## Understanding variations of basic concepts on profit and loss gives you clarity and speed of problem solving

We have already treated an interesting type of profit and loss problem with equal sale prices and equal profit and loss figures in two transactions. This session we will be solving such a problem with equal sale prices but with unequal profit and loss percentages very fast and in two ways. We will use basic and rich profit and loss concepts, fraction concepts and mathematical reasoning to achieve the speed and elegance.

### Basic concepts on Profit and loss

$\text{Profit amount}=\text{Sale price}-\text{Cost price}$.

When the $\text{Profit amount}$ is expressed as a percentage, the * Percentage reference value* is the $\text{Cost price}$, that is,

$\text{Profit percentage}=\displaystyle\frac{\text{Profit amount}\times{100}}{\text{Cost price}}$.

When $\text{Sale price}$ is less than the $\text{Cost price}$, * a loss* is incurred with the loss as,

$\text{Loss amount}=\text{Cost price}-\text{Sale price}$.

When the $\text{Loss amount}$ is expressed as a percentage, the * Percentage reference value* again is the $\text{Cost price}$, that is,

$\text{Loss percentage}=\displaystyle\frac{\text{Loss amount}\times{100}}{\text{Cost price}}$.

These are briefly the basic concepts in the topic area of Competitive test Profit an loss. *The concepts are apparently simple.*

Let's describe an important rich concept on profit and loss.

### Rich concept on profit and loss with expression in fraction

The rich concept on profit and loss enables us to express Sale price $\text{SP}$ directly in terms of Cost price $\text{CP}$ if profit or loss percentage is given. This speeds up solution process considerably.

For example, if profit is 25%, sale price is more than cost price by 25% of cost price by definition and so, it can be expressed as,

$\text{SP}=\text{CP}+0.25\text{CP}=1.25\text{CP}$.

We just know this and don't have to think or deduce it.

Similarly, when loss is 30% we know,

$\text{SP}=\text{CP}-0.3\text{CP}=0.7\text{CP}$.

The percentages are automatically converted to equivalent decimals by dividing with 100 and then is added to or subtracted from 1 to form the coefficient of cost price $\text{CP}$.

For solving the chosen problem, this time we would move one step forward and express the decimal equivalent of profit or loss percentages as fractions.

In dealing with this form of expressions of percentage profit and loss, we should be clear about fraction equivalence of percentages as well as interpretations of sale price to cost price relations in terms of profit or loss.

Let us clarify with **examples**.

If sale price $\text{SP}=\displaystyle\frac{4}{5}\text{CP}$,

- First, we would be able to immediately identify that it is a loss situation as sale price is less than the cost price, and,
- Second, we would know that the loss amount is 20% as shortfall of sale price from cost price is one portion out of 5 portions of cost price. This is use of fraction concepts, portion concepts as well as percentage concepts.

Similarly, if sale price $\text{SP}=\displaystyle\frac{5}{4}\text{CP}$, we would know,

- First, it is a profit situation as sale price is more than the cost price, and,
- Second, profit is 25% as sale price is more than the cost price by 1 portion out of 4 portions of cost price.

Remember, percentage profit or loss is invariably on cost price.

In this session, we will solve the chosen problem using these concepts and also mathematical reasoning. Two approaches to solution will be presented, both in a few steps and wholly in mind.

### Chosen Problem

A man sells two articles for Rs.4000 each with no loss or no gain in the two transactions taken together. If one is sold at a gain of 25%, the other is sold at a loss of,

- $25$%
- $20$%
- $16\frac{2}{3}$%
- $18\frac{2}{9}$%

**Problem Solution 1: Mathematical reasoning based on Abstraction, Rich percentage technique and Compensation principle**

In the first case, let us assume the cost price to be Rs.100 so that at 25% profit the sale price is Rs. 125. As sale prices are same on two occasions, the total sale price would be Rs.250 in two transactions.

As no loss or no gain happens taking the two transactions together, the total cost price must also be Rs. 250 with average of Rs. 125.

In the first case, the cost price is less than the sale price by Rs. 25 and so in the second case, the cost price must be more than the sale price of Rs. 125 by Rs. 25, that is, Rs. 150 to compensate for the shortfall.

What is the loss percent in this second case then?

It would simply be loss of Rs.25 as a percent of cost price of Rs.150.

It is then,

$\displaystyle\frac{25}{150}=\frac{1}{6}=16\frac{2}{3}$%.

Here we have not used the given sale price or calculated any of the cost prices because main computation being on percentages or ratios, the actual figures cancelled out and did not need to be considered. This we call **abstraction of the prices**.

This is the reason why we could choose a very convenient figure of Rs. 100 as the first cost price.

Conveniently choosing the figure of 100 as the starting value of the percentage reference variable, in this case the cost price, simplifies and speeds up solution in any such arithmetic problem where percentage increase or decrease of the reference variable takes place.

This is **Rich percentage technique**.

The solution could be reached very fast and wholly in mind, but one must have clarity on the concepts.

**Answer:** Option c: $16\frac{2}{3}$%.

**Key concepts used:** * Profit and loss basic concepts* --

*--*

**Profit and loss compensation***--*

**Percentage reference concept***--*

**Context awareness***--*

**Sequencing of events concept***.*

**Abstraction technique -- Rich percentage technique -- Mathematical reasoning**Overall, this approach to solving the problem is carried out using what we call—**mathematical reasoning**. This is because we have not really used any formal mathematical deductions, using only reasoning based on math concepts instead.

We will now show you a second approach to solve the problem equally elegantly. In this approach, we would use the more familiar profit and loss relations to begin with. But looking at the cost price relations we would form the answer immediately using the rich profit and loss concept, fraction concepts and the main requirement of the problem of equal total cost and sale prices.

#### Problem Solution 2: Mathematical reasoning based on Rich profit and loss concept with fraction expression, Fraction concepts and Compensation principle

Assuming sale price on both occasions as $\text{SP}$ and cost prices on two occasions as $\text{CP1}$ and $\text{CP2}$ for profit and loss, we have for profit,

$\text{SP}=\text{CP1}+ \text{Profit}=1.25\text{CP1}=\displaystyle\frac{5}{4}\text{CP1}$,

Or, $\text{CP1}=\displaystyle\frac{4}{5}\text{SP}$.

As cost price $\text{CP2}$ in the lossy transaction must compensate the shortfall of $\displaystyle\frac{1}{5}\text{SP}$ to make the sum of the two costs equal to $2\text{SP}$, the value of $\text{CP2}$ would have to be equal to, $\left(1+\displaystyle\frac{1}{5}\right)\text{SP}=\displaystyle\frac{6}{5}\text{SP}$.

It follows,

$\text{SP}=\displaystyle\frac{5}{6}\text{CP2}$, which means $\text{SP}$ is less than $\text{CP2}$ by $\displaystyle\frac{1}{6}$th of $\text{CP2}$ or loss is $16\frac{2}{3}$% of $\text{CP2}$.

This solution is also a quick one, but it involves clarity on fraction to percentage relations.

**Note:** Problem understanding, problem solving strategy formulation and use of the basic and rich concepts enabled reaching the solution in a few simple steps avoiding all time consuming calculations. **The calculations were only on the percentages and not on the actual values of the prices at all.**

### Many ways technique

When we get the opportunity to solve a problem in more than one way, we always take the opportunity. In fact, as a general principle we always look for new ways to solve a problem. We call this approach to problem solving as **many ways technique.** If you practice this approach as a habit, your basic skill in finding new ways to the solution of an otherwise difficult situation or problem gets enhanced. It is a powerful problem solving skill enhancement technique.

A more visible outcome of this approach is the chance to compare multiple solutions and implement the optimum one.

**Resources that should be useful for you**

* 7 steps for sure success in SSC CGL tier 1 and tier 2 competitive tests* or

*to access all the valuable student resources that we have created specifically for SSC CGL, but*

**section on SSC CGL****generally for any hard MCQ test.**

### Other related question set and solution set on SSC CGL Profit and loss and Ratio and Proportion

**SSC CGL Tier II level Solution Set 8 on Profit and loss 1**

**SSC CGL Tier II level Question Set 8 on Profit and loss 1**

**SSC CGL level Solution Set 53 on Profit and loss 4**

**SSC CGL level Question Set 53 on Profit and loss 4**

**SSC CGL level Solution Set 34 on Profit and loss 3**

**SSC CGL level Question Set 34 on Profit and loss 3**

**SSC CGL level Solution Set 29 on Profit and loss 2**

**SSC CGL level Question Set 29 on Profit and loss 2**

**SSC CGL level Solution Set 25 on Arithmetic Percentage Ratios**

**SSC CGL level Question Set 25 on Arithmetic Percentage Ratios**

**SSC CGL level Solution Set 24 on Arithmetic Ratios**

**SSC CGL level Question Set 24 on Arithmetic Ratios**

**SSC CGL level Solution Set 6 on Profit and loss**

**SSC CGL level Question Set 6 on Profit and loss**

**SSC CGL level Solution Set 5 on Arithmetic Ratios**

**SSC CGL level Question Set 5 on Arithmetic Ratios**

**SSC CGL level Solution Set 4 on Arithmetic Ratios**

**SSC CGL level Question Set 4 on Arithmetic Ratios**

#### 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 withpermanent skillset improvement.

**The following are the associated links,**

**How to solve SSC CGL level Profit and loss problems by Change analysis in a few steps 6**

**How to solve a difficult SSC CGL level Profit and loss problem with equal sale prices in a few steps 5**

**How to solve a difficult SSC CGL level Profit and Loss problem in a few steps 4**

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

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

**How to solve in a few steps, Profit and loss problem 1**