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71. A policeman goes after a thief who has 250 metres start. If the policeman runs a kilometre in 4 minutes, and the thief a kilometre in 5 minutes, how far will the thief has gone before he is overtaken?
  A.  4 km
  B.  8 km
  C.  5 km
  D.  1 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
Speed of thief =
1 / 5/60
=
60 / 5
= 12 km/hr
Speed of policeman =
1 / 4/60
=
60 / 4
= 15 km/hr
Here, Z = 250/1000 or 1/4, a = 12, b = 15
Using these values in the shortcut, we get:
Required distance =
1 / 4
x
12 / 15 − 12

=
1 / 4
x
12 / 3
= 1
Hence, the thief has gone 1 km before being overtaken.

72. A train travels 260 km in 2.5 hours and 325 km in 4 hours. Find the average speed of the train.
  A.  80 km/hr
  B.  70 km/hr
  C.  90 km/hr
  D.  40 km/hr
     
   
View Answer

Shortcut:
If a moving body travels a1, a2, a3,......an, metres moving with different speeds S1, S2, S3,........Sn metres per second in time T1, T2, T3,......Tn seconds respective, then it is necessary to calculate the average speed of the body throughout the journey. If the average speed is denoted by As, then As =
Total distance travelled / Total time taken

or
Case I :
a1 + a2 + a3 + ....... + an / T1 + T2 + T3 + ....... + Tn

Case II :
S1T1 + S2T2 + S3T3 + ....... + SnTn / T1 + T2 + T3 + ....... + Tn

Case III :
a1 + a2 + a3 + ....... + an / a1/S1 + a2/S2 + a3/S3 + ....... + an/Sn


Here, a1 = 260, a2 = 325, T1 = 2.5, T2 = 4; We use Case I:
Using these values in the shortcut, we get:
Average speed =
260 + 325 / 2.5 + 4
=
585 / 6.5
= 90
Hence, the average speed is 90 km/hr.

73. A car during its journey travels 10 minutes at a speed of 15 km/hr, another 30 minutes at a speed of 30 km/hr and 1 hours at a speed of 40 km/hr. Find its average speed of the car.
  A.  44.5 km/hr
  B.  34.5 km/hr
  C.  24.5 km/hr
  D.  32.5 km/hr
     
   
View Answer

Shortcut:
If a moving body travels a1, a2, a3,......an, metres moving with different speeds S1, S2, S3,........Sn metres per second in time T1, T2, T3,......Tn seconds respective, then it is necessary to calculate the average speed of the body throughout the journey. If the average speed is denoted by As, then As =
Total distance travelled / Total time taken

or Case I :
a1 + a2 + a3 + ....... + an / T1 + T2 + T3 + ....... + Tn

Case II :
S1T1 + S2T2 + S3T3 + ....... + SnTn / T1 + T2 + T3 + ....... + Tn

Case III :
a1 + a2 + a3 + ....... + an / a1/S1 + a2/S2 + a3/S3 + ....... + an/Sn

Here, S1 = 15, S2 = 30, S3 = 40, T1 = 10/60, T2 = 30/60, T3 = 1; We use Case II:
Using these values in the shortcut, we get:
Average speed =
15 x 10/60 + 30 x 30/60 + 40 x 1 / 10/60 + 30/60 + 1

=
5/2 + 15 + 40 / (10 + 30 + 60)/60

=
(5 + 30 + 80)/2 / (10 + 30 + 60)/60

=
115 x 60 / 100 x 2

=
115 x 3 / 10
= 34.5
Hence, the average speed is 34.5 km/hr.

74. A man walks 6 km at a speed of 6 km/hr, runs 8 km at a speed of 8 km/hr and goes by bus another 10 km. Speed of the bus is 10 km/hr. If the speed of the bus is considered as the speed of the man, find the average speed of the man.
  A.  8 km/hr
  B.  6 km/hr
  C.  4 km/hr
  D.  9 km/hr
     
   
View Answer

Shortcut:
If a moving body travels a1, a2, a3,......an, metres moving with different speeds S1, S2, S3,........Sn metres per second in time T1, T2, T3,......Tn seconds respective, then it is necessary to calculate the average speed of the body throughout the journey. If the average speed is denoted by As, then As =
Total distance travelled / Total time taken

or Case I :
a1 + a2 + a3 + ....... + an / T1 + T2 + T3 + ....... + Tn

Case II :
S1T1 + S2T2 + S3T3 + ....... + SnTn / T1 + T2 + T3 + ....... + Tn

Case III :
a1 + a2 + a3 + ....... + an / a1/S1 + a2/S2 + a3/S3 + ....... + an/Sn

Here, a1 = 6, a2 = 8, a3 = 10, S1 = 6, S2 = 8, S3 = 10. We use Case III:
Using these values in the shortcut, we get:
Average speed =
6 + 8 + 10 / 6/6 + 8/8 + 10/10

=
24 / 1 + 1 + 1
=
24 / 3
= 8
Hence, the average speed is 8 km/hr.

75. A train does a journey without stopping in 4 hours. If it had travelled 2.5 km an hour faster, it would have done the journey in 3 hours 20 minutes. What is its original speed?
  A.  14.5 km/hr
  B.  16.5 km/hr
  C.  18.5 km/hr
  D.  12.5 km/hr
     
   
View Answer

Shortcut:
A train does a journey without stopping in H hours. If it had travelled a km/hr faster, it would have done the journey in h hour, then the original speed is
h / H − h
x a km/hr and the length of the journey is
H x h / H − h
x a km
h = 3 hours 20 minutes = 180 + 20 =
200 / 60
=
10 / 3
hours
Here, H = 4, a = 2.5, h =
10 / 3

Using these values in the shortcut, we get:
Original speed =
10/3 / 4 − 10/3
x 2.5
=
10/3 / (12 − 10)/3
x 2.5
=
10 x 3 / 2 x 3
x 2.5 = 5 x 2.5 = 12.5
Hence, the original speed is 12.5 km/hr.

76. A car finished a journey in 10 hours at the speed of 80 km/hr. If the same distance is to be covered in 8 hours how much more speed does the car have to gain?
  A.  25 km/hr
  B.  20 km/hr
  C.  30 km/hr
  D.  32 km/hr
     
   
View Answer

Shortcut:
A train does a journey without stopping in H hours. If it had travelled a km/hr faster, it would have done the journey in h hour, then the original speed is
h / H − h
x a km/hr and the length of the journey is
H x h / H − h
x a km
Here, H = 10, a = ?, h = 8, original speed = 80
Using these values in the shortcut, we get:
80 =
8 / 10 − 8
x a
80 =
8 / 2
x a
80 = 4a
a =
80 / 4
= 20
Hence, the gain in speed will be 20 km/hr.

77. A car travels a distance of 120 km in 2 hours partly at a speed of 75 km/hr and partly at 25 km/hr. Find the distance travelled at a speed of 75 km/hr.
  A.  115 km
  B.  125 km
  C.  105 km
  D.  120 km
     
   
View Answer

Shortcut:
If a car travels a distance of Z km in t hours partly at a speed of a km/hr and partly at b km/hr, then the distance travelled at a speed of a km/hr is (Z − tb ) x
a / a − b
km
Here, Z = 120, a = 75, t = 2, b = 25
Using these values in the shortcut, we get:
Required distance = (120 − 2 x 25 ) x
75 / 75 − 25

= (120 − 50 ) x
75 / 50

= 70 x
3 / 2
= 35 x 3 = 105
Hence, the distance is 105 km.

78. A thief goes away with a car at a speed of 20 km/hr. The theft has been discovered after half and hour and the owner sets off in another car at 25 km/hr. When will the owner overtake the thief form the start?
  A.  2 hours
  B.  3 hours
  C.  1 hours
  D.  4 hours
     
   
View Answer

Shortcut:
A thief goes away with a bike at a speed of a km/hr. If the theift has been find after 't' hours and the owner sets off in another bike at b km/hr, then the owner will overtake the thief from the start in
a / b − a
x t hours.
Here, a = 20, t = 1/2, b = 25
Using these values in the shortcut, we get:
Required time =
20 / 25 − 20
x
1 / 2

=
20 / 5
x
1 / 2
= 2
Hence, the owner overtake the thief in 2 hours.

79. A car travels a distance of 30 km at uniform speed. If the speed of the car is 5 km/hr more, it takes 1 hour less to cover the same distance. Find the original speed of the car.
  A.  12 km/hr
  B.  10 km/hr
  C.  15 km/hr
  D.  18 km/hr
     
   
View Answer

Shortcut:
A car travels a certain distance 'Z' km at uniform speed. If the speed of the car is a km/hr more, it takes 't' hours less to cover the same distane, then the original speed of the car is
√[(at)2 + (4Zat)] − at / 2t
km/hr.
Here, Z = 30, a = 5, t = 1
Using these values in the shortcut, we get:
Required speed =
√[(5 x 1)2 + (4 x 30 x 5 x 1)] − 5 x 1 / 2 x 1

=
√[(5)2 + 600] − 5 / 2

=
√[25 + 600] − 5 / 2

=
√[625] − 5 / 2

=
25 − 5 / 2

=
20 / 2
= 10
Hence, the original speed of the car is 10 km/hr.

80. A car travels a distance of 30 km at uniform speed. If the speed of the car is 5 km/hr less, it takes 1 hours more to cover the same distance. Find the original speed of the car.
  A.  20 km/hr
  B.  25 km/hr
  C.  15 km/hr
  D.  30 km/hr
     
   
View Answer

Shortcut:
A car travels a certain distance 'Z' km at uniform speed. If the speed of the car is a km/hr more, it takes 't' hours less to cover the same distane, then the original speed of the car is
√[(at)2 + (4Zat)] + at / 2t
km/hr.
Here, Z = 30, a = 5, t = 1
Using these values in the shortcut, we get:
Required speed =
√[(5 x 1)2 + (4 x 30 x 5 x 1)] + 5 x 1 / 2 x 1

=
√[(5)2 + 600] + 5 / 2

=
√[25 + 600] + 5 / 2

=
√[625] + 5 / 2

=
25 + 5 / 2

=
30 / 2
= 15
Hence, the original speed of the car is 15 km/hr.

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