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Percentage - General Questions (2)

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  • Percentage - General Questions
9. 

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 of 48
3
80 x = 80
100

 x = 100.

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10. 

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.
100
 b% of 20 = 1-sym-oparen-h1 b x 20 1-sym-cparen-h1 = 1-sym-oparen-h1 20 a x 1 x 20 1-sym-cparen-h1 = 4 a = 4% of a.
100 100 100 100
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11. 

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 = 1-sym-oparen-h1 x x y 1-sym-cparen-h1 = 1-sym-oparen-h1 y x x 1-sym-cparen-h1 = y% of x
100 100

 A = B.

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12. 

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 = 1-sym-oparen-h1 14 x 100 1-sym-cparen-h1% = 20%.
70
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13. 

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 = 1-sym-oparen-h1 420 x 100 1-sym-cparen-h1  = 700.
60
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14. 

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.

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15. 

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 = 1-sym-oparen-h1 50 x 100 1-sym-cparen-h1% = 45 5 %
110 11
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