## 34th SSC CGL level Solution Set, 3rd on topic Profit and loss

This is the 34th solution set of 10 practice problem exercise for SSC CGL exam and 3rd 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 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 level Question Set 34, Profit and loss 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 this, 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.**companion 34th SSC CGL solution set on Profit and loss 3****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

**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

*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.**

**Tutorials that you should refer to**

*Numbers, Number system and basic arithmetic operations*

*Factorizing or finding out factors*

*Basic concepts on fractions and decimals part 1*

**If you like, **you may * subscribe* to get latest content from this place.

### 34th solution set - 10 problems for SSC CGL exam: topic Profit and loss - Answering time 12 mins

**Problem 1.**

A trader bought a few pens at the rate of 5 per Rs.100 and a second time bought the same number of pens at the rate of 4 per Rs.100. He mixed both the lots and then sold the pens at the rate of 9 per Rs.200. In this business he suffered a loss of Rs.300. The total number of pens he bought was,

- 540
- 545
- 1080
- 1090

**Conventional solution**

Let the number of pens he bought on each occasion be $x$. For the first transaction then he bought $x$ number of pens paying Rs.100 for each 5. Applying unitary method to find out the total cost of the $x$ number of pens,

He bought first time 5 pens for Rs.100,

His cost per pen was then $\displaystyle\frac{100}{5}$,

and for $x$ number of pens total cost, $\displaystyle\frac{100x}{5}$.

Similarly for the second purchase his total cost for purchasing $x$ number of pens at the rate of 4 pens per Rs.100 was,

$\displaystyle\frac{100x}{4}$.

For $2x$ number of pens his total cost was then,

$CP=\displaystyle\frac{100x}{5} + \displaystyle\frac{100x}{4} =\displaystyle\frac{900x}{20}$.

He sold it at the rate of 9 for Rs.200, that is at a sale price per pen of Rs.$\displaystyle\frac{200}{9}$.

For $2x$ number of pens sale price is then,

$SP = Rs.\displaystyle\frac{400x}{9} = CP - Rs.300 $, as loss was Rs.300

$\hspace{7mm}= \displaystyle\frac{900x}{20} - 300$,

Or, $\displaystyle\frac{900x}{20} - \displaystyle\frac{400x}{9} = 300$,

Or, $x\displaystyle\frac{8100 - 8000}{180} = 300$

Or, $100x = 300\times{180}$

Or, $x = 540$.

So total number of pens he bought was, $2\times{540}=1080$.

**Answer: **Option c: 1080.

**Key concepts used:** Basic profit and sale concepts -- unitary method.

#### Elegant approach

We may use the per pen cost and per pen loss as variables thus simplifying deductions considerably. In this case also let's make the same assumption that number of pens purchased on each occasion was $x$.

Per pen cost first time was,

$\displaystyle\frac{100}{5}=20$.

Second lot per pen cost,

$\displaystyle\frac{100}{4}=25$.

Both lot together per pen cost is the average of the two as both lot had the same number of pens, that is, effective per pen purchase price,

$\displaystyle\frac{20+ 25}{2}= \frac{45}{2}$.

Per pen sale cost,

$\displaystyle\frac{200}{9}$.

So per pen loss,

$\displaystyle\frac{45}{2} - \displaystyle\frac{200}{9} = \displaystyle\frac{300}{2x}$, we bring in unknown variable $x$ at the last stage only,

Or, $\displaystyle\frac{5}{18} = \displaystyle\frac{150}{x}$,

Or, $x = \displaystyle\frac{150\times{18}}{5} = 540$.

This approach should be easier to carry out and can be done mostly mentally.

**Note:** If the purchase, sale and loss money figures are in same proportion or ratio, the answer will be same. In other words, if purchase figure to sale figure to loss figure are in the ratio of 1 : 2 : 3, the answer will be same. Instead of Rs.100, Rs.200 and Rs.300 we could have used Rs.1, Rs.2 and Rs.3 and would have got the same answer.

**Exercise:** Explore and understand why this happens.

**Problem 2.**

A shopkeeper bought 15 kg of rice at the rate of Rs.29/kg and 25kg of rice at the rate of Rs.20/kg. He sold the mixture of both types of rice at the rate of Rs.27/kg. His profit (in Rs) was,

- Rs.145
- Rs.125
- Rs.150
- Rs.140

**Solution 2 - Problem analysis and solving**

Total cost price of $15 + 25=40$ kg of rice,

$15\times{29} + 25\times{20} = 435 + 500 = 935$.

Total sale price of this 40 kgs of mixed rice,

$40\times{27} = 1080$.

Profit on Rs.935 was,

$1080 - 935 = 145$.

**Answer:** Option a : Rs.145.

**Key concepts used:** Simple mixture concepts -- basic profit and loss concepts.

**Problem 3.**

A, B and C are partners of a company. During a particular year A received one-third of the profit, B received one-fourth of the profit and C received remaining Rs.5000. How much did A receive?

- Rs.5000
- Rs.3000
- Rs.1000
- Rs.4000

**Solution 3 - Problem analysis and solving**

As the two portions of the whole profit are given in terms of statements that A received one-third of the profit and B received one-fourth of the profit, together they received, $\displaystyle\frac{1}{3} + \displaystyle\frac{1}{4} = \displaystyle\frac{7}{12}$ portion of the profit.

It automatically gives us the remaining portion as,

$1 - \displaystyle\frac{7}{12} = \displaystyle\frac{5}{12} \text { of total profit } = Rs.5000$ which C received as remaining portion of profit

So total profit = Rs.12000 and A received one third of it, that is, Rs.4000.

**Answer:** d: Rs.4000.

**Key concepts used:** Portion of total concept.

**Problem 4. **

A shopkeeper gains 30% while buying the goods and 20% while selling them. Find his total gain percent.

- 50%
- 40%
- 56%
- 36%

**Solution 4.**

By the first statement the shopkeeper gets goods of weight 1.3kg, at the official price CP of 1kg. So his effective cost price $CP_E$ for 1kg of goods was, $\displaystyle\frac{CP}{1.3}$.

By the second statement he sold at selling price per kg,

$SP = 1.2CP$, for making 20% official profit SP should be more than CP by 20% of CP.

Substituting for the actual cost, his actual profit was,

$\text{Profit } = SP - CP_E $

$= CP\left(1.2 - \displaystyle\frac{1}{1.3}\right) $

$= \displaystyle\frac{0.56CP}{1.3}$.

So the percentage profit per kg of sale,

$\text{Profit percent } = \displaystyle\frac{0.56CP}{1.3} \div {\frac{CP}{1.3}}\times{100} $

$\hspace{30mm}= 56$%

**Answer:** Option c: 56%.

**Key concepts used:** Profit and loss basic concepts -- two stage profit concepts.

**Problem 5.**

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

- 16%
- 25%
- 9%
- 5%

**Solution 5.**

For 25% profit his marked price would have been,

$\text{MP } = 1.25\text{ CP}$ where $\text{CP}$ is the cost price.

Discount of 16% is on this marked price and so,

$\text{Discount } = \text{16% of } 1.25\text{ CP} $

$\hspace{21mm}= \text{20% ofÂ CP} $

$\hspace{21mm}= 0.2\text{ CP}$.

Thus his selling price is,

$\text{SP } = 1.25\text{ CP} - 0.2\text{ CP} = 1.05\text{ CP}$.

Finally then the profit is 5%.

**Answer:** Option d: 5%.

**Key concept used:** Discount is on marked price -- profit is on cost price.

**Problem 6.**

If a person sells an article for Rs.720 he would lose 25%. To gain 25% he would have to sell it for,

- Rs.1200
- Rs.1000
- Rs.900
- Rs.960

**Solution 6.**

Loss of 25% on selling an article costing say $\text{CP}$ at Rs.720 means,

$720 = \text{CP}(1 - 0.25) = 0.75\text{CP}$,

Or, $\text{CP}=960$.

To gain 25% on this cost price the sale price should then be,

$\text{Sale price } =\text{CP} + \text{25% of CP} $

$\hspace{22mm}= 960 + 0.25\times{960} $

$\hspace{22mm}= 960 + 240 $

$\hspace{22mm}= Rs.1200$.

**Answer:** Option a : Rs.1200.

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

**Problem 7.**

Ganesh purchased 200 pens at the rate of Rs.12 per piece. He sold 100 pens at a gain of 10%. The percentage gain at which he must sell the remaining pens so as to gain 15% on the whole lot is,

- $17$%
- $21\frac{1}{2}$%
- $20$%
- $17\frac{1}{2}$%

**Solution 7: **

Total cost price of the 200 pens is, $TCP = 2400$.

Selling 100 pens at a gain of 10% means he sold it for $1200 + \text{10% of } 1200 = 1320$.

To get a total gain of 15% on the cost price of Rs.2400, the sale price should be,

$\text{Total sale price } = 2400 + \text{15% of } 2400 = 2760$.

So on the sale of rest 100 pens costing Rs.1200, Ganesh should get, $2760 - 1320 = 1440$, that is an extra of Rs.240. This will be his gain in selling the second lot.

$\text{Rs.240 as a percent of Rs.1200 } = \displaystyle\frac{240\times{100}}{1200} = 20$%.

**Answer: **c: 20%.

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

**Note:** There is a much simpler way to get to the solution of this problem. Can you say how? Take this as an exercise for gaining clarity on concepts.

We will explore this aspect later.

**Problem 8.**

By selling 4 articles for Rs.1000 a man loses 4%. Had he sold 3 articles for Rs.1000, his profit would have been,

- 28%
- 12%
- 30%
- 16%

**Solution 8. **

When the man sells 4 articles for Rs.1000, at a loss of 4%,

$SP = 0.96 CP = 1000$,

Or, $\text{CP for 4 articles } = \displaystyle\frac{1000}{0.96}$,

Or, $\text{CP of 1 article } = \displaystyle\frac{1000}{4\times{0.96}}$.

So cost of 3 articles is,

$\text{CP of 3 articles } = \displaystyle\frac{3\times{1000}}{4\times{0.96}}=1000\displaystyle\frac{0.75}{0.96}$

$\text{Total profit in the second sale }=1000 - 1000\displaystyle\frac{0.75}{0.96}=1000\displaystyle\frac{0.21}{0.96}$

This total profit will be on the cost price of these three articles and the percent profit is then,

$\text{Percent profit } = 1000\displaystyle\frac{0.21}{0.96} \div{1000\displaystyle\frac{0.75}{0.96}}$

$\hspace{30mm}=\displaystyle\frac{0.21\times{100}}{0.75}$

$\hspace{30mm}=28$%

**Answer:** Option a: gain of 28%.

**Key concepts used:** Basic profit loss concepts -- unitary metod.

**Problem 9.**

A retailer purchased articles at the rate of Rs.400 each from a wholesaler. He raised the price by 30% and then allowed a discount of 8% on each article. His profit is then,

- 19%
- 78.4%
- 19.6%
- 22%

**Solution 9 - Problem analysis and solving**

After 30% rise, raised price is=400(1.3),

Giving a discount of 8% on this raised price means reducing it by 8%, that is, reducing it to 92% of this raised price.

The reduced price $= 400(1.3)(0.92)$.

His cost price being Rs.400,

$\text{Profit} = 400(1.3)(0.92) - 400$

$\hspace{15mm}= 400(1 - 1.196) $

$\hspace{15mm}=400(0.196)$.

So percent profit is,

$\text{Percent profit }=\displaystyle\frac{400\times{0.196}\times{100}}{400} = 19.6$%.

**Answer:** Option c: 19.6%.

**Key concepts used:** Price rise percent on original cost price -- discount percent is on raised price -- profit is a percent of cost price.

**Note:** Cost price Rs.400 is unimportant. It could have been any price and still the same answer would been obtained. Can you say why?

This is because all figures are in terms of cost price where finally the percent of cost price is wanted so that cost price is canceled out.

**Problem 10.**

Arun bought a microwave oven at a discount of $16\frac{2}{3}$%. Had he bought the oven at a discount of 25% he would have saved Rs.600. At what price did he buy the oven?

- Rs.7200
- Rs.6000
- Rs.5000
- Rs.7500

**Solution 10. **

Assuming that the labelled price as $P$ Arun saved Rs.600 as a difference of two discounts on the labelled price which is,

$=\text{(25%} - 16\displaystyle\frac{2}{3}\text{%) of labelled price } $

$=(0.25 - \displaystyle\frac{1}{6})\text{ of labelled price } $

$= \displaystyle\frac{1}{12} \text{ of labelled price }$

$= Rs.600$.

So the labelled price = Rs.7200.

The price at which Arun bought the oven is 25% less than the labelled price, that is, 75% of Rs.7200 = Rs.6000.

**Answer:** b: Rs.6000.

**Key concepts used:** From two discount scenarios the all important linear equation on unknown labelled price can be formed.

**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 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**