## 21st SSC CGL Tier II level Solution Set, 4th on topic Profit loss discount

This is the 21st solution set of 10 practice problem exercise for SSC CGL Tier II exam and 4th on topic Profit loss discount. Students should complete the corresponding question set in prescribed time first and then only refer to this solution set for extracting maximum benefits from this resource.

We will repeat here the method of taking a 10 problem test if you have not gone through it already. And if you have not taken the test yet, you can refer to * SSC CGL Tier II level Question Set 21, Profit loss discount 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:

**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,*or any other short but good material to refresh your concepts if you so require.*Ratio and proportion***Answer the questions**in an undisturbed environment with no interruption, full concentration and alarm set at 12 minutes.**When the time limit of 12 minutes is over,**mark up to which point you have answered,**but go on to complete the set.****At the end,**refer to the answers given in the end 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.**Identify and analyze**the problems that**you couldn't do**to learn how to solve those problems.**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.**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.**Give a gap**before you take a 10 problem practice test again.

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

### 21st solution set—10 problems for SSC CGL Tier II exam: topic Profit loss discount 4—Answering time 12 mins

**Problem 1.**

The difference between a discount of 30% and two successive discounts of 20% and 10% on the marked price of an article is Rs.144. The marked price of the article is,

- Rs.7500
- Rs.7000
- Rs.7400
- Rs.7200

**Solution 1: Problem analysis and solving in mind**

From given information, sale price after 30% discount is,

$S_1=0.7M$, where $S_1$ is the first selling price and $M$ the marked price,

In the second case of two successive discounts, after the first 20% discount, the reduced selling price is,

$S_{21}=0.8M$.

The second discount of 10% is applied on this reduced selling price so that the final second selling price after two successive discounts is,

$S_{22}=0.9(0.8M)=0.72M$.

Difference between these two selling prices being Rs.144,

$0.72M-0.7M=0.02M=144$,

Or, $M=7200$.

The problem could easily be solved in mind.

**Answer: **Option d: Rs.7200.

**Key concepts used:** * Basic profit and loss concepts* --

*--*

**Rich percentage concepts, if a variable $M$ is reduced by 40%, the reduced value is, $M-0.4M=0.6M$***in the second of two successive discounts, the discount percentage is applied on the reduced selling price after applying the first discount on the marked price -- Percentage reference concept -- Solving in mind.*

**Successive discounts:****Problem 2.**

A shopkeeper offers 15% discount on all plastic toys. He offers a further discount of 4% on the reduced price to the customers who pay by cash. What does a customer has to pay for a toy of Rs.200?

- Rs.153.3
- Rs.133.7
- Rs.163.2
- Rs.129.8

**Solution 2: Problem analysis and solving in mind**

After the first discount of 15% on marked price $M$, the reduced selling price is,

$S_1=0.85M$.

4% second discount is applied on this reduced selling price to get the final selling price,

$S=0.96(0.85M)$

$=0.96\times{0.85\times{200}}$, the marked price being Rs.200,

$=0.96\times{170}=163.2$.

**Answer:** Option c: Rs.163.2.

**Key concepts used:** Basic profit and loss concepts -- Successive discounts -- Solving in mind.

**Problem 3.**

A dinner set is quoted for Rs.1500. A customer pays Rs.1173 for it. If the customer got a series of discounts and the rate of the first discount is 15%, then the rate of the second discount was,

- 15%
- 8%
- 9%
- 7%

**Solution 3: Problem analysis and solving by successive discount concept**

After the first discount of 15% on the marked price of Rs.1500, the sale price reduced to,

$S_1=1500-225=1275$.

The difference between this intermediate sale price and final sale price is,

$1275-1173=102$.

This difference is the actual second discount amount. Its percentage equivalent amount on the intermediate sale price of Rs.1275 is,

$\displaystyle\frac{102}{1275}\times{100}=\displaystyle\frac{102}{51}\times{4}=8$%.

**Answer:** Option b: $8$%.

**Key concepts used:** *Basic profit and loss concept -- Second discount value is the percentage of the first discounted sale price -- Successive discounts -- Efficient simplification, the steps of quick factor cancellation with minimum calculation is shown.*

**Problem 4. **

A dealer is selling an article at a discount of 5% on marked price. if the marked price is 12% above the cost price and the article was sold for Rs.532, then the cost price is,

- Rs.520
- Rs.500
- Rs.505
- Rs.525

**Solution 4: Problem analysis and solving in mind**

Selling price is after 5% discount of marked price so that,

$532=0.95M$,

Or, $M=\displaystyle\frac{532}{0.95}=\displaystyle\frac{2800}{5}=560$, decimal elimination technique of multiplying numerator by 100 and denominator of 0.05 by 100. At the second stage, both numerator and denominator are multiplied by 2 to generate a 10 in the denominator for ease of cancellation.

Again marked price being 12% above the cost price with percentage this time on cost price $C$, we have,

$M=1.12C=560$,

Or, $C=\displaystyle\frac{560}{1.12}=\displaystyle\frac{56}{112}\times{1000}=500$.

**Answer:** Option b: Rs.500.

**Key concepts used:** Discount concepts -- Markup concept -- Percentage concepts -- Solving in mind -- Efficient simplification: the steps of quick factor cancellation and decimal elimination technique resulting in minimum calculation is shown.

**Problem 5.**

On a television of brand A the discount is 40% and on another brand B, the discount is 25%. The price of A after discount is Rs.2250 more than the price of B after discount. What is the marked price of B, if the marked price of A is Rs.35000?

- Rs.25000
- Rs.21000
- Rs.17850
- Rs.18750

**Solution 5: Problem analysis and solving in mind**

By the given information and discount concepts,

$0.6M_A-0.75M_B=2250$, where $M_A$ and $M_B$ are the marked prices of brand A and brand B,

As $M_A=35000$,

$0.75M_B=0.6\times{35000}-2250=21000-2250$,

Or, $M_B=28000-3000=25000$.

0.75 is converted to $\displaystyle\frac{3}{4}$ and taken to RHS as a multiplying factor of $\displaystyle\frac{4}{3}$. Fraction arithmetic is easier.

**Answer:** Option a: Rs.25000.

**Key concept used:** Discount concepts --- Markup concepts -- Percentage reference concept, markup is on cost price to result in marked price, while discount is on marked price which is reduced by the discount to the sale price -- **Discount concept -- percentage concepts -- Efficient simplification -- Fraction arithmetic -- Solving in mind.**

**Problem 6.**

A person purchased a saree for Rs.7710 after availing a discount of Rs.1285. The percent discount the saree shop offers is,

- $14\displaystyle\frac{4}{7}$%
- $14\displaystyle\frac{2}{7}$%
- $14\displaystyle\frac{3}{7}$%
- $14\displaystyle\frac{1}{7}$%

**Solution 6: Problem analysis and solution by profit and loss and successive discount concepts**

By the given information we get the marked price by adding discount Rs.1285 with sale price Rs.7710, getting marked price as, Rs.8995.

So the discount percentage is,

$\displaystyle\frac{1285}{8995}\times{100}$

$=\displaystyle\frac{257}{1799}\times{100}$

$=\displaystyle\frac{100}{7}$

$=14\displaystyle\frac{2}{7}$%

**Answer:** Option b: $14\displaystyle\frac{2}{7}$%.

**Key concepts used:** * Discount concepts*: actual discount added to sale price gives marked price --

*is actual discount as a percentage of marked price --*

**Percentage discount***--*

**Percentage reference concept***.*

**Percentage concepts**Though conceptually the problem is straightforward, for accuracy we needed to write a few steps.

**Problem 7.**

A man sells an item for a profit of 10%. If he had bought the item at a price 10% less and sold it for Rs.360 less, he would have gained 20%. What is the original cost price of the item?

- Rs.19000
- Rs.17000
- Rs.18500
- Rs.18000

**Solution 7: Problem analysis and solving in mind**

From the given information, in the first case,

$S_1=1.1C_1$,

In the second case,

$S_2=1.2C_2=1.2(0.9C_1)=1.08C_1$,

Or, $360=0.02C_1$.

So, original cost price was 50 times Rs.360, that is, Rs.18000.

**Answer: **Option** **d: Rs.18000.

**Key concepts used:** Basic profit and loss concepts -- Rich prifit and loss concept -- *Percentage concepts -- Percentage reference concept -- Solving in mind***.**

**Problem 8.**

Pens are bought at 15 for Rs.100. How many pens must be sold for Rs.100 to gain 25%?

- 10
- 8
- 11
- 12

**Solution 8: Problem analysis and solving in mind using unitary method**

To get 25% profit by selling at Rs.100, the cost price must be $\displaystyle\frac{100}{1.25}=80$. We actually convert 1.25 to fraction and take $\displaystyle\frac{4}{5}$th of 100. Fraction arithmetic is faster.

**Applying three step unitary method**:

At 100 rupees 15 pens were bought,

So, at 1 rupee $\displaystyle\frac{15}{100}$ pens were bought,

So, at Rs.80, $15\times{\displaystyle\frac{4}{5}}=12$ pens were bought and then sold at Rs.100 to gain 25%.

Solved easily in mind.

**Answer:** Option d: 12.

**Key concepts used:** * Basic profit loss concepts* --

**Rich profit and loss concept, profit of 25% means cost price will be four-fifths of sale price -- Unitary method -- Percentage concept -- Solving in mind.****Problem 9.**

Oil equal to 20% of the weight of groundnut is extracted in an oil mill.The residue left after extraction is sold as cattle feed at the rate of Rs.12.5 per kg. The groundnuts are bought at Rs.20 per kg, while processing cost is Rs.5 per kg. At what price per kg should the oil be sold to earn 20% profit on total of cost of groundnut and cost of processing?

- Rs.100
- Rs.200
- Rs.150
- Rs.250

**Solution 9: Problem analysis and solving in mind**

By the given information, 0.8 kg residue of 1 kg groundnut was sold at Rs.12.5 per kg, that is at Rs.10.

To have 20% profit on total cost of Rs.25 for 1kg, the oil and its residue for 1kg groundnut must be sold at a total price of $1.20=\displaystyle\frac{6}{5}$th of Rs.25, that is, Rs.30.

As the residue 0.8 kg of 1 kg groundnut is sold at Rs.10, the oil of 0.2 kg must be sold for Rs.20, that is at Rs.100 per kg.

**Answer:** Option a: Rs.100.

**Key concepts used:** Basic profit and loss concepts -- Tracking of price and profit of components: * Price profit component tracking* -- Fraction arithmetic -- Percentage concepts -- Solving in mind.

**Problem 10.**

A sells a watch to B at a loss of 12%, but B makes a profit if $12\frac{1}{2}$% by selling the watch to C. If A sells the watch to B at the cost at which C purchased it in the first case, the percentage loss or profit of A now would be,

- 1% loss
- 2% profit
- 1% profit
- 2% loss

**Solution 10: Problem analysis and solving careful tracking of prices and profits of series of transactions**

By the given statement applying basic and rich profit and loss concepts along with percentage concepts, sale price for A is,

$S_A=0.88C_A$, $S_A$ is sale price of A which is cost price of B, $C_B$ and $C_A$ is cost price of A.

And for the transaction of B,

$S_B=\displaystyle\frac{9}{8}\times{C_B}$, where $S_B$ is the sale price of B which is the cost price of C, $C_C$.

At the second instance, sale price for A,

$S_{A2}=C_C=S_B=\displaystyle\frac{9}{8}\times{C_B}$.

Backtracking, as the first cost price B was equal to first sale price of A, that is, $0.88C_A$,

$S_{A2}=\displaystyle\frac{9}{8}\times{0.88C_A}=0.99C_A$.

For the second sale than, A suffered a loss of 1%.

**Answer:** Option a: 1% loss.

**Key concepts used:** Basic profit and loss concept -- * Rich profit and loss concept:* 12% loss means sale price is 0.88 times cost price -- Careful tracking of sale price, cost price and profit in a

**Series of transactions.**Overall, for computational correctness the problems 3 and 6 needed a few steps in writing whereas, for careful tracking of series of transactions, problem 10 also needed jotting down the steps, though the problems could be solved comfortably within a minute's time each. Rest of the seven problems could be solved easily in mind using basic and rich profit and loss, percentage and discount concepts.

Just remember, understanding and applying basic and rich concepts on profit and loss, discount, ratio and percentage will enable you to solve any such problem easily under a minute with no dependence on varieties of formulas.

**Resources that should be useful for you**

* 7 steps for sure success in SSC CGL tier 1 and tier 2 competitive tests* or

*to access all the valuable student resources that we have created specifically for SSC CGL, but*

**section on SSC CGL****generally for any hard MCQ test.**

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#### How to solve difficult SSC CGL Math problems at very high speed using efficient problem solving strategies and techniques

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 withpermanent skillset improvement.

**The following are the associated links,**

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

**How to solve a difficult SSC CGL level Profit and Loss problem in a few steps 4**

**How to solve difficult SSC CGL Profit and loss problems in a few steps 3**

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

**How to solve in a few steps, Profit and loss problem 1**