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.
A. 63 kmph
B. 74 kmph
C. 60 kmph
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?
A.10 km
B. 6.5 km
C. 4 km
D. 6 km
3. Walking at ¾ of his usual speed ,a man is late by 2 ½ hr. The usual time is.
A. 7 ½
B. 7
C. 8 ½
D. 10
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.
A. 36
B. 40
C. 80
D.41
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.
A. 15
B. 10
C. 9.5
D. 4
6.
Two boys
starting from the same place walk at a rate of 5kmph and 5.5kmph respectively. What
time will they take to be 8.5km apart, if they walk in the same direction.
A. 17 hrs
B. 18 hrs
C. 15 hrs
D. 10hrs
7. 2 trains starting at the same time from 2 stations 200km apart and going in opposite direction cross each other at a distance of 110km from one of the stations. What is the ratio of their speeds.
A. 11:6
B. 6:11
C. 11:9
D. 9:11
8. Two trains start from A & B and travel towards each other at speed of 50kmph and 60kmph respectively. At the time of the meeting the second train has traveled 120km more than the first. The distance between them.
A. 1200 km
B. 1300 km
C. 1321 km
D. 1320 km
9. A thief steals a ca r at 2.30pm and drives it at 60kmph.the theft is discovered at 3pm and the owner sets off in another car at 75kmph when will he overtake the thief.
A. 5 pm
B. 4 pm
C. 5.5 pm
D. 9 pm
10. In covering distance, the speed of A & B are in the ratio of 3:4. A takes 30min more than B to reach the destination. The time taken by A to reach the destination is.
A. 2.5 hr
B. 2 hr
C. 3 hr
D. 9 hr
Solutions –
1. 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. Solution
Let
total distance be S
total
time=1hr24min
A to T
:: speed=4kmph
distance=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. 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.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.Solution
Due to
stoppages, it covers 9km less.
time
taken to cover 9 km is [9/54 *60] min = 10min
6.Solution
The
relative speed of the boys = 5.5kmph – 5kmph = 0.5 kmph
Distance
between them is 8.5 km
Time= 8.5km
/ 0.5 kmph = 17 hrs
7. Solution
In
same time ,they cover 110km & 90 km respectively
so
ratio of their speed =110:90 = 11:9
8. Solution
Let
the distance traveled by the first train be x km
then
distance covered by the second train is x + 120km
x/50 =
x+120 / 60
x= 600
so the
distance between A & B is x + x + 120 = 1320 km
9. Solution
Let
the thief is overtaken x hrs after 2.30pm
distance
covered by the thief in x hrs = distance covered by
the
owner in x-1/2 hr
60x =
75 ( x- 1/2)
x= 5/2
hr
thief
is overtaken at 2.30 pm + 2 ½ hr = 5 pm
10. Solution
Ratio
of speed = 3:4
Ratio
of time = 4:3
let A
takes 4x hrs, B takes 3x hrs
then
4x-3x = 30/60 hr
x =
1/2 hr
Time taken by
A to reach the destination is 4x = 4 * 1/2 = 2 hr
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