How to solve a hard CAT level Time and Work problem in a few confident steps 3

Hard problems also can be solved in a confident few steps using your number domain concepts

How to solve a hard CAT level time work problem in a few confident steps3

The conventional approach to math problem solving relies heavily on manipulation of terms using low level mathematical constructs without using the problem solving abilities of the student. Following only this approach to solving problems, students may tend to become used to mechanical and procedural thinking suppressing their inherent creative and innovative out-of-the-box thinking abilities.

On the other hand, conceptual reasoning without firm mathematical base leads to confusion. In solving hard problems you need to strike a balance. In this third session on hard problem solving we have shown again how to do it.

In this series of dealing with hard problems we will discuss,

  • What makes a problem hard,
  • The importance of problem definition and problem modelling in dealing with a hard problem,
  • Use of basic concepts drawn from a number of topic areas to solve a hard problem in one topic area,
  • General importance of grasp of basic algebraic concepts and techniques in solving a hard problem, and
  • How a hard problem can be solved in a few confident steps rather than left alone or spending too much time on it.

Most of the SSC CGL or SBI PO level problems we have dealt with till now we classify as 1 minute problems. By use of basic and rich concepts and suitable problem solving skills, strategies and techniques all these problems can be solved generally under a minute. The examiners also expect the problems solved using proper skills so that the average 1 minute time set for a problem in most tests stands as a just estimate for solving a problem.

A hard problem on the other hand, should take on an average one and half a minute to 2 minutes. These problems are structured and formed in such a way that problem understanding and definition itself are significant hurdles to many. And the examiners will provide you a longer average time of 1.5 to 2 minutes to solve one of these hard problems on an average, again a just average time. You will get more time to solve these type of problems in an actual exam.

From the point of view of exposing various aspects of problem solving, we consider the hard problems more suitable because of the richness of the barriers to the solution in these problems. Ultimately though, once you have analyzed, solved and dissected a few hard problems, the hardness reduces and the problems do not seem to be hard any more, just like any other problem.

Usually in Common Admission Test or CAT for entrance to IIMs or other reputed management institutions, the average time to solve a problem being in the range of 1.5 to 2 minutes, some of the problems (not all) may be of hard problem class.

We will now discuss how you can solve a hard problem in a few confident steps through a third suitable problem example taken from the topic of Time and Work.

Before going ahead, you may refer to the earlier sessions on How to solve a hard CAT level time and work problem in a few confident steps 1 and 
How to solve a hard CAT level time and work problem in a few confident steps 2.

Problem example 1.

Three persons A, B, and C were doing a job. The second person B could finish the job on his own taking 3 days more than the time taken by the other two working together. But when the first person A worked alone he could complete the job taking the time exactly same as the other two working together. Also the first person A's completion time alone was 8 days less than twice that of B's completion time alone. How long would the three have taken to complete the job working together?

  1. 4 days
  2. 5 days
  3. 2 days
  4. 3 days

Problem analysis and modelling - first stage

As most of the given information involve time taken to complete the job, we won't adopt the efficient Work rate technique, with assumption of variables in terms of individual work rates. Instead, in keeping with the problem data, we will assume the work capacities of the three in classical method in terms of number of days taken by each to finish the job working alone as $A$days, $B$days, and $C$days for A, B and C respectively.

This is the first major decision in conformity with the problem statements.

As a next step in dealing with hard problems, we won't analyze all relations expressed in the three statements. We would instead analyze only the simplest ones first. This is problem breakdown, alternative evaluation and easy first principle in action.

The second statement thus gives us the relation,

$A=\displaystyle\frac{1}{\displaystyle\frac{1}{B}+\displaystyle\frac{1}{C}}$, as inverting the sum of inverses of $B$ and $C$ gives the total duration of completition of the job by B and C working together. This is according to the most basic concept of Time and Work problems.

Inverting the two sides of the equation,

$\displaystyle\frac{1}{A}=\displaystyle\frac{1}{B}+\displaystyle\frac{1}{C}$.

The relation corresponding to the third statement is simpler,

$A=2B-8$.

With confidence gathered now we will formulate the third expression corresponding to the first statement but won't analyze it till it is necessary. The expression is,

$B-3=\displaystyle\frac{1}{\displaystyle\frac{1}{A}+\displaystyle\frac{1}{C}}$.

Inverting again,

$\displaystyle\frac{1}{B-3}=\displaystyle\frac{1}{A}+\displaystyle\frac{1}{C}$.

With a brief look at the three expression we know that solving these equations will involve dealing with a quadratic relationship and this was expected.

Leaving the path of direct evaluation we decide to explore the possibilities hidden in the first two simpler equations.

Problem analysis - second stage

The first equation is,

$\displaystyle\frac{1}{A}=\displaystyle\frac{1}{B}+\displaystyle\frac{1}{C}$.

Adding inverted $A$ to both sides we have,

$\displaystyle\frac{2}{A}=\displaystyle\frac{1}{A}+ \displaystyle\frac{1}{B}+\displaystyle\frac{1}{C}$.

we recognize the right side of the equation to be the inverted form of the target number of days all three together will take. Thus this target number of days is,

$\displaystyle\frac{A}{2}=D$, we invert the left side.

Also from the expression,

$\displaystyle\frac{1}{A}=\displaystyle\frac{1}{B}+\displaystyle\frac{1}{C}$,

the first major conclusion we make is, $A$ must be less than $B$ as well as $C$,

$A \lt {B}$, and

$A \lt {C}$.

This follows from Inversion principles in number domain,

If inverted variable $A$ is a sum of two other inverted variables, in inverted form $A$ musy be greater than inverted forms of both the other two variables, and consequently, in normal non-inverted form $A$ must be less than both the other two variables.

An example will make this principle clear,

$\displaystyle\frac{1}{3}=\displaystyle\frac{1}{4}+\displaystyle\frac{1}{12}$.

Problem solving execution - use of free resource of choice values and enumeration

Now we look at the choice values that is our free resource. As all the choice values are integers, it follows that $A$ is an even number. This is the second conclusion.

Now we will explore the possibilities in the second simple expression,

$A=2B-8$,

Or, $B=\displaystyle\frac{A}{2}+4=D+4$.

We will test out each choice value of $D$, the target number of days, and validate value of $A$ and $B$ to check whether $A \lt B$.

Out of the values, 4, 5, 2 and 3 of $D$, only 2, and 3 turn out to be valid.

If $D=4$, $A=8$, and $B=8$, an invalid value.

If $D=5$, $A=10$, and $B=9$, an invalid value.

If $D=2$, $A=4$, and $B=6$, a valid value.

If $D=3$, $A=6$, and $B=7$, a valid value.

For $A=4$ and $B=6$, $C=12$, from $\displaystyle\frac{1}{A}=\displaystyle\frac{1}{B}+\displaystyle\frac{1}{C}$, and

for $A=6$ and $B=7$, $C=42$.

Problem solving final execution - evaluation of third expression

We test these two valid set of values of $A$, $B$ and $C$ with the third expression,

$\displaystyle\frac{1}{B-3}=\displaystyle\frac{1}{A}+\displaystyle\frac{1}{C}$.

With $A=4$, $B=6$, and $C=12$,

$\text{LHS}=\displaystyle\frac{1}{3}$, and

$\text{RHS}=\displaystyle\frac{1}{4}+\displaystyle\frac{1}{12}=\displaystyle\frac{1}{3}$.

So this must be the answer if we assume the problem as not a malformed problem with two valid choice values.

Still testing the second set, $A=6$, $B=7$ and $C=42$,

$\text{LHS}=\displaystyle\frac{1}{4}$, and

$\text{RHS}=\displaystyle\frac{1}{6}+\displaystyle\frac{1}{42}=\displaystyle\frac{4}{21}$, an invalid set of values.

Answer: Option c: 2 days.

Note: Though the mathematical reasoning seems involved and time consuming, with a good sense of number estimation and manipulation (that is expected out of you), and a good grip on basic time and work concepts as well as problem solving strategies, you should reach the solution within a share of 2 minutes by this approach.

Let us see now how to reach the solution by an academic exercise of conventional mathematical deduction.

Solution by mathematical deduction

We have three equations,

$\displaystyle\frac{1}{A}=\displaystyle\frac{1}{B}+\displaystyle\frac{1}{C}$,

$A=2B-8$, and

$\displaystyle\frac{1}{B-3}=\displaystyle\frac{1}{A}+\displaystyle\frac{1}{C}$.

Eliminating $C$ straightaway by subtracting the third from the first we get,

$\displaystyle\frac{1}{A}-\displaystyle\frac{1}{B-3}=\displaystyle\frac{1}{B}-\displaystyle\frac{1}{A}$,

Or, $\displaystyle\frac{2}{A}=\frac{1}{B-4}=\displaystyle\frac{1}{B}+\displaystyle\frac{1}{B-3}$, by using second equation we eliminate $A$,

Or, $B(B-3)=(B-4)(2B-3)$, by transposing,

Or, $B^2-3B=2B^2-11B+12$,

Or, $B^2-8B+12=0$,

Or, $(B-6)(B-2)=0$.

$B$ cannot be equal to 2 as that would make $A$ negative.

So, $B=6$, $A=4$ and $D=2$ days, the answer.

Answer: Option c: 2 days.

Note: This path also seems to be not too imposing if you are quite adept in fast algebraic expression manipulation using efficient simplification concepts.


Deductive reasoning sum up

The first statement formed the Anchor expression, and even gave us at a very early stage, the target value of number of days in terms of  single variable $A$. Additionally from inverse principles we could deduce an important inequality in the form of $A \lt B$.

The second expression also being a simple one and between only two variables $A$ and $B$, it was a situation ripe for testing out the free resource values of the choices against the target $D=\displaystyle\frac{A}{2}$, the second equation $B=\displaystyle\frac{A}{2}+4$ and the inequality, $A \lt B$.

We could eliminate two of the four choice values with 2 and 3 remaining as possibilities. This forced us to use the third expression.

As we have the values of all three variables for each possible set, it was easy to test out each set of values of three variables against the RHS and the LHS of the third equation. Finally only the value of $A=4$ turned out to the valid one fully satisfying all problem sonditions.

While doing the academic exercise of mathematical deduction we encountered the hardness in the form of a quadratic equation in $B$, as we found it easy to eliminate $C$ first then $A$ second.

The quadratic equation though was simple enough to evaluate with ease. Alll in all this academic path seemed to be easier and more assured, though the richness of the path of mathematical reasoning was quite satisfying.

End note: The first time you encounter such a problem, it might take you more time to solve than allowed. But that is your first time. When you go into your final exam, by then you have gone through enough preparations by solving sufficient number of hard problems with an analytical approach so that you won't find any problem hard at all.

An important aside: The same problem can easily be converted to a pipes and cisterns problem with three persons transformed to three pipes, say, $P_1$, $P_2$ and $P_3$ and instead of doing work the pipes will fill a tank taking certain times in doing so. Every other aspect of the problem and its solution will remain unchanged. Check for yourself.

This happens because Time and Work problems are very similar (but not exactly same, there are differences) in nature to the Pipes and Cisterns problems.

Our recommendation: Go through the above process of solution more than once to understand and absorb the concepts fully.


Useful resources to refer to

7 steps for sure success in SSC CGL Tier 1 and Tier 2 competitive tests

How to solve Arithmetic problems on Work-time, Work-wages and Pipes-cisterns

Basic concepts on Arithmetic problems on Speed-time-distance Train-running Boat-rivers

How to solve a hard CAT level Time and Work problem in a few confident steps 3

How to solve a hard CAT level Time and Work problem in a few confident steps 2

How to solve a hard CAT level Time and Work problem in few confident steps 1

How to solve Work-time problems in simpler steps type 1

How to solve Work-time problem in simpler steps type 2

SSC CGL Tier II level Solution Set 10 on Time-work Work-wages Pipes-cisterns 1

SSC CGL Tier II level Question Set 10 on Time-work Work-wages Pipes-cisterns 1

SSC CGL level Solution Set 67 on Time-work Work-wages Pipes-cisterns 6

SSC CGL level Question Set 67 on Time-work Work-wages Pipes-cisterns 6

SSC CGL level Solution Set 66 on Time-Work Work-Wages Pipes-Cisterns 5

SSC CGL level Question Set 66 on Time-Work Work-Wages Pipes-Cisterns 5

SSC CGL level Solution Set 49 on Time and work in simpler steps 4

SSC CGL level Question Set 49 on Time and work in simpler steps 4

SSC CGL level Solution Set 48 on Time and work in simpler steps 3

SSC CGL level Question Set 48 on Time and work in simpler steps 3

SSC CGL level Solution Set 44 on Work-time Pipes-cisterns Speed-time-distance

SSC CGL level Question Set 44 on Work-time Pipes-cisterns Speed-time-distance

SSC CGL level Solution Set 32 on work-time, work-wage, pipes-cisterns

SSC CGL level Question Set 32 on work-time, work-wages, pipes-cisterns