## Hard CAT level Work and Time problem 3 solved step by step

Define and structure the problem clearly, use your basic time and work concepts and apply math reasoning. Solve breaking down the whole into smaller parts.

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 3: Hard time and work question for CAT Quant

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?

- 4 days
- 5 days
- 2 days
- 3 days

### How to solve a hard CAT level Work and Time problem Example 3

#### 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

*. 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.*

**individual work rates**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*,

*and*

**alternative evaluation***in action.*

**easy first principle**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 ofÂ 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 conditions.

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.

Overall, this systematic problem solving approach ensures assured and confident solution in steps.

To go through **more such selected hard CAT Quant problems** with **step by step easy solutions,** click * here*.