Home ≫ Arithmetic Aptitute ≫ Time and Distance ≫ Page 9
81. A man covers one-third of his journey at 18 km/hr and the remaining journey at 36 km/hr. If the total journey is of 160 km, what is his average speed for the whole journey?
  A.  17 km/hr
  B.  25 km/hr
  C.  30 km/hr
  D.  27 km/hr
     
   
View Answer

Shortcut:
If a man covers
a / b
th part of the total journey at speed S1 km/hr and the remaining journey at speed S2 km/hr then his average speed for the whole journey is
S1 x S2 / (1 − a/b)x S1 + a/b x S2
km/hr.
Here, S1 = 18, a/b = 1/3, S2 = 36
Using these values in the shortcut, we get:
Average speed =
18 x 36 / (1 − 1/3)x 18 + 1/3 x 36

=
18 x 36 / [(3 − 1)/3] x 18 + 12

=
18 x 36 / 2/3 x 18 + 12

=
18 x 36 / 2 x 6 + 12

=
18 x 36 / 12 + 12

=
18 x 36 / 24
= 9 x 3
Hence, the average speed is 27 km/hr.

82. A road is of 500 km length. A contractor wants to plant some pillars on the road at every 5 km of distance. Find the total no. of pillars that the contractor has to plant.
  A.  111
  B.  105
  C.  101
  D.  108
     
   
View Answer

Shortcut:
If we want to set pillar at every certain distance, then there are 2 cases, we can set: Case I: When the system is open, i.e road, path etc. If pillars are to be set at every a km distance on the road or the path of b km length, then the toal number of pillars are
b / a
+ 1 or
Total number of pillars =
Total length of the road / Distance between two adjacent pillars
+ 1

Case II: When the system is cloased ie. circle , square, rectangle etc, then If pillars are to be set at every a km distance on a closed system, then the toal number of pillars are
Total perimeter / Distance between two adjacent pillars

Here, a = 5, b = 500
Using these values in the shortcut, we get:
Total Number of pillars =
500 / 5
+ 1 = 100 + 1 = 101
Hence, the total number of pillar are 101.

83. A rat sees a cat 200 metres away from her and scuds off in the opposite direction at a speed of 24 km/hr. Two minutes later the cat perceives her and gives chase at a speed of 32 km/hr. How soon will the cat overtake the rat, and at what distance from the spot where the rat started running?
  A.  Time: 7
1 / 2
minutes; Distance: 2000 metres
  B.  Time: 7
1 / 2
minutes; Distance: 3000 metres
  C.  Time: 5
1 / 2
minutes; Distance: 2500 metres
  D.  Time: 9
1 / 2
minutes; Distance: 4000 metres
     
   
View Answer

Shortcut:
A rat sees a cat z km away from him and scud off in the opposite direction at a speed of a km/hr. If t hurs later the cat perceives him and gives chase at speed of b km/hr, then the cat will overtake rat in
at + z / b − a
hours and at
at + z / b − a
x a km distance from the spot where the cat took flight from.
Here, z =
200 / 1000
=
1 / 5
, a = 24, t =
2 / 60
=
1 / 30
, b = 32
Using these values in the shortcut, we get:
Required time =
24 x 1/30 + 1/5 / 32 − 24

=
4/5 + 1/5 / 8

=
(4 + 1)/5 / 8

=
5/5 / 8

=
1 / 8
hours
or,
1 / 8
x 60 =
15 / 2
= 7
1 / 2
minutes
Hence, the cat will overtake the rat in 7
1 / 2
minutes.
Required distance =
24 x 1/30 + 1/5 / 32 − 24
x 24
=
4/5 + 1/5 / 8
x 24
=
(4 + 1)/5 / 8
x 8
=
5/5 / 8
x 24 = 1 x 3 = 3
Hence, the distance is 3 km or 3000 metres.

84. A rat pursued by a dog, is 100 of her own leaps ahead of him. While the rat takes 8 leaps the dog takes 5 leaps. In one leap the rat goes 1
1 / 4
metres and the dog 2
1 / 4
metres. In how many leaps will the dog overtake the rat?
  A.  500 leaps
  B.  580 leaps
  C.  540 leaps
  D.  520 leaps
     
   
View Answer

Shortcut:
A rat pursued by a dog is "D' of her own leaps ahead of him. While the rat take 'i' leaps the dog takes 'j' leaps. If in one leap the rat goes z1 metres and the dog z2 metres, then the number of leaps in which the dog will overtake the rat are
D / z2/z1 − i/j

Here, D = 100, i = 8, j = 5, z1 =
5 / 4
, z2 =
9 / 4

Using these values in the shortcut, we get:
Required leaps =
100 / (9/4)/(5/4) − 8/5

=
100 / 9/5 − 8/5

=
100 / (9 − 8)/5

=
100 / 1/5
= 100 x 5 = 500
Hence, the dog will overtake the rat in 500 leaps.

85. A hare, pursued by a dog is 80 of her own leaps ahead of him. While the hare takes 4 leaps, the dog takes 3 leaps. 2 leaps of dog is equal to 3 leaps of hare. In how many leaps will the dog overtake the hare?
  A.  380 leaps
  B.  580 leaps
  C.  460 leaps
  D.  480 leaps
     
   
View Answer

Shortcut:
A rat pursued by a dog is "D' of her own leaps ahead of him. While the rat take 'i' leaps the dog takes 'j' leaps. If i1 leaps of rat is equal to j1 leaps of dog, then the number of leaps in which the dog will overtake the hare are
D / i1/j1 − i/j

Here, D = 80, i = 4, j = 3, i1 = 3, j1 = 2
Using these values in the shortcut, we get:
Required leaps =
80 / 3/2 − 4/3

=
80 / (9 − 8)/6

=
80 / 1/6
= 80 x 6 = 480
Hence, the dog will overtake the hare in 480 leaps.

86. A monkey tries to ascend a greased pole 28 metres high. He ascends 4 metres in first minute and slips down 1 metre in the alternate minute. If he continues to ascend in this fashion, how long does he take to reach the top?
  A.  15 minutes
  B.  17 minutes
  C.  19 minutes
  D.  27 minutes
     
   
View Answer

Shortcut:
A monkey tries to ascend a pole. If he ascends in first minute and slips down in the second minute, then the multiple of ascend and descend is
Length of the pole − Distance of ascent / Distance of ascent − Distance of slip down


Now, consider the following two cases:
Case I: If the result is a whole number, then this whole number will be multiple. and the required answer will be (2 x multiple + 1)

Case II: If the result is not the whole number (i.e a fraction) then the multiple will be th whole number of that fraction and the required answer will be 2 x Multiple +
Length of the pole − (Distance of ascent − Distance of slip down) x Multiple / Distance of ascent

Here, Length of pole = 28, Distance of ascent = 4, Distance of slip down = 1
Using these values in the shortcut, we get:
Multiple =
28 − 4 / 4 − 1
=
24 / 3
= 8
Since, multiple is a whole number, the required time to reach the top will be (2 x 8 + 1) = 17
Hence, the monkey will take 17 minutes to reach the top of pole.

87. A monkey tries to ascend a greased pole 51 metres high. He ascends 12 metres in first minute and slips down 4 metres in the alternate minute. If he continues to ascend in this fashion, how long does he take to reach the top?
  A.  12 minutes 5 seconds
  B.  13 minutes 15 seconds
  C.  10 minutes 55 seconds
  D.  15 minutes 5 seconds
     
   
View Answer

Shortcut:
A monkey tries to ascend a pole. If he ascends in first minute and slips down in the second minute, then the multiple of ascend and descend is
Length of the pole − Distance of ascent / Distance of ascent − Distance of slip down


Now, consider the following two cases:

Case I: If the result is a whole number, then this whole number will be multiple. and the required answer will be (2 x multiple + 1)

Case II: If the result is not the whole number (i.e a fraction) then the multiple will be th whole number of that fraction and the required answer will be 2 x Multiple +
Length of the pole − (Distance of ascent − Distance of slip down) x Multiple / Distance of ascent

Here, Length of pole = 51, Distance of ascent = 12, Distance of slip down = 4
Using these values in the shortcut, we get:
Multiple =
51 − 12 / 12 − 4
=
39 / 8
= 5 (approx)
Since, multiple is a fraction, the required time to reach the top will be 2 x 5 +
51 − (12 − 4) x 5 / 12

= 10 +
51 − (8) x 5 / 12

= 10 +
51 − 40 / 12

= 10 +
11 / 12

Hence, the monkey will take 10 minutes 55 seconds to reach the top of pole.

88. A monkey climbing up a greased pole ascends 10 m and slips down 4 m in alternatie minutes. If the pole is 62 m high, how long will it take him to reach the top?
  A.  18 minutes 48 seconds
  B.  12 minutes 28 seconds
  C.  16 minutes 38 seconds
  D.  20 minutes 18 seconds
     
   
View Answer

Shortcut:
A monkey tries to ascend a pole. If he ascends in first minute and slips down in the second minute, then the multiple of ascend and descend is
Length of the pole − Distance of ascent / Distance of ascent − Distance of slip down


Now, consider the following two cases:

Case I: If the result is a whole number, then this whole number will be multiple. and the required answer will be (2 x multiple + 1)

Case II: If the result is not the whole number (i.e a fraction) then the multiple will be th whole number of that fraction and the required answer will be 2 x Multiple +
Length of the pole − (Distance of ascent − Distance of slip down) x Multiple / Distance of ascent

Here, Length of pole = 62, Distance of ascent = 10, Distance of slip down = 4
Using these values in the shortcut, we get:
Multiple =
62 − 10 / 10 − 4
=
52 / 6
=
26 / 3
= 9 (Approx)
Since, multiple is a fraction, the required time to reach the top will be
2 x 9 +
62 − (10 − 4) x 9 / 10

= 18 +
62 − (6) x 9 / 10

= 18 +
62 − 54 / 10

= 18 +
62 − 54 / 10
= 18 +
8 / 10

Hence, the monkey will take 18 min 48 seconds to reach the top of pole.

89. Ravi travelled 600 km by air which formed
3 / 5
of his trip. The part of his trip which was one-fifth of the whole trip, he travelled by car. The rest of the journey was performed by train. The total distance travelled by train was:
  A.  205 km
  B.  200 km
  C.  220 km
  D.  250 km
     
   
View Answer

Let the total distance is y km.
According to the question:
3 / 5
y = 600
y = 600 x
5 / 3
= 200 x 5 = 1000 km
Distance travelled by car = 1000 x
1 / 5
= 200 km
∴ Distance travelled by train = [1000 − (200 + 600)] = 1000 − 800 = 200
Hence, the total distance travelled by train was 200 km.

90. Ajay started cycling along the boundaries of a square field from corner point P. After half an hour, he reached the corener point R, diagonally opposite to P. if his speed was 16 km/hr, the area of the field is:
  A.  14 km2
  B.  32 km2
  C.  16 km2
  D.  26 km2
     
   
View Answer

Distance covered in half an hour = 8 km
2 x (side of the square) = 8 km
or side of the square = 4 km
∴ Area of the square field = 16 km2
Hence, the area of the field is 16 km2.

Copyright © 2020-2022. All rights reserved. Designed, Developed and content provided by Anjula Graphics & Web Desigining .