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

Abstraction technique and use of Rich concepts reduce calculations to a minimum

How to solve in a few steps profit and loss problem 1

Problem

A man sells two wrist watches one at a profit of 30% and another at a loss of 30%, but each at a same selling price of Rs.400. The net profit or loss is,

  1. 6%
  2. 0%
  3. 9%
  4. 1%

Conventional solution

Basic concepts of Profit and loss

Profit of 30% in selling the first watch at Rs.400 or the loss of 30% in selling the second watch at Rs.400 are calculated on the basis of their cost prices.

$\text{Profit} = \text{Selling price} - \text{Cost price}$, and,

$\text{Loss} = \text{Cost price} - \text{Selling price}$.

Thus, the selling price of the first watch is 30% (of its cost price) more than its cost price, and the selling price of the second watch is 30% (of its cost price) less than its cost price.


Initial analysis

Let Cost prices of the two watches be CP1 and CP2 and the common selling price SP.

So total cost price TCP = CP1 + CP2, and

Total sale price is TSP = 2SP = Rs.800.

If TCP is greater than TSP there is a loss otherwise a gain.


Finding the two cost prices in terms of the given sale price

For the first watch sale at a profit of 30%,

$SP = CP1 + .3\times{CP1} = 1.3\times{CP1}$

Or, $1.3\times{CP1} = SP = Rs.400$.

So, $CP1 = \displaystyle\frac{Rs.400}{1.3}$.

For the second watch sale at a loss of 30%,

$SP = CP2 - .3\times{CP2} = 0.7\times{CP2}$

Or, $0.7\times{CP2} = SP = Rs.400$

Or, $CP2 = \displaystyle\frac{Rs.400}{0.7}$.


Finding the loss or gain at the last stage

So, Total cost of the two watches is,

$TCP = \displaystyle\frac{Rs.400}{1.3} + \displaystyle\frac{Rs.400}{0.7}  = \displaystyle\frac{Rs.800}{0.91}$,

whereas total sale price was $TSP = Rs.400 + Rs.400 = Rs.800$. This is less than the total cost price.

Thus there has been a net loss in the whole transaction,

\begin{align} Loss &= \displaystyle\frac{Rs.800}{0.91} - Rs.800 \\ & = Rs.800\times{\displaystyle\frac{.09}{0.91}} \\ & =\frac{Rs.800\times{.09}}{0.91} \end{align}

This loss is on the total cost of $\displaystyle\frac{Rs.800}{0.91}$,

The percentage loss is,

$\frac{\displaystyle\frac{Rs.800\times{.09}}{0.91}}{\displaystyle\frac{Rs.800}{0.91}}= .09=9{\%}$.

Answer: c: 9%.


Abstraction technique and rich concepts for faster solution

If you observe carefully, we have never resorted to calculating the the actual values of the two fractional cost prices, we kept those in fraction form only. Finally there was no need to calculate any awkward fraction because, in percentage ratio calculation the prices were canceled out leaving only the ratio of proportions.

This characteristic enables us to observe in general,

In profit and loss problems, wherever the selling prices are same in two sale situations and final requirement is a percentage (in terms of whatever, sale or cost price), we may ignore the same selling price altogether. The price values will finally cancel out.

This is an example of direct application of abstraction technique, which says,

When in two entities or situations there is a common set of properties, you can focus on only the specific special properties of each leaving out the common properties or vice versa.

In simple terms, abstraction means generalization, focusing only on the core of the problem that is important, shedding details and thereby decreasing unwanted clutter of information.

Sensing this property we delayed fraction calculation till the last stage and achieved a clean solution, even though the method followed was conventional.

Reason for the abstraction to work in this case

The reason why abstraction works in this case is the Rich concept that,

Profit or loss percentage both are in terms of cost prices, and as the final requirement is also a percentage which essentially is a ratio, the sale price or the cost price, any of the prices will finally be canceled out. A corollary of this concept is, whatever be the prices, if the percentages reamain same, the answer will always be the same.

Rich concepts

In any topic area, problems can be solved using the very basic concepts. We have illustrated the basic concepts in profit and loss problems already.

Additionally, when you are well conversant with the basic concepts you can form what we call Rich concepts from the basic concepts itself and use the rich concepts sometimes to solve problems in the specific area in a few steps only.

Rich concepts in Profit and Loss

If CP = Cost price and $x{\%}$ = Profit, in terms of Sale price SP,

$CP =  \displaystyle\frac{SP}{1 + 0.01\times{x}}$.

Similarly, if $x{\%}$ = Loss, in terms of Sale price SP,

$CP =  \displaystyle\frac{SP}{1 - 0.01\times{x}}$.

These relationships are so simple and can so easily be derived from the basic concepts of Profit and Loss topic, it is useful to remember and use these rich or derived concepts when SP and profit are given but CP is unknown.


Efficient solution

Being aware now of these simple but rich concepts, we apply the concepts on our problem. Additionally as we are also aware of the fact that only percentages are important, prices are not, we think and express in terms of the terms like CP1, CP2 and SP not considering their numerical values.

With these two powerful new concepts under the belt, we form our problem expression directly moving to the final stage,

$CP1 + CP2 = SP\left(\displaystyle\frac{1}{1.3} + \displaystyle\frac{1}{0.7}\right) = 2SP\times{\displaystyle\frac{1}{0.91}}$.

This total cost is larger than total sale price of 2SP.

So the loss is,

$2SP\left(\displaystyle\frac{1}{0.91} - 1\right)=2SP\left(\displaystyle\frac{.09}{.91}\right)$.

Or, $Loss = \displaystyle\frac{2SP}{0.91}\times{.09}=9{\%}$ of total cost price.

2SP is part of the total cost and loss both, thus canceled out or abstracted out in our terminology.

This result can be reached in a minute comfortably if you follow this path with full awareness of the rich concepts of Profit and Loss and the Abstraction technique.


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 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