Hard problems can be solved in a few confident steps if you use efficient problem solving techniques
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 second session on hard problem solving we have shown 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 second suitable problem example taken from the topic of Time and Work.
Before going ahead, you may refer to the earlier session on How to solve a hard CAT level time and work problem in a few confident steps 1.
Problem example 1.
On the first occasion, two persons completed a job of 425 work units with the first person working for 5 days more than the second person. If the second person worked for the period the first person worked in the first occasion the work amount would have been double that of the work done if the first person worked for the period the second person worked on the first occasion. On a second occasion, working together the two persons completed the job in 17 days. How long did the second person worked on the first occasion?
- 15 days
- 10 days
- 12 days
- 18 days
Problem analysis and modelling
Adopting our efficient Time and Work problem solving approach, we will assume the work rates in amount of work done per day by the first person as $A$ and by the second person by $B$, with $W=425$ as the total work amount and $x$ the number of days the second person worked on the first occasion. With these assumptions, the first statement translates to,
$x(A+B) +5A=W=425$, no fractions involved.
Analyzing the problem further, we find the third statement directly provides the amount of work done together by the two persons in 1 day,
The second statement is a complex one that we won't convert to algebraic expression at the moment and explore further the two expressions we have got.
Problem solving execution in a few confident steps
Or, $25x +5A=425$, as $(A+B)=25$,
From the choice values, applying the free resource use principle, we find $x$ to be an integer. This implies $A$ to be a factor of 5. Trying out 5, 10, 15, 20 and 25 as the value of A in our enumeration process, we quickly home in to two values of A, namely 10 and 25 that result in integer $x$ value as 15 and 12 that are in the set of choice values. But as, $(A+B)=25$, If $A=25$, $B$ would have been 0, an infeasible situation.
So the answer should be $x=15$ when $A=10$.
By this reasoning there is no real need for stating the algebraic expression corresponding to the second complex statement. The apparently hard problem stands solved at this point.
Expression for second complex statement and verification of the answer
This being an academic analytical session, we will not stop here but go on to form the expression for the second statement as,
The first person worked on the first occasion for $(x+5)$ days and the second person for $x$ days. In the second statement these periods were reversed.
Putting $x=15$ and $A=10$, $B=15$, (as $A+B=25$), both the sides of the equation evaluate to 300. So the answer is verified.
But can we not derive this answer arrived at by mathematical reasoning following mathematical deduction? Let us see how that can be done and what results we get.
Solution by mathematical deduction
We have three equations,
From the second equation,
Combining this with the first equation we have,
Substituting values of $A$ nd $B$ in the third equation,
Eliminating factor 5 and transposing,
As $x$ can't be negative, $x=15$ days.
Answer: Option a: 15 days.
Deductive reasoning sum up
The last statement formed the Anchor expression, and the first statement gave us an opportunity to evaluate possible values of $x$ by enumeration and free resource (the choice values) use principle. We got the answer as $x=15$ days, bypassing the complexity of the second statement completely.
While doing the academic exercise of mathematical deduction we encountered the hardness in the form of a quadratic equation in $x$, but on factoring one of the roots of $x$ being infeasible, the answer was duly found to be the value we arrived at earlier by mathematical reasoning.
Though this problem seems to be hard, its solution may finally take less than a minute and a few steps only.
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 two persons transformed to two pipes, say, $P_1$ and $P_2$ and instead of doing work the pipes will fill a tank of capacity 425 litres or gallons of liquid. 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
Guidelines, Tutorials and Quick methods to solve Work Time problems
How to solve a hard CAT level Time and Work problem in a few confident steps 2