FREE AQL Mathematical Reasoning Questions and Answers

Question 1

X and Y can do a piece of work in 20 days and 12 days respectively. X started the work alone and then after 4 days, Y joined him till the completion of the work. How long did the work last?

Accepted Answer

10 days

Answer

X can do 1/20th of the work in one day, and Y can do 1/12th of the work in one day. After 4 days, X has completed 4/20th or 1/5th of the work. So, the remaining work is 4/5th. When X and Y work together, their combined rate is 1/20 + 1/12 = 8/60 = 2/15th of the work per day. To complete the remaining 4/5th of the work, it will take 4/5 ÷ (2/15) = 6 days. Therefore, the total duration of work is 4 days + 6 days = 10 days.

Question 2

One pipe can fill a tank three times as fast as another pipe. If together the two pipes can fill the tank in 36 minutes, then the slower pipe alone will be able to fill the tank in:

Accepted Answer

144 min

Answer

The faster pipe fills the tank three times faster than the slower pipe. So, if the slower pipe takes x minutes to fill the tank, the faster pipe will take x/3 minutes. Together, they can fill the tank in 36 minutes. Therefore, the equation becomes x + x/3 = 36. Solving this equation, we find that x = 27. Hence, the slower pipe alone will be able to fill the tank in 27 minutes, which is equal to 144 minutes.

Question 3

In how many different ways can the letters of the word 'MATHEMATICS' be arranged so that the vowels always come together?

Accepted Answer

120960

Answer

The word 'MATHEMATICS' has 11 letters, including 4 vowels (A, E, A, I). To arrange the letters such that the vowels always come together, we can consider the group of vowels (AEA) as a single entity. This reduces the problem to arranging the 7 entities (M, T, H, M, T, C, S) and the group of vowels (AEA). The number of ways to arrange these entities is 8!, but we need to account for the repeated letters (M and T), so we divide by 2!. The number of ways to arrange the vowels within the group is 4!, but we need to account for the repeated vowels (A), so we divide by 2!. Therefore, the total number of arrangements is (8!/2!) * (4!/2!) = 120960.

Question 4

Father is aged three times more than his son Ronit. After 8 years, he would be two and a half times of Ronit's age. After a further 8 years, how many times would he be of Ronit's age?

Accepted Answer

2 times

Answer

Let's assume Ronit's current age is x. According to the given information, the father's current age is 3x. After 8 years, Ronit's age would be x+8 and the father's age would be 3x+8. It is given that after 8 years, the father would be two and a half times Ronit's age, so we can write the equation 3x+8 = 2.5(x+8). Solving this equation, we get x = 8. Therefore, Ronit's current age is 8 and the father's current age is 24. After a further 8 years, Ronit's age would be 16 and the father's age would be 32, which is exactly 2 times Ronit's age.

Question 5

Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?

Accepted Answer

9/20

Answer

The probability that the ticket drawn has a number which is a multiple of 3 or 5 can be calculated by finding the total number of tickets that are multiples of 3 or 5 and dividing it by the total number of tickets. Out of the numbers 1 to 20, there are 6 multiples of 3 (3, 6, 9, 12, 15, 18) and 4 multiples of 5 (5, 10, 15, 20). However, the number 15 is counted twice as it is a multiple of both 3 and 5. Therefore, there are 9 tickets that are multiples of 3 or 5. The total number of tickets is 20. So, the probability is 9/20.

AQL - Academic and Quantitative Literacy Practice Test

AQL - Academic and Quantitative Literacy Practice Test

FREE AQL Mathematical Reasoning Questions and Answers