SSC CGL level Solution Set 53, Profit and loss 4

53rd SSC CGL level Solution Set, 4th on topic Profit and loss

ssc cgl level profit loss solution set 4 cover

This is the 53rd solution set of 10 practice problem exercise for SSC CGL exam and 4th on topic Profit and loss. Students should complete the corresponding question set in prescribed time first and then only refer to this solution set.

If you have not taken the test yet, you can refer to SSC CGL level Question Set 53, Profit and loss 4, and then after taking the test come back to this solution.

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 15 minutes.
  3. When the time limit of 15 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 the companion 53rd SSC CGL solution set on Profit and loss 4, to mark your score at 15 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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53rd solution set - 10 problems for SSC CGL exam: topic Profit and loss - Answering time 15 mins

Problem 1.

A trader marks up a product he is selling for 20% profit. If he sells it at a discount of 15%, his net gain percentage is,

  1. 2%
  2. 2.5%
  3. 3.5%
  4. 4%

Solution 1 - Problem analysis and solving

We will use the most basic concepts and primarily descriptive mathematically sound conceptual reasoning. The advantage of this approach is, this process can be carried out wholly in mind and so is very fast.

Profit $P=20\text{% of }CP$ which is extra to $CP$, where $CP$ is the cost price and $P$ is the profit. This is by definition.

The trader marks up the cost price $CP$ in such a way that the profit in excess of CP is 20% of CP.

Thus the marked price, $MP=120\text{% of }CP$.

The discount is on this $MP$.

So discount 15% on $MP = 15\text{% of }(120\text{% of }CP) = 18\text{% of }CP$.

The $MP$ reduces by this amount after discount and reaches the Sale price $SP$.

So,

$SP = MP - \text{discount} $

$= (120\text{% of }CP) - (18\text{% of }CP) $

$= 102\text{% of }CP$.

So the Sale price $SP$ is in excess of Cost price by 2% of Cost price which is then the profit by definition.

Answer: Option a: 2%.

Key concepts used: Basic profit and loss concepts -- Basic discount concepts -- Conceptual reasoning -- Base equalization technique -- Percent reference concept -- Sequencing of events concept.

Note: We start by Profit percentage on anchor or reference value as the Cost price on which the percentage profit is applied to evaluate the actual profit. So profit is always a fraction of the Cost price.

This profit adds up to the Cost price and lands onto the Sale price. On the other hand, the Marked price is discounted by a percentage of itself and reduced by this amount to converge into Sale price again. This is sequencing of events concept which is of prime importance in complex profit and loss problem solving.

Just to remember,

Discount percentage is on the Marked price. So for discount reference value is Marked price.

On the other hand, for Profit, percentage reference value is Cost price and profit percentage is calculated on this cost price.

By Base equalization technique we bring all values in terms of Cost price for final solution.

Problem 2.

A shopkeeper gives a discount of 7% while selling an item. If he had given a discount of 9% he would have got Rs.15 less as profit. The marked price for the item is then,

  1. Rs.750
  2. Rs.720
  3. Rs.600
  4. Rs.712.50

Solution 2 - Problem analysis and solving

We will again use very basic percentage concepts, basic discount concepts and basic profit and loss concepts.

As in both the cases the trader offers discount on the Marked price, say, $MP$, with Cost price, say, $CP$ remaining same, we will first apply the discount on the $MP$ to get the Sale price in the two cases.

Let the Sale price in the first case be $SP_1$ and in the second case $SP_2$.

So in first case profit,

$P_1=SP_1-CP=0.93MP-CP$.

We have converted the Sale price in terms of the fixed Marked price, $MP$.

Similarly in the second case profit,

$P_2=SP_2-CP=0.91MP-CP$.

Thus subtracting the second equation from the first, Cost price $CP$ is eliminated and we get the reduction in profit as,

$\text{Rs.}15=0.02MP$,

Or, Marked price, $MP=\text{Rs.}750$.

We have used first base equalization technique to convert the sale prices to the fixed marked price. Finally we have used Change analysis technique to get the value of reduction in profit (which is the change) in terms of the change in sale price converted to the form of change in terms of marked price.

Answer: Option a : Rs.750.

Key concepts used: Basic percentage concepts -- Basic discount concepts -- Basic profit and loss concepts -- Base equalization technique -- Change analysis technique.

Problem 3.

A dishonest trader declares that he sells his goods at cost price, but actually he uses a false weight of 850 gm for 1 kg. His gain percent is,

  1. $17\frac{12}{17}$%
  2. $17\frac{11}{17}$%
  3. $11\frac{11}{17}$%
  4. $71\frac{11}{17}$%

Solution 3 - Problem analysis

This is a typical problem that has inbuilt in it a capacity for creating confusion in the problem solver's mind. But if you think actually what is happening, which we call Context awareness and use the very basic concepts, this problem should be solved well within 30 seconds.

Solution 3 - Problem solving execution

What is actually happening?

The trader declares that he sells 1 kg of goods at the cost price of 1 kg, say, $CP$. But the 1 kg he sells actually weighs 850 gm for which he paid, $0.85CP$ only. This is his actual cost price.

So by definition his profit is,

$CP-0.85CP=0.15CP$.

Don't confuse the cost price to be $CP$ in this case for profit calcualtion. He actually makes this profit on a cost price of $0.85CP$. So the profit is,

$\displaystyle\frac{0.15CP}{0.85CP}=\frac{300}{17}=17\frac{11}{17}$%.

Answer: Option b: $17\frac{11}{17}$%.

Key concepts used: Context awareness -- Basic profit and loss concepts.

Problem 4.

A man sells two articles for Rs.4000 each with no loss or no gain in the two transactions taken together. If one is sold at a gain of 25%, the other is sold at a loss of,

  1. $25$%
  2. $20$%
  3. $16\frac{2}{3}$%
  4. $18\frac{2}{9}$%

Solution 4.

Let us assume the sale price be Rs.125.

In the first case then, the cost price would be Rs.100 so that 25% profit is made.

We have to find out the cost price in the second case with sale price pegged still at Rs.125 so that by the loss incurred, the profit of Rs.25 is cancelled out. To cancel out the profit of Rs.25, the cost price must be then Rs.150 in the second case so that an actual loss amounting to same Rs.25 is made.

What is the loss percent in this second case then?

It is simply be loss of Rs.25 as a percent of cost price of Rs.150.

This is,

$\displaystyle\frac{25}{150}=\frac{1}{6}=16\frac{2}{3}$%.

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

Key concepts used: Profit and loss basic concepts -- Profit and loss balancing concept -- Percentage reference concept -- Context awareness -- Sequencing of events concept -- Abstraction technique.

Note: By using abstraction technique, instead of sale value of Rs.4000, we have conveniently assumed a sale value of Rs.125. This we could do because all increases or decreases are in terms of percentages so that actual values may be anything if the percentage values in problem description remain unchanged.

Verify yourself.

This is a powerful technique that can simplify many a profit and loss problem considerably.

Problem 5.

A dishonest trader using a faulty balance makes a profit of 5% while buying as well as while selling his goods. His actual gain percent in the whole process amounts to,

  1. 10.25%
  2. 11%
  3. 10%
  4. 10.5%

Solution 5.

We will again stick to the most basic concepts and sequentially follow the events.

While buying the trader makes 5% profit. That means he buys Rs.105 worth of goods by paying Rs.100.

While selling he makes again a profit of 5%. But this time the profit percentage is to be applied to Rs.105 to get an idea of net profit with cost price as the original cost price.

Applying the second profit of 5% on Rs.105, we get a sale price of,

$105(1+0.05)=105\times{1.05}$.

So the net profit is,

$105\times{1.05}-100=5+5\times{1.05}=10.25$.

In the second case reference value for the profit percentage changes from 100 to 105.

Answer: Option a: 10.25%.

Key concept used: Basic profit and loss concept -- Sequence of event awareness -- Percent reference value concept.

Problem 6.

A loss of 19% gets converted to a profit of 17% when selling price is increased by Rs.162. The cost price of the article is,

  1. Rs.360
  2. Rs.450
  3. Rs.600
  4. Rs.540

Solution 6.

We will apply here change analysis technique.

For the loss of 19%, the sale price is,

$0.81CP$, where $CP$ is the cost price that remains fixed.

In the second instance for the profit of 17%, the selling price would be,

$1.17CP$.

The difference in the two selling prices is,

$\text{Rs.}162=0.36CP$.

So,

$CP=\displaystyle\frac{162}{0.36}$

$=\displaystyle\frac{16200}{36}$

$=\displaystyle\frac{1800}{4}$

$=\text{Rs.}450$

Answer: Option b : Rs.450.

Key concepts used: Basic profit and loss concepts -- Change analysis technique -- Base equalization technique in converting the two selling prices to the fixed cost price -- Basic percentage concepts.

Problem 7.

A fruit-seller makes a profit of 20% by selling oranges. If he charges Rs.1 more for each orange, he would be able to make a profit of 40%. The selling price of an orange in the first case is then,

  1. Rs.7
  2. Rs.5.5
  3. Rs.5
  4. Rs.6

Solution 7:

By the problem description, Rs.1 is equivalent to additional profit of 20% of the cost price.

So 100% of the cost price is Rs.5.

With an additional 20%, that is, Rs.1 increase in the first case, the desired selling price is then, Rs.6.

Answer: d: Rs.6.

Key concepts used: Basic profit and loss concepts -- Change analysis technique -- Basic percentage concepts.

Problem 8.

After 4 years of purchasing a house, the houseowner sold it for a price 25% more than the purchase price, but had to pay a tax of 50% of the gain. The tax amount as a percentage of the original purchase price is,

  1. $50$%
  2. $25$%
  3. $4\frac{1}{6}$%
  4. $12.5$%

Solution 8.

The sale price is more than the original purchase price, say, $P$ by,

$0.25P=\displaystyle\frac{1}{4}P$.

The tax of 50% is paid on this gain.

So the tax is,

$0.5\times{\displaystyle\frac{1}{4}P}=\displaystyle\frac{1}{8}P$.

This is equivalent to 12.5% of P.

Answer: Option d: 12.5%.

Key concepts used: Basic profit loss concepts -- Basic percent concepts -- Basic decimal concepts -- Basic fraction concepts.

Problem 9.

The marked price of a watch is Rs.1600. The shopkeeper gives two successive discounts to the customer, the first being 10%. If the customer pays Rs1224 for the watch, the second discount is,

  1. 5%
  2. 10%
  3. 20%
  4. 15%

Solution 9 - Problem analysis and solving

After the first discount of 10% on Rs.1600, the sale price drops to,

$\text{Rs.}1600 - \text{Rs.}160=\text{Rs.}1440$.

As the customer pays finally after the second discount, Rs.1224, the second discount amounts to,

$\text{Rs.}1440 - \text{Rs.}1224=\text{Rs.}216$.

This discount is on the sale price of $\text{Rs.}1440$. So the percentage discount is,

$\displaystyle\frac{21600}{1440}=\frac{2400}{160}=15$%.

Answer: Option d: 15%.

Key concepts used: Successive discount concept -- Sequence of events awareness -- Basic percentage concepts -- Percentage reference concept.

Problem 10.

How much percent more than the cost price should a shopkeeper mark his goods so that after allowing a discount of 25% on the marked price, the gain is 20%?

  1. 60%
  2. 55%
  3. 50%
  4. 70%

Solution 10.

Assuming cost price to be 100 sale price will be 120 for 20% gain.

To give a discount of 25% on the marked price reducing the marked price to the sale price, the sale price must then be, 75% of marked price.

If $120=0.75MP$, where $MP$ is the marked price, the marked price will be,

$MP=\displaystyle\frac{120}{0.75}=120\times{\frac{4}{3}}=160$.

This is 60 more than the cost price of 100, that is 60% more than the cost price.

Answer: a: 60%.

Key concepts used: Basic profit and loss concepts -- Basic discount concepts -- Sequence of events awareness -- Basic percentage concepts -- Percentage reference concept.


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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