SSC CGL level Solution Set 6, Profit and Loss

Sixth SSC CGL level Solution Set, topic Profit and Loss

SSC CGL level Arithmetic Profit and Loss Solution Set 6

This is the sixth solution set of 10 practice problem exercise for SSC CGL exam on topic Profit and Loss. Students must complete the corresponding question set in prescribed time first and then only refer to the solution set.

It is emphasized here that answering in MCQ test is not at all the same as answering in a school test where you need to derive the solution in perfectly elaborated steps.

In MCQ test instead, you need basically to deduce the answer in shortest possible time and select the right choice. None will ask you about what steps you followed.

Based on our analysis and experience we have seen that, for accurate and quick answering, the student

  • must have complete understanding of the basic concepts of the topics
  • is adequately fast in mental math calculation
  • should try to solve each problem using the most basic concepts in the specific topic area and
  • does most of the deductive reasoning and calculation in his head rather than on paper.

Actual problem solving happens in items 3 and 4 above. But how to do that?

You need to use your problem solving abilities only. There is no other recourse. Your problem solving skill can be significantly increased by intelligent and systematic practice.

This solution set will give you a number of powerful approaches to solve lengthy problems in a very short time. You need to understand the concepts and apply in other problems to really be able use these valuable concepts during actual tests with confidence.

Sixth solution set- 10 problems for SSC CGL exam: topic Profit and Loss - time 12 mins

Q1. 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

Solution: 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 = $120{\%}(400 + x) = 400 + 80 + 1.2x$, where $80 + 1.2x$ = Profit.

As profit doubles after selling price increase, $80 + 1.2x = 2x$,

Or, $0.8x = 80$,

Or, $x = 100$.

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

Alternative approach:

Change in selling price = $0.2(400 + x) = $ change in profit $ = x$.

Or, $0.8x = 80$.

In this approach the steps upto finding the cost price is same. As a next step only the changes in selling price and thereby this change causing a change in profit are considered.

This approach is faster, if you are comfortable with 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.

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.

Q2. A trader marked the price of a commodity so as to include a profit of 25% but allowed a discount of 16% on marked price. His actual profit will be,

  1. 16%
  2. 5%
  3. 25%
  4. 9%

Solution: Assuming cost price as Rs.100, the marked price = Rs.125 as it includes a profit of 25%.

Actual selling price is less than this marked price by the discount of $16{\%}\times{125} = 20$.

Thus selling price is, Rs.105 and actual profit is $5{\%}$.

Answer: Option b: $5{\%}$.

Key concepts used: Using basic concepts finding marked price first with assumed cost price Anchored at 100. All subsequent values will be in terms of this anchor and hence will reflect the percentages -- next taking care to apply the $16{\%}$ discount on this marked price and deducting the discount from the marked price to obtain the selling price and hence the profit -- all along for ease of deduction we have assumed the anchor value of cost price as 100.

Q3. A shopkeeper allows 23% discount on his advertised price and still makes a profit of 10%. If he gains Rs.56 on one article his advertised price in Rs. is,

  1. 780
  2. 820
  3. 790
  4. 800

Solution: $10{\%}$ profit is on cost price and equals Rs. 56. So, cost price of an item is 10 times Rs.56, that is, Rs.560.

It follows then, selling price = $560 + 56 = \text{Rs.}616$.

As the reduction of $23{\%}$ by discounting is on the advertised price, and after reduction we land onto the selling price, we have,

$0.77x = \text{Rs.}616$, where $x$ = advertised price.

Or, advertised price $= \displaystyle\frac{\text{Rs.}616}{0.77} = \text{Rs.}800$.

Answer: Option d: 800.

Key concepts used: Clear concept regarding on which quantity to apply the percentages -- profit percentage is applied as a convention on cost price and discount percentage (which is a reduction) on marked price.

Q4. If there is a profit of $20{\%}$ on the cost price of an article, the percentage of profit calculated on its selling price will be,

  1. 20
  2. $8\frac{1}{3}$
  3. $16\frac{2}{3}$
  4. 24

Solution: Assuming cost price as 100 selling price is 120, as profit = $20{\%}$ adds up to cost price.

Then the profit percentage of 20 to 120 is,

$\displaystyle\frac{20}{120}\times{100} = 16\frac{2}{3}$.

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

Key concepts used: First getting the profit and selling price and then changing the base of percentage to selling price instead of cost price.

Q5. A shopkeeper gains $20{\%}$ while buying the goods and $30{\%}$ while selling them. His total gain is,

  1. $36{\%}$
  2. $50{\%}$
  3. $40{\%}$
  4. $56{\%}$

Solution: To gain $20{\%}$ while buying, the shopkeeper buys goods worth Rs.120 at Rs.100 (this is potential profit, where his investment is Rs.100).

To make a profit of another $30{\%}$, his selling price will be an additional $30{\%}$ on the official buying price, that is,

$30{\%}$ of Rs.120 = Rs.36.

Thus his profit is = Rs.20 + Rs.36 = Rs.56 on his actual cost price of Rs.100. This is equivalent to a total profit of $56{\%}$.

Answer: Option d: $56{\%}$.

Key concepts used: Taking care to get the actual cost price with the gain at the buying stage -- at the second stage profit percentage is on the official cost price rather than the actual lower cost price, so that due to two stage gain, his actual gain is added up.

Q6. A discount of $15{\%}$ on one article is same as discount of $20{\%}$ on another article. The costs of two articles can be,

  1. 60, 40
  2. 85, 60
  3. 80, 60
  4. 40, 20

Solution: From the given statement,

0.15 of CP1 = 0.20 of CP2, where CP1 and and CP2 are the cost prices of two articles.

Or, $\displaystyle\frac{CP1}{CP2} = \displaystyle\frac{0.20}{0.15}=\displaystyle\frac{4}{3}$.

Only the option c with costs 80, 60 satisfies this condition.

Answer: Option c : 80, 60.

Key concepts used: Careful formation of the basic relationship between the two costs -- discount is on the costs as the selling price or marked price is not in the picture -- ratio equalization using the choice values.

Q7. A trader sells two bullocks for Rs.8400 each neither losing nor gaining in total. If he sells one of the bullocks at $20{\%}$ profit, the other is sold at a loss of,

  1. $21{\%}$
  2. $20{\%}$
  3. $14\frac{2}{7}{\%}$
  4. $18\frac{2}{9}{\%}$

Solution: Profit for the first bullock being $20{\%}$ on its cost price, its selling price,

Rs.8400 = 1.2 of Cost price,

Or, Cost price of first bullock = $\displaystyle\frac{\text{Rs.}8400}{1.2} = \text{Rs.}7000$, and so the profit $=\text{Rs.}8400 - \text{Rs.}7000=\text{Rs.}1400$.

By selling the second bullock at selling price Rs.8400, the trader must have suffered a loss of this Rs.1400 thus nullifying his gain in the first sale.

So cost price of second bullock $=\text{Rs.}8400 + \text{Rs.}1400=\text{Rs.}9800$, and loss percentage with respect to this cost price is,

Loss $=\displaystyle\frac{1400}{9800}\times{100}{\%}=14\frac{2}{7}{\%}$.

Answer: Option c : $14\frac{2}{7}{\%}$.

Key concepts used: Using the basic concepts of profit as a cost percent and using the given information, deriving the final cost value and the profit value in a step by step approach.

Alternate method: As ultimately percent profit was required with given profit also as a percent, Rs.8400 as equal sale value is a superfluous information. It could have been any value but equality of both sale values would have given the same result. Let us see how.

Let CP1, CP2, SP and are the cost and sale prices. So from the statement of the first sale we have,

1.2 of CP1 = SP. We need to find out the profit $(SP - CP1)$ in terms of SP.

With this goal in mind, we further proceed to get,

1.2 of (SP - CP1)=0.2 of SP

Or, Profit $=\frac{0.2}{1.2}$ of SP.

This being the loss amount,

CP2 = SP + $\frac{1}{6}$ of SP = $\frac{7}{6}$ of SP,

Loss in second sale = CP2 - SP = CP2 - $\frac{6}{7}$ of CP2=$\frac{1}{7}$ of CP2 = $14\frac{2}{7}{\%}$ of second bullock cost.

Note: This alternate method is mathematically precise and avoids practically any calculation. But this is an abstract method. Unless the student is well versed in dealing with such abstraction without any numerical values, she should follow the first, more usual method as the calculations here are very simple.

Nevertheless adoption and practice of this abstract approach should surely improve the speed of problem solving of the student.

Q8. Ramu buys a plot of land for Rs.96000. He sells $\frac{2}{5}$ths of the land at a loss of $6{\%}$. He wants now to make a profit of $10{\%}$ on the whole transaction by selling the remaining land. At what percentage profit should he sell his remaining land?

  1. $7{\%}$
  2. $20{\%}$
  3. $14{\%}$
  4. $20\frac{2}{3}{\%}$

Solution: The loss of his first sale was,

$0.06\times{\frac{2}{5}}$ of 96000 = 0.024 of 96000.

His target gain is $10{\%}$ or 0.1 of 96000. So in his next sale he has to gain 0.124 of 96000.

As he has to gain this on value of the rest of the land, that is, three fifths or 0.6 of 96000, his target profit percent should be,

$\frac{0.124}{0.6}$ of $100{\%} = 20\frac{2}{3}{\%}$.

Answer: Option d: $20\frac{2}{3}{\%}$.

Key concepts used: Converting all percents into decimal and expressing these as a fraction of the total cost price simplifies the whole deduction. The total cost value in this case acts as a common factor that was not needed to be brought into any calculation finally. But continued reference to it added meaning to the statements -- dealing directly with initial loss, target gain and needed gain kept the focus on the proceedings -- target gain must make up the intial loss, all in terms of the common factor of Cost price.

Note: From the problem solving technique point of view, we have used all through only the proportional changes in terms of the whole, without needing to calculate the actual values of the changes. First change was the loss amount, second was the target profit amount and the third was the required profit amount.

Along with the Change analysis technique though the use of modified Base equalization technique, converting all the changes as proportion of common base factor of cost price was necessary for this simple deduction to be possible.

Q9. A trader marked his goods at $20{\%}$ above the cost price of Rs.245346. He sold half the stock at the marked price, one quarter at a discount of $20{\%}$ on the marked price and the rest at a discount of $40{\%}$ on the marked price. His total gain is,

  1. $2{\%}$
  2. $15{\%}$
  3. $4.5{\%}$
  4. $13.5{\%}$

Solution: First sale price = Half of 1.2 of cost price = 0.6 of cost price.

Second sale price = One fourth of (1.2 of cost price - 0.2 of 1.2 of cost price)= 0.25 of 0.96 of cost price= 0.24 of cost price.

Third sale price = One fourth of (1.2 of cost price - 0.4 of 1.2 of cost price)= 0.25 of 0.72 of cost price=0.18 of cost price.

Total sale price = (0.6 + 0.24 + 0.18) of cost price = 1.02 of cost price. In total he then gained a net profit of $2{\%}$.

Answer: Option a: $2{\%}$.

Key concepts used: Conversion of each percent to decimal portions of the cost price -- expressing all values in terms of portions of cost price -- directly expressing each sale value -- finally elimination of the need to do any difficult multiplication and associated arithmetic calculation on large values. Cost price information becomes superfluous.

Note: This is another example of using a modified form of base equalization technique in which all values are expressed in terms of proportion of common base or reference value. The whole series of calculations are done on the proportions only keeping the base value out of the calculations. Effectively the base value is altogether eliminated from the consideration because the final result is to be expressed as a proportion of the same reference base value.

Any type of mathematical problem having these characteristics can be solved quickly and efficiently using this powerful problem solving technique.

Q10. 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

Solution: From the problem statement,

The difference in profit percentage, that is, $1{\%}$ equals Rs.15. As profit percentage is usually expressed as a percent of the cost price, the cost price here would be $100{\%}$, that is, Rs.1500.

Answer: Option a : Rs.1500.

Key concepts used: Direct change analysis -- equating it to the given amount -- concept of profit as a percent of cost price -- cost price is taken as $100{\%}$.


Resources that should be useful for you

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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.

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