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61. A bus leaves the bus-terminal 1 hour before the scheduled time. The driver decreases its speed by 8 km/hr. At the next bus-terminal 240 km away, the train reached on time. Find the original speed of the bus.
  A.  42 km/hr
  B.  38 km/hr
  C.  24 km/hr
  D.  48 km/hr
     
   
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Shortcut:
A train leaves the station h hours before the schedule time. The driver decreases its speed by a km/hr. At the next station Z km away, the train reached on time. Then the train 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 = 1, Z = 240, a = 8
Using these values in the shortcut, we get:
Required speed =
2 x 240 / √[(1)2 + 4 x 240 x1/8] − 1

=
2 x 240 / √[1 + 4 x 30] − 1

=
2 x 240 / √[1 + 120] − 1

=
2 x 240 / √[121] − 1

=
2 x 240 / 11 − 1

=
2 x 240 / 10
= 2 x 24 = 48
Hence, the speed is 48 km/hr.

62. When a man travels equal distance at S1 and S2 km/hr, his average speed is 9 km/hr. But when he travels at these speeds for equal times his average speed is 12 km/hr. Find the difference of the two speeds and also find the values of S1 and S2.
  A.  Difference: 12 km/hr; S1: 18 km/hr; S2: 10 km/hr
  B.  Difference: 14 km/hr; S1: 12 km/hr; S2: 6 km/hr
  C.  Difference: 12 km/hr; S1: 18 km/hr; S2: 6 km/hr
  D.  Difference: 15 km/hr; S1: 18 km/hr; S2: 6 km/hr
     
   
View Answer

Shortcut:
When a man travels equal distance at speed S1 and S2 km/hr, his average speed is a km/hr. But when he travels at these speeds for equal times his average speed is b km/hr, then the values of S1 and S2 are [b + √{b(b − a)}] km/hr and [b − √{b(b − a)}] km/hr respectively. And the difference of the two speeds is 2√{b(b − a)}] km/hr.
Here, a = 9, b = 12
Using these values in the shortcut, we get:
Difference of two speed = 2√{12(12 − 9)}] = 2√[12 x 3] = 2√[36] = 2 x 6 = 12
Hence, the difference in speed is 12 km/hr.
S1 = [12 + √{12(12 − 9)}] = [12 + √{12 x 3}] = [12 + √{36}] = 12 + 6 = 18
Hence, S1 is 18 km/hr
S2 = [12 − √{12(12 − 9)}]
= [12 − √{12 x 3}]
= [12 − √{36}] = [12 − 6] = 6
Hence, S2 is 6 km/hr

63. Ravi travels for 6 hours at the speed of 80 km/hr and for 9 hours at the speed of 120 km/hr. At the end of it, he finds that he has convered
3 / 5
of the total distance. At what average speed should he travel to cover the remaining distance in 8 hours?
  A.  120 km/hr
  B.  130 km/hr
  C.  140 km/hr
  D.  136 km/hr
     
   
View Answer

Shortcut:
A man travels for h1 hours at the speed of a km/hr and for h2 hours at the speed of b km/hr. At the end of it, if he finds that he has covered 'd' of the total distance, then his average speed, to cover the remaining distance in h hours
(ah1 + bh2)(1/d − 1) / h

Here, h1 = 6, a = 80, h2 = 9, b = 120, d =
3 / 5
, h = 8
Using these values in the shortcut, we get:
Required average speed =
(80 x 6 + 120 x 9)(1/3/5 − 1) / 8

=
(480 + 1080)(5/3 − 1) / 8

=
1560 x (5 − 3)/3 / 8

=
1560 x 2/3 / 8

=
520 x 2 / 8
= 65 x 2 = 130

Hence, the average speed is 130 km/hr.

64. If Amit takes 4 hours to cover a distance and Biru is two times faster than Amit, then what time will Biru take to cover the same distance?.
  A.  2 hours
  B.  4 hours
  C.  1 hours
  D.  6 hours
     
   
View Answer

Shortcut:
If a man M1 walking at the rate of S1 km/hr, takes t1 hours to cover a distance and another person M2, walking at the rate of S2 km/hr, takes t2 hours to cover the same distance, then S1t1 = S2t2 = Distance or, S1t1 = S2t2 = constant.
Thus we see that both speed and time are inversely proportional to each other. That is, if the speed increases to 4 times, the time will decrease of
1 / 4
times.
We know, S1t1 = S2t2
Since, S2 = 2S1 (As given)
So, S1t1 = 2S1t2
⇒ t1 = 2t2
t2 =
t1 / 4
=
4 / 2
= 2
Hence, Biru will cover the same distance in 2 hours.

65. If Amit takes 4 hours to cover a distance and he is 2 times faster than Biru, then what time will Biru take to cover the same distance?
  A.  4 hours
  B.  6 hours
  C.  5 hours
  D.  8 hours
     
   
View Answer

Shortcut:
If a man M1 walking at the rate of S1 km/hr, takes t1 hours to cover a distance and another person M2, walking at the rate of S2 km/hr, takes t2 hours to cover the same distance, then S1t1 = S2t2 = Distance or, S1t1 = S2t2 = constant.
Thus we see that both speed and time are inversely proportional to each other. That is, if the speed increases to 4 times, the time will decrease of
1 / 4
times.
We know, S1t1 = S2t2
Since, S1 = 2S2 (As given)
So, 2S2 x t1 = S2 x t2
⇒ 2 x 4 = t2
or, t2 = 8
Hence, Biru will cover the same distance in 8 hours.

66. If Biru is 25% faster than Amit, then what time will he take to travel the distance which Amit travels in 25 minutes?
  A.  25 minutes
  B.  20 minutes
  C.  30 minutes
  D.  35 minutes
     
   
View Answer

Shortcut:
If a man M1 walking at the rate of S1 km/hr, takes t1 hours to cover a distance and another person M2, walking at the rate of S2 km/hr, takes t2 hours to cover the same distance, then S1t1 = S2t2 = Distance or, S1t1 = S2t2 = constant.
Thus we see that both speed and time are inversely proportional to each other. That is, if the speed increases to 4 times, the time will decrease of
1 / 4
times.
We know, S1t1 = S2t2
Since, S2 =
125 / 100
x S1 (As given)
So, S1 x 25 =
125 / 100
x S1 x t2
25 x 4 = 5 x t2
⇒ t2 =
100 / 5
= 20
Hence, Biru will cover the distance in 20 minutes.

67. If Biru takes 25% less time than Amit, to cover the same distance. What should be the speed of Biru if Amit walks at a rate of 5 km/hr?
  A.  2
2 / 3
km/hr
  B.  8
2 / 3
km/hr
  C.  6
2 / 3
km/hr
  D.  4
2 / 3
km/hr
     
   
View Answer

Shortcut:
If a man M1 walking at the rate of S1 km/hr, takes t1 hours to cover a distance and another person M2, walking at the rate of S2 km/hr, takes t2 hours to cover the same distance, then S1t1 = S2t2 = Distance or, S1t1 = S2t2 = constant.
Thus we see that both speed and time are inversely proportional to each other. That is, if the speed increases to 4 times, the time will decrease of
1 / 4
times.
We know, S1 x t1 = S2 x t2
Since, t2 =
100 − 25 / 100
x t1 (As given)
So, S1 x t1 = S2 x
75 / 100
x t1
5 x t1 = S2 x
3 / 4
x t1
⇒ S2 =
5 x 4 / 3
=
20 / 3
= 6
2 / 3

Hence, Biru speed is 6
2 / 3
km/hr.

68. Ram travelled 60 km by steamer, 270 km by train and 30 km by horse. It took 10 hours 30 minutes. If the rate of the train is 6 times that of the horse and 3 times that of the steamer, find the rate of the train per hour.
  A.  60 km/hr
  B.  50 km/hr
  C.  70 km/hr
  D.  65 km/hr
     
   
View Answer

Shortcut:
A man travels a km by steamr, b km by train and c km by horse. It took T hours. If the rate of train is 'n' times that of the horse and 'm' times that of the steamer, then the rate of the train is
ma + b + nc / T
km/hr
Rate of horse is
1 / n
x
ma + b + nc / T
km/hr
Rate of steamer is
1 / m
x
ma + b + nc / T
km/hr
T = 10 hours 30 minutes = 10.5 hours
Here, a = 60, b = 270, c = 30, T = 10.5, n = 6, m = 3
Using these values in the shortcut, we get:
Rate of train =
3 x 60 + 270 + 6 x 30 / 10.5

=
180 + 270 + 180 / 10.5

=
180 + 270 + 180 / 10.5

=
630 / 10.5
= 60
Hence, the rate of the train is 60 km/hr.

69. Ajay covers a certain distance on bike. Had he moved 6 km/hr faster, he would have taken 20 minutes less. If he had moved 4 km/hr slower, he would have taken 20 minutes more. Find the distance and original speed.
  A.  Distance: 25 km; Speed: 34 km/hr
  B.  Distance: 40 km; Speed: 24 km/hr
  C.  Distance: 35 km; Speed: 20 km/hr
  D.  Distance: 45 km; Speed: 25 km/hr
     
   
View Answer

Shortcut:
A man covers a certain distance on scooter. Had he moved a km/hr faster, he would have taken h1 hours less. If he had moved b km/hr slower, he would have taken h2 hours more, then the original speed (S) is
ab(h1 + h2) / h2a − h1b
km/hr and the distance is
h1 x S(S + a) / a
km
Here, a = 6, h1 =
20 / 60
=
1 / 3
, b = 4, h2 =
20 / 60
=
1 / 3

Using these values in the shortcut, we get:
Original speed =
6 x 4 x (1/3 + 1/3) / 1/3 x 6 − 1/3 x 4

=
6 x 4 x 2/3 / 2 − 4/3

=
2 x 4 x 2 / (6 − 4)/3

=
2 x 4 x 2 / 2/3

=
2 x 4 x 2 x 3 / 2
= 2 x 4 x 3 = 24
Hence, its original speed is 24 km/hr.
Distance =
1/3 x 24(24 + 6) / 6
=
8 x 30 / 6
= 8 x 5 = 40
Hence, the distance is 40 km/hr.

70. A thief is spotted by a policeman from a distance of 300 metres. When the policeman starts the chase, the thief also starts running. Assuming the speed of the thief 11 km/hr, and that of the policeman 14 km/hr, how far will have the thief run before he is overtaken?
  A.  5.2 km
  B.  2.5 km
  C.  1.1 km
  D.  1.5 km
     
   
View Answer

Shortcut:
A criminal is spotted by a policeman from a distance of 'Z' km. When the policeman starts the chase, the criminal also starts running. Assuming the speed of the criminal 'a' km/hr, and that of the policeman 'b' km/hr, then the criminal will run before he is overtaken is Z x
a / b − a
km
or The distance covered by the thief before he gets caught is Speed of criminal x
Lead of distance / Relative speed
km/hr
Here, Z =
300 / 1000
or
3 / 10
, a = 11, b = 14
Using these values in the shortcut, we get:
Required distance =
3 / 10
x
11 / 14 − 11

=
3 / 10
x
11 / 3
=
11 / 10
= 1.1
Hence, the thief ran 1.1 km before being overtaken.

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