- If a person can do a piece of work in
**‘n’**days, then in one day, the person will do**‘1/n’**work. Conversely, if the person does**‘1/n’**work in one day, the person will require**‘n’**days to finish the work … More on Work and Wages

Question 1 |

Two friends A and B were employed to do a work. Initial deadline was fixed at 24 days. Both started working together but after 20 days, A left the work and the whole work took 30 days to complete. In how much time can B alone can do the work?

40 | |

50 | |

60 | |

70 |

**Work and Wages**

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Question 1 Explanation:

Let the total work be 24 units. It is given that A and B together can do the work in 24 days.
=> Combined efficiency of A and B = 24/24 = 1 unit / day
=> Work done in 20 days = 20 units
=> Work left = 24 - 20 = 4 units
Now, this remaining 4 units of work was done by B alone in 10 days.
=> Efficiency of B = 4/10 = 0.4
Therefore, time required by B alone to do the work = 24/0.4 = 60 days

Question 2 |

A and B took a job to be completed in 20 days. They started working together and after 12 days, C joined them and the whole job finished in 15 days. How much time would C require to complete the job if only C was hired?

15 | |

12 | |

10 | |

8 |

**Work and Wages**

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Question 2 Explanation:

Let the total job be 20 units. It is given that A and B took the job to be completed in 20 days.
=> Combined efficiency of A and B = 20/20 = 1 unit / day
Now, job done in 12 days = 12 units
=> Job Left = 8 units
Now, this remaining 8 units of job has been done by all A, B and C together.
Let the efficiency of C be 'x'.
=> Combined efficiency of A, B and C = 1+x units/ day
Now, with this efficiency, the job got completed in 3 more days.
=> Job done in 3 days = 3 x (1+x) = 8 units
=> x = 5/3
Therefore, efficiency of C = x = 5/3 units / day
Hence, time required by C alone to do the job = 20/(5/3) = 12 days

Question 3 |

Three people A, B and C working individually can finish a job in 10, 12 and 20 days respectively. They decided to work together but after 2 days, A left the work and after another one day, B also left work. If they got two lacs collectively for the entire work, find the difference of the highest and lowest share.

70000 | |

60000 | |

10000 | |

20000 |

**Work and Wages**

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Question 3 Explanation:

Let the total work be LCM(10, 12, 20) = 60 units
=> Efficiency of A = 60/10 = 6 units / day
=> Efficiency of B = 60/12 = 5 units / day
=> Efficiency of C = 60/20 = 3 units / day
Since the number of working days are different for each person, the share of each will be calculated in the ratio of the units of work done.
Now, A works for 2 days and B works for 3 days.
=> Work done by A = 2 x 6 = 12 units
=> Work done by B = 3 x 5 = 15 units
=> Work done by C = 60 - 12 - 15 = 33 units
Therefore, ratio of work done = 12:15:33 = 4:5:11
So, A's share = (4/20) x 2,00,000 = Rs 40,000
B's share = (5/20) x 2,00,000 = Rs 50,000
C's share = (11/20) x 2,00,000 = Rs 1,10,000
Therefore, difference of the highest and lowest share = Rs 1,10,000 - 40,000 = Rs 70,000

Question 4 |

A alone and B alone can do a work in respectively 18 and 8 days more than both working together. Find the number of days required if both work together.

12 | |

8 | |

16 | |

36 |

**Work and Wages**

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Question 4 Explanation:

Let the time required to complete the work by A and B together = n days
=> Time required by A alone = n + 18 days
=> Time required by B alone = n + 8 days
Therefore, n

^{2}= 18 x 8 = 144 => n = 12 Hence, A and B require 12 days to complete the work if they work together.Question 5 |

Three friends A, B and C are employed to make pastries in a bakery. Working individually, they can make 60, 30 and 40 pastries respectively in an hour. They decided to work together but due to lack of resources, they had to work on shifts of 30 minutes. Find the time taken to make 185 pastries.

4 hours | |

3 hours 45 minutes | |

4 hours 15 minutes | |

5 hours |

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Question 5 Explanation:

It is given that A, B and C make 60, 30 and 40 pastries respectively in an hour.
=> In 30 minutes, they will make 30, 15 and 20 pastries respectively.
So, in one cycle of 1 hour 30 minutes where each works for 30 minutes, pastries made = 30 + 15 + 20 = 65
Now, in 2 cycles (3 hours), 130 pastries would be made.
In the next 30 minutes, A would make 30 pastries.
So, total time elapsed = 3 hours 30 minutes and pastries made = 130 + 30 = 160
In the next 30 minutes, B would make 15 pastries.
So, total time elapsed = 4 hours and pastries made = 160 + 15 = 175
In the next 15 minutes, C would make 10 pastries.
So, total time elapsed = 4 hours 15 minutes and pastries made = 175 + 10 = 185
Therefore, total time taken = 4 hours 15 minutes

Question 6 |

A person employed a group of 20 men for a construction job. These 20 men working 8 hours a day can complete the job in 28 days. The work started on time but after 18 days, it was observed that two thirds of the work was still pending. To avoid penalty and complete the work on time, the employer had to employ more men and also increase the working hours to 9 hours a day. Find the additional number of men employed if the efficiency of all men is same.

40 | |

44 | |

64 | |

80 |

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Question 6 Explanation:

Let the total work be 3 units and additional men employed after 18 days be 'x'.
=> Work done in first 18 days by 20 men working 8 hours a day = (1/3) x 3 = 1 unit
=> Work done in last 10 days by (20 + x) men working 9 hours a day = (2/3) x 3 = 2 unit
Here, we need to apply the formula

**M**, where M_{1}D_{1}H_{1}E_{1}/ W_{1}= M_{2}D_{2}H_{2}E_{2}/ W_{2}_{1}= 20 men D_{1}= 18 days H_{1}= 8 hours/day W_{1}= 1 unit E_{1}= E_{2}= Efficiency of each man M_{2}= (20 + x) men D_{2}= 10 days H_{2}= 9 hours/day W_{2}= 2 unit So, we have 20 x 18 x 8 / 1 = (20 + x) x 10 x 9 / 2 => x + 20 = 64 => x = 44 Therefore, additional men employed = 44Question 7 |

6 men and 10 women were employed to make a road 360 km long. They were able to make 150 kilometres of road in 15 days by working 6 hours a day. After 15 days, two more men were employed and four women were removed. Also, the working hours were increased to 7 hours a day. If the daily working power of 2 men and 3 women are equal, find the total number of days required to complete the work.

19 | |

35 | |

34 | |

50 |

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Question 7 Explanation:

We are given that the daily working power of 2 men and 3 women are equal.
=> 2 Em = 3 Ew
=> Em / Ew = 3/2, where 'Em' is the efficiency of 1 man and 'Ew' is the efficiency of 1 woman.
Therefore, ratio of efficiency of man and woman = 3 : 2.
If 'k' is the constant of proportionality, Em = 3k and Ew = 2k.
Here, we need to apply the formula

**∑(M**, where ∑(M_{i}E_{i}) D_{1}H_{1}/ W_{1}= ∑(M_{j}E_{j}) D_{2}H_{2}/ W_{2}_{i}E_{i}) = (6 x 3k) + (10 x 2k) ∑(M_{j}E_{j}) = (8 x 3k) + (6 x 2k) D_{1}= 15 days D_{2}= Number of days after increasing men and reducing women H_{1}= 6 hours H_{2}= 7 hours W_{1}= 150 km W_{2}= 210 km So, we have 38k x 15 x 6 / 150 = 36k x D_{2}x 7 / 210 => 38k x 6 = 12k x D_{2}=> D_{2}= 19 days Therefore, total days required to complete the work = 15 + 19 = 34 daysQuestion 8 |

A stadium was to be built in 1500 days. The contractor employed 200 men, 300 women and 750 robotic machines. After 600 days, 75% of the work was still to be done. Fearing delay, the contractor removed all women and 500 robotic machines. Also, he employed some more men having the same efficiency as earlier employed men. This led to a speedup in work and the stadium got built 50 days in advance. Find the additional number of men employed if in one day, six men, ten women and fifteen robotic machines have same work output.

1100 | |

1340 | |

1300 | |

1140 |

**Work and Wages**

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Question 8 Explanation:

Let the total work be 4 units.

=> Work done in first 600 days = 25% of 4 = 1 unit

=> Work done in next 850 days = 75% of 4 = 3 unit

Also, we are given that the daily work output of 6 men, 10 women and 15 robotic machines are same.

=> 6 Em = 10 Ew = 15 Er

=> Em : Ew : Er = 5 : 3 : 2, where 'Em' is the efficiency of 1 man, 'Ew' is the efficiency of 1 woman and 'Er' is the efficiency of 1 robotic machine.

Therefore, ratio of efficiency of man, woman and robotic machine = 5:3:2.

If 'k' is the constant of proportionality, Em = 5k, Ew = 3k and Er = 2k.

Here, we need to apply the formula

∑(M

∑(M

D

D

H

W

W

So, we have

3400k x 600 / 1 = (1500 + 5m)k x 850 / 3

=> 3400k x 1800 = (1500 + 5m)k x 850

=> 1500 + 5m = 7200

=> 5m = 5700

=> m = 1140

Therefore, additional men employed = 1140

=> Work done in first 600 days = 25% of 4 = 1 unit

=> Work done in next 850 days = 75% of 4 = 3 unit

Also, we are given that the daily work output of 6 men, 10 women and 15 robotic machines are same.

=> 6 Em = 10 Ew = 15 Er

=> Em : Ew : Er = 5 : 3 : 2, where 'Em' is the efficiency of 1 man, 'Ew' is the efficiency of 1 woman and 'Er' is the efficiency of 1 robotic machine.

Therefore, ratio of efficiency of man, woman and robotic machine = 5:3:2.

If 'k' is the constant of proportionality, Em = 5k, Ew = 3k and Er = 2k.

Here, we need to apply the formula

**∑(M**

(M, where_{i}E_{i}) D_{1}H_{1}/ W_{1}= ∑(M

_{j}E_{j}) D_{2}H_{2}/ W_{2}∑(M

_{i}E_{i}) = (200 x 5k) + (300 x 3k) + (750 x 2k)∑(M

_{j}E_{j}) = (200 x 5k) + (m x 5k) + (250 x 2k), where 'm' is the additional men employedD

_{1}= 600 daysD

_{2}= 850 daysH

_{1}= H_{2}= Daily working hoursW

_{1}= 1 unitW

_{2}= 3 unitsSo, we have

3400k x 600 / 1 = (1500 + 5m)k x 850 / 3

=> 3400k x 1800 = (1500 + 5m)k x 850

=> 1500 + 5m = 7200

=> 5m = 5700

=> m = 1140

Therefore, additional men employed = 1140

Question 9 |

3 men and 4 women can complete a work in 10 days by working 12 hours a day. 13 men and 24 women can do the same work by working same hours a day in 2 days. How much time would 12 men and 1 women working same hours a day will take to complete the whole work?

4 | |

6 | |

8 | |

10 |

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Question 9 Explanation:

Here, we need to apply the formula

**∑(M**, where ∑(M_{i}E_{i}) D_{1}H_{1}/ W_{1}= ∑(M_{j}E_{j}) D_{2}H_{2}/ W_{2}_{i}E_{i}) = (3 x m) + (4 x w) ∑(M_{j}E_{j}) = (13 x m) + (24 x w), where 'm' is the efficiency of each man and 'w' is the efficiency of each woman D_{1}= 10 days D_{2}= 2 days H_{1}= 12 hours H_{2}= 12 hours W_{1}= W_{2}= Work to be done So, we have (3m + 4w) x 10 x 12 = (13m + 24w) x 2 x 12 => 15m + 20w = 13m + 24w => 2m = 4w => m = 2w => m : w = 2 : 1 Therefore, ratio of efficiency of man and woman = 2 : 1 If the constant of proportionality be 'k', Efficiency of each man = m = 2k Efficiency of each woman = w = k Now, we re-apply the same formula.**∑(M**, where ∑(M_{i}E_{i}) D_{1}H_{1}/ W_{1}= ∑(M_{j}E_{j}) D_{2}H_{2}/ W_{2}_{i}E_{i}) = (3 x m) + (4 x w) ∑(M_{j}E_{j}) = (12 x m) + (1 x w) D_{1}= 10 days D_{2}= Days requires by 12 men and 1 woman H_{1}= 12 hours H_{2}= 12 hours W_{1}= W_{2}= Work to be done So, we have (3m + 4w) x 10 x 12 = (12m + w) x D_{2}x 12 => 30m + 40w = (12m + w) x D_{2}=> 60k + 40k = (24k + k) x D_{2}=> 100k = 25k x D_{2}=> D_{2}= 4 Therefore, 12 men and 1 woman would require 4 days to complete the work.Question 10 |

600 men can make a road in 500 days. They start working together but after every hundred days, 50 men leave the work. Find the total time (in days) it takes to make the road.

600 | |

550 | |

650 | |

750 |

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Question 10 Explanation:

600 men can make a road in 500 days.
=> Total work = 600 x 500 = 3,00,000 units
Now, 600 men start working but after every hundred days, 50 men leave the work.
=> Work done in first 100 days = 600 x 100 = 60,000 units and total work done = 60,000 units
=> Work done in next 100 days = 550 x 100 = 55,000 units and total work done = 1,15,000 units
=> Work done in next 100 days = 500 x 100 = 50,000 units and total work done = 1,65,000 units
=> Work done in next 100 days = 450 x 100 = 45,000 units and total work done = 2,10,000 units
=> Work done in next 100 days = 400 x 100 = 40,000 units and total work done = 2,50,000 units
=> Work done in next 100 days = 350 x 100 = 35,000 units and total work done = 2,85,000 units
=> Work done in next 50 days = 300 x 50 = 15,000 units and total work done = 3,00,000 units
Therefore, total time (days) taken to make the road = 650 days

Question 11 |

10 men working 9 hours a day can complete a work in 24 days. How much time will it take to complete the work if 15 men are employed for 6 hours a day?

18 | |

20 | |

24 | |

30 |

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Question 11 Explanation:

Here, we need to apply the formula

**M**, where M_{1}D_{1}H_{1}E_{1}/ W_{1}= M_{2}D_{2}H_{2}E_{2}/ W_{2}_{1}= 10 men D_{1}= 24 days H_{1}= 9 hours/day W_{1}= W_{2}= Work to be done E_{1}= E_{2}= Efficiency of each man M_{2}= 15 men D_{2}= Days required by 15 men H_{2}= 6 hours/day So, we have 10 x 24 x 9 = 15 x 6 x D_{2}=> D_{2}= 24 Therefore, 15 men working 6 hours / day would require 24 days to complete the work.**Note :**Here, 10 men working 9 hours a day is equivalent to 90 working hours by 1 man. 15 men working 6 hours / day is also equivalent to 90 working hours by 1 man. Hence, the number of days required to complete the work would not change.Question 12 |

32 Bakers working 6 hours a day can make 400 cakes in 25 days. If 30 such bakers are given a contract to make 300 cakes in 24 days, how many hours a day should they work?

4 | |

5 | |

6 | |

8 |

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Question 12 Explanation:

Here, we need to apply the formula

**M**, where M_{1}D_{1}H_{1}E_{1}/ W_{1}= M_{2}D_{2}H_{2}E_{2}/ W_{2}_{1}= 32 bakers D_{1}= 25 days H_{1}= 6 hours/day W_{1}= 400 cakes E_{1}= E_{2}= Efficiency of each baker M_{2}= 30 bakers D_{2}= 24 days H_{2}= Daily working hours for 30 bakers W_{2}= 300 cakes So, we have 32 x 25 x 6 / 400 = 30 x 24 x H_{2}/ 300 => H_{2}= 5 Therefore, 30 bakers should work 5 hours / day to make 300 cakes in 24 days.Question 13 |

Two friends A and B were hired to paint a room. A alone can paint the room in 10 days and is twice as efficient as B. Due to lack of resources, they decided to work alternatively with A starting first. Find the days it will take to paint the room.

13 | |

15 | |

16 | |

18 |

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Question 13 Explanation:

We are given that A alone can paint the room in 10 days and A is twice as efficient as B.
=> B alone would take double the time than A alone ,i.e., 20 days.
Now, let the total work of painting be LCM (10, 20) = 20 units.
=> Efficiency of A = 20/10 = 2 units / day
=> Efficiency of B = 20/20 = 1 unit / day
Since they work alternatively with A starting first, room painted in 1 cycle of 2 days = 3 units
=> Room painted in 6 cycles (total 12 days) = 18 units
Now, we are left with 20-18 = 2 units of painting and it is A's turn to paint, who can paint 2 units in a day.
Therefore, total number of days required to paint the room = 12 + 1 = 13

Question 14 |

Two friends A and B take a job for Rs. 10000. Had they worked alone, A would have taken 20 days while B would have taken 30 days. They started working together but after 10 days, A left and B completed the remaining work alone. Find the difference between their share.

0 | |

1000 | |

2000 | |

5000 |

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Question 14 Explanation:

Let the total job be LCM (20, 30) = 60 units
=> Efficiency of A = 60/20 = 3 units / day
=> Efficiency of B = 60/30 = 2 units / day
Now, since they work for different number of days, the amount of Rs 10,000 would be divided in the ratio of units of work done.
Work done by A in 10 days = 30 units
Work done by B in 10 days = 20 units
Work left = 60 - 30 - 20 = 10 units
This 10 units of leftover work was done by B alone.
Therefore, total work done by A = 30 units
Total work done by B = 20 + 10 = 30 units
Hence, ratio of share in amount = 30 : 30 = 1 : 1
=> Each would be getting Rs 5000.
Thus, difference in share = 0

Question 15 |

If 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do the same in 2 days, the time taken by 15 men and 20 boys in doing the same type of work will be:

4 days | |

5 days | |

6 days | |

7 days |

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Question 15 Explanation:

Let the amount of work 1 man can do in 1 day be x and the amount of work 1 boys can do in 1 day be y.
Then, 6x + 8y = 1/10 and 26x + 48y = ½.
Solving these two equations, we get: x = 1/100 and y = 1/200.
Amount of work done by 15 men and 20 boys in 1 day = 15/100 + 20/200 = ¼.
Therefore, the answer is 4 days.

There are 15 questions to complete.