Aptitude

[Triton Networking Academy]

Dear Reader,
Below are three different problems dealing with time, speed and distance calculations.

Question 1

A car driver starts from home and takes 345 minutes in driving to office and returning home. The driver would have gained 120 minutes if he travels at the speed of return journey in both ways. Then the time he would take if he use his forward journey's speed both ways, is:

a)7 hrs & 45mins b)6 hrs & 12min c)7 hrs d)8 hrs & 15mins

Answer : a) 7 hrs & 45 mins

Solution :

Let the distance between the office and home be X km.
Note,
Time taken to reach office = time taken for his forward journey of X km.
Time taken to reach home = time taken for his return journey of X km.

Given, he takes 345 minutes for his forward and return journey.
i.e., he takes 345 minutes = 345/60 hours = 23/4 hours
Then, time taken for his forward journey of X km + time taken for return journey of X km = 23/4 hours
Multiply the above by 2 on both sides,
we have, time taken for his forward journey of 2X km + time taken for return journey of 2X km = 23/2 hours...(1)

It is given in question that he would gain 2 hours (120 mins) if he travels at return journey's speed on both ways.
Then he takes 345 - 120 = 255 minutes for his forward and backward journey
That is, he takes 255 minutes (15/4 hours) for his forward journey of 2X km
From (1), 15/4 + time taken to return journey of 2X km = 23/2 hours
Time taken to return journey of 2X km = 23/2 hours - 15/4 hours = 31/4 hours = 31/4 x 60 minutes = 465 minutes = 7 hours 45 minutes.

Hence he takes 7 hours 45 mins if he use his forward journey's speed on both ways.

Question 2

A man leaves office daily at 7pm. A driver with car starts at 4.30pm, comes from the man's home to pick the man at 7 pm from office and bring back home. One day, the man leaves office at 5.30 pm ,instead of waiting for driver he takes auto and reaches home at 7.30 pm. In the way he meets the car at:

a) 5.40 pm b) 6.30 pm c) 5.50 pm d) 6.20 pm

Answer : a) 5.40 pm

Solution :

Let the distance between the office and home be X km.
Time taken by the driver to cover X km = 4.30 to 7 pm = 5/2 hours
Time taken by the man to cover X km by auto = 5.30 pm + 7.30 pm = 2 hours
Therefore, speed of the driver = X/(5/2) km/hr = 2X/5 (distance/time)
And, the speed of the auto = X/2 km/hr.

Let them meet y hours after 4.30 pm
Then, the distance travelled by the driver in y hours = 2X/5 x y = 2Xy/5 km
And, the distance travelled by the man in y-1 hours = X/2 x y-1 = X(y-1)/2 km
2Xy/5 + X(y-1)/2 = X
4Xy + 5X(y-1) = 10X
9Xy = 15X
y = 15/9 = 5/3 = 1 + 2/3
That is, y = 1 hr 40 mins.

Since the driver starts at 4.30pm, they meets at 5.40 pm.

Question 3

A man leaves office daily at 7 pm. A driver with car starts at 5 pm and moves at 30km/hr, comes from the man's home to pick the man from office and bring back home. One day he gets free at 6 pm and instead of waiting for his driver, he starts walking towards home at a speed of 6 km/hr. In the way he meets the car and returns home on car. At what time he reaches home if the distance between the office and home is 60 km?

a) 8.40 pm b) 7.30 pm c) 7.50 pm d) 8.20 pm

Answer : a) 8.40 pm

Solution :

The driver starts at 5 pm at 30 km/hr
The man starts at 6 pm and walks at 6km/hr
Distance between office and home = 60km

Let they meet X hours after 5 pm.
Distance covered by driver in X hours= 30X km = (speed x time)
Distance covered by man in X-1 hours = 6(X-1) km (since he starts at 6 pm, we calculate for X-1 hours)

Therefore, 30X + 6(X-1) = 60
30X + 6X - 6 = 60
30X + 6X = 66
X = 66/36 = 11/6 hours = 1 hour 50 minutes.

Hence, they meet 1 hr 50 mins after 5 pm, i.e., they meet at 6.50 pm.
Now, the driver takes 1 hr 50 mins to reach X km.
Therefore, he picks the man at 6.50 pm and returns to home by taking 1 hr 50 mins after 6.50 pm.
Hence the required time = 6.50 pm + 1 hr + 50 mins = 8.40 pm.