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Usual time

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by liferocks » Wed May 05, 2010 7:32 am
Since distance is constant time is inversely propositional to speed
since the person is going at 5/6 of its usual speed time taken will be 6/5 of usual time
i.e 1/5th of usual time is the extra time taken

hence 1/5 th of usual time is 10min
so the usual time to cover the journey is 5*10=50minutes

Ans option C
"If you don't know where you are going, any road will get you there."
Lewis Carroll
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by Ashish8 » Wed May 05, 2010 9:09 am
I used a combination of backward substitution and picking numbers.

V = d/t

I picked 100 as my distance since all the times were also divisible by 10.

Lets start with C (Since its in the middle)

V = 100/50 = 2

If the train is going 5/6 its usual speed then its speed becomes 5/3.

5/3 = 100/t --> t = 60, which is 10 mins late. Answer = C
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by gmatjedi » Sat May 15, 2010 12:20 pm
algebraic solution:

d=distance
r=rate
t=time

distance will be constant

d=rt
d=(5/6)r(t+10)

rt=(5/6)rt +(50/6)r
t=50 min
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by indiantiger » Sat May 15, 2010 1:23 pm
solution:

speed = distance/time
Lets consider normal speed to be s
distance to be d
time taken in the first case to be t

Now in both the cases, one with speed s and other with 5/6s speed the distance is the same (d)

first case time (t) = distance(d)/speed(s)
second case time t' = distance (d)/(5/6s)

now t' = t+10
t+10 = d/s*6/5
t=d/s-----given
d/s+10 = 6d/5s
10 = 6d/5s-d/s
10 = d/s(6/5-1)
10 = t(1/5)
t=50(C)
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