Latest TCS Aptitude Question SOLUTION: John leaves Town A. at 12:02 PM and drives to Town B. a distance of 120 miles. His average speed for the first 60 miles was 50 mph and for the final 60 miles it was 40 mph. Jim lea

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John leaves Town A. at 12:02 PM and drives to Town B. a distance of 120 miles. His average speed for the first 60 miles was 50 mph and for the final 60 miles it was 40 mph. Jim leaves Town A. at 12:20 PM and also drives to Town B. What must Jim's average speed be in order to arrive at Town B at the same time that John arrives?

Asked In TCS padmasri (10 years ago)

Solved Vivek Jain Read Solution (11)

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Read Solution (11) :

suneetha  (10 years ago)

john avg speed=(2*40*50)/(40+50)=44.44 mph
so, john will take 120/44.44=2.7 hours
so jim will take 2.7-0.3=2.4 hours
so his avg speed=120/2.4=50 mph

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ANIKET BAG  (10 years ago)

We know, Time= Distance/speed
Now here time is constant as both of them require the same time
Let Jim's speed be x
Therefore, John's total time= Jim's total time
or, (60/50)+(60/40)=(18/60)+(120/x)
or, x= 50 mph

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Vivek Jain  (10 years ago)

If a body covers part of the journey at speed p and the remaining part of the journey at a speed q and the distances of the two parts of the journey are in the ratio m : n, then the average speed for the entire journey is:
= (m+n) pq / (mq+np)


here m:n:1:1
p=50mph and q=40mph

so in order to arrive at Town B at the same time that John arrives, jim has to maintain a speed of 44.44mph or 44 4/9 mph

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Vivek Jain  (10 years ago)

Jim also has to consider 18 min lag. so yes 120/2.4 = 50 mph is the answer

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Bhargavi  (10 years ago)

y minus 0.3 ?? Suneetha

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Div  (10 years ago)

@Bhargavi
18/60=0.3

Is this solution Helpfull? Yes (3) | No (2)  

saipraveen  (10 years ago)

@bhargavi
john leaves frm town a at 12:02 and jim leavs same at 12:20 so dere is a time diff betwn dem is abot 18min
sin order to conver th min to hours we have 18/60=13/10 which is =0.3
by eliminatng th 0.3 hours we get the tym with respect to the john tym there by we can calculate jim speed i.e (spd=dist/tym)..

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premal  (9 years ago)

According to me the solution should be----
Total time taken by John is = 60/50 + 60/40 = 2.7h
and then we calculate the average speed of JIM we assume the speed of JIM be xmph
2.7= 120/x
x=44.4mph
I HOPE IT CAN BE HELPFUL TO U .......

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renuka  (9 years ago)

Bhargavi 0.3 is diff b/w 12:02&12:20 is 18 minutes by converting into hours i.e
18/60=0.3

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Siva Prasad Madugula  (9 years ago)

John=(12:02-2:44)
Jim should travel 120miles n 124 mnts
therefore 61.0333 mph

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RM ARUNACHALAM  (4 years ago)

John takes 72 minutes to drive the first 60 miles and 90 minutes to drive the next 60 miles or a total of 162 minutes (2.7 hours) for the trip. John arrives at Town B. at 2:44 PM. Jim leaves 18 minutes later and thus must drive the 120 miles in 2 hrs and 24 minutes (2.4 hours)in order to arrive at 2:44 PM. Thus, 120 miles divided by 2.4 hours equals 50 mph.

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