Question

X, Y and Z walks at 6, 12 and 18 km/h respectively. They start from Jhumritalaiya towards Delhi at 2, 5 and 7 p.m. respectively, when Y catches up with X, Y sends X back with a message to Z, when will Z get the message?
(a) 9.45 pm (b) 9 pm (c) 8.45 pm (d) 9.15 pm

Answer 1

Answer:

8 : 45 pm

Step-by-step explanation:

We analyze the question as follows :

X : speed = 6 km/h.

Time of departure = 2 pm

Y : speed = 12 km/h

Time of departure = 5pm

Z : speed = 18 km/h

Time of departure = 7 pm

By the time Y leaves the distance covered by x will be :

Time = 5 pm - 2 pm = 3 hrs

Distance covered = 3 × 6 = 18 km

Let the point where x meets y be C and the distance from the point where x is to that point be d then:

Y covers = (d + 18) km

x covers = d km

They take the same time to reach C and therefore :

d/6 = (18 + d) / 12

12d = 6(18 + d)

12d = 108 + 6d

6d = 108

d = 18 km

The time taken for y to meet x is therefore :

18/6 = 3 hrs

The time will be :

5pm + 3hrs = 8 pm.

At this time Z has traveled for 1 hour covering a distance of = 1 × 18 = 18 km

The distance between X and the starting point is : 18 + 18 = 36 km

But Z has already covered 18 km.

The distance between them is therefore 18 km.

Let the meeting point of Z and C be R and the distance between Z and the meeting point be s then the distance between X and the meeting point will be (18 - s) km.

Now :

s/18 = (18 - s) / 6

6s = 18(18 - s)

6s = 324 - 18s

6s + 18s =324

24s = 324

s = 324/24

s = 13.5 km

Time taken for them to meet =

13.5/18 = 0.75 hrs

This equals to = 45 minutes

The time they meet is :

8.00 + 45 = 8:45 pm