How to Prepare for Quantitative Aptitude for CAT (2014)

Block III: Arithmetic and Word-based Problems

Chapter 9. TIME AND WORK

INTRODUCTION

The concept of time and work is another important topic for the aptitude exams. Questions on this chapter have been appearing regularly over the past decade in all aptitude exams. Questions on Time and Work have regularly appeared in the CAT especially in its online format.

Theory

In the context of the CAT, you have to understand the following basic concepts of this chapter:

If A does a work in a days, then in one day A does Æ of the work.

If B does a work in b days, then in one day B does Æ of the work.

Then, in one day, if A and B work together, then their combined work is + .

or

In the above case, we take the total work to be done as “1 unit of work”. Hence, the work will be completed when 1 unit of work is completed.

For example, if A can do a work in 10 days and B can do the same work in 12 days, then the work will be completed in how many days.

One day’s work = 1/10 + 1/12 = (12 + 10)/120

[Taking LCM of the denominators]

= 22/120

Then the number of days required to complete the work is 120/22.

Note that this is a reciprocal of the fraction of work done in one day. This is a benefit associated with solving time and work through fractions. It can be stated as—the number of time periods required to complete the full work will be the reciprocal of the fraction of the work done in one time period.

ALTERNATIVE APPROACH

Instead of taking the value of the total work as 1 unit of work, we can also look at the total work as 100 per cent work. In such a case, the following rule applies:

If A does a work in a days, then in one day A does Æ % of the work.

If B does a work in b days, then in one day B does Æ % of the work.

Then, in one day, if A and B work together, then their combined work is

This is often a very useful approach to look at the concept of time and work because thinking in terms of percentages gives a direct and clear picture of the actual quantum of work done.

What I mean to say is that even though we can think in either a percentage or a fractional value to solve the problem, there will be a thought process difference between the two.

Thinking about work done as a percentage value gives us a linear picture of the quantum of the work that has been done and the quantum of the work that is to be done. On the other hand, if we think of the work done as a fractional value, the thought process will have to be slightly longer to get a full understanding of the work done.

For instance, we can think of work done as 7/9 or 77.77%. The percentage value makes it clear as to how much quantum is left. The percentage value can be visualised on the number line, while the fractional value requires a mental inversion to fully understand the quantum.

An additional advantage of the percentage method of solving time and work problems would be the elimination of the need to perform cumbersome fraction additions involving LCMs of denominators.

However, you should realise that this would work only if you are able to handle basic percentage calculations involving standard decimal values. If you have really internalised the techniques of percentage calculations given in the chapter of percentages, then you can reap the benefits for this chapter.

The benefit of using this concept will become abundantly clear by solving through percentages the same example that was solved above using fractions.

Example: If A can do a work in 10 days and B can do the same work in 12 days, then the work will be completed in how many days.

One day’s work = 10% + 8.33% = 18.33% (Note, no LCMs required here)

Hence, to do 100% work, it will require: 100/18.33.

This can be solved by adding 18.33 mentally to get between 5–6 days. Then on you can go through options and mark the closest answer.

The process of solving through percentages will yield rich dividends if and only if you have adequate practice on adding standard percentage values. Thus, 18.33 × 5 = 91.66 should not give you any headaches and should be done while reading for the first time.

Thus a thought process chart for this question should look like this.

If A can do a work in 10 days (Æ means 10% work) and B can do the same work in 12 days (Æ 8.33% work Æ 18.33% work in a day in 5 days 91.66% work Æ leaves 8.33% work to be done Æ which can be done in 8.33/18.33 of a day = 5/11 of a day (since both the numerator and the denominator are divisible by 1.66), then the work will be completed in 5days.

The entire process can be done mentally.

The Concept of Negative Work

Suppose, that A and B are working to build a wall while C is working to break the wall. In such a case, the wall is being built by A and B while it is being broken by C. Here, if we consider the work as the building of the wall, we can say that C is doing negative work.

Example: A can build a wall in 10 days and B can build it in 5 days, while C can completely destroy the wall in 20 days. If they start working at the same time, in how many days will the work be completed.

Solution: The net combined work per day here is:

A’s work + B’s work – C’s work = 10% + 20% – 5% = 25% work in one day.

Hence, the work will get completed (100% work) in 4 days.

The concept of negative work commonly appears as a problem based on pipes and cisterns, where there are inlet pipes and outlet pipes/leaks which are working against each other.

If we consider the work to be filling a tank, the inlet pipe does positive work while the outlet pipe/leak does negative work.

Application of Product Constancy Table o Time and Work

The equation that applies to Time and Work problems is

Work Rate × Time = Work done

This equation means that if the work done is constant, then Æ

Work rate is inversely proportional to time. Hence, the Product Constancy Table will be directly applicable to time and work questions.

[Notice the parallelism between this formula and the formula of time speed and distance, where again there is product constancy between speed and time if the distance is constant.]

Time is usually in days or hours although any standard unit of time can be used. The unit of time that has to be used in a question is usually decided by the denominator of the unit of work rate.

Here, there are two ways of defining the Work rate.

(a)In the context of situations where individual working efficiencies or individual time requirements are given in the problem, the work rate is defined by the unit: Work done per unit time.

In this case, the total work to be done is normally considered to be 1 (if we solve through fractions) or 100% (if we solve through percentages).

Thus, in the solved problem above, when we calculated that A and B together do 18.33% work in a day, this was essentially a statement of the rate of work of A and B together.

Then the solution proceeded as:

18.33% work per day × No. of days required = 100% work

Giving us: the no. of days required = 100/18.33 =

(b)In certain types of problems (typically those involving projects that are to be completed), where a certain category of worker has the same rate of working, the Work rate will be defined as the number of workers of a particular category working on the project.

For instance, questions where all men work at a certain rate, the work rate when 2 men are working together will be double the work rate when 1 man is working alone. Similarly, the work rate when 10 men are working together will be 10 times the work rate when 1 man is working alone.

In such cases, the work to be done is taken as the number of man-days required to finish the work.

Note, for future reference, that the work to be done can also be measured in terms of the volume of work defined in the context of day-to-day life.

For example, the volume of a wall to be built, the number of people to be interviewed, the number of chapattis to be made and so on.

WORK EQUIVALENCE METHOD
(To Solve Time and Work Problems)

The work equivalence method is nothing but an application of the formula:

Work rate × Time = Work done (or work to be done)

Thus, if the work to be done is doubled, the product of work rate × time also has to be doubled. Similarly, if the work to be done increases by 20%, the product of work rate × time also has to be increased by 20% and so on.

This method is best explained by an example:

A contractor estimates that he will finish the road construction project in 100 days by employing 50 men.

However, at the end of the 50th day, when as per his estimation half the work should have been completed, he finds that only 40% of his work is done.

(a)How many more days will be required to complete the work?

(b)How many more men should he employ in order to complete the work in time?

Solution:

(a)The contactor has completed 40% of the work in 50 days.

If the number of men working on the project remains constant, the rate of work also remains constant. Hence, to complete 100% work, he will have to complete the remaining 60% of the work.

For this he would require 75 more days. (This calculation is done using the unitary method.)

(b)In order to complete the work on time, it is obvious that he will have to increase the number of men working on the project.

This can be solved as:

50 men working for 50 days Æ 50 × 50 = 2500 man-days.

2500 man-days has resulted in 40% work completion. Hence, the total work to be done in terms of the number of man-days is got by using unitary method:

Work left = 60% = 2500 × 1.5 = 3750 man-days

This has to be completed in 50 days. Hence, the number of men required per day is 3750/50 = 75 men.

Since, 50 men are already working on the project, the contractor needs to hire 25 more men.

[Note, this can be done using the percentage change graphic for product change. Since, the number of days is constant at 50, the 50% increase in work from 40% to 60% is solely to be met by increasing the number of men. Hence, the number of men to be increased is 50% of the original number of men = 25 men.]

The Specific Case of Building a Wall (Work as volume of work)

As already mentioned, in certain cases, the unit of work can also be considered to be in terms of the volume of work. For example, building of a wall of a certain length, breadth and height.

In such cases, the following formula applies:

where L, B and H are respectively the length, breadth and height of the wall to be built, while m, t and d are respectively the number of men, the amount of time per day and the number of days. Further, the suffix 1 is for the first work situation, while the suffix 2 is for the second work situation.

Consider the following problem:

Example: 20 men working 8 hours a day can completely build a wall of length 200 meters, breadth 10 metres and height 20 metres in 10 days. How many days will 25 men working 12 hours a day require to build a wall of length 400 meters, breadth 10 metres and height of 15 metres.

This question can be solved directly by using the formula above

Then we get (200 × 10 × 20)/(400 × 10 × 15) = (20 × 8 × 10)/(25 × 12 × d₂)

\ d₂ = 5.333/0.6666 = 8 days

Alternatively, you can also directly write the equation as follows:

d₂ = 10 × (400/200) × (10/10) × (20/15) × (20/25) × (8/12)

This can be done by thinking of the problem as follows:

The number of days have to be found out in the second case. Hence, on the LHS of the equation write down the unknown and on the RHS of the equation write down the corresponding knowns.

d₂ = 10 × ....

Then, the length of the wall has to be factored in. There are only two options for doing so. viz:

Multiplying by 200/400 (< 1, which will reduce the number of days) or multiplying by 400/200 (>1, which will increase the number of days).

The decision of which one of these is to be done is made on the basis of the fact that when the length of the wall is increasing, the number of days required will also increase.

Hence, we take the value of the fraction greater than 1 to get

d₂ = 10 × (400/200)

We continue in the same way to get

No change in the breadth of the wall Æ hence, multiply by 10/10 (no change in d₂)

Height of the wall is decreasing Æ hence, multiply by 15/20 (< 1 to reduce d₂)

Number of men working is increasing Æ hence, multiply by 20/25 (< 1 to reduce d₂)

Number of hours per day is increasing Æ hence, multiply by 8/12 (< 1 to reduce the number of days)

The Concept of Efficiency

The concept of efficiency is closely related to the concept of work rate.

When we make a statement saying A is twice as efficient as B, we mean to say that A does twice the work as B in the same time. In other words, we can also understand this as A will require half the time required by B to do the same work.

In the context of efficiency, another statement that you might come across is A is two times more efficient than B. This is the same as A is thrice as efficient as B or A does the same work as B in 1/3rd of the time.

Equating Men, Women and Children This is directly derived from the concept of efficiencies.

Example: 8 men can do a work in 12 days while 20 women can do it in 10 days. In how many days can 12 men and 15 women complete the same work.

Solution: Total work to be done = 8 × 12 = 96 man-days.

or total work to be done = 20 × 10 = 200 woman-days.

Since, the work is the same, we can equate 96 man-days = 200 woman-days.

Hence, 1 man-day = 2.08333 woman-days.

Now, if 12 men and 15 women are working on the work we get

12 men are equal to 12 × 2.08333 = 25 women

Hence, the work done per day is equivalent to 25 + 15 women working per day.

That is, 40 women working per day.

Hence, 40 × no. of days = 200 woman days

Number of days = 5 days.

WORKED-OUT PROBLEMS

Problem 9.1 A can do a piece of work in 10 days and B in 12 days. Find how much time they will take to complete the work under the following conditions:

(a)Working together

(b)Working alternately starting with A.

(c)Working alternately starting with B.

(d)If B leaves 2 days before the actual completion of the work.

(e)If B leaves 2 days before the scheduled completion of the work.

(f)If another person C who does negative work (i.e. works against A and B and can completely destroy the work in 20 days) joins them and they work together all the time.

Solution

(a)1 day’s work for A is 1/10 and 1 day’s work for B is 1/12.

Then, working together, the work in one day is equal to:

= of the work. Thus working together they need 60/11 days to complete the work Æ 5.45 days.

Alternately, you can use percentage values to solve the above question:

A’s work =10%, B’s work = 8.33%. Hence, A + B = 18.33% of the work in one day.

Hence, to complete 100% work, we get the number of days required = 100/18.33 Æ 5.55 days.

This can be calculated as

@ 18.33% per day in 5 days, they will cover 18.33 × 5 = 91.66%. (The decimal value 0.33 is not difficult to handle if you have internalised the fraction to percentage conversion table of the chapter of percentages).

Work left on the sixth day is: 8.33%, which will require: 8.33/18.33 of the sixth day.

Since, both these numbers are divisible by 1.66 we get 5/11 of the sixth day will be used Æ 0.45 of the sixth day is used.

Hence, 5.45 days are required to finish the work.

Note: Although the explanation to the question through percentages seems longer, the student should realise that if the values in the fraction-to-percentage table is internalised by the student, the process of solution through percentage will take much lesser time because we are able to eliminate the need for the calculation of LCMs, which are often cumbersome. (if the numbers in the problem are those that are covered in the fraction to percentage conversion table). In fact, the percentage method allows for solving while reading.

(b)Working alternately: When two people are working alternately the question has to be solved by taking 2 days as a unit of time instead of 1 day.

So in (a) above, the work done in 1 day will be covered in 2 days here.

Thus, in 2 days the work done will be 18.33%. In 10 days it will be 91.66%. On the 11th day A works by himself.

But A’s work in 1 day is 10%. Therefore, he will require 4/5 of the 11th day to finish the work.

(c)Working alternately starting with B: Here, there will be no difference in work completed by the 10th day. On the 11th day, B works alone and does 8.33% of the work (which was required to complete the work). Hence, the whole of the 11th day will get used.

(d)If B leaves 2 days before the actual completion of the work: In this case, the actual completion of the work is after 2 days of B’s leaving. This means, that A has worked alone for the last 2 days to complete the work. But Adoes, 10% work in a day. Hence, A and B must have done 80% of the work together (@18.33% per day).

Then, the answer can be found by

80/18.33 + 20/10 days.

Note: For calculation of 80/18.33, we can use the fact that the decimal value is a convenient one. If they worked together they would complete 73.33% of the work in 4 days and the work that they would have done on the 5th day would be 6.66%.

At the rate of 18.33% work per day while working together, they would work together for 6.66/18.33 of the 5th day. Since both the numerator and denominator are divisible by 1.66 the above ratio is converted into 4/11 = 0.3636.

Hence, they work together for 4.3636 days after which B leaves and then A completes the work in 2 more days. Hence, the time required to finish the work would be = 6.3636 days.

(e)If B leaves 2 days before the scheduled completion of the work: Completion of the work would have been scheduled assuming that A and B both worked together for completing the work (say, this is x days). Then, the problem has to be viewed as x – 2 days was the time for which A and B worked together. The residual amount of work left (which will be got by 2 days work of A and B together) would be done by A alone at his own pace of work.

Thus we can get the solution by:

Number of days required to complete the work = [(100/18.33) – 2] +

(f)If C joins the group and does negative work, we can see that one day’s work of the three together would be

A’s work + B’s work – C’s work = 10% + 8.33% – 5% = 13.33%

Hence, the work will be completed in (100/13.33) days.

[Note: This can be calculated by 13.33 × 7 = 13 × 7 + 0.33 × 7 = 93.33.

Then, work left = 6.66, which will require half a day more at the rate of 13.33% per day.

Advantage of Solving Problems on Time and Work through Percentages Students should understand here, that most of the times the values given for the number of days in which the work is completed by a worker will be convenient values like: 60 days, 40 days, 30 days, 25 days, 24 days, 20 days, 16 days, 15 days, 12 days, 11 days, 10 days, 9 days, 8 days, 7 days, 6 days, 5 days, 4 days, 3 days and 2 days. All these values for the number of days will yield convenient decimal values. If your fraction to percentage table is internalised, you can use the process of solving while reading by taking the percentage of work done per day process rather than getting delayed by the need to find LCM’s while solving through the process of the fraction of work done per day.]

Problem 9.2 A contractor undertakes to build a wall in 50 days. He employs 50 people for the same. However, after 25 days he finds that the work is only 40% complete. How many more men need to be employed to:

(a)complete the work in time?

Solution In order to complete the work in time, the contractor has to finish the remaining 60% of the work in 25 days.

Now, in the first 25 days the work done = 50 × 25 =1250 man-days Æ 40% of the work.

Hence, work left = 60% of the work = 1875 man-days.

Since, 25 days are left to complete the task, the number of people required is 1875/25 = 75 men.

Since, 50 men are already working, 25 more men are needed to complete the work.

Thought process should go like: 1250 Æ 40% of work. Hence, 1875 man-days required to complete the work.

Since there are only 25 days left, we need 1875/25 = 75 men to complete the work.

(b)Complete the work 10 days before time?

For this purpose, we have to do 1875 man-days of work in 15 days. Hence, men = 1875/15 = 125 men.

Problem 9.3 For the previous problem, if the contractor continues with the same workforce:

(a)how many days behind schedule will the work be finished?

Solution He has completed 40% work in 25 days. Hence, to complete the remaining 60% of the work, he would require 50% more days (i.e. 37.5 days) (Since, 60% is 1.5 times of 40%)

Hence, the work would be done 12.5 days behind schedule.

(b)how much increase in efficiency is required from the work force to complete the work in time?

Solution If the number of men working is kept constant, the only way to finish the work in time is by increasing the efficiency so that more work is done every man-day.

This should be mathematically looked at as follows:

Suppose, that 1 man-day takes care of 1 unit of work.

Then, in the first 25 days, work done = 25 (days) × 50 (men) × 1 (work unit per man-day) = 1250 units of work.

Now, this 1250 units of work is just 40% of the work.

Hence, work left = 1875 units of work.

Then, 25 (days) × 50 (men) × z (work units per man-day) = 1875 Æ z = 1.5

Thus, the work done per man-day has to rise from 1 to 1.5, that is, by 50%. Hence, the efficiency of work has to rise by 50%.

Problem 9.4 A is twice as efficient as B. If they complete a work in 30 days find the times required by each to complete the work individually.

Solution When we say that A is twice as efficient as B, it means that A takes half the time that B takes to complete the same work.

Thus, if we denote A’s 1 day’s work as A and B’s one day’s work as B, we have

A = 2B

Then, using the information in the problem, we have: 30 A + 30 B = 100% work

That is, 90 B = 100% work Æ B = 1.11 % (is the work done by B in 1 day) Æ B requires 90 days to complete the work alone.

Since, A = 2B Æ we have A = 2.22 % Æ A requires 45 days to do the work alone.

You should be able to solve this mentally with the following thought process while reading for the first time:

= 3.33%. = 1.11%. Hence, work done is 1.11% per day and 2.22% per day Æ 90 and 45 days.

Problem 9.5 A is two times more efficient than B. If they complete a work in 30 days, then find the times required by each to complete the work individually.

Solution Interpret the first sentence as A = 3B and solve according to the process of the previous problem to get he answers. (You should get A takes 40 days and B takes 120 days.)

LEVEL OF DIFFICULTY (I)

LEVEL OF DIFFICULTY (II)

Directions for Questions 6 to 10: Read the following and answer the questions that follow.

A set of 10 pipes (set X) can fill 70% of a tank in 7 minutes. Another set of 5 pipes (set Y) fills 3/8 of the tank in 3 minutes. A third set of 8 pipes (set Z) can empty 5/10 of the tank in 10 minutes.

LEVEL OF DIFFICULTY (III)

Directions for Questions 1 to 10: Study the following tables and answers the questions that follow.

Darbar Toy Company has to go through the following stages for the launch of a new toy:

The profile of the company’s manpower is

Directions for Questions 11 to 20: Read the following and answer the questions that follow.

A fort contains a granary, that has 1000 tons of grain. The fort is under a siege from an enemy army that has blocked off all the supply routes.

The army in the fort has three kinds of soldiers:

Sepoys Æ 2,00,000.

Mantris Æ 1,00,000

Footies Æ 1,00,000

100 Sepoys can hold 5% of the enemy for one month.

100 Mantris can hold 10% of the enemy for 15 days.

50 Footies can hold 5% of the enemy for one month.

A sepoy eats 1 kg of food per month, a Mantri eats 0.5 kg of food per month and a footie eats 3 kg of food. (Assume 1 ton = 1000 kg).

The king has to make some decisions based on the longest possible resistance that can be offered to the enemy.

If a king selects a soldier, he will have to feed him for the entire period of the resistance. The king is not obliged to feed a soldier not selected for the resistance.

(Assume that the entire food allocated to a particular soldier for the estimated length of the resistance is redistributed into the king’s palace in case a soldier dies and is not available for the other soldiers.)

Directions for Questions 21 to 25: Study the following and answer the questions that follow.

A gas cylinder can discharge gas at the rate of 1 cc/minute from burner A and at the rate of 2 cc/minute from burner B (maximum rates of discharge). The capacity of the gas cylinder is 1000 cc of gas.

The amount of heat generated is equal to 1 kcal per cc of gas.

However, there is wastage of the heat as per follows:

@(Include higher extremes)

Level of Difficulty (I)

Hints

Level of Difficulty (II)

2.They drill at planned rate for 3 days (say x)

3x + 5(x + 8) = 280

5.Work left = 1/7th of original work. A would have done it in z/3 hours.

6–10.First set of ten pipes operate at 10%/min (filling) i.e. Filling done by 1 pipe = 1%/minute

Second set of five pipes operate at 12.5%/minute (filling)

i.e. Filling done by 1 pipe = 2.5%/minute

Set Z (Emptying) = 5%/minute.

Emptying per pipe = 0.625%/minute.

11.8A = 12B = 15D.

13.Since (rate of filling or emptying) a diameter, the emptying rate is 10 gallons/minute.

14.The net flow in the original situation is 5 litres/minutes. The required answer will be such that when it is doubled, the increase in emptying rate (in litres/minute) should be more than 5 litres/minute.

15.Required answer = .

17.Ditch dug per day = = 54 metres

\ x + (x + y) + (x + 2y) = 54

10(x + 2y) = 14x

Use options to solve.

18.Work done on the first three days is 37%. Hence, work done on the next 7 days is 63%.

Since, this is A + B’s work we get

One day’s work of (A + B) = 9%

Also, 5A = 4B.

Hence A = 4% and B = 5%

B, turns out to be the fastest worker.

19.Solve through options.

20.Work done by the first typist in 2 minutes = 20%.

Hence, the first typist’s work rate = 10% per minute.

22.Solve through options.

23.Solve through options.

24.Let x hours be required to complete the work together.

Then, = . Check the options to see which one fits the equation.

27.A + B = = 41.66%

(where A and B represent the work per hour of pipes A and B respectively).

Solve using options to see which one fits the remaining conditions of the problem.

For example, if we check option b (4 hrs), then we get that the work of the faster pipe (say A) = 25%. Then B = 16.66%.

Then B was open for 4/2 = 2 hours and A was open for 6/3 = 2 hours.

Work done = 25% × 2 + 16.66% × 2 = 83.33% = of work.

Hence, this option is correct.

28.Let the discharges per pipe be A and B litres/hr.

Then 17(A + B) = 425

i.e. A + B = 25

From this point of the solution, proceed to check the conditions of the question through options.

29.Difference in times required by the first man (A) and the second man (B) = 3 hours. Also, if t_A and t_B are the respective times, then

2t_B – t_A = 8 and t_B – t_A = 3

30.Let the work/minute of the four pipes be A, B, C and D respectively.

Then, D – A = 1.66% and D + A = 5%

Level of Difficulty (III)

1–10.Interpretation of the first row of the first table in the question:

Design and Development requires 30 expert man- days or 60 non-expert man-days.

Hence, work done in 1 expert man-day = 3.33% and work done in 1 non-expert man-day = 1.66%. Further, from the second table, it can be interpreted that: A is an expert at design and development. Hence, his work rate is 3.33% per day and B, D and E are ready to work as non-experts on design and development, hence their work rate is 1.66% per day each.

Thus, in 1 day the total work will be

A + B + D + E = 3.33 + 1.66 + 1.66 + 1.66 = 8.33% work.

Thus, 12 days will be required to finish the design and development phase.

1. + + + + = 40.5

2.Increase of number of days Æ = 4 days.

[This happens since the work rate will drop from 16.66% to 10% due to A and C’s refusal to work.]

3–4.Find out the work done by each of the 5 workers.

11–20.The resistance offered is equal for 100 numbers of all types of soldiers.

11–13.If all sepoys are chosen, the food requirement will be 200 tons/month. The resistance will last for 5 months.

If footies are chosen, the food will last for = 3.33 months.

If mantris are chosen, the food will last for = 20 months.

Hence, all mantris must be chosen.

19–20.For these questions, since food is no longer a constraint, the constraint then becomes the number of lives. Then, the assumption will be that the resistance lasts for one month with a loss of either 2000 sepoys, 2000 mantris or 1000 footies.

19.Length of resistance = = 250 months.

20.In 6 months, the resistance will have lost 12000 mantris. He would also have lost all other soldiers since he has not fed them.

21. = 200.

23.At 1 cc/minute, the loss of heat is 20%. Hence, when 1000 cc of the gas is used, out of the 1000 kcal of heat generated 200 kcal will be lost.

Solutions and Shortcuts

Level of Difficulty (I)

1.He will complete the work in 20 days. Hence, he will complete ten times the work in 200 days.

2.6 men for 12 days means 72 mandays. This would be equal to 4 men for 18 days.

3.A’s one day work will be 5%, while B will do 6.66 % of the work in one day. Hence, their total work will be 11.66% in a day.

In 8 days they will complete Æ 11.66 × 8 = 93.33%

This will leave 6.66% of the work. This will correspond to 4/7 of the ninth day since in 6.66/11.66 both the numerator and the denominator are divisible by 1.66.

4.A’s work = 5% per day

B’s work = 6.66% per day

C’s work = 4% per day.

Total no. of days = 100/15.66 = 300/47 = 6(18/47)

5.N + A = 10%

N = 8.33%

Hence A = 1.66% Æ 60 days.

6.The ratio of the wages will be the inverse of the ratio of the number of days required by each to do he work. Hence, the correct answer will be 3:2 Æ ` 30

7.24 man days + 18 women days = 20 man days + 28 woman days

Æ 4 man days = 10 woman days.

Æ 1 man day = 2.5 woman days

Total work = 24 man days + 18 woman days = 60 woman days + 18 woman days = 78 woman days.

Hence, 1 man + 1 woman = 3.5 women can do it in 78/3.5 = 156/7 = 22(2/7) days.

8.The data is insufficient, since we only know that the work gets completed in 200 boy days and 300 women days.

9.A = 10%, B = 5% and Combined work is 20%. Hence, C’s work is 5% and will require 20 days.

10.In 5 days, A would do 25% of the work. Since, B finishes the remaining 75% work in 10 days, we can conclude that B’s work in a day = 7.5%

Thus, (A + B) = 12.5% per day.

Together they would take 100/12.5 = 8 days.

11.A = 20%, B = 10% and A + B + C = 50%. Hence, C = 20%. Thus, in two days, C contributes 40% of the total work and should be paid 40% of the total amount.

12.Total man days required = 600 man days. If 5 workers leave the job after ‘n’ days, the total work would be done in 35 days. We have to find the value of ‘n’ to satisfy:

20 × n + (35 – n) × 15 = 600.

Solving for n, we get

20n – 15n + 35 × 15 = 600

5n = 75

n = 15.

13.Let the time taken by Arun be ‘t’ days. Then, time taken by Vinay = 2t days.

1/t + 1/2t = 1/7 Æ t = 10.5

14.Subhash can copy 200 pages in 40 hours (reaction to the first sentence). Hence, Prakash can copy 100 pages in 40 hours. Thus, he can copy 30 pages in 30% of the time: i.e. 12 hours.

15.30X = 20 (X + 6) Æ 10X = 120 Æ X = 12.

16.Sashi = 4%, Rishi = 5%. In five days, they do a total of 45% work. Rishi will finish the remaining 55% work in 11 more days.

17.Raju = 10%, Vicky = 8.33% and Tinku = 6.66%. Hence, total work for a day if all three work = 25%.

In 2 days they will complete, 50% work. On the third day onwards Raju doesn’t work. The rate of work will become 15%. Also, since Vicky leaves 3 days before the actual completion of the work, Tinku works alone for the last 3 days (and must have done the last 6.66 × 3 = 20% work alone). This would mean that Vicky leaves after 80% work is done. Thus, Vicky and Tinku must be doing 30% work together over two days.

Hence, total time required = 2 days (all three) + 2 days (Vicky and Tinku) + 3 days (Tinku alone)

18.Sambhu requires 16 days to do the work while Kalu requires 18 days to do the work.

(1/16 + 1/18) × n = 1

Æ n = 288/34 = 144/17

19.Let Anjay take 3t days, Vijay take 2t days and Manoj take 6t days in order to complete the work. Then we get:

1/3t + 1/2t + 1/6t = 1 Æ t = 1. Thus, Manoj would take 6t = 6 days to complete the work.

20.After 100 days and 4500 mandays, only 1/6^th of the work has been completed. You can use the product change algorithm of PCG to solve this question.

100 × 45 = 16.66% of the work. After this you have 200 days (i.e. 100% increase in the time available) while the product 200 × no. of men should correspond to five times times the original product.

This will be got by increasing the no. of men by 150% (300/200).

21.Since the ratio of money given to Apurva and Amit is 2:3, their work done would also be in the same ratio. Thus, their time ratio would be 3:2 (inverse of 2:3). So, if Apurva takes 12 days, Amit would take 8 days and the total number of days required (t) would be given by the equation:

(1/12 + 1/8)t = 1 Æ t = 24/5 = 4.8 days

22.Raju being twice as good a workman as Vijay, you can solve the following equation to get the required answer:

1/R + 1/2R = 1/14.

Solving will give you that Vijay takes 42 days.

23.40n = 30 (n + 5) Æ n = 15

24.12 × 5 man days + 16 × 5 Boy days

= 13 × 4 man days + 24 × 4 Boy days

Æ 8 man days = 16 Boy days

1 man day = 2 Boy days.

Required ratio of man’s work to boy’s work = 2 : 1.

25.A’s rate of working is 10 per cent per day while B’s rate of working is 5 per cent per day. In 5 days they will complete 75 per cent work. Thus the last 25 per cent would be done by B alone. Working at the rate of 5 per cent per day, B would do the work in 5 days.

26.Work equivalence method:

30 × 5 × 16 = 20 × 6 × n

Gives the value of n as 20 days

27.Ajay’s daily work = 4.1666%, Vijay’s daily work = 3.33% and the daily work of all the three together is 8.33%. Hence, Pradeep’s daily work will be 0.8333%. Hence, he will end up doing 10% of the total work in 12 days. This will mean that he will be paid ` 20.

28.Total work = 15 × 210 = 3150 mandays.

After 100 days, work done = 15 × 100 = 1500 mandays.

Work left = 3150 – 1500 = 1650 mandays.

This work has to be done with 30 men working each day.

The number of days (more) required = 1650/30 = 55 days.

29.A + V + S = 1(1)

A + V = 19/23

V + S = 8/23

Æ A + 2V + S = 27/23(2)

(2)–(1) gives us: V = 4/23.

30.Interpret the starting statement as: Anmol takes 30 days and Vinay takes 90 days. Hence, the answer will be got by:

(1/30 + 1/90) * n = 1

Alternatively, you can also solve using percentages as: 3.33 + 1.11 = 4.44% is the daily work. Hence, the no. of days required is 100/4.44 = 22.5 days.

31.After 27 days, food left = 4 × 200 = 800 soldier days worth of food. Since, now there are only 80 soldiers, this food would last for 800/80 = 10 days. Number of extra days for which the food lasts = 10 – 4 = 6 days.

32.Total work of Anju, Manju and Sanju = 16.66%

Anju’s work = 10%

Manju’s work = 4.166%

Sanju’s work = 2.5%

So Sanju can reap the field in 40 days.

33.Ajay + Vijay = 1/28 and Ajay + Vijay + Manoj = 1/21.

Hence, Manoj = 1/21 – 1/28 = 1/84.

Hence, Manoj will take 84 days to do the work.

34.A + M = 8.33, M + B = 6.66 and A = 2B Æ A’s 1 days work = 3.33%, M’s = 5% and B’s = 1.66%. Thus, Mohan would require 100/5 = 20 days to complete the work if he works alone.

35.A + V = 16.66% and A = 10% Æ V = 6.66%. Consequently Vijay would require 100/6.66 = 15 days to do it alone.

36.The rate of filling will be 20% and the net rate of filling (including the leak) is 16.66%. Hence, the leak accounts for 3.33% per hour. i.e. it will take 30 hours to empty the tank.

37.A + B = 16.66%. From here solve this one using the options. Option (c) fits the situation as it gives us A’s work = 10%, B’s work = 6.66% as also that B takes 5 minutes more than A (as stipulated in the problem).

38.A + B = 5.55 + 11.11 = 16.66. In two days, 33.33% of the work will be done. C adds 16.66% of work to that of A and B. Hence, the rate of working will go to 33.33%. At this rate it would take 2 more days to complete the work.

Hence, in total it will take 4 days to complete the entire work.

39.24 × 8 × 10 = N × 10 × 6 Æ N = 32

40.n × 20 = (n – 12) × 32 Æ n = 32.

41.12 × 18 = 12 × 6 + 16 × t Æ t = 9

42.(A + B)’s work = C’s work.

Also if A takes ‘a’ days

B would take ‘a – 5’ days

and C would take ‘a – 9’ days.

Solving through options, option ‘c’ fits.

A (15 days) Æ A’s work = 6.66%

B (10 days) Æ B’s work = 10%

C (6 days) Æ C’s work = 16.66%

43.The cistern fills in 6 hours normally, means that the rate of filling is 16.66% per hour. With the leak in the bottom, the rate of filling becomes 10% per hour (as it takes 10 hours to fill with the leak).

This means that the leak drains out water at the rate of 6.66% per hour. This in turn means that the leak would take 100/6.66 = 15 hours to drain out the entire cistern.

44.Since the net work of the three taps is 10% and the first and second do 20% + 10% = 30%. Hence, the third pipe must be a waste pipe emptying at the rate of 20% per hour. Hence, the waste pipe will take a total of 5 hours to empty the tank.

45.A’s work = 10%

B’s negative work = 6.66%

(A + B)’s work = 3.33%

To fill a half empty tank, they would take 50/3.33 = 15 hours.

46.The work rate would be 10% on the first day, 5% on the second day and 2.5% on the third day. For every block of 3 days there would be 17.5% work done. In 15 days, the work completed would be 17.5 × 5 = 87.5%. On the sixteenth day, work done = 10% Æ 2.5% work would be left after 16 days. On the 17^th day the rate of work would be 5% and hence it would take half of the 17^th day to complete the work. Thus, it would take 16.5 days to finish the work in this fashion.

47.(A + B) = 2C.

Also,(A + C) = 3B

36(A + B + C) = 1

Solving for C, we get:

36 (2C + C) = 1 Æ 108C = 1

C = 1/108

Hence, C takes 108 days.

48.A + B + C = 19%. In the first two hours they will do 38 % of the work. Further, for the next two hours work will be done at the rate of 15% per hour. Hence, after 4 hours 68% of the work will be completed, when tap B is also closed. The last 32% of the work will be done by A alone. Hence, A does 40% (first 4 days) + 32% = 72% of the work.

49.Without the leak:

Rate of work = 20% + 5% = 25%. Thus, it would have taken 4 hours to complete the work.

Due to the leak the filling gets delayed by 1 hour. Thus, the tank gets filled in 5 hours. This means that the effective rate of filling would be 20% per hour. This means that the rate at which the leak empties the tank is 5% per hour and hence it would have taken 20 hours to empty a filled tank.

50.In 6 days A would do 25% of the work and in 8 days B would do 25% of the work himself. So, C has to complete 50% of the work by himself.

In all C would require 30 days to do 50% of the work. So, he would require 22 more days.

Level of Difficulty (II)

1.25 (n – 12) = 21 n + 300. Solving this equation, n = 150. Hence, the first division harvest 3150 tons.

2.Let n be the number of meters planned per day. Start from the options to find the number of planned days. In the options the 2 feasible values are 30 meters and 27 meters (as these divide 270). Suppose we check for 30 meters per day, the work would have got completed in 9 days as per the original plan. In the new scenario:

3n + 5(n + 8) = 280 Æ n = 30 too. Hence, this option is correct.

Note that if we tried with 27 meters per day the final equation would not match as we would get:

3n + 6(n + 8) = 280 Æ which does not give us the value of n as 27 and hence this option is rejected.

3.To solve this question first assume the values of x and y (such that x < y). If you take x as 10 hours and y as 15 hours, you will get a net work of 3.33% per hour. At this rate it will take 20 hours to fill the tank from one third full. Using this condition try to put these values of x and y into the options to check the values.

For instance option (a) gives the value as 3 × 10 × 15/10 = 45 which is not equal to 20.

4.n(1/45 + 1/40) + 23/40 = 1 Æ n = 9.

5.Since A finishes 6/7^th of the work in 2z hours .

B would finish 12/7 of the work in 2z hours.

Thus, to do 1/7^th of the work (which represents the remaining work), B would require 2z/12 = z/6 hours. Option (d) is correct.

6–10.

Set X can fill 10% in a minute. Hence, every Pipe of set X can do 1% work per minute. Set Y has a filling capacity of 12.5% per minute (or 2.5% per minute for each tap in set Y). Set Z has a capacity of emptying the tank at the rate of 5% per minute and each tap of set Z can empty at the rate of 0.625% per minute.

6.If all the 23 pipes are opened the per minute rate will be:

10 + 12.5 – 5 = 17.5% Æ Option b is correct.

7.Set X will do 5 % per minute and Set Y will do 6.25% per minute, while set Z will do 5% per minute (negative work). Hence, Net work will be 6.25% per minute. To fill 49% it will take slightly less than eight minutes and the value will be a fraction. None of the first three options matches this requirement. Hence, the answer will be (d).

8.If 4 of the taps of set Z are closed, the net work done by Set Z would be –2.5% while the work done by Sets X and Y would remain 10% and 12.5% respectively. Thus, the total work per minute would be 20% and hence the tank would take 5 minutes to fill up.

9.Again if we close 4 taps of set Z, the rate of emptying by set Z would be 2.5% per minute. A half filled tank would contain 50% of the capacity and hence would take 50/2.5 = 20 minutes to empty.

10.The rate per minute with the given changes (in percentage terms) would be:

Set X =11%, Set Y = 15% and Set Z = –6%.

Hence, the net rate = 11 + 15 – 6 = 20% per minute and it would take 5 minutes for the tank to fill. If all pipes are opened at 7:58, the tank would get filled at 3:03.

11.Let Ajit’s rate of work be 100/2 = 50 work units per day. Baljit would do 100/3 = 33.33 work units per day and Diljit does 133.33/5 = 26.66 units of work per day. Their 1 days work = 50 + 33.33 + 26.66 = 110 units of work per day. In 20 days, the total work done would be 2200 units of work and hence for Baljit to do it alone it would take: 2200/33.33 = 66 days to complete the same work.

12.The 32 minutes extra represents the extra time taken by the pipes due to the leak.

Normal time for the pipes Æ n × (1/14 + 1/16) = 1

Æ n = 112/15 = 7 hrs 28 minutes.

Thus, with 32 minutes extra, the pipes would take 8 hours to fill the tank.

Thus, 8(1/14 + 1/16) – 8 × (1/L) = 1 Æ 8/L

= 8(15/112) – 1

1/L = 15/112 – 1/8

= 1/112.

Thus, L = 112 hours.

13.The outlet pipe will empty the tank at a rate which is double the rate of filling (Hence, 10 gallons per minute). If the inlet is shut off, the tank will get emptied of 100 gallons of water in ten minutes.

14.The net inflow when both pipes are opened is 5 litres a minute.

The outlet flow should be such that if its rate is doubled the net inflow rate should be negative or 0.

Only an option greater than or equal to ‘5’ would satisfy this condition.

Option (b) is the only possible value.

15.XÆ12 days Æ 8.33% of the work per day.

Y Æ 18 days Æ 5.55% of the work per day

Z Æ 10 days Æ 10% of the work per day.

In three days, the work done will be 25 + 16.66 + 30 = 71.66%. The remaining work will get done by Y in 28.33/5.55 = 5.1 days.

[Note: You need to be fluent with your fraction to percentage conversions in order to do well at these kinds of calculations.]

16.A takes 9 days to complete the work

B takes 24 days to complete the work

C takes 16 days to complete the work

In 4e days, work done by all three would be:

4 × (1/9 + 1/24 + 1/16)

= 4 × = 124/144

= 31/36 of the work.

Work left for B would be 5/36 of the work.

B would require: (5/36) × 24 = 3.33 days.

17.The per day digging of all three combined is 54 meters. Hence, their average should be 18. This means that the first should be 18 – x, the second, 18 & the third 18 + x.

The required conditions are met if we take the values as 15,18,21 meters for the first, second and third diggers respectively. Hence, (a) is the correct answer.

18.The equations are:

3(A + B + C) = 37/100 = 37% of the work.

7(A + B) = 63/10 Æ A + B = 9/100 = 9%

(Where A, B and C are 1 day’s work of the three respectively).

Further, 5A= 4B gives us

A = 4% and B = 5% work per day.

In 3 days (A + B + C) do 37% of the work.

Out of this A and B would do 27% (= 3 × 9%) of the work. So, C would do 3.33% of the work per day.

Thus, B is the fastest and he would require 20 days to complete the work.

19.A + B = 20% of the work. Use trial and error with the options to get the answer.

Checking for option (a), A = 10% and B = 10%. If A doubles his work and B halves his work rate, the total work in a day would become A = 20, B = 5. This would mean that the total work would get completed in 4 days which is the required condition that needs to be matched if the option is to be correct. Hence, this option is correct.

20.Since the first typist types for 4 minutes and the second typist types for exactly 6 minutes, the work left (which is given as 1/5 of the total work) would be the work the first typist can do in 2 minutes. Thus, the time taken by the first typist to do the work would be 10 minutes and his rate of work would be 10% per minute. Also, since both the typists can do the work together in 6 minutes, their combined rate of work would be 100/6 = 16.66% per minute.

Thus, the second typist’s rate of work would be 16.66 – 10 = 6.66% per minute.

He would take 100/6.66 = 15 minutes to complete the task alone.

21.From the condition of the problem and a little bit of trial and error we can see that the first cook worked for 4 minutes and the 2^nd and 3^rd cooks also worked for 4 minutes. As 4(A) + 4(B + C) = 4(A + B + C) and we know that A + B + C = 20 idlis per minute.

Thus, the first cook make 20 idlis in 4 minutes. To make 160 idlis he would take 32 minutes.

22.Solve this using options. If we check for option (c) i.e. the work of a man exceeds the work of a woman by 5 times, we would get the following thought process:

Total work = 6 days × (3 women + 2 men) = 18 woman days + 12 man days = 18 woman days + 60 woman days = 78 woman days.

Thus, 9 women would take 78/9 days = 8.66 days and hence 3 men should do the same work in 3.66 days. This translates to 3 × 3.66 = 10 man days or 50 woman days which is incorrect as the number of woman days should have been 78.

Thus, we can reject this option.

If we check for option (d) i.e. the work of a man exceeds the work of a woman by 6 times, we would get the following thought process:

Total work = 6 days × (3 women + 2 men) = 18 woman days + 12 man days = 18 woman days + 72 woman days = 90 woman days.

Thus, 9 women would take 90/9 days = 10 days and hence 3 men should do the same work in 5 days. This translates to 3 × 5 = 15 man days or 90 woman days which is correct as the number of woman days should be 90.

Thus, we select this option.

23.0.5(A + B + C) = 50% of the work.

Means Æ A, B and C can do the full work in 1 hour.

Thus, (A + B + C) = 100%

From this point it is better to solve through options. Option (c) gives the correct answer based on the following thought process.

If c = 50% work per hour, it means C takes 2 hours to complete the work.

Consequently, A would take 3 hours and hence do 33.33% work per hour.

Since, A + B + C = 100%, this gives us B’s hourly work rate = 16.66%.

For this option to be correct these nos. should match the second instance and the information given there.

According to the second condition:

A + 4B should be equal to 100%. Putting A = 33.33% and B = 16.66% we see that the condition is satisfied. Hence, this option is correct.

24.Option (a) is correct because: 1/10.5 + 1/14 = 1/6 which matches all the conditions of the problem.

25.Solve by trial and error by putting values for x and y in the options.

26.Use options for this question as follows:

If discharging delivery is 40, filling delivery will be 16.66% less (this will give a decimal value right at the start and is unlikely to be the answer. Hence, put this option aside for the time being.)

Option (c) gives good values. If discharging delivery is 60, filling delivery will be 50. Also, time taken for discharge of 3600 cu m will be 60 minutes and the time taken for delivery will be 72 minutes (12 minutes more – which is the basic condition of the problem).

27.The interpretation of the first statement is that (a) & (b) do 41.66 percent of the work per hour. From this point if we go through the options, option (b) fits the situation as 4 hours per one person means 25 percent work per hour per person. Consequently this means 16.66 percent per work per hour per other person.

28.From the last statement we know that since both the pipes would require 17 hours to fill the tank together, they would discharge 425/17 = 25 liters per hour together.

From this point try to fit the values from the options in order to see which one satisfies all the conditions.

In the case of option (a): Second pipe open for 10 hours, first pipe open for 15 hours.

When the interchange occurs: Second pipe open for 15 hours, first pipe open for 10 hours Æ gives us that the respective rates of the two pipes would be 3:4 (as the first pipe delivers half the amount of the second pipe- if it delivers 3 liters per minute the second pipe would need to deliver 4 liters per minute).

Thus, if the delivery of the first pipe is 3n liters per minute, the delivery of the second pipe would be 4n liters per minute. Then, in 10 hours of the second pipe and 15 hours of the first pipe, the total water would be 85n, which should be equal to the total water of the two pipes in 17 hours each. But in 17 hours each, the two pipes would discharge 17 × 7n = 119n. Thus, we reject this option.

In the case of option (c): Second pipe open for 15 hours, first pipe open for 20 hours.

When the interchange occurs: Second pipe open for 20 hours, first pipe open for 15 hours Æ gives us that the respective rates of the two pipes would be 2:3 (as the first pipe delivers half the amount of the second pipe- if it delivers 2 liter per minute the second pipe would need to deliver 3 liters per minute).

Thus, if the delivery of the first pipe is 2n liters per minute, the delivery of the second pipe would be 3n liters per minute. Then, in 15 hours of the second pipe and 20 hours of the first pipe, the total water would be 85n, which should be equal to the total water of the two pipes in 17 hours each. In 17 hours each, the two pipes would discharge 17 × 5n = 85n. Thus, we realize that this is the correct option.

29.In order to solve this question, if we look at the first statement, we could think of the following scenarios:

If the time taken by the first man and the woman is 1 hour (100% work per hour) , the time taken by the second man would be 4 hours (25% work per hour). In such a case, the total time taken by all three to complete the task would be 100/125 = 0.8 hours. But this value is not there in the options. Hence, we reject this set of values.

If the time taken by the first man and the woman is 2 hours (50% work per hour) , the time taken by the second man would be 5 hours (20% work per hour). In such a case, the total time taken by all three to complete the task would be 100/70 = 10/7 hours. But this value is not there in the options. Hence, we reject this set of values.

If the time taken by the first man and the woman is 3 hours (33.33% work per hour) , the time taken by the second man would be 6 hours (16.66% work per hour). In such a case, the total time taken by all three to complete the task would be 100/50 = 2 hours. Since this value is there in the options we should try to see whether this set of values meets the other conditions in the question.

In this case, it is given that the first man working alone takes as much time as the second man and the woman. Since, the work of all three is 50%, this means that the work of the first man is 25%. Consequently the work of the woman is 8.33%.

Looking at the third condition given in the problem – the time taken by the first man to do the work alone (@ 25% per hour he would take 4 hours) should be 8 hours less than double the time taken by the second man. This condition can be seen to be fulfilled here because the second man would take 6 hours to complete his work (@ 16.66% per hour) and hence, double his time would be 12 hours- which satisfies the difference of 8 hours.

Thus, the total time taken is 2 hours.

30.Let the inlets be A, B, C and D.

A + B + C = 8.33%

B + C + D = 6.66%

A + D = 5%

Thus, 2A + 2B + 2C + 2D = 20%

and A + B + C + D = 10%

Æ 10 minutes would be required to fill the tank completely.