## Hard problems can be solved quickly using math problem solving techniques

Define the hard CAT level time and work problem very clearly. Use basic time and work concepts for more clarity. Now apply your problem solving skills.

*Use math concepts of course, but give a structure to the problem, analyze, apply math reasoning, break down the whole into smaller parts and move ahead.*

Before going ahead through the problem followed by step by step solution,, 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: Hard time and work question for CAT Quant

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

### Solution to the CAT level hard time and work problem 2: CAT Quant

#### Problem analysis and modelling

Adopting our * efficient Time and Work problem solving approach*, we will assume the

*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,*

**work rates in amount of work done per day**$x(A+B) +5A=W=425$, 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,

$17(A+B)=425$,

Or, $(A+B)=25$.

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

We have,

$x(A+B) +5A=425$,

Or, $25x +5A=425$, as $(A+B)=25$,

Or, $x=17-\displaystyle\frac{A}{5}$

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

*, 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.*

**enumeration process**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,

$(x+5)B=2xA$.

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

*? Let us see how that can be done and what results we get.*

**mathematical deduction**#### Solution by mathematical deduction

We have three equations,

$A+B=25$,

$5x+A=85$, and

$(x+5)B=2xA$.

From the second equation,

$A=85-5x$.

Combining this with the first equation we have,

$B=25-A=5x-60$.

Substituting values of $A$ nd $B$ in the third equation,

$(x+5)B=2xA$,

Or, $(x+5)(5x-60)=2x(85-5x)$,

Or, $5x^2-35x-300=170x-10x^2$.

Eliminating factor 5 and transposing,

$3x^2-41x-60=0$,

Or, $(3x+4)(x-15)=0$

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

*and*

**enumeration***(the choice values) use principle. We got the answer as $x=15$ days, bypassing the complexity of the second statement completely.*

**free resource**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.

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 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.*

*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*.