## 20th SSC CGL Tier II level Solution Set, 3rd on topic Profit loss discount

This is the 20th solution set of 10 practice problem exercise for SSC CGL Tier II exam and 3rd 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 20, Profit loss discount 3*, 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

### 20th solution set—10 problems for SSC CGL Tier II exam: topic Profit loss discount 3—Answering time 12 mins

**Problem 1.**

A man sells an item 5% above its cost price. If he had bought it 5% less than what he had paid for it and sold it at Rs.2 less he would have gained 10%. The original cost price of the item is,

- Rs.300
- Rs.200
- Rs.100
- Rs.400

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

From given information, initial profit is,

$S_1=1.05C_1$, where $S_1$ is the original selling price and $C_1$ the original cost price,

In the second case,

$S_2=S_1-2=1.1C_2=1.1(0.95C_1)$.

Or, $1.05C_1-2=1.045C_1$,

Or, $C_1=\displaystyle\frac{2}{0.005}=400$.

Problem solved mentally using basic concepts. Just remember the rich concept of profit and loss—if profit is say 25%, sale price $S=C+0.25C=1.25C$, where $C$ is the cost price.

**Answer: **Option d: Rs.400.

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

*-- Correct variable selection -- Solving in mind.*

**Rich profit and loss concepts**#### Solution 1: Conceptual solution by reasoning and use of change analysis

The following would be the reasoning based on profit and loss concepts and change analysis.

In the first case profit was 5% because the sale price was 1.05 times the original cost price. The second time, the cost price reduced by 5% from the original cost price to become 0.95 times original cost price. Considering 10% profit on this new cost price, the new sale price becomes $1.1\times{0.95}=1.045$ times the original cost price.

The difference between these two sale prices, 0.005 times original cost price is 2. 200 times both sides gives original cost price as Rs.400.

**Problem 2.**

A trader bought 200 eggs out of which 38 eggs were broken. He sold the remaining eggs at the rate of Rs.4.80 per dozen and thus gained 8%. His total investment is,

- Rs.45
- Rs.120
- Rs.60
- Rs.80

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

Reducing 38 from 200, the number of eggs sold was 162 which is 13.5 dozen. These eggs were sold at Rs.4.80 per dozen at a total sale price of,

$S=4.80\times{13.5}=64.80$.

For 8% profit this sale price would equal 1.08 times total cost price. So total cost price is,

$\displaystyle\frac{64.80}{1.08}=60$, multiplying numerator and denominator by 100 and eliminating 108 with numerator.

**Answer:** Option c : Rs.60.

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

**Problem 3.**

A book seller allowed 10% discount on printed price. He gets 30% commission from the publisher. His profit percent is,

- $26\displaystyle\frac{3}{7}$
- $28\displaystyle\frac{4}{7}$
- $20$
- $25$

**Solution 3: Problem analysis and solving in mind**

If $P$ is the printed price, after 30% commission, the book seller gets a book at $0.7P$ which is his cost price, while after 10% discount on printed price the sale price reduces to $0.9P$ so that profit is,

$0.9P-0.7P=0.2P$.

And percentage profit is,

$\displaystyle\frac{0.2P}{0.7P}\times{100}=28\displaystyle\frac{4}{7}$%.

**Answer:** b: $28\displaystyle\frac{4}{7}$.

**Key concepts used:** *Basic profit and loss concept -- Marked price concept -- Discount is on marked or printed price, commission is also on printed price while profit percentage is on cost price. Cost price is after commission. Sale price is after discount. -- Solving in mind.*

**Problem 4. **

The cost price of an article is marked up by 200% and the article is sold at Rs.540 after giving a discount of 25%. What is the original cost price?

- Rs.360
- Rs.250
- Rs.240
- Rs.300

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

Because of 25% discount on marked price, the sale price is $\displaystyle\frac{3}{4}$th of marked price. So marked price is $\displaystyle\frac{4}{3}$rd of sale price Rs.540, that is, Rs.720.

As cost price is marked up by 200% or in fact tripled to form the marked price, it is one-third of marked price Rs.720, that is, Rs.240.

After 200% markup on cost price $C$, the marked price becomes,

$M=C+2C=3C$.

**Answer:** Option c: Rs.240.

**Key concepts used:** Profit and loss basic concepts -- fraction arithmetic -- percentage concepts -- discount concept.

**Problem 5.**

If the discount on an article is 15%, its selling price is Rs.816. What would be its selling price if the discount on it were 25%?

- Rs.800
- Rs.750
- Rs.720
- Rs.200

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

With 15% discount the selling price $S$ is 0.85 times the marked price $M$,

$S=816=0.85M$,

So marked price is,

$M=\displaystyle\frac{816}{0.85}=\frac{81600}{85}=\displaystyle\frac{4800}{5}=960$

With 25% discount then, the selling price would be $\displaystyle\frac{3}{4}$th of 960, that is, Rs.720.

**Answer:** Option c: Rs.720.

**Key concept used:** Basic profit and loss concepts -- **Discount concept -- percentage concepts.**

**Problem 6.**

A trader allows a trade discount of 20% and a cash discount of $6\frac{1}{4}$% on the marked price of the goods and gets a net gain of 20% on the cost. By how much above the cost price would the goods have been marked for sale?

- 40%
- 70%
- 50%
- 60%

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

After 20% trade discount on marked price $M$, it is reduced to,

$0.8M$.

The cash discount of $6\displaystyle\frac{1}{4}=\displaystyle\frac{25}{4}$% is equivalent to $\displaystyle\frac{1}{16}$ after dividing with 100.

Final selling price after two successive discounts will then be less than 0.8M by,

$0.8M\times{\displaystyle\frac{1}{16}}=0.05M$, the percentage reference in this case is the first discounted selling price of 0.8M.

So the final selling price, considering 20% profit on cost price $C$ is then,

$S=1.2C=(0.8-0.05)M=0.75M=\displaystyle\frac{3}{4}M$,

Or, Marked price $M=1.2C\times{\displaystyle\frac{4}{3}}=1.6C$.

Markup is 60% on cost price.

**Answer:** Option d : 60%.

**Key concepts used:** Basic profit and loss concepts -- Rich profit and loss concepts -- Successive discount concept -- Percentage reference concept -- Percentage to fraction conversion, fraction arithmetic is easier -- Rich percentage concept—when a reducing percentage of say 10% is applied on a variable A, its reduced value becomes, $A-0.1A=0.9A$—this concept is applied frequently in expressing sale price in terms of cost price taking care of profit or loss as well as sale price in terms of marked price after taking care of discount.

**Problem 7.**

A store offers a variety of discounts that range between 20% and 25% inclusive. If a book is discounted to a price of Rs.270, then its maximum possible original price was,

- Rs.360
- Rs.345.5
- Rs.337.5
- Rs.334

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

For maximum possible original price or marked price, the highest of the discount range, that is 25% discount must have been applied on it to get the given discounted price. Any discount less than the maximum of the range would result in a smaller increase from given value of Rs.270.

To explain further, after discount of say 10% on marked price M, the discounted sale price would be,

$S=M-0.1M$,

Or, $M=\displaystyle\frac{S}{1-0.1}$.

With a given value of $S$, the larger the discount is, the smaller the denominator will be, and consequently larger the marked price $M$ will be.

So the discounted price would in this case be 0.75 times or, $\displaystyle\frac{3}{4}$th of the original price or the marked price.

The original price would then be $\displaystyle\frac{4}{3}$rd of Rs.270, or Rs.360.

**Answer: **a: Rs.360.

**Key concepts used:** Basic profit and loss concepts -- *Percentage concepts -- Discount concepts -- Decimal conversion to fraction which is easier to deal with -- Maximum marked price concept***.**

**Problem 8.**

Two blends of tea costing Rs.350 and Rs.400 per kg were mixed in the ratio of 2 : 3 by weight. If one-fifth of the mixture is sold at Rs.460 per kg and the remaining at the rate of Rs.550 per kg, the profit percent will be,

- 50%
- 20%
- 30%
- 40%

**Solution 8: Problem analysis and solving using rich ratio and mixing concepts**

By given information, every 5 kg of mixture will have 2kg of tea costing Rs.350 per kg and 3kg of tea costing Rs.400 per kg. Total cost of this amount of 5kg of mixed tea would be,

$2\times{350}+3\times{400}=1900$ with per kg cost of Rs.380.

For every 5kg, one-fifth of mixed tea, that is, 1 kg costing Rs.380 is sold at Rs.460 per kg getting a profit of Rs.80. Rest 4kg of this 5kg of mixture is sold at a selling price of Rs.550 per kg, that is, profit gained will be, Rs.170 per kg, or, Rs.680 for 4 kg.

Total profit would then be, Rs.760 for this 5kg of mixed tea.

Per kg profit would then be, Rs.152, and profit percentage,

$\displaystyle\frac{152}{380}\times{100}=40$%.

**Answer:** Option d: 40%.

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

**Rich ratio concept by which we deal with every 2+3=5 kg of mixture -- mixing concept -- Percentage concept -- Percentage profit concept.****Problem 9.**

A shopkeeper sold an article at a loss of 10% after giving a discount equal to half of the marked price. Then the cost price is,

- $\displaystyle\frac{7}{9}$th of marked price
- $\displaystyle\frac{5}{9}$th of marked price
- $\displaystyle\frac{1}{9}$th of marked price
- $\displaystyle\frac{4}{9}$th of marked price

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

By the given statement sale price is,

$S=0.5M=0.9C$.

Because of 10% loss, sale price is,

$S=C-0.1C=0.9C$, $C$ being the cost price and $M$ marked price.

So cost price is,

$C=\displaystyle\frac{5}{9}$th of marked price $M$.

**Answer:** Option b: $\displaystyle\frac{5}{9}$th of marked price.

**Key concepts used:** Basic profit and loss concepts -- Discount concepts -- Marked price concept -- Percentage concepts.

**Problem 10.**

If 60% discount is offered on the marked price so that the selling price becomes equal to the cost price, then what was the percentage mark up?

- 40%
- 100%
- 150%
- 250%

**Solution 10: Problem analysis and solving in mind**

By the given statement applying basic and rich profit and loss concepts along with percentage concepts,

$S=C=0.4M$, where $S$ is the sale price, $C$ cost price and $M$ marked price.

So,

$M=\displaystyle\frac{C}{0.4}=2.5C$.

Percentage mark up is then 150%,

$M=C+\text{150% of }C=C+1.5C=2.5C$.

**Answer:** c: 150%.

**Key concepts used:** Basic profit and loss concept -- Percentage concepts, 2.5 times means 250% and 1.5 times means 150% -- Rich profit and loss concepts -- Discount concepts.

Overall, excluding problem 6 on successive discounts and problem 8 on mixture, all other eight problems could be solved in mind easily using basic and rich concepts. These two problems also can be solved in mind with a little practice, but for assured correctness, it is recommended to use writing for these two solutions.

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**