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A Problem on Time and Distance

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by beshtesh » Mon Oct 06, 2008 3:13 am
Can anyone solve this problem for me?

To travel 600 Km , train x takes 8 hours more than train y. If however, the speed of the train x is doubled, it takes 2 hours less than train y. What is the speed of the train y?
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by vvnaveen » Mon Oct 06, 2008 5:46 am
ans: speed of y = 50 km/hr

speed of x be s1 and speed of y be s2
let 't' be the time taken by x at speed s1.

s1 is proportional to 1/t

so if s1 is doubled then, time t will be halved.

With s1 speed , it takes 8 hrs more than y and with 2s1 speed it takes 2 hrs less.

So the diff in time = 10 hrs = 0.5t (as time will be halved - this is the key idea)
therefore, t= 20 hrs

[S1 = D/t = 600/20 = 30 km/hr ](not needed to calculate. but 2 cross check we need)
s2(speed of y) = D/(t - 8) = 600/(20 - 8) = 50 km/hr.

Check:
Speed -Distance- Speed - Time
S1 - 600 km - 30 km/hr - 20 hrs
S2 - 600 km - 50 km/hr - 12 hrs
diff - 8 hrs
S1(doubled) - 600km - 60 km/hr - 10 hrs
diff - (12-10= 2hrs)
Done..
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Thank You

by beshtesh » Mon Oct 06, 2008 8:36 am
Thanks Naveen

beshtesh
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Explanation

by SYim » Tue Oct 07, 2008 10:36 am
Here's another way to solve this problem.

1. Set up the rate, time, distance:
rate/ time/ distance
X1 (slower): Rx (rate for x) /8+h (8 hrs more than y) / 600
Y : Ry (rate for y) / h / 600
X2 :(faster) 2Rx (when rate doubles) / h-2 (2 hrs less) / 600

The distance will be all the same because they are all traveling the same distance; 600 miles.

2. Since the distance is the same for both X1 and X2, you can write an equation: X1 = X2 --> Rx(8+h) = 2Rx(h-2) --> divide both by Rx -->
8+h = 2(h-2) --> You can find h --> h=12

3. Knowing that h=12, you can solve the formula D=RT for Y
600=Ry x 12 --> 600/12 --> Ry=50

Hope this helps-
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by vivek.kapoor83 » Tue Oct 07, 2008 11:13 am
i think more simpler if we do...
speed of x =x
speed of y =y
Acc to 1st cond. (600/x)+8 = 600/y
Acc to 2nd cond. (600/2x)-2=600/y
Solving for x & y ...above eqn
u will get y=50.
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