How to solve SSC CGL level Profit and loss problems by change analysis in a few steps 6 | SureSolv

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

Instead of calculating the changed values step by step, just equate the change and its effect

How to solve profit and loss problems in a few steps by change analysis technique 6

In profit and loss problems when price or profit changes by certain percentages, without calculating the actual changed values and just by using the change and its effect, solution can be achieved very fast. We would solve two problems on profit and loss to highlight the power of this technique.

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 set of concepts involving marked price and discount.

Basic concepts on marked price and discount

It is a standard market practice to mark or print the price on a label attached to a product for sale. The seller and the buyer both know that there will be a discount on this marked price so that the buyer can purchase the item happily at the discounted price less than the marked price.

The discounted price is the actual sale price and difference between the sale price and the cost price of the item is the profit.

To formally express these ideas, let's assume marked price $\text{MP}$ is marked above the cost price $\text{CP}$ by 30% so that,

$\text{MP}=1.3\text{CP}$.

Let us also assume that the seller has decided to offer a discount of 20% to the buyer. This percentage will be on the marked price to reduce it to the level of sale price $\text{SP}$ so that,

$\text{SP}=\text{MP}-0.2\text{MP}$

$=0.8\text{MP}$

$=0.8\times{1.3}\text{CP}$

$=1.04\text{CP}$.

The profit accrued in this case is thus, $0.04\text{CP}$ and in percentage of cost price, 4% of cost price.

These form briefly the basic concepts on cost price, sale price, marked price, discount and profit.

Remember the core concept on profit and loss—percentage profit or loss is invariably on cost price.

In this session, we will solve two chosen problems using change analysis technique. For both, we will also solve each of the problems in usual method also to showcase the speed and power of change analysis technique.

The main theme of this session is,

Wherever you find change, look for analyzing and using the change for reaching the solution.

Chosen Problem 1

The marked price of an article is 50% above its cost price. When marked price is increased by 20% and selling price is increased by 20%, the profit is doubled. If original marked price was Rs.600, the original selling price was,

  1. Rs.400
  2. Rs.580
  3. Rs.500
  4. Rs.550

Conventional solution to chosen problem 1

From first statement,

Original Marked price = 150% of original cost price = Rs.600.

So, Original cost price = $\frac{2}{3}$ of Rs.600 = Rs.400.

If original selling price $= 400 + x$, where, $x$ was original profit, after 20% increase in selling price,

The changed selling price becomes,

$\text{SP}_{changed}= 120{\%} \text{ of }(400 + x)$

$= 400 + 80 + 1.2x$, where $80 + 1.2x$ = New Profit.

As profit doubles after selling price increase,

$80 + 1.2x = 2x$,

Or, $0.8x = 80$,

Or, $x = 100$.

So, original selling price = $400 + x = \text{Rs.}500$.

Answer: Option c: Rs.500.

Faster solution to Chosen problem 1 using Change Analysis Technique

As original marked price of Rs. 600 was 50% more than the cost price $\text{CP}$, it is,

$1.5=\displaystyle\frac{3}{2}$ times the cost price, and so,

Cost price $\text{CP}=\frac{2}{3}\times{\text{Rs.}600}=\text{Rs.}400$ just as before.

Note that cost price doesn't change, the changes are in the sale price and the marked price. Both of these are increased by 20% to double the profit.

If $x$ is the original profit amount, the original sale price was, $(400+x)$ and after increasing this sale price by 20%, the profit becomes $2x$, that is, it increases by $x$.

So,

Change in selling price is

$0.2(400 + x) = $ change in profit $ = x$, cost price remaining unchanged,

Or, $80 +0.2x=x$,

Or, $0.8x = 80$,

Or, $x=100$.

So, original sale price was Rs.500.

In this approach, the steps up to finding the cost price are same as before.

At the next step only, the change in selling price and its effect on changing the profit are equated.

This approach is faster, if you are comfortable using Change Analysis Technique which is more conceptual and thereby abstract. Once you get habituated though, this technique may enable you to solve large number of problems much faster in many areas.

Answer: Option c: Rs.500.

Key concepts used: The very first step is to find out the value of cost price. This is the natural first step as on this step only hinges further progress towards the solution and more so, because of the given marked price value, you are urged towards finding this value in a few seconds in your mind -- as profit is directly related to selling price and cost price only, you disregard change in marked price. This finally turns out to be superfluous information -- for quickest solution just equate the change in selling price to change in profit.

We won't evaluate the changed sale price. We would use just the change in sale price thus reducing one step and a few valuable seconds.

Chosen problem 2.

While selling a watch a shopkeeper gives a discount of $5{\%}$. If he gives a discount of $6{\%}$ he earns Rs.15 less as profit. What is the marked price of the watch?

  1. Rs.1500
  2. Rs.1200
  3. Rs.1400
  4. Rs.750

Conventional solution to Chosen problem 2

Let's assume the cost price, original sale price, changed sale price and the marked price to be, $\text{CP}$, $\text{SP1}$, $\text{SP2}$ and $\text{MP}$ respectively.

With 5% discount profit was,

$\text{Profit}_1=\text{SP1}-\text{CP}=0.95\text{MP}-\text{CP}$.

The second time after 6% discount profit decreased by Rs.15. 

So for the second instance,

$\text{Profit}_2=0.94\text{MP}-\text{CP}$

Or, $\text{Profit}_1-15=0.94\text{MP}-\text{CP}$.

Subtracting this second profit relation from the first,

$15=0.01\text{MP}$,

Or, $\text{MP}=\text{Rs.}1500$.

Answer: Option a: Rs.1500.

This is the usual step by step solution. We will now use change analysis technique to solve the problem.

Faster solution to Chosen problem 2 using Change Analysis Technique

From the problem statement,

The difference in discount percentage, that is, $1{\%}$ equals reduced profit of Rs.15, because the anchor value of cost price remains unchanged. The reduction in profit is solely due to $1{\%}$ increase in discount causing corresponding decrease in sale price and hence profit.

As discount percentage is always expressed as a percent of the marked price, the $100{\%}$ of marked price would be,

$100\times{\text{Rs.15}}=\text{Rs.1500}$.

Answer: Option a : Rs.1500.

Key concepts used: Direct change analysis -- Equating change in discount to the reduction in profit amount as anchor value of cost price remains unchanged -- concept of discount as a percent of marked price -- Marked price is taken as $100{\%}$.

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 section on SSC CGL to access all the valuable student resources that we have created specifically for SSC CGL, but generally for any hard MCQ test.

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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,

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