago = P 1 – R n 100

 3. If A is R% more than B, then B is less than A by R x 100 %. (100 + R)

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

1

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%

Explanation:

Number of runs made by running = 110 – (3 x 4 + 8 x 6)

= 110 – (60)

= 50. Required percentage = 50 x 100 % = 45 5 % 110 11

2.

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

Explanation:

Let their marks be (x + 9) and x.

 Then, x + 9 = 56 (x + 9 + x) 100 25(x + 9) = 14(2x + 9) 3x = 99 x = 33

So, their marks are 42 and 33.

3.

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

Explanation:

Suppose originally he had x apples.

Then, (100 – 40)% of x = 420. 60 x x = 420 100 x = 420 x 100 = 700. 60

4.

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

nswer: 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 x 100 % = 20%. 70

5.

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

Explanation:

 x% of y = x x y = y x x = y% of x 100 100 A = B.

6.

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

Explanation:

 20% of a = b 20 a = b. 100 b% of 20 = b x 20 = 20 a x 1 x 20 = 4 a = 4% of a. 100 100 100 100

7.

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

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 of 48 3 80 x = 80 100 x = 100.

8.

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

Explanation:

 5% of A + 4% of B = 2 (6% of A + 8% of B) 3 5 A + 4 B = 2 6 A + 8 B 100 100 3 100 100 1 A + 1 B = 1 A + 4 B 20 25 25 75  1 – 1 A = 4 – 1 B 20 25 75 25 1 A = 1 B 100 75

 A = 100 = 4 . B 75 3 Required ratio = 4 : 3

9.

 A student multiplied a number by 3 instead of 5 . 5 3

What is the percentage error in the calculation?

 A. 34% B. 44% C. 54% D. 64%

Explanation:

Let the number be x.

 Then, error = 5 x – 3 x = 16 x. 3 5 15

 Error% = 16x x 3 x 100 % = 64%. 15 5x

10.

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

Explanation:

Number of valid votes = 80% of 7500 = 6000. Valid votes polled by other candidate = 45% of 6000

 = 45 x 6000 = 2700. 100

11

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%

Explanation:

Total number of votes polled = (1136 + 7636 + 11628) = 20400. Required percentage = 11628 x 100 % = 57%. 20400

12.

Two tailors 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

Explanation:

Let the sum paid to Y per week be Rs. z.

Then, z + 120% of z = 550. z + 120 z = 550 100 11 z = 550 5 z = 550 x 5 = 250. 11

13.

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

Explanation:

Let the amount taxable purchases be Rs. x.

 Then, 6% of x = 30 100 x = 30 x 100 = 5. 100 6 Cost of tax free items = Rs. [25 – (5 + 0.30)] = Rs. 19.70

14.

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

Explanation:

 Rebate = 6% of Rs. 6650 = Rs. 6 x 6650 = Rs. 399. 100

 Sales tax = 10% of Rs. (6650 – 399) = Rs. 10 x 6251 = Rs. 625.10 100 Final amount = Rs. (6251 + 625.10) = Rs. 6876.10

15.

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%

Explanation:

Increase in 10 years = (262500 – 175000) = 87500.

 Increase% = 87500 x 100 % = 50%. 175000 Required average = 50 % = 5%. 10