Latest TCS Aptitude Question SOLUTION: A and B start from their house at 1Oam. They travel from their house on MG road at 2OKmph and 4Okmph These is a T junction on their path. A turns left at the T junction at 12:00

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A and B start from their house at 1Oam. They travel from their house on MG road at 2OKmph and 4Okmph These is a T junction on their path. A turns left at the T junction at 12:00 noon B reaches the T
junction earlier and turns right. Both of them continue travelling till 2 :OOpm What is the distance between A and B at 2:OOpm


150 km /120 km /160 km/ 140 km

Asked In TCS dhritiman (8 years ago)

Unsolved Read Solution (8)

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Read Solution (8) :

Naresh P  (8 years ago)

By the data Distance from starting point to T is : 20*2 = 40 kms
Therefore time taken by B to reach T : 40km/40hmph= 1hr
So from 11:00AM to 2:00 PM B travels the distance : 40*3 = 120kms
from 12:00noon to 2:00PM A travels the distance 20*2 = 40kms
So total distance is 160 kms

Is this solution Helpfull? Yes (11) | No (1)  

gec  (8 years ago)

easiest way with no confusion..
1) A started at 10 am at 12 noon he covers 40 k.ms (i.e covers 20 km per hour)
B started at 10 am at 12 noon he covers 80 k.ms (i.e covers 40 km per hour)
so at 12 noon distance between them is 80-40= 40 k.m ( since same direction)
2) now A started at 12 p.m at 2 p.m he covers 40 k.m
B started at 12 p.m at 2 p.m he covers 80 k.m
since they are travelling in opposite direction there distance increases i.e added
so distance b/n them is 40+80=120.
so total distance between them is ( distance between them at 12p.m + distance between them at 2 p.m)
finally it becomes 40+120=160

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SURAJ  (8 years ago)

160km......both started at 10 o clock then b reches at junction T at 11.00....hence B cover 40*3= 120 distance to its right upto 2.00........and A reach junction T at 12.00 cover 40 km....after that he turn left and cover 20*2=40 km distance....hence both are 120+40=160km from each other.

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devi vara prasad mudili  (8 years ago)

Ans) 160km
because A speed is 20km per 1hr so he would be reached T jn. at 12.00 noo ie A travels 40km for 2 hours
similarly B speed is 40km per hour so he reached T jn. at 11.00 am so he reached the total distance is from T jn. is 120km and A traveled distance is from T jn. is 40km so total distance between A & B is (20+20)+(40+40+40)=160km

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brijesh borad  (8 years ago)

Ans is 160km.
A reaches T junction at 12:00 noon at speed of 20kmph but B has 40kmph(2*20) so B will reach in half time of A at T junction so at 11:00 noon B reach at T junction. Both are travels till 2 pm so A travel aat left for 2 hours and B travel at right for 3 hours.
So that A travels 40km (2 hours at 20kmph) at left of T junction and B travels 120km (3 hours at 40kmph) at right.
So total distance 40km+120km=160km.

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Pankaj Kumar  (8 years ago)

A,B(A's speed= 20 KMPH & B's speed= 40KMPH) at 10am-------------------12am-------------------
at 12 noon A turn left , and distance covered 40km
B also cover 80km , 40km on straight and 40km on right way and another 2 hr also B cover 80km ,,,,, total is 40+80=120
A cover 40km on left way
hence on straight line total distance is 120+40= 160 ans

Is this solution Helpfull? Yes | No (2)  

Vigneswaran  (7 years ago)

A starts from starting point at 10 am.both travelled in same road.after 2 hours A could have taken left..in 2 hours A would have completed 40 KMS..so the distance between A and junction is 40..so B also travelled 40 KMS..and extra for 3 hours and completed 120 KMS after turn..now A is 40KMS apart from junction..and B is 120 KMS from junction.. toatal distance will be 40+120=160kms

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Anusha  (4 years ago)

from 12 to 2,the time diff is 2hrs.
d1=t1*s1=2*40=120
d2=t2*s2=2*20=40
D=d1+d2=120+40=160km.

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