1.

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.


2.

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.

60/100 x x = 420

x = (420 x 100)/60 = 700.


3.

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%.


4.

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.
Answer: Option E
Explanation: x% of y = (x/100 x y) = (y/100 x x) = y% of x

A = B.


5.

In a certain school, 20% of students are below 8 years of age. The number of students above 8 years of age is 2/3 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
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

80/100x = 80

x = 100.