1363.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.
than B to reach the destination. The time taken by A to reach the destination is.
5 hrs
3 hrs
2 hrs
4 hrs
Explanation:
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 = ½ hr
Time taken by A to reach the destination is 4x = 4 * ½ = 2 hr
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 = ½ hr
Time taken by A to reach the destination is 4x = 4 * ½ = 2 hr
1364.A person covers a certain distance at 7kmph .How many meters does he cover in 2
minutes
minutes
1000m
1500m
2400m
2000m
Explanation:
speed=72kmph=72*5/18 = 20m/s
distance covered in 2min =20*2*60 = 2400m
speed=72kmph=72*5/18 = 20m/s
distance covered in 2min =20*2*60 = 2400m
1365.If a man runs at 3m/s. How many km does he run in 1hr 40 min
19km
17km
18km
16km
Explanation:
speed of the man = 3*18/5 kmph = 54/5kmph
Distance covered in 5/3 hrs=54/5*5/3 = 18km
Distance covered in 5/3 hrs=54/5*5/3 = 18km
1366.Walking at the rate of 4knph a man covers certain distance in 2hr 45 min. Running at a
speed of 16.5 kmph the man will cover the same distance in.
speed of 16.5 kmph the man will cover the same distance in.
20mins
25mins
30mins
40mins
Explanation:
Distance=Speed* time 4*11/4=11km
New speed =16.5kmph
therefore Time=D/S=11/16.5 = 40min
Distance=Speed* time 4*11/4=11km
New speed =16.5kmph
therefore Time=D/S=11/16.5 = 40min
1368.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.
40 min less. If he had moved 2kmph slower he would have taken 40min more. the
distance is.
20
25
30
40
Explanation:
Let distance = x m
Usual rate = y kmph
x/y – x/y+3 = 40/60 hr
2y(y+3) = 9x (equation 1)
x/y-2 – x/y = 40/60 hr y(y-2) = 3x (equation 2)
divide 1 & 2 equations
by solving we get x = 40
Let distance = x m
Usual rate = y kmph
x/y – x/y+3 = 40/60 hr
2y(y+3) = 9x (equation 1)
x/y-2 – x/y = 40/60 hr y(y-2) = 3x (equation 2)
divide 1 & 2 equations
by solving we get x = 40
1369.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.
45kmph.for how many min does the bus stop per hr.
5mins
25mins
20mins
10mins
Explanation:
Due to stoppages, it covers 9km less.
time taken to cover 9 km is [9/54 *60] min = 10min
Due to stoppages, it covers 9km less.
time taken to cover 9 km is [9/54 *60] min = 10min
1370.A motorist covers a distance of 39km in 45min by moving at a speed of xkmph for the
first 15min.then moving at double the speed for the next 20 min and then again moving at
his original speed for the rest of the journey .then x=?
first 15min.then moving at double the speed for the next 20 min and then again moving at
his original speed for the rest of the journey .then x=?
36kmph
44kmph
42kmph
55kmph
Explanation:
Total distance = 39 km
Total time = 45 min
D = S*T
x * 15/60 + 2x * 20/60 + x * 10/60 = 39 km
x = 36 kmph
Total distance = 39 km
Total time = 45 min
D = S*T
x * 15/60 + 2x * 20/60 + x * 10/60 = 39 km
x = 36 kmph
1371.A & B are two towns. Mr.Fara covers the distance from A to B on cycle at 17kmph
and returns to A by a tonga running at a uniform speed of 8kmph.his average speed
during the whole journey is.
and returns to A by a tonga running at a uniform speed of 8kmph.his average speed
during the whole journey is.
11.88kmph
10.88kmph
12kmph
10kmph
Explanation:
When same distance is covered with different speeds, then the average speed = 2xy / x+y
=10.88kmph
When same distance is covered with different speeds, then the average speed = 2xy / x+y
=10.88kmph
1372.A car covers 4 successive 3km stretches at speed of 10kmph,20kmph,30kmph &
60kmph resp. Its average speed is.
60kmph resp. Its average speed is.
17kmph
20 kmph
16kmph
18kmph
Explanation:
Average speed = total distance / total time
total distance = 4 * 3 = 12 km
total time = 3/10 + 3/20 + 3/30 + 3/60
= 36/60 hr
speed =12/36 * 60 = 20 kmph
Average speed = total distance / total time
total distance = 4 * 3 = 12 km
total time = 3/10 + 3/20 + 3/30 + 3/60
= 36/60 hr
speed =12/36 * 60 = 20 kmph
1373.A person walks at 5kmph for 6hr and at 4kmph for 12hr. The average speed is.
3 mph
2 1/3mph
4 1/3 mph
3 1/6mph
Explanation:
avg speed = total distance/total time
= 5*6 + 4*12 / 18
=4 1/3 mph
avg speed = total distance/total time
= 5*6 + 4*12 / 18
=4 1/3 mph
1374.A bullock cart has to cover a distance of 80km in 10hrs. If it covers half of the journey
in 3/5th time.wht should be its speed to cover the remaining distance in the time left.
in 3/5th time.wht should be its speed to cover the remaining distance in the time left.
10 kmph
12 kmph
15 kmph
16 kmph
Explanation:
Time left = 10 – 3/5*10
= 4 hr
speed =40 km /4 hr
=10 kmph
Time left = 10 – 3/5*10
= 4 hr
speed =40 km /4 hr
=10 kmph
1376.A is twice as fast as B and B is thrice as fast as C is. The journey covered by B in?
17 mins
18 mins
20 mins
15 mins
Explanation:
speed& ratio
a : b = 2: 1
b : c = 3:1
Time& ratio
b : c = 1:3
b : c = 18:54
(if c covers in 54 min i..e twice to 18 min )
speed& ratio
a : b = 2: 1
b : c = 3:1
Time& ratio
b : c = 1:3
b : c = 18:54
(if c covers in 54 min i..e twice to 18 min )
1377.A man performed 3/5 of the total journey by ratio 17/20 by bus and the remaining
65km on foot.wht is his total journey.
65km on foot.wht is his total journey.
110km
120km
130km
111km
Explanation:
Let total distance is x
x-(3/5x + 17/20 x) =6.5
x- 19x/20 = 6.5
x=20 * 6.5
=130 km
Let total distance is x
x-(3/5x + 17/20 x) =6.5
x- 19x/20 = 6.5
x=20 * 6.5
=130 km
1378.A train M leaves Meerat at 5 am and reaches Delhi at 9am . Another train N leaves
Delhi at 7am and reaches Meerut at 1030am At what time do the 2 trains cross one
Another
Delhi at 7am and reaches Meerut at 1030am At what time do the 2 trains cross one
Another
50mins
25mins
56mins
50mins
Explanation:
Let the distance between Meerut & Delhi be x
they meet after y hr after 7am
M covers x in 4hr
N covers x in 3 ½ i.e 7/2 hr
speed of M =x/4
speed of N = 2x/7
Distance covered by M in y+2 hr + Distance covered by N in
y hr is x
x/4 (y+2) +2x/7(y)=x
y=14/15hr or 56 min
Let the distance between Meerut & Delhi be x
they meet after y hr after 7am
M covers x in 4hr
N covers x in 3 ½ i.e 7/2 hr
speed of M =x/4
speed of N = 2x/7
Distance covered by M in y+2 hr + Distance covered by N in
y hr is x
x/4 (y+2) +2x/7(y)=x
y=14/15hr or 56 min
1379.An aeroplane covers a certain distance at a speed of 240 kmph in 5 hours. To cover
the same distance in 1 hours, it must travel at a speed of:
the same distance in 1 hours, it must travel at a speed of:
300 kmph
360 kmph
600 kmph
720 kmph
Explanation:
Distance = (240 x 5) = 1200 km.
Speed = Distance/Time
Speed = 1200/(5/3) km/hr. [We can write 1 hours as 5/3 hours]
Required speed =(1200 * 3/5) km/hr= 720 km/hr.
Distance = (240 x 5) = 1200 km.
Speed = Distance/Time
Speed = 1200/(5/3) km/hr. [We can write 1 hours as 5/3 hours]
Required speed =(1200 * 3/5) km/hr= 720 km/hr.
3148.The ratio between the speeds of two trains is 7 : 8. If the second train runs 400 km in 4 hours, then the speed of the first train is:
70 km/hr
75 km/hr
84 km/hr
87.5 km/hr
Explanation:
Let the speed of two trains be 7$ x $ and 8$ x $ km/hr.
| Then, 8$ x $ =$ \left(\dfrac{400}{4} \right) $= 100 |
| $\Rightarrow x $ =$ \left(\dfrac{100}{8} \right) $= 12.5 |
$\therefore$ Speed of first train = $(7 \times 12.5)$ km/hr = 87.5 km/hr.
3151.Walking 6/7th of his usual speed, a man is 12 minutes too late. What is the usual time taken by him to cover that distance?
1 hr 42 min
1 hr
2 hr
1 hr 12 min
Explanation:
New speed = 6/7 of usual speed
Speed and time are inversely proportional.
Hence new time = 7/6 of usual time
Hence, 7/6 of usual time - usual time = 12 minutes
=> 1/6 of usual time = 12 minutes
=> usual time = 12 x 6 = 72 minutes = 1 hour 12 minutes
3154.A car travelling with $ \dfrac{5}{7} $ of its actual speed covers 42 km in 1 hr 40 min 48 sec. Find the actual speed of the car.
17$ \dfrac{6}{7} $km/hr
25 km/hr
30 km/hr
35 km/hr
Explanation:
| Time taken = 1 hr 40 min 48 sec = 1 hr 40$ \dfrac{4}{5} $min = 1$ \dfrac{51}{75} $hrs =$ \dfrac{126}{75} $hrs. |
Let the actual speed be $ x $ km/hr.
| Then,$ \dfrac{5}{7} x \times \dfrac{126}{75} $= 42 |
| $\Rightarrow x $ =$ \left(\dfrac{42 \times 7 \times 75}{5 x 126} \right) $= 35 km/hr. |
3156.A man on tour travels first 160 km at 64 km/hr and the next 160 km at 80 km/hr. The average speed for the first 320 km of the tour is:
35.55 km/hr
36 km/hr
71.11 km/hr
71 km/hr
Explanation:
| Total time taken =$ \left(\dfrac{160}{64} +\dfrac{160}{80} \right) $hrs.=$ \dfrac{9}{2} $hrs. |
| $\therefore$ Average speed =$ \left(320 \times\dfrac{2}{9} \right) $km/hr= 71.11 km/hr. |
3157.An aeroplane covers a certain distance at a speed of 240 kmph in 5 hours. To cover the same distance in 1$ \dfrac{2}{3} $ hours, it must travel at a speed of:
300 kmph
360 kmph
600 kmph
720 kmph
Explanation:
Distance = $(240 \times 5)$ = 1200 km.
Speed = Distance/Time
Speed = 1200/$\dfrac{5}{3}$ km/hr. [We can write 1$\dfrac{2}{3}$ hours as 5/3 hours]
| $\therefore$ Required speed =$ \left(1200 \times\dfrac{3}{5} \right) $km/hr= 720 km/hr. |