SSC CGL level Solution Set 29, Profit and loss 2

29th SSC CGL level Solution Set, 2nd on topic Profit and loss

SSC CGL Solution Set 29 profit and loss

This is the 29th solution set of 10 practice problem exercise for SSC CGL exam and 2nd on topic Profit and loss. Students should complete the corresponding question set in prescribed time first and then only refer to this solution set.

We will repeat the method of taking the test and if you have not taken the test yet, will expect you to refer to the SSC CGL level Question Set 29 on Profit and loss to take the test before you go through its solution here.

Method for taking this 10 problem test and get the best results from the test set:

  1. Before start, go through the tutorials Numbers, Number system and basic arithmetic operations, Factorizing or finding out factors, HCF and LCM, Basic concepts on fractions and decimals part 1, Ratio and proportion or any other short but good material to refresh your concepts if you so require.
  2. Answer the questions in an undisturbed environment with no interruption, full concentration and alarm set at 12 minutes.
  3. When the time limit of 12 minutes is over, mark up to which point you have answered, but go on to complete the set.
  4. At the end, refer to the answers given in this companion 29th SSC CGL solution set on Profit and loss 2, to mark your score at 12 minutes. For every correct answer add 1 and for every incorrect answer deduct 0.25 (or whatever is the scoring pattern in the coming test). Write your score on top of the answer sheet with date and time.
  5. Identify and analyze the problems that you couldn't do to learn how to solve those problems.
  6. Identify and analyze the problems that you solved incorrectly. Identify the reasons behind the errors. If it is because of your shortcoming in topic knowledge improve it by referring to only that part of concept from the best source you get hold of. You might google it. If it is because of your method of answering, analyze and improve those aspects specifically.
  7. Identify and analyze the problems that posed difficulties for you and delayed you. Analyze and learn how to solve the problems using basic concepts and relevant problem solving strategies and techniques.
  8. Give a gap before you take a 10 problem practice test again.

Important: both practice tests and mock tests must be timed, analyzed, improving actions taken and then repeated. With intelligent method, it is possible to reach highest excellence level in performance.

Resources that should be useful for you

You may refer to:

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.

Tutorials that you should refer to

Numbers, Number system and basic arithmetic operations

Factorizing or finding out factors

HCF and LCM

Basic concepts on fractions and decimals part 1

Ratio and proportion

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29th solution set - 10 problems for SSC CGL exam: topic Profit and loss - Answering time 12 mins

Problem 1.

A manufacturer sells an article to a wholesale dealer at a profit of 10%. The wholesale dealer sells it to a shopkeeper at 20% profit. Ultimately, the shopkeeper sells it to a customer at a price of Rs. 56100, suffering a loss of 15%. The cost price of the article for the manufacturer is then,

  1. Rs. 10000
  2. Rs. 25000
  3. Rs. 50000
  4. Rs. 55000

Solution 1 - Problem analysis and solving

This is a multistage sale and purchase situation in which at each stage profit and loss may need to be considered thus increasing the complexity. Let us understand the mechanism of this multistage process in brief.

Let $C_1$, $S_1$, and $P_1$ be the cost, sale price and the profit in the first stage, $C_2$, $S_2$, and $P_2$ be the cost, sale price and the profit in the second stage and  $C_3$, $S_3$, and $P_3$ be the cost, sale price and the profit in the third stage.

For our first transaction then,

$S_1 = C_1 + 0.1C_1=1.1C_1$,

This is because of the 10% profit (of cost price) to the manufacturer.

As a strategy, we transform the profit in terms of the cost price to reduce the number of variables to two in the profit loss relationship.

Furthermore, in multistage transactions, the sale price of one stage becomes the cost price of the next stage. This is the only relationship between the two stages - profit remains pertinent to only one individual stage of transaction.

For the second transaction,

$S_2 = 1.2C_2$ because of 20% profit in this stage.

But cost price of this stage two is the sale price of the previous stage, that is, $C_2= S_1$. We have then,

$S_2 = 1.2\times{S_1} = 1.2\times{1.1C_1}$

$\hspace{7mm}=1.32C_1$.

Similarly for the third transaction, as at 15% loss, the sale price is less than cost price by 15% of cost price,

$S_3 = 56100 = 0.85C_3 $

$\hspace{7mm}= 0.85S_2 = 0.85\times{1.32C_1}$,

Or, $C_1 = \displaystyle\frac{56100}{1.32\times{0.85}} = \displaystyle\frac{51000}{1.2\times{0.85}}$

$\hspace{13mm}= \displaystyle\frac{1000000}{4\times{5}}$

$\hspace{13mm}= 50000$.

Rich concept of multistage sale purchase transactions

Though we have proceeded to establish relationship between the adjoining stages of sale and purchases, examining the whole process above we can see that the initial cost is related to the final stage variables through a series of multiplicative terms,

$S_3 = (1+P_3)(1 +P_2)(1+P_1)C_1$


$S_3 = (1+P_3)C_3$,

Or, $S_3 = (1+P_3)S_2$, as $S_2=C_3$,

Or, $S_3 = (1+P_3)(1 +P_2)C_2$, as $S_2=(1 +P_2)C_2$,

Or, $S_3 = (1+P_3)(1 +P_2)S_1$, as $S_1=C_2$,

Or, $S_3 = (1+P_3)(1 +P_2)(1+P_1)C_1$, as $S_1=(1 +P_1)C_1$


Alternative process of solution using rich concept of multistage profit and loss

In any such problem of multistage transactions, only one of the variables will be the final unknown with all other variables in all stages known. Without going through the detailed steps we can straightaway employ this rich concept of relationship between the profits, costs and sale prices.

Using this approach in our problem directly,

$561000 = 1.1\times{1.2}\times{0.85}\times{C_1}$,

Or, $C_1 = \displaystyle\frac{56100}{1.1\times{1.2}\times{0.85}}=50000$

Answer: Option c: Rs. 50000.


Key concepts used: Multistage sale and purchase transactions -- rich concept of profit and loss in a multistage transaction.

Problem 2.

The ratio in which a variety of Darjeeling tea at Rs. 32/kg is mixed with a variety of Assam tea costing Rs. 25/kg to get a gain of 20% by selling the blended tea at Rs. 32.40/kg is,

  1. 4 : 3
  2. 2 : 5
  3. 3 : 4
  4. 5 : 2

Solution 2 - Problem analysis and solving

As ratio is desired we would assume a single unknown variable $x$ as the amount of Darjeeling tea in the desired mixture as a proportion of 1 kg mixture with a cost contribution of $32x$. Obviously $x$ is a proper fraction with value between 0 and 1.

Similarly cost contribution of $(1-x)$ portion of Assam tea in a mixture of 1 kg is, $25(1-x)$, giving a total cost of 1 kg of mixture as,

$32x + 25(1-x)=7x + 25$ per kg.

To get a profit of 20% out of selling this mixture at the selling price of Rs. 32.40 per kg, we get the cost and selling price relationship as,

$SP = 1.2CP$,

Or, $CP = \displaystyle\frac{32.4}{1.2} = 7x + 25$

Or, $7x + 25 = 27$,

Or, $x=\displaystyle\frac{2}{7}$,

and, $1-x= \displaystyle\frac{5}{7}$.

So the desired ratio of mixture is, 2: 5.

Answer: Option b : 2:5.

Key concepts used: Mixture concepts -- profit and loss concepts -- ratio concepts.

Problem 3.

As a marketing campaign a company decides to offer an extra amount of 20% of a toothpaste free with no change in price. If per gm total cost of the toothpaste to the company is same and original profit were 80% on this particular product, what would be the new profit after the free offer?

  1. 40%
  2. 50%
  3. 60%
  4. 70%

Solution 3 - Problem analysis and solving

By the problem definition before the free offer,

$SP = 1.8CP_1$, where $SP$ and $CP_1$ were the selling price and cost price of one unit of toothpaste of say, $x$ gm.

As cost price varies directly with the amount of toothpaste per gm, after 20% increase in amount as free, the cost of the toothpaste tube containing now $1.2x$ toothpaste increases to $CP_2=1.2CP_1$. 

But as the selling price remains same as before we have now,

$Profit = SP - CP_2 $

$\hspace{13mm}= 1.8CP_1 - 1.2CP_1 $

$\hspace{13mm}= 0.6CP_1$

$\hspace{13mm}=\displaystyle\frac{0.6CP_2}{1.2}$

$\hspace{13mm}=0.5CP_2$.

Thus profit in terms of increased cost is now 50%.

Answer: b: 50%.

Key concepts used: Basic profit and loss concepts -- marketing concepts..

Problem 4.

A man sells 320 mangoes at the cost price of 600 oranges when prices of mango to orange are in the ratio of 3 : 2. What is his gain percent?

  1. 10%
  2. 15%
  3. 20%
  4. 25%

Solution 4.

As mango price to orange price ratio is 3 : 2, cost price of 600 oranges will be equal to cost price of $\frac{2}{3}\times{600} =400$ mangoes.

So the man sells 320 mangoes at the cost price of 400 mangoes.

If cost of 1 mango is $x$, cost price of 400 mangoes $=400x$ which is the sale price of 320 mangoes. As cost price of 320 mangoes is nothing other than $=320x$, the profit is,

$P=400x - 320x=80x$ on the cost of $320x$, that is, the profit percent is, 25%.

Answer: Option d: 25%.

Key concepts used: Profit and loss basic concepts -- ratio concepts -- percent concepts.

Problem 5.

Ramu started business with a capital of Rs. 100000. Bishu joined the business one year later with a capital contribution of Rs. 200000. At the end of three years from the start of the business, the profit earned was Rs. 84000. The share of Bishu in the profit exceeded that of Ramu by,

  1. Rs. 15000
  2. Rs. 14000
  3. Rs. 12000
  4. Rs. 10000

Solution 5.

First year end profit for Ramu, $P_{r1}= P_1$ and no profit for Bishu.

For second and third year together profit for Ramu, $P_{r2}=\frac{1}{3}P_2$ and for Bishu, $P_{b2} = \frac{2}{3}P_2$.

Given, Total profit,

$P = P_1 + P_2 = 84000$.

Total profit for Ramu in 3 years,

$P_r = P_1 + \frac{1}{3}P_2$,

Total profit for Bishu in 3 years,

$P_b = \frac{2}{3}P_2$.

So, $P_b - P_r = \frac{1}{3}P_2 - P_1$

Here we need to assume that the profits in each year was in proportion of the investments.

By this assumption we get, first year profit one portion of total profit of 7 portions and next two years 6 portions (1 + 2= 3 portions of capital invested for two years). Thus $P_2=6$ portions of total profit, $P_1=1$ portion of total profit and, Bishu's excess share of profit as,

$P_b - P_r = \frac{1}{3}P_2 - P_1 = 1$ portion of total profit out of 7 portions of 84000, that is, Rs12000..

Answer: Option c: Rs. 12000.

Key concept used: Ratio -- share of profit in relation to investment.

Alternate solution

Total investment in terms of Rs-year,

$I = 100000 + 300000 + 300000 = 700000$.

Out of this, Ramu's investment in terms of Rs-year was,

$I_r = 100000 + 100000 + 100000 = 300000$.

Similarly Bishu's investment in terms of Rs-year was,

$I_b=200000 + 200000 = 400000$.

Thus Bishu will be getting a share of profit of 4 portions and Ramu 3 portions out of total 7 portions.

Their difference is thus 1 portion out of 7 total in favor of Bishu.

If the total profit were Rs.84000, Bishu will get one seventh, that is, Rs, 12000 extra.

Rule applicable:

Profit share of a person will be proportional to his investment Rs.-year ratio to the total Rs.-year investment.

Problem 6.

An article was sold at a profit of 4%, but due to general price rise, the cost of the article to the company rose by 30%. As a temporary measure, to watch the market trend, the company decided not to increase the price of the article for three months. What percent of loss the company would suffer on the article during this three month period?

  1. 40%
  2. 20%
  3. 50%
  4. 20%

Solution 6.

Before cost rise the relation of sale price and cost price was,

$SP = 1.04CP_1$, because of 4% profit.

After price rise, sale price is still $SP$ but the cost price is now, $CP_2 = 1.3CP_1$

So the loss after the cost price rise is,

$Loss = CP_2 - SP$

$\hspace{10mm}=1.3CP_1 - 1.04CP_1$

$\hspace{10mm}=0.26CP_1$

$\hspace{10mm}=\displaystyle\frac{0.26CP_2}{1.3}$

$\hspace{10mm}=0.2CP_2$

Thus loss is 20%.

Answer: Option b : 20%.

Key concepts used: Basic profit and loss concepts.

Alternate solution

Increased cost price, $CP_2 = 1.3CP_1$, and

Unchanged sale price, $SP=1.04CP_1$ giving a ratio,

$\displaystyle\frac{CP_2}{SP} = \frac{1.3CP_1}{1.04CP_1}=\frac{5}{4}$.

It means for every 5 unit cost sale loss will be 1 unit, that is, loss will be 20%.

This is a conceptual solution using ratio concept in between.

It is a direct application of Many ways technique.

Problem 7.

Due to market pressure a tradesman first discounted an article by 5% on its marked price and then again decided to offer a further discount of 4%. The selling price as a percent of original marked price after two discounts is finally,

  1. 93.2%
  2. 92.8%
  3. 91.2%
  4. 91.8%

Solution 7.

On first discount the selling price became 95% of marked price.

The next discount was 4% on this 95% of marked price $=0.95\times{0.04} = 3.8$% of marked price.

So the total discount on marked price was $ = 5 + 3.8 = 8.8$%, and the final selling price as percent of original marked price $=100 - 8.8=91.2$%.

Answer: c: 91.2%.

Key concepts used: Successive discounts and final selling price as a percent of original marked price.

Problem 8.

A dealer buys an article marked at Rs. 25000 on two successive discounts of first 20% and then 5%. He spends Rs. 1000 for its repair and then sells it for Rs. 25000. What is his gain or loss percent?

  1. loss of 25%
  2. gain of 25%
  3. loss of 10%
  4. gain of 10%

Solution 8.

The first discount was 20% on marked price resulting in 80% of marked price as the cost price.

Then the next discount of 5% was on this 80% and resulted in an effective discount of 4% of additional discount on the original marked price.

So the total effective discount was 24% on the original marked price.

His discount was thus, $25000\times{24}$% $=6000$ and cost price $=25000 - 6000 = 19000$.

He spent Rs 1000 for repair making the effective cost price to Rs. 20000.

Selling it at Rs. 25000 he made a profit of Rs 5000 on Rs. 20000 which was 25%.

Answer: Option b: gain of 25%.

Key concepts used: Discount -- effective discount on two discounts -- profit and loss.

Problem 9.

To attract more visitors zoo authority announces a 20% discount on every ticket which costs Rs. 5. Due to this discount the sale of this ticket increases by 28%. The percentage increase in number of visitors due to the discount is then,

  1. 50%
  2. 40%
  3. 60%
  4. no change

Solution 9 - Problem analysis and solving

As the ticket value denomination is only one and rest of the values are in percentages, actual ticket value is not important in this problem.

The basic relationship of total sale value is,

$T = V\times{p}$, where $T$ is the total sale value, $V$ is the number of visitors and $p$ is the price of ticket.

Before the discount we had, say,

$T_1 = V_1\times{p_1}$,

And after discount,

$T_2 = V_2\times{p_2}$.

By the problem definition, $T_2 = 1.28T_1$ and $p_2=0.8p_1$.

So taking the ratio we get,

$1.28 = 0.8\displaystyle\frac{V_2}{V_1}$,

Or, $V_2 = 1.6V_1$, a 60% increase.

Answer: Option c: 60% increase.

Key concepts used: Proportionality concepts -- ratio concepts.

Problem 10.

During annual sale in July, a shopkeeper offers his goods at a discount of 50% except for the last week when he offers an additional discount of 40%. The prices of goods sold during the last week, as a percent of marked price, is then,

  1. 10%
  2. 30%
  3. 70%
  4. 90%

Solution 10.

During the month the discount offered was 50% of the marked price. So the sale price was 50% of the marked price.

Duriing the last week the additional discount of 40% was on this sale price which effectively amounted to, $0.4\times{0.5} = 0.2 = 20$% of additional discount on the original marked price.

During the last week total effective discount on the marked price was then $=50 + 20=70$%.

Thus the sale price during the last week was,

$\text{Marked price} - \text{70% of Marked price} = \text{30% of Marked price}$.

Answer: b: 30%.

Key concepts used: Discount on discount concept.


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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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