Dear Reader,
Given are three moderately difficult aptitude questions to practice for Wipro. Hope you find these questions interesting and helpful.

Question 1

Trisha and Priya talked over certain important issues for nearly one hour. After that they went to Coffee Day shop—Trisha had cold coffee and Priya asked for hot coffee. Both the drinks were charged at the same rate of Rs.75 per cup. Trisha started walking at 6 kmph towards eastern direction and Priya started walking at 5 kmph towards western direction. They started at 11 am. At what time will they be 44 km apart ?

a) 3 am b) 2 pm c) 3pm d) 4 pm

Answer : c) 3 pm

Solution :

Trisha and Priya started walking in eastern and western direction respectively at 11 am.
Trisha walked at 6 kmph and Priya started walking at 5 kmph. So the relative speed is 11 kmph.
Time required to be 44 Kms apart when the relative speed is 11 Km/hr = Distance Between Them / Relative Speed = 44/11 = 4.
They will be 44 km apart after four hours from the time of starting. i.e. 3 pm. (candidates must be alert to not to mark a) 3 am as answer)

Question 2

Sundaramurthy can row at 8.5 kmph in still water and he finds it takes him thrice as long to row up as to row down the river. Find the speed of river in kmph.

a) 3.25 km/hr b) 5.25 km/hr c) 2.75 km/hr d) 4.25 km/hr

Answer : d) 4.25 km/hr

Solution :

Let the speed of the river be c.
His speed when rowing down = 8.5 + c
His speed when rowing up = 8.5 - c
In the question it has been given that "it takes him thrice as long to row up as to row down the river". This means the time taken for him to row up is 3 times the time taken for him to row down. In other words, his overall speed when rowing down will be 3 times greater than his overall speed when rowing up.

Let the time taken for upstream be Tup and for downstream be Tdown.
Tup = 3 x Tdown
Tdown / Tup = 1 / 3
Speed Up = Distance / Time Up
Speed Down = Distance / Time Down
Speed Up / Speed Down = Time Down / Time Up

8.5 - c / 8.5 + c = 1 / 3
8.5 + c = (8.5 - c) 3
8.5 + c = 25.5 - 3c
c + 3c = 25.5 – 8.5 = 17
4c = 17
c = 4.25 km/hr

Question 3

In a software company employees arranged for a picnic. At the planning stage it was found that per employee share of the total cost would be same as the number of enlisted employees. However, on the day of picnic the number of employees who were present was 20 less than the enlisted number. Consequently, per employee share shot up by 20%. What was the number of employees who had enlisted initially? How much did each employee pay finally?

a) 120, Rs.172.80 b) 100, Rs.172.80 c) 120, Rs.144 d) none of these.

Answer : c) 120, Rs.144

Solution :

Let the number of employees initially be n
From the question, the Cost per Employee = n (as it is given that "employee share of the total cost would be same as the number of enlisted employees")
Hence total cost = Cost per Employee x Number of Employees = n x n = n²
Twenty employees did not turn up and hence those participating in picnic = (n – 20). Hence the cost of n² has to be shared by the present (n - 20) employees.
Total cost per employee will be n² / (n-20)
This is to be given to be 20% more than the original figure.

So, n² / (n-20) = (120/100) n
n² = 1.20 n (n-20)
Dividing by n on both sides,
n = 1.2 (n - 20)
n = 1.2n - 24
24 = 0.2n
120 = n
Initial amount planned to be paid by each employee = n = Rs.120
Final and actual amount paid by each employee = n² / (n-20) = 14400/100 = Rs. 144.