How to solve difficult SSC CGL Profit and loss problems in a few steps 3

Use of rich Ratio and Algebraic concepts leads us to the solution in no time

How to solve difficult SSC CGL profit loss problems in a few steps 3

Usually profit and loss problems do not come difficult, variations are few and with a bit of grounding on concepts you can solve any of this category of problems easily. That's why this one surprised us. At first glance there seemed to be too many variables and the problem unsolvable. But when we looked deeper we discovered the clues to elegant solution one by one.

Not surprisingly the new crucial concept that provided the breakthrough belonged to the category of rich algebraic concepts as well as rich ratio concepts both apparently different from the profit and loss topic. This again shows how at the basic level, the concepts of basic topic areas can be intertwined with each other. 

We will now see how.


A trader bought a number of pens at the rate of 5 per Rs. $P$, and in a second occasion he purchased the same number of pens at the rate of 4 per Rs. $P$. Then he mixed the two lots together and sold the pens at the rate of 9 per Rs. $Q$ suffering thus a loss of Rs. $R$. If $P : Q : R = 1 : 2 : 3$, find how many pens the trader bought.

  1. 540
  2. 545
  3. 1080
  4. 1090

Problem analysis and first stage simplification using basic ratio concept

At first glance unknown variables seem to be too many for any feasible solution. But when we find the answer choice values to be simple numbers we know for sure that all except the single desired variable will be eliminated and the breakthrough technique lies within the problem.

When we examine the problem more closely, we notice the ratio and immediately find the way to reduce number of unknown variables from 3 of $P$, $Q$ and $R$ to 1, the HCF of the three. We transform then the ratio terms to their actual values by introducing as a product factor, their HCF as an unknown variable $x$,

Thus the ratio is transformed as,

$P : Q : R = x : 2x : 3x$,

where $x$ is the HCF of the three quantities $P$, $Q$ and $R$ and the actual values of the variables are, $P=x$, $Q=2x$ and $R=3x$. This is direct application of the most basic concept of ratios.

Rich algebraic ratio concept of Ratio variable elimination property

We state the new Ratio variable elimination property as,

When a number of variables appear in a problem related with each other as a numeric ratio, the ratio variable quantities will be eliminated from consideration if all the variables appear in a linear equation together.

This is primarily a rich algebraic concept of rich ratio concept variety, it is a powerful hybrid concept.

In our case till this point, in transforming the three price/cost varibales $P$, $Q$ and $R$ into $x$, $2x$ and $3x$ as their actual values, we have reduced the number of variables from 3 to 1. This is significant simplification.

But we also foresee that, these three variables will certainly appear in a future linear equation all together, as we know, in basic Profit and loss all relations are of linear form. Here we are concerned with such a one in,

$CP - SP = Loss$, a perfectly linear equation.

Implementation of first stage simplification

After converting the three price/cost ratio variables to their actual values we face two choices of paths for further progress,

  • to proceed with per pen purchase cost, per pen sale price and per pen loss, or,
  • total purchase cost, total sale price and the total loss.

As purchase quantities on both occasion are equal we found it possible to choose the per pen cost/price deduction path thus simplifying the computations to an extent. If we had taken conventional total price/cost calculation path, the deductive load would have increased. Let us see how.

Conventional total cost/price deduction

Let the number of pens purchased on each occasion be $N$.

The total cost price on the first occasion is then,

$CP_1 = \displaystyle\frac{NP}{5}=\frac{Nx}{5}$.

And the total cost price on the second occasion is,

$CP_2 = \displaystyle\frac{NP}{4}=\frac{Nx}{4}$.

So that the grand total of the cost price is,

$CP =Nx\left(\displaystyle\frac{1}{5} + \displaystyle\frac{1}{4}\right)=\displaystyle\frac{9Nx}{20}$

Similarly total sale price,

$SP = 2N\displaystyle\frac{Q}{9} = \displaystyle\frac{4Nx}{9}$.

Finally loss is,

$Loss = R = 3x = CP - SP$

$=\displaystyle\frac{9Nx}{20} - \displaystyle\frac{4Nx}{9}$


Eliminating $x$,

$N =3\times{180} = 540$,

So the total pens purchased was,

$2N = 1080$, our answer as option c.

In this path of solution we had to carry the variable $N$ through a few steps. Instead in the elegant solution we will take up the per pen price/cost path and will bring in $N$ at the last possible step.

Elegant solution along the per pen deduction path

In the first purchase per pen cost is,

$CPP_1 = \displaystyle\frac{P}{5} = \frac{x}{5}$.

In the second purchase per pen cost is,

$CPP_2 = \displaystyle\frac{P}{4} = \frac{x}{4}$.

Number of pens purchased on both occasions being same, average per pen cost after mixing the lots together is,

$CPP = \displaystyle\frac{1}{2}\left(\displaystyle\frac{x}{5} + \displaystyle\frac{x}{4}\right) = \displaystyle\frac{9x}{40}$

Similarly the per pen sale price of the mixed lot of pens is,

$SPP = \displaystyle\frac{Q}{9} = \frac{2x}{9}$

Bringing in the number of pens purchased each time as $N$ now we have the per pen loss as,

$\text{Per pen loss } = \displaystyle\frac{R}{2N} = \displaystyle\frac{3x}{2N}$

$=CPP - SPP$

$=\displaystyle\frac{9x}{40} - \displaystyle\frac{2x}{9}$


Eliminating $x$ we get,

$N = 360\times{\displaystyle\frac{3}{2}}=540$.

Total number of pens purchased is then,

$2N = 1080$.

Answer: c: 1080.


First we simplified the complexity of the problem by introducing HCF $x$ as a factor for each of the ratio variables to get their actual values in terms of the single variable $x$ thus reducing the number of these variables from 3 to 1.

Second, avoiding the conventional path of dealing with the total values of cost price, sale price and loss, we took up the per item values thus simplifying the deduction. This was possible because of same number of items purchased on both occasions and we could work on the average per item purchase cost. 

In the third stage, using the basic relation of profit or loss, we formed the linear equation in per item cost, per item sale price and per item loss, bringing in the number of items purchased at the last possible stage.

Because of the linear relationship, the variable HCF $x$ was eliminated leaving only the desired unknown variable, the number of items purchased on each occasion.

This new variable elimination technique we named as Ratio variable elimination rich concept that stated (we repeat),

When a number of variables appear in a problem related with each other as a numeric ratio, the ratio variable quantities will be eliminated from consideration if all the variables appear in a linear equation together.

In our problem, this powerful hybrid algebraic and ratio rich concept could be applied as basic profit and loss relations are linear.

Inherent abstraction and further exploration

This is an example of direct application of abstraction of simpler problems of this type when the cost value, sale value and loss value appear as simple numeric terms of 1, 2, 3 or 100, 200, 300, but always in ratio of 1 : 2 : 3 giving us the idea that actual values of these variables are not important at all, only their relative values are of significance. Thus it could imagined that these three values can even be algebraic variables but still in a ratio of 1 : 2 : 3.

You can tune the ratio values or the rate of purchase or sale values to see how the profit or loss behave.

You may ask us, why so much trouble?

Our answer will be, with this explorative experimentative approach you will gain total control of the profit and loss concept domain for arriving at quick elegant solution of any problem in the domain.

How to solve difficult SSC CGL Math problems at very high speed using efficient problem solving strategies and techniques

These resources should be extremely useful for you to speed up your in-the-exam-hall SSC CGL math problem solving. You will find these under the subsection Efficient Math Problem Solving.

This is a collection of high power strategies and techniques for solving apparently tricky looking problems in various topic areas usually within a minute. These are no bag of tricks but are based on concepts and strategies that are not to be memorized but to be understood and applied with ease along with permanent skillset improvement.

The following are the associated links,

How to solve similar problems in a few seconds, Profit and loss problem 2, Domain modeling

How to solve difficult SSC CGL Profit and loss problems in a few steps 1