Arithmetic - Time & Work - Previous Year CAT/MBA Questions
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Renu would take 15 days working 4 hours per day to complete a certain task whereas Seema would take 8 days working 5 hours per day to complete the same task. They decide to work together to complete this task. Seema agrees to work for double the number of hours per day as Renu, while Renu agrees to work for double the number of days as Seema. If Renu works 2 hours per day, then the number of days Seema will work, is
Answer: 6
Text Explanation :
Workspace:
Amal and Vimal together can complete a task in 150 days, while Vimal and Sunil together can complete the same task in 100 days. Amal starts working on the task and works for 75 days, then Vimal takes over and works for 135 days. Finally, Sunil takes over and completes the remaining task in 45 days. If Amal had started the task alone and worked on all days, Vimal had worked on every second day, and Sunil had worked on every third day, then the number of days required to complete the task would have been
Answer: 139
Text Explanation :
Workspace:
Sam can complete a job in 20 days when working alone. Mohit is twice as fast as Sam and thrice as fast as Ayna in the same job. They undertake a job with an arrangement where Sam and Mohit work together on the first day, Sam and Ayna on the second day, Mohit and Ayna on the third day, and this three-day pattern is repeated till the work gets completed. Then, the fraction of total work done by Sam is
- (a)
1/5
- (b)
3/10
- (c)
1/20
- (d)
3/20
Answer: Option B
Text Explanation :
Workspace:
The amount of job that Amal, Sunil and Kamal can individually do in a day, are in harmonic progression. Kamal takes twice as much time as Amal to do the same amount of job. If Amal and Sunil work for 4 days and 9 days, respectively, Kamal needs to work for 16 days to finish the remaining job. Then the number of days Sunil will take to finish the job working alone, is
Answer: 27
Text Explanation :
Let the time taken by A and K to complete a work is x and 2x days respectively.
Work done in a day is the efficiency of a person.
Hence, if efficiencies of A, S and K are in Harmonic Progression, time taken by them to finish a work will be in Arithmetic Progression.
∴ Time taken by S alone is arithmetic mean of time taken by A and K alone.
⇒ Time taken by S alone = (x + 2x)/2 = 1.5x
Using unitary method
⇒ 1 = + +
⇒ x = 4 + 6 + 8
⇒ x = 18
∴ S will take 1.5 × 18 = 27 days to finish the job alone.
Hence, 27.
Concept:
Workspace:
Pipes A and C are fill pipes while Pipe B is a drain pipe of a tank. Pipe B empties the full tank in one hour less than the time taken by Pipe A to fill the empty tank. When pipes A, B and C are turned on together, the empty tank is filled in two hours. If pipes B and C are turned on together when the tank is empty and Pipe B is turned off after one hour, then Pipe C takes another one hour and 15 minutes to fill the remaining tank. If Pipe A can fill the empty tank in less than five hours, then the time taken, in minutes, by Pipe C to fill the empty tank is
- (a)
60
- (b)
90
- (c)
75
- (d)
120
Answer: Option B
Text Explanation :
Let time taken by A and C alone to fill the tank is A and C hours respectively.
∴ Time taken B alone = (A - 1) hours.
Case 1: All three can fill the tank in 2 hours.
⇒ - + = ...(1)
Case 2: B works for 1 hour and C works for 5/4 hours.
⇒ - + = 1 ...(2)
(1) × 9/4 - (2)
⇒ - + = - 1
⇒ - =
⇒ - = 1
⇒ A(A - 1) = 18(A - 1) - 10A
⇒ A2 - A = 8A - 18
⇒ A2 - 9A + 18 = 0
⇒ A = 3 or 6 hours.
Since A should be less than 5, hence we accept only A = 3 hours.
Now, from (1), we have
⇒ - + =
⇒ - + =
⇒ C = 3/2 hours = 90 minutes.
Hence, option (b).
Concept:
Workspace:
Rahul, Rakshita and Gurmeet, working together, would have taken more than 7 days to finish a job. On the other hand, Rahul and Gurmeet, working together would have taken less than 15 days to finish the job. However, they all worked together for 6 days, followed by Rakshita, who worked alone for 3 more days to finish the job. If Rakshita had worked alone on the job then the number of days she would have taken to finish the job, cannot be
- (a)
21
- (b)
17
- (c)
16
- (d)
20
Answer: Option A
Text Explanation :
Let the total work to be done = LCM(7, 15, 6) = 630 units.
⇒ Combined efficiency of Rahul, Rakshita and Gurmeet < 630/7 = 90 units/day
All 3 worked for 6 days and then Rakshita worked for 3 days. Calculating total work done
∴ 630 = (combined efficiency of all 3) × 6 + (efficiency of Rakshita) × 3
⇒ (efficiency of Rakshita) × 3 = 630 - (combined efficiency of all 3) × 6
⇒ (efficiency of Rakshita) = 210 - (combined efficiency of all 3) × 2
⇒ (efficiency of Rakshita) > 210 - 90 × 2
⇒ (efficiency of Rakshita) > 30
∴ Rakshita's efficiency is greater than 30 units/day.
⇒ Time taken by Rakshita < 630/30 = 21 days.
∴ Rakshita will take less than 21 days to finish the job. Hence, she cannot take 21 days to finish the job.
Hence, option (a).
Workspace:
Gautam and Suhani, working together, can finish a job in 20 days. If Gautam does only 60% of his usual work on a day, Suhani must do 150% of her usual work on that day to exactly make up for it. Then, the number of days required by the faster worker to complete the job working alone is
Answer: 36
Text Explanation :
Let work done per day (efficiency) of Gautam and Suhani are 'g' and 's' units.
A shortfall of 40% for Gautam is compensated by 50% extra work done by Suhani.
⇒ 0.4 × g = 0.5 × s
⇒ g/s = 5/4
⇒ Let g = 5x and s = 4x
Together they can complete the work in 20 days.
⇒ Total work to be done = 20 × (5x + 4x) = 180x units
The faster among the two is Gautam whose efficiency is 5x.
∴ Time required by Gautam alone = 180x / 5x = 36 days.
Hence, 36.
Workspace:
Working alone, the times taken by Anu, Tanu and Manu to complete any job are in the ratio 5 : 8 : 10. They accept a job which they can finish in 4 days if they all work together for 8 hours per day. However, Anu and Tanu work together for the first 6 days, working 6 hours 40 minutes per day. Then, the number of hours that Manu will take to complete the remaining job working alone is:
Answer: 6
Text Explanation :
Ratio of time taken by Anu, Tanu and Manu is 5 : 8 : 10.
⇒ Ratio of efficiencies of Anu, Tanu and Manu = : : = 8 : 5 : 4
Let their efficiencies be 8x, 5x and 4x respectively per hour.
Total work done by them in 4 days = (8x + 5x + 4x) × 4 × 8 = 17x × 32 = 544x
Now, Anu and Tanu worked by 6 days working 6 hours 40 minute i.e., 6 hours daily.
∴ Worked completed by Anu and Tanu = 13x × 6 × = 520x
⇒ Time taken by Manu to complete the remaining work = = 6 hours.
Hence, 6.
Workspace:
A group of N people worked on a project. They finished 35% of the project by working 7 hours a day for 10 days. Thereafter, 10 people left the group and the remaining people finished the rest of the project in 14 days by working 10 hours a day. Then the value of N is
- (a)
23
- (b)
140
- (c)
36
- (d)
150
Answer: Option B
Text Explanation :
Let the efficiency of each person be ‘e’ units/hour.
Given,
35% work is done by N men in 10 days working for 7 hours/day, while
65% work is done by N-10 men in 14 days working for 10 hours/day
⇒ =
⇒ =
⇒ N =
⇒ 13N = 14N – 140
⇒ N = 140
Hence, option (b).
Workspace:
Bob can finish a job in 40 days, if he works alone. Alex is twice as fast as Bob and thrice as fast as Cole in the same job. Suppose Alex and Bob work together on the first day, Bob and Cole work together on the second day, Cole and Alex work together on the third day, and then, they continue the work by repeating this three-day roster, with Alex and Bob working together on the fourth day, and so on. Then, the total number of days Alex would have worked when the job gets finished, is
Answer: 11
Text Explanation :
Bob takes 40 days to finish the work. Alex takes is twice as fast as Bob and hence will take half the time taken by Bob i.e., 20 days. Alex is also thrice as fast as Cole, hence Cole will take thrice the time taken by Alex, i.e., 60 days.
Time take by
Alex = 20 days
Bob = 40 days
Cole = 60 days
Let the total work to be done = 120 units.
∴ Efficiency of
Alex = 6 units/day
Bob = 3 units/day
Cole = 2 units/day
Work done/cycle = 22 units.
∴ Work done in 5 cycles = 5 × 22 = 110 units
Work left after 5 cycles (15 days) = 120 – 110 = 10 units.
On 16th day Alex and Bob will together complete 9 units of work while remaining 1 unit of work will be completed by Bob and Cole on 17th day.
∴ Alex worked for 10 days in 5 complete cycles + on 16th day i.e., total 11 days.
Hence, 11.
Workspace:
Anu, Vinu and Manu can complete a work alone in 15 days, 12 days and 20 days, respectively. Vinu works everyday. Anu works only on alternate days starting from the first day while Manu works only on alternate days starting from the second day. Then, the number of days needed to complete the work is
- (a)
8
- (b)
6
- (c)
5
- (d)
7
Answer: Option D
Text Explanation :
Let the total work to be done = LCM (15, 12, 20) = 60 units.
Efficiency of Anu = 60/15 = 4 units/day
Efficiency of Vinu = 60/12 = 5 units/day
Efficiency of Manu = 60/20 = 3 units/day
Let us calculate the work done every 2 days.
Vinu works on both the days, hence work done by Vinu = 2 × 5 = 10 units
Anu works on alternate days, hence work done by Anu = 4 units
Manu works on alternate days, hence work done by Manu = 3 units
∴ Total work done in 2 days = 10 + 4 + 3 = 17 units.
Hence total work done in 6 days = 3 × 17 = 51 units.
On 7th day Vinu and Anu will work, hence work done = 5 + 4 = 9 units.
∴ Work done in total 7 days = 51 + 9 = 60 units [Work gets completed now]
∴ The number of days needed to complete the work is 7 days.
Hence, option (d).
Workspace:
Amar, Akbar and Anthony are working on a project. Working together Amar and Akbar can complete the project in 1 year, Akbar and Anthony can complete in 16 months, Anthony and Amar can complete in 2 years. If the person who is neither the fastest nor the slowest works alone, the time in months he will take to complete the project is
Answer: 32
Text Explanation :
Let the total work to be done = LCM (12, 16, 24) = 48 units.
Let the efficiency of Amar, Akbar and Anthony be ‘x’, ‘y’ and ‘z’ units/month
⇒ x + y = 48/12 = 4 …(2)
⇒ y + z = 48/16 = 3 …(1)
⇒ z + x = 48/24 = 2 …(3)
Adding all these equations
⇒ x + y + z = 9/2 = 4.5 …(4)
Solving these four equations we get,
x = 1.5, y = 2.5 and z = 0.5
∴ The person who is neither slowest not fastest is Amar with efficiency of 1.5 units/month.
∴ Time required by Amar to complete the task alone = 48/1.5 = 32 months.
Hence, 32.
Workspace:
Anil can paint a house in 60 days while Bimal can paint it in 84 days. Anil starts painting and after 10 days, Bimal and Charu join him. Together, they complete the painting in 14 more days. If they are paid a total of ₹ 21000 for the job, then the share of Charu, in INR, proportionate to the work done by him, is
- (a)
9100
- (b)
9000
- (c)
9150
- (d)
9200
Answer: Option A
Text Explanation :
Let the total work to be done = LCM (60, 84) = 420 units.
Efficiency of Anil = 420/60 = 7 units/day
Efficiency of Bimal = 420/84 = 5 units/day
Work done by Anil in 10 + 14 days = 24 × 7 = 168 units
Work done by Bimal in 14 days = 14 × 5 = 70 units
∴ Work done by Charu = 420 – 70 - 168 = 182 units
Fraction of work done by Charu = 182/420 = 91/210
∴ Payment received by Charu = 91/210 × 21,000 = Rs. 9,100
Hence, option (a).
Workspace:
Two pipes A and B are attached to an empty water tank. Pipe A fills the tank while pipe B drains it. If pipe A is opened at 2 pm and pipe B is opened at 3 pm, then the tank becomes full at 10 pm. Instead, if pipe A is opened at 2 pm and pipe B is opened at 4 pm, then the tank becomes full at 6 pm. If pipe B is not opened at all, then the time, in minutes, taken to fill the tank is:
- (a)
140
- (b)
120
- (c)
144
- (d)
264
Answer: Option C
Text Explanation :
Let the filling and emptying capacity of A and B be ‘a’ and ‘b’ units/hour respectively.
Case 1: A is opened at 2 pm and B at 3 pm
Total work done till 10 pm = 8a - 7b
Case 2: A is opened at 2 pm and B at 4 pm
Total work done till 6 pm = 4a - 2b
Since work done is same in both cases, we have
8a – 7b = 4a – 2b
⇒ 4a = 5b
Now time taken by A alone to fill the tank = (Total work)/a = (4a – 2b)/a = (5b – 2b)/(5b/4) = 12/5 hours = 144 minutes.
Hence, option (c).
Workspace:
One day, Rahul started a work at 9 AM and Gautam joined him two hours later. They then worked together and completed the work at 5 PM the same day. If both had started at 9 AM and worked together, the work would have been completed 30 minutes earlier. Working alone, the time Rahul would have taken, in hours, to complete the work is:
- (a)
10
- (b)
12
- (c)
12.5
- (d)
11.5
Answer: Option A
Text Explanation :
Let the efficiency of Rahul be ‘r’ units per hour and for Gautam be ‘g’ units per day.
Initially, Rahul works for 8 hours and Gautam works for 6 hours.
∴ Work done by them = 8r + 6g …(1)
If both of them worked starting from 9 AM, they would have completed the work in 7.5 hours.
∴ Work done by them = 7.5r + 7.5g …(2)
⇒ 8r + 6g = 7.5r + 7.5g
⇒ 0.5r = 1.5g
⇒ r = 3g
∴ Work to be done = 8r + 6g = 8r + 2r = 10r
∴ Time taken by Rahul to complete the work alone = 10r/r = 10 hours.
Hence, option (a).
Workspace:
Anil can paint a house in 12 days while Barun can paint it in 16 days. Anil, Barun, and Chandu undertake to paint the house for ₹ 24000 and the three of them together complete the painting in 6 days. If Chandu is paid in proportion to the work done by him, then the amount in INR received by him is
Answer: 3000
Text Explanation :
Let the area to be painted = LCM(12, 16) = 48 units.
⇒ Efficiency of Anil = 48/12 = 4 units/day
and Efficiency of Barun = 48/16 = 3 units/day
Work done by Anil in 6 days = 6 × 4 = 24 units
Work done by Anil in 6 days = 6 × 3 = 18 units
∴ Remaining work done by Chandu = 48 – 24 – 18 = 6 units.
⇒ Payment received by Chandu = 24000/48 × 6 = Rs. 3,000
Alternately,
Fraction of work done by Anil in 6 days = 6/12 = ½
Fraction of work done by Barun in 6 days = 6/16 = 3/8
⇒ Fraction of work done by Chandu in 6 days = 1 - ½ - 3/8 = 1/8
Since, Chandu completes 1/8th of the work, he will receive 1/8th of the payment.
∴ Chandu’s payment = 1/8 × 24,000 = Rs. 3,000
Hence, 3000.
Workspace:
John takes twice as much time as Jack to finish a job. Jack and Jim together take one-thirds of the time to finish the job than John takes working alone. Moreover, in order to finish the job, John takes three days more than that taken by three of them working together. In how many days will Jim finish the job working alone?
Answer: 4
Text Explanation :
Let John’s efficiency be 1 unit/day.
∴ Jack’s efficiency is 2 units/day.
Jack and Jim’s efficiency is thrice John’s efficienc.
∴ 2 + eJim = 3 × 1
⇒ eJim = 1
John takes three days more than that taken by three of them working together
Let the time taken by all three together is d.
∴ Total work to be done = 4 × d = 1 × (d + 3)
⇒ d = 1
∴ John finished the work in d + 3 = 4 days.
Since, John and Jim have same efficiency, Jim will also finish the work in 4 days.
Hence, 4.
Workspace:
A contractor agreed to construct a 6 km road in 200 days. He employed 140 persons for the work. After 60 days, he realized that only 1.5 km road has been completed. How many additional people would he need to employ in order to finish the work exactly on time?
Answer: 40
Text Explanation :
In 60 days 1.5 kms out of 6 kms is built.
∴ In 60 days 1.5/6 = 1/4th work is done.
To complete the whole work, it would take 4 × 60 = 240 days, i.e., 180 more days.
Days remaining now is 200 – 60 = 140 days.
∴ To complete the remaining work in 140 days, contractor will have to hire more people. Assuming he hires x more men.
The work which 140 men would take 180 more days, now need to be completed by 140 + x men in 140 days.
∴ 180 × 140 = 140 × (140 + x)
⇒ x = 40
∴ Contractor hires 40 more people.
Hence, 40.
Workspace:
At their usual efficiency levels, A and B together finish a task in 12 days. If A had worked half as efficiently as she usually does, and B had worked thrice as efficiently as he usually does, the task would have been completed in 9 days. How many days would A take to finish the task if she works alone at her usual efficiency?
- (a)
36
- (b)
24
- (c)
12
- (d)
18
Answer: Option D
Text Explanation :
Let total work be the LCM of 12 and 9 = 36 units.
Let the efficiency of A and B be a and b respectively.
Work done per day when A and B are working together = 36/12 = 3 units.
∴ a + b = 3 ...(1)
Work done per day when A is working at half efficieny and B is working at thrice efficiency = 36/9 = 4 units.
∴ (a/2) + 3b = 4 ...(2)
Solving (1) and (2), we get; a = 2.
Time taken by A alone to complete the work = 36/2 = 18 days.
Hence, option (d).
Workspace:
Three men and eight machines can finish a job in half the time taken by three machines and eight men to finish the same job. If two machines can finish the job in 13 days, then how many men can finish the job in 13 days?
Answer: 13
Text Explanation :
Let the work done by one man and one machine per day be x and y respectively.
Three men and eight machines can finish a job in half the time taken by three machines and eight men to finish the same job.
Since efficiency is inversely proportional to the time taken, so the efficiency of 3 men and 8 machines is twice that of 8 men and 3 machines.
∴ (3x + 8y) = 2(8x + 3y)
∴ 13x = 2y.
So, work done by 13 men in a day = work done by 2 machines in a day.
∴ If two machines can finish the job in 13 days, same work will be done by 13 men in 13 days.
Hence, 13.
Workspace:
Anil alone can do a job in 20 days while Sunil alone can do it in 40 days. Anil starts the job, and after 3 days, Sunil joins him. Again, after a few more days, Bimal joins them and they together finish the job. If Bimal has done 10% of the job, then in how many days was the job done?
- (a)
13
- (b)
12
- (c)
15
- (d)
14
Answer: Option A
Text Explanation :
Anil in one day can do 1/20th of the work
Sunil in one day can do 1/40th of the work
Anil starts the job and Sunil joins him after three days.
So, Anil would have done 3/20th of the work by the time Sunil joins
After Sunil joins, they both would be doing 3/40th of work everyday
Now, it is known that Bimal joins them after some days and finishes 10% of the work (i.e. 1/10th of the work).
Now, Anil alone had done 3/20th of the work in first 3 days and Bimal completes 1/10th of the work
So, in total they would have done 3/20 + 1/10 = 1/4th of the work.
Remaining work = 3/4th, which would be done by Anil and Sunil together.
Anil and Sunil together comple 1/20th + 1/40th = 3/40th work in 1 day.
Hence, time taken by them to comple 3/4th of the work = 10 days.
Combining everything,
Total number of days = 3 + 10 = 13 days.
Hence, option (a).
Workspace:
John gets Rs 57 per hour of regular work and Rs 114 per hour of overtime work. He works altogether 172 hours and his income from overtime hours is 15% of his income from regular hours. Then, for how many hours did he work overtime?
Answer: 12
Text Explanation :
If John works the same number of regular and over-time hours say 'p'
The income would be 57p and 114p
Let's say that he works 'x' hours regular and 'y' hours overtime...
So, the income would be 57x and 114y
we are told that 114y is 15% of 57x
114y = 0.15 × 57x
y = 0.075x
we also know that x + y = 172
therefore, x + 0.075x = 1.075x = 172
x = 160
y = 172 - 160 = 12
Therefore, the number of hours he worked overtime is 12 hours.
Hence, 12.
Workspace:
A tank is fitted with pipes, some filling it and the rest draining it. All filling pipes fill at the same rate, and all draining pipes drain at the same rate. The empty tank gets completely filled in 6 hours when 6 filling and 5 draining pipes are on, but this time becomes 60 hours when 5 filling and 6 draining pipes are on. In how many hours will the empty tank get completely filled when one draining and two filling pipes are on?
Answer: 10
Text Explanation :
Let a filling pipe fills the tank at ‘a’ liters per hour and a draining pipe drains at ‘b’ liters per hour.
Work done by 6 filling and 5 emptying pipes in 6 hours = 6(6a - 5b)
Work done by 5 filling and 6 emptying pipes in 60 hours = 60(5a - 6b)
6(6a – 5b) = 60(5a – 6b)
∴ 6a – 5b = 50a – 60b
∴ 44a = 55b
∴ a = 5b/4
Let the tank gets filled completely in ‘m’ hours when one draining pipe and two filling pipes are on.
Work done by 2 filling and 1 emptying pipes in m hours = m(2a - b)
∴ m(2a – b) = 6(6a – 5b)
∴ =
∴ =
Solving this, we get m = 10
Hence, 10.
Workspace:
Humans and robots can both perform a job but at different efficiencies. Fifteen humans and five robots working together take thirty days to finish the job, whereas five humans and fifteen robots working together take sixty days to finish it. How many days will fifteen humans working together (without any robot) take to finish it?
- (a)
36
- (b)
32
- (c)
45
- (d)
40
Answer: Option B
Text Explanation :
Let work-done by a human and a robot in a day are ‘h’ and ‘r’ respectively.
Work done by 15 human and 5 robot in 30 days = 30(15h + 5r)
Work done by 5 human and 15 robot in 60 days = 60(5h + 15r)
∴ Total work = 30(15h + 5r) = 60(5h + 15r)
⇒ 15h + 5r = 10h + 30r
⇒ h = 5r
Let fifteen humans take ‘y’ days to finish the job.
Work done by 15 humans in y days = y × 15
∴ 30(15h + 5r) = y × 15h
⇒ 30(15h + h) = y × 15h
⇒ y = 32
Hence, option (b).
Workspace:
When they work alone, B needs 25% more time to finish a job than A does. They two finish the job in 13 days in the following manner: A works alone till half the job is done, then A and B work together for four days, and finally B works alone to complete the remaining 5% of the job. In how many days can B alone finish the entire job?
- (a)
18
- (b)
22
- (c)
16
- (d)
20
Answer: Option D
Text Explanation :
Ratio of time taken by A and B to complete a certain task = 4 : 5.
∴ Ratio of their efficiencies = 5 : 4.
Now, A completes 50% of the work alone and B completes 5% of the work alone. That means, A and B together complete 45% of the work together in 4 days.
Out of this 45%, work done by A and B individually will be in the ratio of their efficiencies.
∴ Work done by B in 4 days = 4/9th of 45% = 20%. [Since the ratio of their efficiencies is 5 : 4]
⇒ B completes 20% (i.e., 1/5th) of the work in 4 days.
Thus, B alone can finish the entire job in 20 days.
Hence, option (d).
Workspace:
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