Q1. Two friends, Akash & Beenu had some candies each. One of them had 15 candies more than the other. The candies with Akash was 60% of the total candies with them. How many candies did each have?

A. 40, 25

B. 47, 32

C. 45, 30

D. 49, 34

Sol : Option C Explanation Let the candies with be (x + 15) and x.
Therefore, x + 15 = 60/100(x + 15 + x)
(x + 15) = 3/5(2x + 15)
5x + 75 = 6x + 45
x = 30
So, the marks of two students are 45 and 30.

Q2. A fruit seller had some oranges. He sells 30% oranges and still has 140 mangoes. Originally, he had:

A. 288 oranges

B. 300 oranges

C. 672 oranges

D. 200 oranges

Sol : Option D Explanation: Suppose originally he had x oranges.
Then, (100 - 30)% of x = 140.
70/100 x = 140
x = (140 x 100)/70 = 200.

Q3. What percentage of numbers from 1 to 30 has 1 or 9 in the unit's digit?

A. 12

B. 15

C. 20

D. 22

Sol : Option C Explanation: Such numbers from 1 to 30 are 1, 9, 11, 19, 21, 29
Number of such numbers =6
Required percentage is (6/20 * 100) % = 20%

Q4. If M = y% of z and N = z% of y, then which of the following must be true?

A. M is lesser than N.

B. M is more than N

C. Relation between M and N cannot be determined.

D. None of these

Sol : Option D Explanation: y% of z = (y/100 x z) = (z/100 x y) = z% of y
M = N.

5. In ABC College, 65% of students are less than 20 years of age. The number of students more than 20 years of age is 2/3 of number of students of 20 years of age which is 42. What is the total number of students in the ABC College?

A. 75

B. 90

C. 130

D. 200

Sol : Option D Explanation: Let total number of students is x.
Then, number of students more than or equal to 20 years of age = (100 - 65)% of x = 35% of x.
As per the question, 35% of x = 42 + 2/3 of 42. So 35x/100 = 70. Hence x = 200.

Q.6. A student erroneously multiplied a number by 2/5 instead of 5/2.What is the percentage error in the calculation?

A. 24%

B. 54%

C. 74%

D. 84%

Sol : Option D Explanation: Let the number be 100. 2/5 of 100 is 40 while 5/2 of 100 is 250.
Now the difference is 210 on a base of 250. Therefore, percentage difference is 210/250 into 100 = 84%.

Q.7. The sum of the first six terms of an AP is 48 and the common difference is 2. What is the 4th term?

A. 9

B. 11

C. 13

D. 7

Sol : Option A Explanation: Let the 1st number of an A.P is x,
As the sum of 1st 6 numbers is 48, so x+ x+2+x+4+x+6+x+8+x+10 = 48, so x= 3
So 4^{th} term is 9.

Q8. Three candidates, Ajay, Bijoy & Chandu contested an election and received 1800, 3300 and votes 3900 respectively. What percent of the total votes did A get?

A. 20%

B. 40%

C. 45%

D. 70%

Sol : Option A Explanation: Total no. of votes polled = (1800 + 3300 + 3900) = 9000.
Required percentage = (1800/9000 * 100)% = 20%.

Q9. A Stationery seller had some Pens, Sharpeners, Erasers & Pencils. He sells 65% of the total units and still has 175 units. Originally, he had:

A. 588 units

B. 400 units

C. 272 units

D. 500 units

Sol : Option D Explanation: Suppose originally he had x units.
Then, (100 - 65)% of x = 175.
35/100 x = 175
x = 500

Q10. The total population of a village increased from 1,80,00 to 22, 500 in a decade. The average percentage increase of population per year of that village is:

A. 2.37%

B. 2.5%

C. 3. 6%

D. 6.75%

Sol : Option B Explanation: Population increase in 10 years = (22500- 18000) = 4500.
Increase% = (4500/18000 x 100)% = 25%
Required average = (25/10)% = 2. 5%