percentage

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

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.

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

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