time and work for CAT

Time and Work Concepts for CAT Questions [LCM Method]

Time and Work Concepts for CAT Questions [LCM Method]
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Problems based on time and work which are asked in CAT can easily be solved by applying an alternate concept which is the method of LCM. Using LCM method, Time and Work problems which appear in CAT are solved quickly and efficiently. In this article, I will be focusing on the application of this method. Let us take some CAT like question on Time and Work to understand this concept.

Question 1: Rakesh alone can do a work in 10 days. Brijesh can do the same job in 15 days. If both Rakesh and Brijesh work together, then in how many days the work will get completed?

Explanation:

Assume that the total work to be LCM of 10 and 15 =30 units.

Rakesh can complete 30 units of work in 10 days. That means in one day Rakesh can complete 30/10 = 3 units of work.

Similarly, Brijeh can complete 30 units of work in 15 days. Or, we can say that in one day Brijesh can complete 30/15 = 2 units of work.

Now, if both Rakesh and Brijesh work together the amount of work completed by them in one day =3 + 2 =5 units of the work.

Therefore, the number of days taken by them together to complete the work = 30/5 = 6 days.

From the above example, it is clear that the method of LCM is not only simple to understand but also easy to apply.

A student who is very new to this topic must deliberately apply this method wherever possible to get hold of the approach.

Shortcut: If two people take X and Y days respectively to complete a work alone, then together they will complete it in XxY/(X+Y) days.

Let us take more examples to see how we apply the method of LCM to get the answer.

Question 2: Arun, Barun and Tarun can do a work alone in 10, 12 and 15 days respectively. If all three work together then in how many days does the job get completed?

Explanation:
LCM of 10,12 and 15 = 60.

Assume that the total work is of 60 units.

Amount of work done by Arun in one day = 60/10 = 6 units per day.

Similarly, the amount of work done by Barun in one day = 60/12 = 5 units per day.

And the amount of work done by Tarun 60/15 = 4 units per day.

So if all three together work then the amount of work done in one day = 6 + 5 + 4 = 15 units per day

Therefore, the number of days required to complete the work = 60/15 = 4 days.

Question 3: Tilak, Mukesh, and Sonali can complete a work in 10, 12 and 15 days respectively. All three agree to complete the work together. After 2 days, Tilak leaves the work. Mukesh left the job 3 days before the completion of the work. Sonali alone continues until the work got completed. How many days did it take to complete the task?

Explanation:
LCM of 10,12 and 15 = 60.

Assume that the total work is of 60 units and let us assume that it took X days for it to complete.

Work done by Tilak= 60/10 = 6 units per day.

Similarly, the work done by Mukesh =60/12 = 5 units per day.

And, the work done by Sonali = 60/50 = 4 unit per day.

We know all three didn’t contribute till the end of the work.

As Tilak worked only for 2 days, his contribution= 6 x 2 = 12 units.

Also, Mukesh left the work 3 days before the completion, so he worked for X-3 days. Therefore, his contribution = 5(X-3) units.

Similarly, the contribution of Sonali = 4X.

If we add the contribution of all three, then the total unit has to be equal to 60.

Or, 12 +5(X-3) +4X=60.

Or, X=7 days.

From the above three examples, it is clear that the method of LCM is quite helpful in solving elementary questions based on time and work.

There are other varieties of questions which are framed from time and work, and it is always advisable to solve all the variations to make oneself comfortable with this topic.

Let us discuss such types of questions from time and work, which frequently appear in competitive exams.

Time and Work Concept of efficiency for CAT

If the same work is given to two people, say A and B, and if A is more efficient than B, then it is evident that A will take less number of days than B to complete the same work. We can also deduce that efficiency is inversely proportional to the number of days taken to finish the job.

Important: If A can finish a work in X days and B is K times as efficient as A, then the time required by both A and B working together to finish the job will be X/(1+K).

Question 4: Mayank can finish a work in 10 days. Mahesh is 2 times as efficient as Mayank. If they work together, in how many days will the work be completed?

Normal method

Let us assume that Mayank complete 1 unit of work per day. Also, the total work =10 units.

Since Mahesh is two times as efficient as Mayank, then Mahesh can do 2 units of work per day.

Now, if they both work together, then the unit of work they can complete in one day equals one plus two i.e. three units.

Therefore, the number of days taken by them together =10/3 days

Applying the direct shortcut

Comparing with the formula we have X =10 and K=2.

Therefore, the number of days taken by them together =10/3 days

Important: A is K times as good a worker as B, and takes X days less than B to finish the work. Then the amount of time required by A and B working together=KX/(K^2-1) days.

Question 5: Rahul is three times as good a worker as Durgesh and takes 8 days less than Durgesh to finish the work; find the amount of time required by Rahul and Durgesh working together to do the job?

Normal method

Let us assume that Rahul alone takes x days to complete the work.

As we know, efficiency is inversely proportional to days, the number of days taken by Durgesh to complete the same work = 3x.

Therefore, Rahul is taking 3x-x = 2x days less than Durgesh; thus, 2x = 8 or x = 4.

Hence we can say that Rahul and Durgesh alone take 4 and 12 days respectively to complete the work.

Therefore together they will be taking 4 x 12/(4+12)=3 days.

Shortcut:  using the formula we get = 3×8/(3^2-1)=3 days.

Let us take more questions based on the above concepts which will strengthen our understanding in solving problems related to time and work.

Solved questions on time and work for CAT Exam

Question 1: Ram and Shyam working together can finish a work in 12 days. If Ram alone can do the same job in 20 days, then in how many days Shyam alone can complete the task?

Explanation:

LCM of 12 and 20 equal 60.

Assume that the total work is of 60 units.

Work done by Ram in one day = 60/20 = 3 units.

Similarly, work done by both Ram and Shyam in one-day = 60/12 =5 units.

Therefore, work done by Shyam in one day equals 5-3 =2 units.

Hence, the time taken by Shyam to complete the work alone = 60/2 = 30 days.

Question 2: Three friends, Akash, Bhairav, and Charan can do a piece of work in 12, 18 and 24 days respectively. They work at it together. Akash stops the job after 4 days, and Bhairav is called off two days before the work is finished. In how many days was the work completed?

Explanation:

LCM of 12, 18 and 24 =72.

Total Work= 72 units.

Akash can complete 72/12 = 6 units of work per day.

Bhairav and complete 72/18 = 4 units of work per day.

Charan can complete 72/24 = 3 units of work per day.

Let us assume that it took X days to complete the work.

As Akash worked only for 4 days, the units of work contributed by him= 4×6 =24 units.

Similarly, contribution of Bhairav =4(X-2) units and contribution of Charan = 3X units.

Adding the contributions of all 3 we get

24+4(X-2) +3X=72.

Or, X=8 days.

Question 3: Two candles of the same length are lighted at the same time. The first is consumed in 6 hours and the second in 4 hours. Assuming that each candle burns at a constant rate in how many hours after being lighted was the first candle twice the length of the second?

Explanation:

LCM of 6 and 4 = 12. Let us assume that both the candles are of length 12 cm each.

The rate of burning of first candle = 12/6=2 cm per hour.

Similarly, the rate of burning of second candle = 12/4 = 3 cm per hour.

Also, let us assume that it takes x hours after which the first candle is twice the length of the second.

The length of the first candle burned in x hours=2x cms. So, the length remaining = (12-2x) cm

The length of the second candle burned in x hours = 3x cms. So, the length remaining = (12-3x) cm

As per the question,

12-2x=2(12-3x)

x=3 hours.

Work Equivalent Concept

This is a based on the universal concept of measuring work in two parameters; Agent who is doing work and the duration the work takes to get completed. Let us understand with the help of an example.

Suppose the time taken to complete a task by 4 men all of the equal efficiency is 8 days. But say we need to get the work done in 4 days only. The logical deduction is that we need to increase the manpower. But how much more men required? Yes, everyone can answer that it will require 8 men to complete the work in 4 days.  BY applying simple logic we can solve this question, but we will see the mathematical calculation behind this reasoning.

From the first scenario, i.e. 4 men can complete the task in 8 days; we measured the work in terms of effort and time required. That is,

Work = Number of Men \( \times \) Number of days (we will keep the unit of measurement of work as man-days)

So, when the number of days decreases to half, to the keep the measurement of the work same, we have to double the manpower.

Some Standard Results:

  1. When work (W) is constant then, the number of men (M) is inversely proportional to the number of days (D) required to complete the work. i.e., \(M \propto \frac{1}{D}\).

Example: As we saw in the above example, in both cases work remained the same, Number of men was inversely proportional to the number of days

  1. If the number of days (D) is constant, then work is directly proportional to the number of men working. i.e., more men more work and vice versa.
  1. If the number of men (M) is constant, then work is directly proportional to the number of days. i.e., more days more work and vice versa.

Let us see some simple solved examples based on work equivalence:

Question 1: If 5 men take 90 days to complete a piece of work then how many days will 15 men take to complete the same work?

Solution:

Here the total work can be simply expressed as 5\( \times \)90 man-days

Now to finish the same job, the measurement of the work should be the same, i.e. 5\( \times \)90 man-days.

Let 15 men take x days to finish then by using the concept of work equivalence

15x = 5\( \times \). 90 or x = 30 days.

Alternate approach: As work is constant, \(M \propto \frac{1}{D}\). Here the number of men going up three times (from 5 to 15), hence the number of days required will be 1/3rd of 90 = 30 days.

Question 2: It takes 6 technicians a total of 10 hours to build a new server from Direct Computer, with each working at the same rate. If six technicians start to build the server at 11 am and one technician per hour is added beginning at 5 pm, at what time will the server be complete? (CAT 2002)

Solution:

Application of work equivalence concept:

Total work = 6\( \times \)10 = 60 man-hours

From 11 am to 5 pm, i.e. for 6 hours 6 technicians worked. The portion of work completed by them = 6\( \times \)6 = 36 man-hours. So, 60-36 = 24 man-hours work is pending.

From 5 pm onwards, at every interval of one hour, one more technician joins them.

Hence amount of work done between 5 pm and 6 pm = 7 men\( \times \) 1 hour = 7 man-hours.

Similarly,

Amount of work done between 6 pm and 7 pm = 8 men\( \times \) 1 hour = 8 man-hours and between 7 pm and 8 pm = 9 men\( \times \) 1 hour = 9 man-hours.

Observe that 7 + 8 +9 = 24 man-hours. Therefore, the task gets completed by 8 pm.

Different persons with different efficiencies working together

There is a type of problem in which different persons working together have different efficiencies. In such questions, the approach to solve them varies slightly. Let us take an example to understand.

Question 3: 2 men and 3 women can complete a work in 16 days, while 3 men and 7 women can complete it in 10 days. In how many days will 5 women complete it?

Solution:

Let each man and each woman work x and y units of work per day respectively.

From the first condition, units of work done in one day are 2x + 3y.

So the total units of work = (2x + 3y)\( \times \)16

Similarly, from the second condition, total units of work = (3x + 7y)\( \times \)10.

Since both are same work we get

(2x + 3y)\( \times \)16 = (3x + 7y)\( \times \)10

Or 32x + 48y = 30x + 70y

=> 2x = 22y

=> \(\frac{x}{y} = \frac{{11}}{1}\) so let us assume that x = 11 and y= 1.

Therefore, the total units of work = (2x + 3y)\( \times \)16 = (2\( \times \)11 + 3\( \times \)1)\( \times \)16=400 units.

Hence 5 women can complete the work in 400/5 = 80 days.

Constantly increasing work with the increase in time

Question 4: 29 horses can graze a uniformly growing grass field in 7 days, 25 horses can do the same in 9 days. In how many days can 32 horses graze the same field?

Solution:

Let the amount of grass on day one was x units and y units of grass grow per day.

From the first condition, the field was completely grazed in 7 days. Therefore, amount of work done = x + 7y. And to complete this work, 29 horses took 7 days, hence the relation we get is:

x + 7y = 29\( \times \)7 …. eq (1)

Similarly, from the second condition we get:

x + 9y = 25\( \times \)9 …. eq (2)

Solving both the equations for x and y we get, x = 126 and y = 11.

Now, let us take the number of days in which 32 horses graze the same field is N.

Therefore, the required relation is:

x + Ny = 32N

Or 126 + 11N = 32N

Or N = 6 days

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Time and Work

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