Hard problems can be solved in few confident steps if you can define the problem well
To solve hard math, instead of thinking conventionally and mechanically, use your problem solving skills, inventive thinking and basic math concepts.
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.
To skip the next section and move on directly to the hard time and work problem click here.
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 recedes 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 suitable problem example taken from the topic of Time and Work.
Hard CAT Quant Time and Work Problem 1
Ramu and Shyam can complete a job in 48 hours working together. On an occasion, Ramu worked alone for half the time Shyam takes to complete the work, and then Ramu left and Shyam came on to work for the period that Ramu takes to complete one-third of the work. When Shyam stopped, five-sixth of the work was done. What is the least time that will be taken either by Ramu or Shyam to complete the work alone?
- 96 hours
- 80 hours
- 60 hours
- 72 hours
How to solve a hard CAT level Time and Work problem in a few confident steps
Problem analysis and modelling
As is our efficient Time and Work problem solving approach, we will assume the work rates in amount of work done per hour by Ramu as $R$ and by Shyam by $S$, with $W$ as the total work amount. With these assumptions, the first statement translates to,
$48(R+S)=W$, no fractions involved.
But going through the next two statements we find it gives us only one more relationship which is not linear in R and S. We felt, it is best to represent this seemingly complex relationship straightway for getting some control on the problem.
By second and third statement we find, two portions of work done sum of which is $\frac{5}{6}W$. The first portion is done by Ramu and the second by Shyam, but each have done their share of work for durations that are expressed in terms of work rate of the other.
This is, "Interdependence between variables", which being inverse introduces quadratic (or square) terms. This is the first element of hardness in the problem. Let us see its exact nature.
Shyam takes 1 hour to do $S$ amount of work by definition. So to complete the work of amount $W$ he will take (by unitary method),
$\displaystyle\frac{W}{S}$ hours.
In the first stage Ramu works for half of this time, that is,
$\displaystyle\frac{W}{2S}$ hours.
As Ramu does $R$ amount of work in 1 hour, during the time he works in the first stage, he will complete an work amount,
$\displaystyle\frac{RW}{2S}$.
Similarly in the second stage, Shyam will complete an work amount,
$\displaystyle\frac{SW}{3R}$.
So sum of these two will give our second equation,
$\displaystyle\frac{RW}{2S}+\displaystyle\frac{SW}{3R}=\displaystyle\frac{5W}{6}$,
Or, $\displaystyle\frac{R}{2S}+\displaystyle\frac{S}{3R}=\displaystyle\frac{5}{6}$, $W$ eliminated.
It turns out to be a quadratic equation between two main variables $R$ and $S$ and here lies the complexity.
At this point, the only option that we see is to straighten out the quadratic equation quickly to see what it can offer.
Problem solving execution in a few confident steps
$\displaystyle\frac{R}{2S}+\displaystyle\frac{S}{3R}=\displaystyle\frac{5}{6}$,
Or, $3R^2 -5SR+2S^2=0$, multiply both sides by $6SR$ and take all terms to one side of the equation. This is simple algebra.
We use an important technique in this step, eliminate the denominator.
This is the point where we will stop for a moment and point out the simplicity of the barrier in front. Though in time and work problems we have never faced a quadratic equation, this additional barrier may not be so difficult to cross.
This is an important approach in solving any hard problem,
If you reach an unknown pattern of a problem state and only one path ahead exists, you go ahead along the path using known concepts without any hesitation or misgving. You will surely find the solution lying around the corner.
The equation is factored as,
$(3R-2S)(R-S)=0$.
This equation turns out to be intended to be factored easily. But again we have used another skill, factoring algebraic expressions.
This gives two possibilities,
$3R=2S$, or
$R=S$.
Again this barrier seems confusing at first.
We don't have any information by which we can negate one of the possibilities. So we decide to go ahead testing each possibility, but using a new concept, that is, the basic concept of ratio and proportion. This is how in time and work problem, concept of factoring a quadratic equation and basic ratio concept played invaluable roles.
Taking the first possibility,
$3R=2S$,
Or, $R : S = 2n : 3n$, where $2n$ and $3n$ are actual values of $R$ and $S$ respectively, $n$ having been assumed as the cancelled out HCF.
From the first condition we had,
$48(R+S)=W$.
So,
$W=240n$, and Ramu and Shyam will take respectively, $120$ hours and $80$ hours to complete the job working alone.
Checking against the choices we find 80 hours as one of the choices.
Let us now check the second possibility, $R=S$, or, $R : S=m : m$, where actual values of $R$ and $S$ both are $m$.
This gives from first equation,
$48(R+S)=W$,
Or, $W=98m$, and Ramu and Shyam both take 96 hours to complete the job.
Finally then Option b or 80 hours is the answer.
Deductive reasoning sum up
The first equation obtained from the first statement is what we call the Anchor statement. Finally this will generate first the total work amount and then the time taken by each to complete the work. We are used to making form of such a statement in time and work problems as simple as possible by the use of agent work rate concept.
We got the first taste of hardness in forming the second expression itself. This is problem modelling and the process and formation of the model should be based on clear understanding of basic concepts and methods.
Use of agent work rate concept together with unitary method and breaking up two portions of work in two stages made the process an easy walk. The last approach is the problem breakdown technique. When you find a statement too complex to deal with at one go, break it up into manageable parts.
Though we could form the expression easily, it turned out to be an equation with inverses. Employing efficient simplification, and denominator elimination, the awkward looking expression has been straightened out into a quadratic equation, a new barrier.
The quadratic equation though looked to be an equation that could be instantly factored into two valid expressions. The action of factoring is automatic.
At this stage we are faced with the next hardness barrier. There are two possibilities and both can be valid. Enumeration has been decided to be the only way forward by checking out each possibility.
Question is, how to check out? Actual values of variables are not known. This is identified as an ideal situation to apply basic concepts of ratio and proportion as, linear relationship between the two variables are known.
Alternative approach without using ratio concept at this stage
We have,
$48(R+S)=W$, and first possibility as,
$3R=2S$,
Or, $R=\displaystyle\frac{2}{3}S$.
Substituting,
$48\times{\left(\displaystyle\frac{2}{3}S +S\right)}=W$,
Or, $80S=W$.
This means Shyam will take 80 hours to complete the job of amount $W$.
Overall, this systematic problem solving approach ensures assured and confident solution in steps.
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 finals, you will make enough preparations by solving enough number of such problems with an analytical approach that, any such problem you won't find hard at all.
An important aside:
The same problem can easily be converted to a pipes and cisterns problem with Ramu and Shyam transformed to two pipes, say, $P_1$ and $P_2$ and instead of doing work the pipes will fill a tank of capacity $W$. 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.
To go through more such selected hard CAT quant problems with step by step easy solutions, click here.