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51. A man takes 8 hours to walk to a certain place and ride back. However, if he walks both ways he needs 2 hours more. How long would he have taken to ride both ways?
  A.  6 hours
  B.  3 hours
  C.  7 hours
  D.  5 hours
     
   
View Answer

Shortcut:
A man takes 't' hours to walk to a certain place and ride back. However, If he walks both ways heneeds t1 hours more, then the time taken by him to ride both ways is (t − t1) hours. or Both ways walking = one way walking and one way riding time + gain in time.
Here, t = 8, t1 = 2
Using these values in the shortcut, we get:
Required time = 8 − 2 = 6
Hence, he would take 6 hours.

52. A man leaves a point X and reaches the point Y in 8 hours. Another man leaves the point Y, 4 hours later and reaches the point X in 8 hours. Find the time in which first man meets to the second man.
  A.  5 hours
  B.  6 hours
  C.  7 hours
  D.  8 hours
     
   
View Answer

Shortcut:
A man M1 leaves a point X and reaches Y in h1 hours. If another man M2 leaves the point Y, t hours later than M1 and reaches the point X in h2 hours, then the time in which M1 meets to M2 is
(h2 + t)
h1 / h1 + h2

Here, h1 = 8, h2 = 8, t = 4
Using these values in the shortcut, we get:
Required time = (8 + 4)
8 / 8 + 8
= 12 x
8 / 16
= 6
Hence, he would take 6 hours.

53. A man leaves a point X and reaches the point Y in 8 hours. Another man leaves the point Y, 4 hour earlier and reaches the point X in 7 hours. Find the time in which the first man meets to the second man.
  A.  1
1 / 15
hr
  B.  2
1 / 15
hr
  C.  2
2 / 15
hr
  D.  3
2 / 15
hr
     
   
View Answer

Shortcut:
A man M1 leaves a point X and reaches Y in h1 hours. If another man M2 leaves the point Y, t hours earlier than M1 and reaches the point X in h2 hours, then the time in which M1 meets to M2 is
(h2 − t)
h1 / h1 + h2

Here, h1 = 8, h2 = 7, t = 4
Using these values in the shortcut, we get:
Required time = (8 − 4)
8 / 8 + 7
= 4 x
8 / 15
=
32 / 15
= 2
2 / 15

Hence, he would take 2
2 / 15
hours.

54. A person covers a distance in 18 minutes if he runs at a speed of 25 km/hr on an average. Find the speed at which he must run to reduce the time of journey to 15 minutes.
  A.  20 km/hr
  B.  25 km/hr
  C.  30 km/hr
  D.  40 km/hr
     
   
View Answer

Shortcut:
Speed and time taken are inversely proportional. Therefore, S1T1 = S2T2 = S3T3 and so on Where S1, S2, S3 ... are speed and T1, T2, T3 ... are the time taken to travel the same distance.
Here, S1 = 25, T1 = 18, S2 = ?, T2 = 15
Using these values in the shortcut, we get:
25 x 18 = S2 x 15
S2 =
25 x 18 / 15
= 30
Hence, the speed is 30 km/hr.

55. Without any stoppage a person travels a certain distance at an average speed of 40 km/hr, and with stoppages he covers the same distance at an average speed of 30 km/hr. How many minutes per hour does he stop?
  A.  15 minutes
  B.  20 minutes
  C.  25 minutes
  D.  18 minutes
     
   
View Answer

Shortcut:
Without any stoppage if a person travels a certain distance at an average speed of a km/hr, and with stoppages he covers the same distance at an average speed of b km/hr, then he stops 60 x
a − b / a
minutes per hour. or Time of rest per hour = 60 x
Difference of speed / Speed without stoppage
minutes.
Here, a = 40, b = 30
Using these values in the shortcut, we get:
Required stops in minutes = 60 x
40 − 30 / 40

= 60 x
10 / 40
= 15
Hence, he will take rest of 15 minutes.

56. Ram has to cover a distance of 40 km in 5 hours. If he covers half of the journey in 3/5th time, what should be his speed to cover the remaining distance in the time left?
  A.  12 km/hr
  B.  14 km/hr
  C.  20 km/hr
  D.  10 km/hr
     
   
View Answer

Shortcut:
A man has to cover a distance of 'a' km in 't' hours. If he covers nth part of the journey in mth of the total time, then his speed should be
a / t
x
1 − n / 1 − m
km/hr to cover the remaining distane in the remaining time.
Here, a = 40, t = 5, n =
1 / 2
, m =
3 / 5

Using these values in the shortcut, we get:
Required speed =
40 / 5
x
1 − 1/2 / 1 − 3/5

= 8 x
(2 − 1)/2 / (5 − 3)/5
= 8 x
1/2 / 2/5

= 8 x
1 x 5 / 2 x 2
= 2 x 5 = 10
Hence, his speed will be 10 km/hr.

57. Deepu has to cover a distance of 50 km in 10 hours. If he covers
2 / 5
of the journey in
4 / 5
th time, what should be his speed to cover the remaiining distance in the time left?
  A.  15 km/hr
  B.  10 km/hr
  C.  25 km/hr
  D.  18 km/hr
     
   
View Answer

Shortcut:
A man has to cover a distance of 'a' km in 't' hours. If he covers nth part of the journey in mth of the total time, then his speed should be
a / t
x
1 − n / 1 − m
km/hr to cover the remaining distane in the remaining time.
Here, a = 50, t = 10, n =
2 / 5
, m =
4 / 5

Using these values in the shortcut, we get:
Required speed =
50 / 10
x
1 − 2/5 / 1 − 4/5

= 5 x
(5 − 2)/5 / (5 − 4)/5
= 5 x
3/5 / 1/5

= 5 x
3 x 5 / 1 x 5
= 5 x 3 = 15
Hence, his speed will be 15 km/hr.

58. A man rode out a certain distance by train at the rate of 20 km/hr and walked back at the rate of 5 km/hr. The whole journey took 5 hours. What distance did he ride?
  A.  20 km
  B.  10 km
  C.  15 km
  D.  25 km
     
   
View Answer

Shortcut:
A person rode out a certain distacne by train at the rate of a km/hr and walked back at the rate of b km/hr. If the whole journey took 't' hours, then the distance he rode is
ab / a + b
x t
Here, a = 20, b = 5, t = 5
Using these values in the shortcut, we get:
Required distance =
20 x 5 / 20 + 5
x 5
=
100 / 25
x 5 = 4 x 5 = 20
Hence, the distance will be 20 km.

59. A man travels 180 km in 3 hours, partly by air and partly by train. If he had travelled all the way by air, he would have saved
2 / 5
of the time he was in train and would have arrived at his destination 1 hour early. Find the distance he travelled by air and train.
  A.  25 km
  B.  45 km
  C.  35 km
  D.  55 km
     
   
View Answer

Shortcut:
A person travels Z km in h1 hours, partly by air and partly by train. If he had travelled all the way by air, he would have saved m/n of the time he was in train and would have arrived at his destination h2 hours early, then the distance he travelled by air is Z x
h1 − (h2)/(m/n) / h1 − h2
km
Here, Z = 180, h1 = 3,
m / n
=
2 / 5
, h2 = 1
Using these values in the shortcut, we get:
Required distance = 180 x
3 − (1)/(2/5) / 3 − 1

= 180 x
3 − 5/2 / 2

= 180 x
(6 − 5)/2 / 2

= 180 x
1/2 / 2

= 180 x
1 / 4
= 45
Hence, the distance he travelled is 45 km.

60. One aeroplance started 30 minutes later than the scheduled time from a place 750 km away from its destination. To reach the destination at the scheduled time the pilot had to increase the speed by 125 km/hr. What was the speed of the aeroplance per hour during the journey?
  A.  325 km/hr
  B.  275 km/hr
  C.  475 km/hr
  D.  375 km/hr
     
   
View Answer

Shortcut:
One aeroplane starts h hours later than the scheduled time from a place Z km away from its destination. To reach the destination at the scheduled time the pilot has to increase the speed by 'a' km/hr. Then the plane takes
√(h2 + 4Zh/a) + h / 2
hours in the normal case.
And the normal speed of the aeroplane is
2Z / √(h2 + 4Zh/a) + h
km/hr.
Here, h =
30 / 60
, Z = 750, a = 125
Using these values in the shortcut, we get:
Required speed =
2 x 750 / √[(1/2)2 + (4 x 750 x 1/2)/(125)] + 1/2

=
2 x 750 / √[(1/4) + (2 x 750)/125] + 1/2

=
2 x 750 / √[(1/4) + (2 x 6)] + 1/2

=
2 x 750 / √[1/4 + 12] + 1/2

=
2 x 750 / √[(1 + 48)/4] + 1/2

=
2 x 750 / √(49/4) + 1/2

=
2 x 750 / 7/2 + 1/2

=
2 x 750 / (7 + 1)/2

=
2 x 750 / 4
= 2 x 187.5 = 375
Hence, the speed is 375 km/hr.

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