Solution by domain modeling of similar problems: Profit loss pt 2

Submitted by Atanu Chaudhuri on Mon, 29/02/2016 - 11:32

How to solve similar profit and loss problem lightning quick pt 2: Domain modeling of similar problems

You Can Solve Similar Problems by Domain Modeling of Profit Loss Questions in Seconds

This is on how to solve similar profit and loss questions lightning quick by domain modeling of similar problems using basic and advanced concepts.

We have already solved a problem on profit and loss in only a few steps. As a recap, we reproduce it here partly.

Previous Problem Recap: The Question

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,

We have started with the conventional solution and then went on to the efficient solution. Briefly the conventional solution was as follows. We are skipping the basic concepts here. If you want, you may refer to our earlier related discourse.

From Efficient Conventional solution to a Faster Solution

Efficient Conventional Solution

Cost prices CP1 and CP2 are calculated in terms of the common selling price SP using profit and loss formula,

The first cost price CP1 when profit is 30%: $CP1 = \displaystyle\frac{Rs.400}{1.3}$.

The second cost price CP2 when loss is 30%: $CP2 = \displaystyle\frac{Rs.400}{0.7}$.

Total cost price $TCP = CP1 + CP2= \displaystyle\frac{Rs.400}{1.3} + \displaystyle\frac{Rs.400}{0.7} = \displaystyle\frac{Rs.800}{0.91}$, when

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

As total cost price TCP is more than total sale price TSP, there is loss,

$Loss = \displaystyle\frac{Rs.800}{0.91} - Rs.800 = 9{\%}$.

This followed conventional method, but postponing calculations till the last stage improved the solution time significantly.

In a More Efficient Solution, an Abstracted Variable SP is used Instead of Actual Value

We have prices CP1 and CP2 and sale price SP not considering their numerical values at all.

The total cost is,

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

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

The loss is,

$Loss = 2SP\left(\displaystyle\frac{1}{0.91} - 1\right)=\displaystyle\frac{2SP}{.91}\times{.09}= 9{\%} \text{ of total cost price}$.

This is a faster solution because you could eliminate the steps of expressing the two costs in individual steps. The reason why you could do this lies in the pattern of common sale prices on both sales leading to the use of a single abstracted variable SP for the selling prices.

The Second Similar Question

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.300. Find the net profit or loss.

Only one change from the first question - the selling price is changed from Rs.400 to Rs.300, percentages remain same.

Already used to the idea of expressing the common Sale Price as a variable SP, the final step can be reached directly,

$Loss = \displaystyle\frac{2SP}{0.91}\times{.09}=9{\%}$ of total cost price. With percentage of loss or gain needed, the common sale price cancels out.

The solution reached in one step (mentally).

The interesting pattern is: Loss of 9% will be the answer whatever be the value of total cost price derived from same sale price SP.

You have effectively abstracted out the sale price from consideration completely along with the need for calculation of cost prices.

You could do this because the final percentage loss is in terms of the total cost. The actual total cost value which in turn depends on the same sale price value, will change but won't change the percent loss value.

This is abstraction technique in action.

Abstraction technique

This technique is described as,

In a problem with two situations, when there is a common set of properties between the two situations, you can focus on only the specific special properties of each situation that are not common ignoring the common properties.

By first level of this abstraction you have effectively factored out the sale value from calculations altogether.

This is the first level of abstraction in this type of problem. Result:

Whatever be the equal sale price in such a profit and loss problem, if other percent values and actions on the values remain unchanged the answer will remain same.

Third similar problem version

A man sells two wrist watches one at a profit of 50% and another at a loss of 50%, but each at a same selling price of Rs.300. The net profit or loss is,

The profit and loss percentages in two sales are still same but the values are different from the earlier problem. The sale prices being same can be ignored.

Mentally calculate the answer,

$CP1 + CP2 = SP\left(\displaystyle\frac{1}{1.5} + \displaystyle\frac{1}{0.5}\right) = 2SP\times{\displaystyle\frac{1}{0.75}}$

Because of the change of percentage loss and gain in the individual transactions to 50%, the denominators of individual CPs changed.

Loss is now 25% of total sale price. To get this, only the denominator of the total cost price needed calculation.

In the same problem, if the individual percent loss or gain is changed to 40%, the denominator of the total cost price will be 0.84,

$CP1 + CP2 = SP\left(\displaystyle\frac{1}{1.4} + \displaystyle\frac{1}{0.6}\right) = 2SP\times{\displaystyle\frac{1}{0.84}}$.

Loss = (1 - 0.84) = 16%. That fast you get it.

Pattern recognition technique

Do you recognize any pattern in these similar type of problems?

As the percentage profit or loss changes, the net result follows a consistent mathematical pattern:

30% loss/gain leads to a 9% net loss.
50% loss/gain leads to a 25% net loss.
40% loss/gain leads to a 16% net loss.

The pattern is: loss/gain is the square of the ten's digit of the percentage. For example, 30% has a ten's digit of 3, and the resulting loss is 9%. Similarly, 40% gives 16%, and 50% gives 25%.

Conjecture: This pattern should work correctly for any same values of loss/gain in a two-transaction profit loss problem with same sale value.

Testing the Conjecture

Let’s test this pattern with a new scenario: a 20% gain on one watch and a 20% loss on the other, with the selling price still Rs. 300.

Using the same method:

$CP1+CP2=SP\left(\displaystyle\frac{1}{1.2}+\displaystyle\frac{1}{0.8}\right)=2SP\times{\displaystyle\frac{1}{0.96}}$

The result is a 4% net loss, confirming our pattern on the change in profit/loss percentage to 20%.

How to solve similar problems in a few seconds

Armed with the knowledge gained from the small discovery of a specific pattern of the result in similar problems, where the defining parameter value of individual loss or gain percentage may take up a specific type of different value (only in multiples of ten), if we face now a similar problem with a loss and gain of 60% in two sale situations, we may select the correct answer of net loss of 36% in a few seconds without going through any derivation at all.

And this will hold good whatever be the common sale value in two sale transactions.

Specifying these similar problems with precision: Domain modeling of similar problems

We have now reached the stage where the questions such as, "What really are the specifics of this similar type of problem?" or "What is the use of this approach at all?" automatically arise in our mind.

To reach the last stage of this step by step long process, we must specify with precision the "similar type of problem" where our capacity to solve such a problem has grown to even solving a new instance of such a problem in seconds. We attempt to define this similar type of problem in abstraction precisely as,

Precise and abstract problem definition

If two similar objects (anything that can be bought and sold, from pens to bullocks) are sold on two separate transactions, one at a loss of $n\times{10{\%}}$, where $n$ may take only values 1, 2, 3, 4, 5, 6, 7, 8 and 9, at the same sale price of $p$, where $p$ is any positive real number, what will be the net loss or gain?

Using the knowledge of useful patterns in such problem, we may directly arrive at the answer as, $n^2{\%}$ net loss.

The precise and abstract problem definition form a small domain model consisting of all profit loss problems conforming to the precise definitions.

What would be the advantages?

If you can create the properties of such a group of similar profit and loss problems, you will get the answer to any problem fitting the domain in seconds.

Conclusion: Rapid Solutions Using Patterns

By recognizing patterns and using abstraction, we can solve these problems in seconds, regardless of the specific selling prices. The key is to focus on percentage changes and apply the pattern of the ten’s digit squared.

Resources that should be useful for you

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