## 8th SSC CGL Tier II level Solution Set, 1st on topic Profit and loss

This is the 8th solution set of 10 practice problem exercise for SSC CGL Tier II exam and 1st 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 Tier II level Question Set 8, Profit and loss 1*, and then after taking the test come back to this solution.

We will repeat the method of taking the test.

### 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 15 minutes.**When the time limit of 15 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 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.**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.

### 8th solution set - 10 problems for SSC CGL Tier II exam: topic Profit and loss - Answering time 15 mins

**Problem 1.**

A dealer purchased a microwave oven for Rs.7660. After giving a discount of 12% on its marked price, he still gained 10%. The marked price was,

- Rs.9755
- Rs.8426
- Rs.8246
- Rs.9575

**Solution 1 - Problem analysis**

The percentage of discount is on the marked price, say, $M$ which is to be found out, whereas the profit percentage of 10% is on the given cost price of Rs.7660. While * adding the profit to the cost price we get the sale price*, the

**same sale price we would get by subtracting the discount from the marked price.**Using * percentage application concept for profit and loss *we can say,

$\text{Sale price}=M-\text{12% of }M=0.88M$

Similarly adding profit percentage to the cost price we would get,

$\text{Sale price}=7660+\text{10% of }7660=1.1\text{ of }7660$.

We have delayed the multiplication of 7660 by 1.1 as we knew 11 will be canceled out later. This is application of * delayed evaluation technique* which is a component of

**efficient simplification.**#### Solution 1 - Problem solving execution

We have arrived at the sale price from marked price by reducing it by discount, and also from the cost price by increasing it by profit. Now we only have to equate the two sale prices,

$0.88M=1.1\times{7660}$,

Or, $M=\displaystyle\frac{76600}{8}$, we have * eliminated all decimals by multiplying by 10s multiples* and canceled out 11,

Or, $M=9575$, dividing directly by 8.

**Answer: **Option d: Rs.9575.

**Key concepts used:** * Basic profit and loss concepts* -- Rich profit and loss concepts -- marked price concept -- discount concept --

*, you need to apply the percentage profit on cost price and percentage discount on marked price --*

**percentage application concept for profit and loss***, the presence of the factor of 11 was detected and utilized --*

**efficient simplification**

**delayed evaluation technique -- decimal elimination technique.**#### Explanation based on Graphical representation

The following is the graphical representation of the problem,

Reduction of marked price by 12% results in the value of Sale price as $0.88M$ which converges with the Sale price arrived at by addition of 10% profit to Cost price, that is, $1.1\text{ of }7660$.

**Note.** One can solve the problem wholly mentally using the rich profit and loss concepts and the technique of delayed evaluation. Division of 76600 by 8 is simple.

**Problem 2.**

If a man were to sell a chair for Rs.720, he would lose 25%. For gaining 25%, he has to sell it for,

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

**Solution 2 - Problem analysis and solving**

Here also by the * anchor value of percentage concept* where anchor is the cost, though the value the cost price on which profit and loss percentages are to be applied, is not given, from the loss percentage of 25% and sale price of Rs.720 of the first transaction we can conclude,

$\text{Sale price }=720=\text{Cost price }-\text{25% of Cost price}$,

Or, $\text{Cost price }=\frac{4}{3}\times{720}$.

If target profit is 25%, the target sale price would have to be,

$\text{Target sale price }=1.25\text{ of Cost price}$

$=\frac{5}{4}\times{\frac{4}{3}}\times{720}$

$=1200$

Again we have used delayed evaluation technique to cancel out factors at the last stage and carry out a simplified calculation just once.

**Answer:** Option c : Rs.1200.

**Key concepts used:** Basic profit and loss concepts -- * anchor value of percentage concept*, for profit or loss the percentage is to be usually applied on the anchor value of the cost price -- delayed evaluation technique.

**Problem 3.**

Purchasing a motorcycle from Surya, Naveen spent Rs.1680 on it for repair and sold it to Pintu for Rs.35910 at a profit of 12.5%. If Surya sold the motorcycle to Naveen at a loss of 28%, the purchase cost of the motorcycle for Surya is,

- Rs.35000
- Rs.38000
- Rs.40000
- Rs.42000

**Solution 3 - Problem analysis and solving**

Usually in Profit and loss problems two events happen, but in this problem we see four events. To deal with the apparent confusion we must first clearly plot out the events in time sequence,

1. **First:** Surya purchased a motorcycle at cost $C$ which is to be found out.

2. **Second:** Surya sold the motorcycle to Naveen at a loss of 28%. So the cost price to Naveen was, $0.72C$, less 28% of $C$. This was the sale price of Surya.

3. **Third:** Naveen spent an additional amount of Rs.1680 for repair making its effective cost to him as, $0.72C + 1680$.

4. **Fourth:** Naveen sold the motorcycle thus repaired to Pintu at a profit of 12.5% which is one-eighth more than his cost price. In other words,

$\text{Naveen's sale price }=\frac{9}{8}\text{ of }(0.72C +1680)=35910$.

**Solution 3 - Problem solving execution final stage**

So,

$0.72C +1680=\frac{8}{9}\times{35910}=8\times{3990}=31920$,

Or, $0.72C=30240$,

Or, $C=\displaystyle\frac{3024000}{72}=42000$.

**Answer:** d: Rs.42000.

**Key concepts used:** * Sequencing of events* -- basic profit and loss concepts --

**percentage conversion techniques -- efficient simplification.****Note:** Without sequencing of the events one after the other, it won't have been possible to get a clear idea of the costs and prices involved. Additionally in profit and loss problems as a part of efficient simplification, we convert percentages in two ways, either as a decimal or as a fraction as the need may be. This is **percentage conversion techniques.**

The * most used percentages and corresponding fractions* are useful to remember,

1. $4\text{%}=\frac{1}{25}$

2. $5\text{%}=\frac{1}{20}$

3. $8\text{%}=\frac{2}{25}$

4. $10\text{%}=\frac{1}{10}$

5. $12.5\text{%}=\frac{1}{8}$

6. $20\text{%}=\frac{1}{5}$

7. $25\text{%}=\frac{1}{4}$

8. $30\text{%}=\frac{3}{10}$

9. $40\text{%}=\frac{2}{5}$

10. $50\text{%}=\frac{1}{2}$

11. $60\text{%}=\frac{3}{5}$

12. $75\text{%}=\frac{3}{4}$

13. $80\text{%}=\frac{4}{5}$

14. $90\text{%}=\frac{9}{10}$.

$72\text{%}$ didn't correspond to any convenient fraction and so we have used instead* its equivalent decimal form*, $0.72$, dividing it by 100.

**Problem 4. **

Praveen bought a plot of land for Rs.96000 and sold $\frac{2}{5}$th of it at a loss of 6%. If he wanted to make a profit of 10% on the whole transaction by selling the remaining land, the gain % of the remaining land must then be,

- $20\text{%}$
- $20\frac{2}{3}\text{%}$
- $7\text{%}$
- $14\text{%}$

**Solution 4 - Problem analysis and solving**

Desired profit on whole land transaction is,

$10\text{% of }96000=\frac{1}{10}\times{96000}$

Loss in the first transaction is $6\text{% of }\frac{2}{5}\text{th of }96000=\frac{3}{125}\text{ of }96000$. This has been the shortfall.

So total desired profit is,

$\frac{1}{10}\times{96000}=\text{profit on}\frac{3}{5}\times{96000} - \frac{3}{125}\times{96000}$,

Or, $P\times{\frac{3}{5}}=\frac{1}{10}+\frac{3}{125}$,

Or, $P=\frac{1}{6}+\frac{1}{25}=\frac{31}{150}=\frac{62}{3}\text{%}=20\frac{2}{3}\text{%}$

**Answer:** Option b: $20\frac{2}{3}\text{%}$.

**Key concepts used:** Profit and loss basic concepts -- * two stage transaction concepts* -- complete bypassing of multiplication of factor 96000 because all of the profits and losses are some portion of this whole --

*--*

**efficient simplification***-- fraction arithmetic -- percentage concepts.*

**percentage conversion techniques****Problem 5.**

A shopkeeper bought two watches at a total cost of Rs.840. He then sold one at a loss of 12% and the other at a profit of 16%. There was no loss or gain in the whole transaction. The cost price of the watch on which the shopkeeper gained was then,

- Rs.390
- Rs.370
- Rs.380
- Rs.360

**Solution 5.**

Let us assume the cost of the watch on which profit was made is $x$. So the cost of the other watch is $(840-x)$.

The profit $=0.16x$, and

the loss $=0.12(840-x)$.

As there is no net loss or gain these two must be equal,

$0.12(840-x)=0.16x$,

Or, $4x=3(840-x)=3\times{840}-3x$,

Or, $7x=3\times{840}$,

Or, $x=3\times{120}=360$.

We have used the technique of equalizing the profit and loss amounts after choosing the variable $x$ judiciously along with delayed evaluation technique.

**Answer:** Option d: Rs.360.

**Key concept used:** Basic profit and loss concepts -- * increment equalization technique*, the profits and losses were the increments and decrements neutralizing each other --

*.*

**delayed evaluation technique****Problem 6.**

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

- Rs.50000
- Rs.10000
- Rs.25000
- Rs.55000

**Solution 6 - Problem analysis and solving**

This is a problem of * multistage transaction *where at each stage the sale price of the previous stage becomes the cost price of the next stage. Profits or losses accordingly increases or decreases the amounts starting from the first cost price. Thus we won't calculate the profits, losses or amounts at each stage separately, doing it only once after establishing the relation between the first cost price and the last sale price.

* Working backwards*, the last sale price is,

$56100=0.85\text{ of last cost price}$, with 15% loss the sale price is less than the cost price by 0.15 times the cost price

Or, $56100=0.85\times{\text{shopkeeper cost price}}$

Thus,

$56100=0.85\times{\text{(shopkeeper cost price)}}$,

$=0.85\times{\text{(wholesale dealer sale price)}}$,

$=0.85\times{1.2}\times{\text{(wholesale dealer cost price)}}$, because of 20% profit by the wholesale dealer

$=0.85\times{1.2}\times{\text{(manufacturer sale price)}}$,

$=0.85\times{1.2}\times{1.1}\times{\text{(manufacturer cost price)}}$, because of the 10% profit by the manufacturer

So,

$\text{Manufacturer cost price }=\displaystyle\frac{56100}{0.85\times{1.2}\times{1.1}}$,

$=\displaystyle\frac{330000}{6.6}$, canceling factor 17 and multiplying 5 by 1.2

Or, $C=50000$.

**Answer:** Option a : Rs.50000.

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

*-- efficient simpliffication --*

**percentage application for profit and loss**

**working backwards approach.****Problem 7.**

A house and a shop were sold for Rs.1 lakh each. The house sale resulted in a loss of 20% whereas the shop sale made a profit of 20%. The entire transaction then resulted in,

- No loss or gain
- Gain of Rs.$\frac{1}{24}$ lakh
- Loss of Rs.$\frac{1}{12}$ lakh
- Loss of Rs.$\frac{1}{18}$ lakh

**Solution 7: **

The sale price are same in two transactions as Rs. 1 lakh.

So in house transaction as the loss is 20%, the sale price of the house is, 80% of cost price, that is $\frac{4}{5}$th of the cost price. Thus,

Cost price of the house is, $\frac{5}{4}\times{\text{1 lakh}}$.

Similarly in the shop transaction as the profit is 20%, the sale price of the shop is, 120% of cost price, that is, $\frac{6}{5}\times{\text{Cost of shop}}$,

Or, $\text{Cost of the shop}=\frac{5}{6}\times{\text{1 lakh}}$.

Together, the total cost of the house and the shop is then,

$\left(\frac{5}{4}+\frac{5}{6}\right)=\frac{25}{12}=2\frac{1}{12}\text{of 1 lakh}$.

As this is more than the total sale price of 2 lakh, there is net loss of $\frac{1}{12}$ of Rs. 1 lakh, that is, loss of Rs. $\frac{1}{12}$ lakh.

**Answer: **c: loss of Rs. $\frac{1}{12}$ lakh.

**Key concepts used:** Basic profit and loss concepts -- * percentage application for profit and loss* --

*, we have kept Rs. 1 lakh as a factor throughout knowing that it will be eliminated in the end; we did only fraction arithmetic --*

**efficient simplification***--*

**delayed evaluation technique**

**fraction arithmetic.****Problem 8.**

A fruit seller sells lemons at the rate of 5 for 3 rupees which he bought at 2 for a rupee. His profit percent is,

- 10%
- 15%
- 25%
- 20%

**Solution 8 - Problem analysis**

If we find out per piece cost and sale prices we can find the profit or loss percent.

#### Solution 8 - Problem solving execution

The fruit seller buys 2 lemons for 1 rupee.

So **per piece lemon cost** is,

$\frac{1}{2}$ rupee.

Then he sells 5 lemons for 3 rupees.

So his **per piece sale price** is,

$\frac{3}{5}$ rupee which is more than the cost price.

So the profit on each $\frac{1}{2}$ rupee is,

$\frac{3}{5}-\frac{1}{2}=\frac{1}{10}$ rupee.

**Profit percent** is then,

$\frac{2}{10}=\frac{1}{5}=20\text{%}$, reverse *conversion of profit as a fraction to profit as a percentage* **multiplying the fraction by 100.**

**Answer:** Option d: 20%.

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

*, calculating the cost and sale prices per piece of lemon --*

**base equalization***--*

**fraction to percent conversion**

**unitary method.****Problem 9.**

The ratio at which tea costing Rs.150/kg is to be mixed with tea costing Rs.192/kg so that the mixed tea when sold for Rs.194.40/kg gives a profit of 20% is,

- 2:5
- 3:5
- 5:3
- 5:2

**Solution 9 - Problem analysis and solving**

Let us assume in 1 kg of mixed tea we have mixed $x$ kg of tea costing 150/kg and so $1-x$ kg of tea costing Rs.192/kg.

Thus the total cost for each kg of tea becomes,

$C=150x +(1-x)192$.

By selling this 1 kg mixed tea at the price of Rs.192.40 a profit of 20% is made.

So,

$194.40=1.2C$,

Or, $C=\frac{194.40}{1.2}=162=150x +(1-x)192$,

Or, $42x=30$,

Or, $x=\frac{5}{7}$, that is, 5 portions of the 7 whole portions.

So the desired ratio is, $5:2$.

**Answer:** Option d: $5:2$.

**Key concepts used:** Basic profit and loss concepts -- mixing concept -- * portions concepts*, at the start itself only one variable in the mixture is assumed the other being in terms of the whole and the assumed variable $x$; this will ultimately result in ratio of variable to whole; subtracting variable portion from whole portion we will get the second mixed portion -- basic ratio concepts -- linear algebraic equations -- percentage application concept.

**Problem 10.**

The percentage loss when an article is sold at Rs.50 is the same as the percentage profit when it is sold for Rs.70. The loss or gain is,

- $20\text{%}$
- $16\frac{2}{3}\text{%}$
- $10\text{%}$
- $22\frac{2}{3}\text{%}$

**Solution 10. **

Cost prices in both situations is same and let us assume it to be $C$.

So, $(1-PL)C=50$, and

$(1+PL)C=70$, where loss and profit percentage is $PL$.

Taking the ratio and simplifying,

$5(1+PL)=7(1-PL)$,

Or, $12PL=2$.

Or, $PL=\frac{2}{12}=\frac{100}{6}\text{%}=16\frac{2}{3}\text{%}$

**Answer:** b: $16\frac{2}{3}\text{%}$.

**Key concepts used:** Percentage application for profit and loss -- assuming one variable for the same cost, two linear equations in two variables formed -- solving of two linear equations -- fraction to percent conversion.

**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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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,**

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