Home ≫ Arithmetic Aptitute ≫ Time and Work ≫ Page 5
41. A and B can together finish a work in 30 days. They worked for it for 20 days and then B left. The remaining work was done by A alone in 20 days. A alone can finish the work in:
  A.  50 days
  B.  55 days
  C.  60 days
  D.  65 days
     
   
View Answer

Shortcut:
X and Y can do a work in 'a' and 'b' days respectively, they began the work together, but X left after some time and Y finished the remaining work in 'c' days, then the number of days after which X left is given by
(
ab / a + b
)(
b − c / b
)

Here,
ab / a+b
= 30
Now as per question,
B left the work, i.e b = time taken by A to complete the whole work, c = 20
Using all these values in the shortcut, we get:
20 = 30 x
b − 20 / b

2b = 3( b − 20)
2b = 3b − 60
b = 60
Hence, A can complete work in 60 days.

42. 1 man or 2 women or 3 boys can do a work in 55 days. Then in how many days will 1 man, 1 woman and 1 boy do the work?
  A.  20 days
  B.  25 days
  C.  35 days
  D.  30 days
     
   
View Answer

Shortcut:
If a1 men or a2 women or a3 boys can do a work in 'Z' days, then the number of days in which 1 man, 1 woman and 1 boy do the same work in
Z x a1 x a2 x a3 / (a1 x a2) + (a1 x a3) + (a2 x a3)

Here, a1 = 1, a2 = 2, a3 = 3, Z = 55
Using all these values in the shortcut, we get:
=
55 x 1 x 2 x 3 / 1x2 + 1x3 + 2x3

=
55 x 1 x 2 x 3 / 2 + 3 + 6

=
55 x 6 / 11
= 5 x 6 = 30
Hence, 1man, 1 woman and 1 boy will do in 30 days.

43. A group of men decided to do a work in 10 days, but five of them became absent. If the rest of the group did the work in 15 days, find the original number of men.
  A.  15 men
  B.  12 men
  C.  18 men
  D.  16 men
     
   
View Answer

Shortcut:
A group of men decided to do a work in 'a' days but 'n' of them became absent. If the rest of the group did the work in 'b' days, then the original number of man is
nb / b − a

Here, a = 10, n = 5, b = 15
Using all these values in the shortcut, we get:
=
5 x 15 / 15 − 10
=
5 x 15 / 5
= 15
Hence, original number of men is 15.

44. A certain number of men can do a work in 50 days. If there were 8 men more it coukld be finished in 10 days less. How many men are there?
  A.  30 men
  B.  32 men
  C.  34 men
  D.  36 men
     
   
View Answer

Shortcut:
A certain number of men can do a work in 'Z' days. If there were 'a' men more it could be finished in 'z' days less, then the number of men originally are
a(Z − z) / z

Here, Z = 50, a = 8, z = 10
Using all these values in the shortcut, we get:
=
8(50 − 10) / 10
=
8 x 40 / 10
= 8 x 4 = 32
Hence, there are 32 men.

45. Ravi decided to buld a house in 35 days. He employed 100 men in the beginning and 100 more after 30 days and completed the construction in stipulated time. If he had not employed the additional men, how many days behind schedule would it have been finished?
  A.  3 days
  B.  4 days
  C.  6 days
  D.  5 days
     
   
View Answer

Shortcut:
A man decided to build a house in 'Z' days. He employed 'a' men in the beginning and 'b' more men after 'z' days and completed the construction in given time. If he had not employed the additional men, then the men in the beginning would have finished the work
Z(a + b) − bz / a
days and it would have been
b(Z − z) / a
days behind the schedule.
Here, Z = 35, a = 100, b = 100, z = 30
Using all these values in the shortcut, we get:
=
100(35 − 30) / 100
= 5 days

46. Ajay, Biru and Chinu can do a work in 8, 16, 24 days respectively. They all begin together. Ajay continues to work till it is finished, Chinu leaving off 2 days and Biru one day before its completion. In what time is the work finished?
  A.  5 days
  B.  7 days
  C.  6 days
  D.  4 days
     
   
View Answer

Shortcut:
X, Y and Z can do a work in a, b and c days respectively. They all begin together. If X contiues to work till it is finished. Z leaves after working d2 days and Y, d1 days before its completion, then the time in which work is finished is
a(b x c + d1 x c + d2 x b) / ab + ac + bc
days
Here, a = 8, b = 16, c = 24, d1 = 1, d2 =2
Using all these values in the shortcut, we get:
=
8(16 x 24 + 1 x 24 + 2 x 16) / 8 x 16 + 8 x 24 + 16 x 24

=
8(384 + 24 + 32) / 128 + 192 + 384

=
8 x 440 / 704
=
10 / 2
= 5 days

47. There is a sufficient food for 300 men for 30 days. After 26 days, 180 men leave the place. For how many days will the rest of the food last for the rest of the men?
  A.  8 days
  B.  9 days
  C.  10 days
  D.  12 days
     
   
View Answer

Shortcut:
There is a sufficient food for 'M' men for 'Z' days. If after 'z' days 'm' men leave the place, then the rest of the food will last for the rest of the men for
[
Z − z / M − m
x M
]
days
Here, M = 300, Z = 30, z = 26, m = 180
Using all these values in the shortcut, we get:
=
[
30 − 26 / 300 − 180
x 300
]
=
4 / 120
x 300 =
1 / 30
x 300 = 10 days

48. Ajay takes as much time as Biru and Chinu together take to finish a job. Ajay and Biru working together finish the job in 15 days. Chinu alone can do the same job in 20 days. In how many days can Biru alone do the same work?
  A.  110 days
  B.  120 days
  C.  115 days
  D.  125 days
     
   
View Answer

Shortcut:
A takes as much time as B and C together take to finish a job. If A and B working together finish the job in 'a' days. C alone can do the same job in 'b' days, then B alone can do the same work in
2ab / b − a
days and A alone can do the same work in
2ab / b + a
days
Here, a = 15, b = 20
Using all these values in the shortcut, we get:
=
2 x 15 x 20 / 20 − 15
=
2 x 15 x 20 / 5
= 120 days

49. A team of 30 men is supposed to do a work in 38 days. After 25 days, 10 more men were employed and the work finished one day earlier. How many days would it have been delayed if 5 more men were not employed?
  A.  2 days
  B.  4 days
  C.  5 days
  D.  3 days
     
   
View Answer

Shortcut:
A team of 'm' persons is supposed to do a work in 'Z' days. After 'z1' days, 'n' more persons were employed and the work was finished 'z2' days earlier, then the number of days it would have been delayed if 'a' more persons were not employed is
n{Z − (z1 + z2)} − z2m / m
days and the number of days in which the work would have been finished is
(m + n)(Z − z2) − z1n / m
days
Here, m=30, n = 10, Z = 38, z1 = 25, z2 = 1, a = 5
Using all these values in the shortcut, we get:
=
10{38 − (25 + 1)} − 1 x 30 / 30

=
10{38 − 26} − 30 / 30

=
10(12) − 30 / 30

=
120 − 30 / 30
=
90 / 30
= 3 days

50. Ajay, Biru and Chinu can do a work in 10 days, 20 days and 30 days respectively. They started the work together but after 5 days Ajay left. Biru left the work 2 days before the completion of the work. In how many days was the work completed?
  A.  5.5 days
  B.  8.5 days
  C.  7.5 days
  D.  9.5 days
     
   
View Answer

Shortcut:
X, Y and Z can do a work in 'a' days, 'b' days and 'c' days respectively. They started the work together but after d1 days X left. If Y left the work d2 days before the completion of the work, then the whole work will be completed in
[
b(a − d1) + d2a / b + c
][
c / a
]
days
Here, a = 10, b = 20, c = 30, d1 = 5, d2 = 2
Using all these values in the shortcut, we get:
=
[
20(10 − 5) + 2 x 10 / 20 + 30
][
30 / 10
]

=
[
20(5) + 20 / 50
][
30 / 10
]
=
[
100 + 20 / 50
]
x 3 =
[
120 / 50
]
x 3 =
12 x 3 / 5
=
36 / 5
= 7.2

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