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Numerical Ability
Time Distance and Speed
A fast train takes 3 hours less than a slow train for a journey of 900 km, If the speed of the slow train is 15 km/hr less than that of the fast train, the speeds of the slow train is (1) 50 km/hr (2) 75 km/hr (3) 60 km/hr (4) 45 km/hr
Read Solution (Total 6)
-
- Let the speeds are =x,x-15
900/x-15 - 900/x= 3
300x-300x+4500=x^2-15x
x^2-15x -4500=0
(x-75)(x+60)=0=>x=75kmph
Slower train speed=75-15=60kmph - 4 years agoHelpfull: Yes(5) No(1)
- Let speed taken by slow train be x
thus, speed taken by fast train be (x+15)
time by slow train=900/x
time by fast train=900/x+15
time taken by slow train-time taken by fast train=3
900/x-900/x+15=3
by cross multiplication we get,
900x
13500-900x/x(x+150)=3
by solving these equation we get,
(x-50)(x-45)=0
x=50 x=45
speed of slow train is x-15= 50-15=45
thus the ans is option 4=45km/hr - 5 years agoHelpfull: Yes(1) No(2)
- Let the time taken by the slow train be x hr
A/c to question,
the time taken by the fast train is 3 hr less than slow train
i.e time taken by the fast train is (x-3) hr
Let the speed of the fast train be y km/hr
then,
A/c to question,
the speed of the slow train is (y-15) km/hr
we are given with the distance covered is 900 km
we know that,
[DISTANCE=SPEED *TIME]
from above formula;
(y-15)*x = (x-3)y
yx-15x = xy-3y
i.e
y=5x
putting the value of y=5x in speed of slow thain ,we get
x(5x-15)=900 [distance=speed*time]
...........
...........
solving the equation ,we get
x(x-15) + 12(x-15)=0
x = -12 (not possible)
x = 15 (possible)
putting the value of x in above equation ,we get
y = 15 * 5
i.e y = 75
now,
putting the value of y in the speed of slow train, we get
speed of slow train = (y - 15) km/hr
=(75 - 15) = 60 km/hr
i.e
speed of slow train is 60 km/hr - 4 years agoHelpfull: Yes(1) No(1)
- simply trail for the each option and check the time taken for slow train and fast train differnce should be 3 hours.(Answers is 60km/hr)
- 4 years agoHelpfull: Yes(0) No(0)
- 3. 60 km/hr
- 3 years agoHelpfull: Yes(0) No(0)
- As per Questions
distance for both=900km
let speed of fast train=x.speed of slow train=x-15
let time for slow train =t.time for fast train=t-3
As distance is same for both so equating we get
x(t-3)=(x-15)t
here we get ratio of x/t=5
as distance is 900km
so substitute in any eequation we get t=15
so x becomes 75 which is speed of fast train.
therefore speed of slow train is 75-15=60 - 3 years agoHelpfull: Yes(0) No(0)
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