Latest TCS Aptitude Question SOLUTION: One day ,Eesha started 30 min late from home and reached her office 50 min late while driving 25% slower than her usual speed.how much time in min Eesha usually take to reach her o

• Placement Papers
• /
• TCS
• /
• Numerical Ability
• /
• Time Distance and Speed

TCS Question Papers: TCS Interview Experience(91) TCS Company Profile

(#M40039391) TCS Numerical Ability question Keep an EYE

One day ,Eesha started 30 min late from home and reached her office 50 min late while driving 25% slower than her usual speed.how much time in min Eesha usually take to reach her office from home.

a. 80min
b. 50min
c. 60min
d. 70 min

Asked In TCS ch.harshitha (10 years ago)

Unsolved Read Solution (10)

Is this Puzzle helpful?   (23)   (5) Submit Your Solution Time Distance and Speed

Read Solution (10) :

manaswini devabhakthuni  (10 years ago)

distance remains constant.hence s*t=3*s*(t+20)/4
hence t=60

Is this solution Helpfull? Yes (27) | No (3)  

shiv kumar sharma  (10 years ago)

home---------------office(time taken =t,speed=u)
one day
time taken=t+20
speed=3u/4
t*u=(t+20)*(3u/4)
solving t=60min

Is this solution Helpfull? Yes (19) | No   

Aritra Das  (10 years ago)

25% decrease in the speed results in 33.33% increase in time due to inverse relationship of them.
so here 33.33%= 50-30 i.e 20 mins
then 100% = 20*3 =60mins

Is this solution Helpfull? Yes (5) | No (1)  

raviteja  (10 years ago)

Let the distance be D and the usual speed S. Then the time taken to travel is D/S=T(1).The new speed is (S- 1/4 S) = ¾ S. The new time taken is D/3/4 S = T + 20 (50 – 30)(20) we get 60 as answer.

Is this solution Helpfull? Yes (4) | No (6)  

PRATIK SINHA  (10 years ago)

c.60 min
100/75=t+20/t
t=60

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

AMRITA  (10 years ago)

Speed is inversely proportional to time
While she drives 25% ,i.e 1/4S ..slower means she drove at 3/4(S)

When speed became 3/4(S) then Time taken should be 4/3(T)
i.e, She has taken 4/3(T) - T extra to cover the distance.
Extra Time = T/3 = 20 min (20 min late due to slow driving)
Actual time T = 60 Minutes

Is this solution Helpfull? Yes (3) | No   

M. srikar  (10 years ago)

d=st s1*t1=s2*t2 100t=75(t+20) t=60min

Is this solution Helpfull? Yes (1) | No (3)  

dinesh  (10 years ago)

speed is 3/4 then time is 4/3 which is 1+1/3
so 1/3 of actual speed =20
speed = 60

Is this solution Helpfull? Yes (1) | No   

sapan kumar paul  (10 years ago)

ans-b because

Is this solution Helpfull? Yes | No (3)  

sandipan  (10 years ago)

25%=20
thn 100=80

Is this solution Helpfull? Yes | No (1)  

Please Login to Read Solution.