Latest TCS Aptitude Question SOLUTION: a student reached his school late by 20mins by travelling at a speed of 9kmph.if he had travelled at a speed of 12kmph,he would have reached his school 20mins early.what is the di

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a student reached his school late by 20mins by travelling at a speed of 9kmph.if he had travelled at a speed of 12kmph,he would have reached his school 20mins early.what is the distance between house and school?

Asked In TCS ratna (11 years ago)

Solved lala Read Solution (6)

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Read Solution (6) :

lala  (11 years ago)

distance = distance either student reaches school before time or after time
9(t+20/60) = 12(t-20/60)
i.e
9(t+1/3) = 12(t-1/3)
9(3t+1) = 12(3t-1)
t=7/3
put t value in any of distance equation
put t= 7/3 in 9(t+1/3)
we get 24 km

Is this solution Helpfull? Yes (58) | No (4)  

NILOY SUTRADHAR  (11 years ago)

let the distance is x,
so,(x/9-x/12)=(20+20)/60 [as 20 min earlier and 20min late]
solving,x=24

Is this solution Helpfull? Yes (17) | No (3)  

Rahul singh  (11 years ago)

ans=24km.
sol:
9=D/(t+20/60): 20 mins converted into hours by dividing 60!
&
12=D/(t-20/60)
now
by solving this we will get
t=7/3 hours
now
D=9*{(7/3)+(20/60)}
D= 24 km

Is this solution Helpfull? Yes (10) | No   

sandeep  (11 years ago)

i dont think it wud be ever asked in tcs.....

Is this solution Helpfull? Yes (7) | No (8)  

alpesh kurhade  (11 years ago)

d/9-d/12=40/60
solving ,3d=2*9*12
so, d=24

Is this solution Helpfull? Yes (6) | No   

Amit Bundela  (11 years ago)

Let, the actual time be x;
student reach 20min late,so time=(x+(20/60))
so,distance=9*(x+(20/60))...(1)
again,at speed rate of 12km/hr we have,
student reach 20min early,so time=(x-(20/60))
so,distane=12*(x-(20/60))...(2)
from eqn 1& 2,
9*(x+1/3)=12*(x-1/3)
27x+9=36x-12
9x=21
x=7/3 hrs...(3),actual time to reach school
thus,distance between house and school is-
d=9*(7/3+20/60)=9*(7/3+1/3)=9*(8/3)=72/3="24 km" answer

so,distance=9*(x-20/60)...(1)

Is this solution Helpfull? Yes (4) | No (3)  

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