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TIME & DISTANCE PROBLEMS - APTITUDE QUESTIONS






 1)A train covers a distance in 50 min ,if it runs at a speed of 48kmph on an average.The  speed at which the train must run to reduce the time of journey to 40min will be.

Answer and Solution :
Time=50/60 hr=5/6hr
Speed=48mph
distance=S*T=48*5/6=40km
time=40/60hr=2/3hr
New speed = 40* 3/2 kmph= 60kmph




2)Vikas can cover a distance in 1hr 24min by covering 2/3 of the distance at 4 kmph and the rest at 5kmph.the total distance is?


Answer and Solution :
 Let total distance be S
total time=1hr24min
A to T :: speed=4kmph
diistance=2/3S
T to S :: speed=5km
distance=1-2/3S=1/3S
21/15 hr=2/3 S/4 + 1/3s /5
84=14/3S*3
S=84*3/14*3
= 6km




3)walking at ¾ of his usual speed ,a man is late by 2 ½ hr. the usual time is.


Answer and Solution :
 Usual speed = S
Usual time = T
Distance = D
New Speed is ¾ S
New time is 4/3 T
4/3 T – T = 5/2
T=15/2 = 7 ½




4)A man covers a distance on scooter .had he moved 3kmph faster he would have taken 40 min less. If he had moved 2kmph slower he would have taken 40min more.the distance is.
 


Answer and Solution :

 Let distance = x m
Usual rate = y kmph
x/y – x/y+3 = 40/60 hr
2y(y+3) = 9x ————–1
x/y-2 – x/y = 40/60 hr y(y-2) = 3x —————–2
divide 1 & 2 equations
by solving we get x = 40




5)Excluding stoppages,the speed of the bus is 54kmph and including stoppages,it is 45kmph.for how many min does the bus stop per hr.


Answer and Solution :
Due to stoppages,it covers 9km less.
time taken to cover 9 km is [9/54 *60] min = 10min





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