percentage

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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%

Answer: Option B

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

Answer: Option C

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

Answer: Option D

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

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 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

Answer: Option E

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%

Answer: Option B

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

Answer: Option C

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

Answer: Option D

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

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 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

Answer: Option E

Explanation:

 

 

x% of y = x x y = y x x = y% of x
100 100

A = B.
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