Quantitative Aptitude

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PROFIT AND LOSS

Cost Price (CP):

Cost price is the price at which a product is purchased

For example,

A person bought a pen for Rs.$10$ and sold it for Rs.$12$. What is the cost price of that pen?

Cost price of the pen $=\;10\;Rs$

Selling Price (SP):

Selling price is the price at which a product is sold

For example,

A person bought a pen for Rs.$10$ and sold it for Rs.$12$. What is the selling price of that pen?

Selling price of the pen $=\;12\;Rs$

Profit (P) or Loss (L):

Profit: If the selling price of a product is greater than the cost price. Then the material is said to be sold with a profit.

Profit = Selling price – Cost Price

Profit $\to$ Selling Price $\gt$ Cost Price

Loss: If the selling price of a material is lesser than the cost price. Then the material is said to be sold with a loss.

Loss = Cost Price – Selling Price

Loss $\to$ Selling Price $\lt$ Cost Price

🌾For example,

A person bought a pen for Rs.$10$ and sold it for Rs.$12$. Is it a profit or loss?

  • Selling price of the pen $=\;12\;Rs$
  • Cost price of the pen $=\;10\;Rs$
  • Since Selling price $\gt$ Cost Price, then we can say that the person had a profit of $Rs.2$

A person bought a pencil for Rs.$3$ and sold it for Rs.$2$. Is it a profit or loss?

  • Selling price of the pen $=\;2\;Rs$
  • Cost price of the pen $=\;3\;Rs$
  • Since Selling price $\lt$ Cost Price, then we can say that the person had a loss of $Rs.1$

Profit or Loss Percentage $(P\%)$ or $(L\%)$:

Profit percentage: How much percentage of amount gained after selling a product $$Profit\%=\frac{Selling \;Price-Cost\;price}{Cost\;Price}×100$$

$$P\%=\frac{SP-CP}{CP}×100$$

Loss percentage: How much percentage of amount lost after selling a product $$Loss\%=\frac{Cost \;Price-Selling\;Price}{Cost\;Price}×100$$

$$L\%=\frac{CP-SP}{CP}×100$$

Practice Question

🌾A man bought a bicycle for $Rs.4000$ and sold it for $Rs.3000$. Answer the following questions

$1)$ Is it profit or loss?

$2)$ Find the amount he gained/lost.

$3)$ Find the percentage of gain/loss.

  1. Since Selling price $\lt$ Cost price, that man incurred a loss.
  2. Amount he lost,$$Loss=CP-SP$$ $$=4000-3000$$ $$=1000$$
  3. Loss percentage, $$Loss\%=\frac{Cost \;Price-Selling\;Price}{Cost\;Price}×100$$ $$=\frac{4000-3000}{4000}×100$$ $$=\frac{1000}{4000}×100$$ $$=25\%$$

How to find the Cost Price?

When SP and profit is given:$$Cost\;Price=Selling\;Price-Profit$$

When SP and profit percentage is given:

From profit percentage formula,$$Profit\%=\frac{Selling \;Price-Cost\;Price}{Cost\;Price}×100$$ Rearranging,$$\frac{Profit\%}{100}=\frac{Selling \;Price}{Cost\;Price}-1$$ $$\frac{Profit\%}{100}+1=\frac{Selling \;Price}{Cost\;Price}$$$$\frac{Profit\%+100}{100}=\frac{Selling \;Price}{Cost\;Price}$$$$Cost\;Price=\frac{100}{Profit\%+100}×Selling\;Price$$

$$CP=\frac{100}{100+P\%}×SP$$

When SP and loss is given:$$Cost\;Price=Selling\;Price+Loss$$

When SP and loss percentage is given:

From loss percentage formula,$$Loss\%=\frac{Cost \;Price-Selling\;Price}{Cost\;Price}×100$$ Rearranging,  $$\frac{Loss\%}{100}=1-\frac{Selling\;Price}{Cost\;Price}$$Then,$$\frac{Selling\;Price}{Cost\;Price}=1-\frac{Loss\%}{100}$$$$Cost\;Price=\frac{100}{100-Loss\%}×Selling\;Price$$

$$CP=\frac{100}{100-L\%}×SP$$

How to find the Selling Price?

When CP and profit is given: $$Selling\;Price=Cost\;Price+Profit$$

When CP and profit percentage is given:

From profit percentage formula,$$Profit\%=\frac{Selling\;Price-Cost\;Price}{Cost\;Price}×100$$ Rearranging, $$\frac{Profit\%×Cost\;Price}{100}={Selling \;Price-Cost\;Price}$$$$Selling\;Price=Cost\;Price+\frac{Profit\%×Cost\;Price}{100}$$$$Selling\;Price=\frac{100+Profit\%}{100}×Cost\;Price$$

$$SP=\frac{100+P\%}{100}×CP$$

When CP and loss is given:$$Selling\;Price=Cost\;Price-Loss$$

When CP and loss percentage is given:

From loss percentage formula,$$Loss\%=\frac{Cost \;Price-Selling\;Price}{Cost\;Price}×100$$ Rearranging, $$\frac{Loss\%×Cost\;Price}{100}={Cost \;Price-Selling\;Price}$$$$Selling\;Price=Cost\;Price-\frac{Loss\%×Cost\;Price}{100}$$$$Selling\;Price=\frac{100-Loss\%}{100}×Cost\;Price$$

$$SP=\frac{100-L\%}{100}×CP$$

Summarizing,$$Profit=SP-CP$$$$Loss=CP-SP$$$$P\%=\frac{SP-CP}{CP}×100$$$$L\%=\frac{CP-SP}{CP}×100$$$$SP=\frac{100+P\%}{100}×CP$$$$SP=\frac{100-L\%}{100}×CP$$$$CP=\frac{100}{100+P\%}×SP$$$$CP=\frac{100}{100-L\%}×SP$$

Practice Question

🌾A shopkeeper purchases six small cold drink bottles for $Rs.100$. For how much should he sell one such bottle to get a profit of $20\%$? (SSC-CGL $2023$)

a) $Rs.22$

b) $Rs.20$

c) $Rs.21$

d) $Rs.23$

Correct Answer: Option B

Given:

  1. Cost price of $6$ bottles $=Rs.100$
  2. $Profit\%=20$

To find:

Selling price of $1$ bottle

Solution:

When CP and profit percentage is given:

$$Selling\;Price=Cost\;Price+\frac{Profit\%×Cost\;Price}{100}$$ $$=100+\frac{20\%×100}{100}$$ $$=120$$

Selling price of $6$ bottles to get a profit of $20\%$ is $Rs.120$

Selling price of $1$ bottle  is $120/6=Rs.20$

Practice Question

🌾By selling a watch for $Rs.2000$, a shopkeeper loses $20\%$. How much would he gain or lose by selling it for $Rs.3000$ (SSC-CGL $2023$)

a) $25\%\;gain$

b) $15\%\;loss$

c) $25\%\;loss$

d) $20\%\;gain$

Correct Answer: Option D

Given:

  1. $SP_1=Rs.2000$
  2. $L_1\%=20$
  3. $SP_2=Rs.3000$
  4. $(P/L)_2\%=??$

To find:

$Profit/Loss\%$ for second selling price.

Solution:

When SP and loss percentage is given:

$$Cost\;Price=\frac{100}{100-Loss\%}×Selling\;Price$$ $$=\frac{100}{100-20}×2000$$ $$=1.25×2000$$ Hence, cost price is $Rs. 2500$

If the selling price is $Rs.3000$, then it will be a profit. 

Hence Profit percentage is, $$P\%=\frac{SP-CP}{CP}×100$$ $$P\%=\frac{3000-2500}{2500}×100$$ $$P\%=\frac{500}{2500}×100$$ $$Profit\%=20\%$$

Two successive profits:

If a product is sold with two successive profits, first-hand at $P_1\%$ and second-hand at $P_2\%$, then total profit percentage will be, $$P\%=P_1+P_2+\frac{P_1 P_2}{100}\%$$

Two successive profits means, the cost price of the product is getting increased $$CP\xrightarrow{\text{P1%}}CP\uparrow \xrightarrow{\text{P2%}}CP\uparrow \uparrow$$

Proof:

Using the formula, when CP and profit percentage is known, the SP will be,$$SP=\frac{100+P\%}{100}×CP$$For first-hand transaction with $P_1\%$ profit, SP will be$$SP_1=\frac{100+P_1}{100}×CP$$

For the second-hand transaction, CP of the product is nothing but the first-hand transaction’s SP,$$SP_2=\frac{100+P_2}{100}×SP_1$$$$SP_2=\frac{100+P_2}{100}×\frac{100+P_1}{100}×CP$$$$SP_2=\frac{100^2+100P_1+100P_2+P_1 P_2}{100^2}×CP$$$$SP_2=\frac{100(100+P_1+P_2+P_1 P_2/100)}{100^2}×CP$$$$SP_2=\frac{(100+P_1+P_2+P_1 P_2/100)}{100}×CP$$ Comparing the above equation with $SP=\frac{100+P\%}{100}×CP$ , we can say, $$P\%=P_1+P_2+\frac{P_1 P_2}{100}\%$$

Two successive profits, $$P\%=P_1+P_2+\frac{P_1 P_2}{100}\%$$

Practice Question

🌾John bought a watch for $Rs.110$ with $10\%$ loss and sells it to Ranjan with $20\%$ profit. How much money did Ranjan lose?

Solution:

Read this question very carefully. Eventhough it says loss and profit, this question is a “Two consecutive profits” problem. Don’t start this problem from John, start it from the shopkeeper. For John it’s a loss but for shopkeeper it’s a profit. $$Shop\xrightarrow{\text{P1%}} John \xrightarrow{\text{P2%}}Ranjan$$

John bought the watch for $Rs.110$ with a loss of $10\%$ (or) shopkeeper sold the watch to John with $10\%$ profit, then the original price of the watch is$$CP=\frac{100}{100+10\%}×110$$ $$=100Rs$$

John sells the watch with a profit of $20\%$, then the amount Ranjan spent will be, $$SP=\frac{100+P\%}{100}×CP$$ $$=\frac{100+20}{100}×110$$ $$=\frac{120}{100}110$$$$=132Rs$$

Hence Ranjan lost $132-100=32$ rupees during these transactions.

Shortcut: This problem involves two transactions, one $10\%$ profit and one $20\%$ profit. Hence total profit $\%$ will be, $$P\%=P_1+P_2+\frac{P_1 P_2}{100}$$ $$P\%=10+20+\frac{200}{100}$$$$=32\%$$ Ranjan paid $32\%$ more money than the original price of the watch. Since the original price is $100$, Ranjan lost $32$ rupees.

Two successive loss:

If a product is sold with two successive loss, first-hand at $L_1\%$ and second-hand at $L_2\%$, then total loss percentage will be, $$L\%=L_1+L_2-\frac{L_1 L_2}{100}\%$$

Two successive loss means, the cost price of the product is getting reduced $$CP\xrightarrow{\text{L1%}}CP\downarrow \xrightarrow{\text{L2%}}CP\downarrow \downarrow$$

Proof:

Using the formula, when CP and loss percentage is known, the SP will be,$$SP=\frac{100-L\%}{100}×CP$$For first-hand transaction with $L_1\%$ loss, SP will be$$SP_1=\frac{100-L_1}{100}×CP$$

For the second-hand transaction, CP of the product is nothing but the first-hand transaction’s SP,$$SP_2=\frac{100-L_2}{100}×SP_1$$$$SP_2=\frac{100-L_2}{100}×\frac{100-L_1}{100}×CP$$$$SP_2=\frac{100^2-100L_1-100L_2+L_1 L_2}{100^2}×CP$$$$SP_2=\frac{100(100-L_1-L_2+L_1 L_2/100)}{100^2}×CP$$$$SP_2=\frac{(100-L_1-L_2+L_1 L_2/100)}{100}×CP$$$$SP_2=\frac{100-(L_1+L_2-L_1 L_2/100)}{100}×CP$$ Comparing the above equation with $SP=\frac{100-L\%}{100}×CP$ , we can say, $$L\%=L_1+L_2-\frac{L_1 L_2}{100}\%$$

Practice Question

🌾John bought a watch for $Rs.90$ with $10\%$ profit and sells it to Ranjan with $20\%$ loss. How much money did Ranjan gain?

Solution:

Read this question very carefully. Eventhough it says profit and loss, this question is a “Two consecutive loss” problem. Don’t start this problem from John, start it from the shopkeeper. For John it’s a profit but for shopkeeper it’s a loss. $$Shop\xrightarrow{\text{L1%}} John \xrightarrow{\text{L2%}}Ranjan$$

John bought the watch for $Rs.90$ with a profit of $10\%$ (or) we can say the shopkeeper sold the watch to John with $10\%$ loss, then the original price of the watch is$$CP=\frac{100}{100-10\%}×90$$ $$=100Rs$$

John sells the watch with a loss of $20\%$, then the amount Ranjan spent will be, $$SP=\frac{100-L\%}{100}×CP$$ $$=\frac{100-20}{100}×90$$ $$=\frac{80}{100}90$$$$=72Rs$$

Hence Ranjan gained $100-72=28$ rupees during these transactions.

Shortcut: This problem involves two transactions, one $10\%$ loss and one $20\%$ loss. Hence total loss $\%$ will be, $$L\%=L_1+L_2-\frac{L_1 L_2}{100}$$ $$L\%=10+20-\frac{200}{100}$$$$=28\%$$ Ranjan paid $28\%$ less money than the original price of the watch. Since the original price is $100$, Ranjan gained $28$ rupees.

Profit and a loss (or) vice-versa:

If a product is sold first-hand with a profit of $P\%$ and second-hand with a loss of $L\%$ or vice-versa, then total profit or loss percentage will be, $$P(or)L\%=P-L-\frac{PL}{100}\%$$

Note: If the answer is positive, then it is a profit. If negative, then it is a loss.

$$CP\xrightarrow{\text{P%}}CP\uparrow \xrightarrow{\text{L%}}CP\downarrow$$

Proof:

For first-hand transaction with $P\%$ profit, SP will be$$SP_1=\frac{100+P}{100}×CP$$

For the second-hand transaction with $L\%$ loss,$$SP_2=\frac{100-L}{100}×SP_1$$$$SP_2=\frac{100+P}{100}×\frac{100-L}{100}×CP$$$$SP_2=\frac{100^2+100P-100L-PL}{100^2}×CP$$$$SP_2=\frac{100(100+P-L-PL/100)}{100^2}×CP$$$$SP_2=\frac{(100+P-L-PL/100)}{100}×CP$$ Comparing the above equation with $SP=\frac{100+P\%}{100}×CP$ , we can say, $$Total\%=P-L-\frac{PL}{100}\%$$

If both profit and loss are at same percentage$(say\;x\%)$, then $$Total\%=x-x-\frac{x^2}{100}\%$$$$Total\%=-\frac{x^2}{100}\%$$ Since the value is always negative, we can conclude that the whole transaction will always incurs a loss.

If a product is sold twice, once with a profit of $x\%$ and then with a loss of $x\%$ or vice-versa, the whole transaction will always result in a loss of $$L\%=\frac{x^2}{100}\%$$ (i.e.,) the value of the product always gets reduced.

Practice Question

🌾John bought a watch for $Rs.110$ with a loss of $10\%$ and sells it to Ranjan with a loss of $10\%$. How much amount did Ranjan pay less than the cost price?

Solution:

Read this question very carefully. Eventhough it says loss and loss, this question is a “Profit and a loss” problem. Don’t start this problem from John, start it from the shopkeeper. For John it’s a loss but for shopkeeper it’s a profit. $$Shop\xrightarrow{\text{P%}} John \xrightarrow{\text{L%}}Ranjan$$John bought the watch for $Rs.110$ with a loss of $10\%$ (or) shopkeeper sold the watch to John with $10\%$ profit, then the original price of the watch is$$CP=\frac{100}{100+10\%}×110$$ $$=100Rs$$

John sells the watch with a loss of $10%$, then the selling price will be, $$SP=\frac{100-L\%}{100}×CP$$ $$=\frac{100-10}{100}×110$$ $$=\frac{90}{100}110$$$$=99Rs$$

Hence Ranjan paid $1$ rupee less than the CP.

Shortcut: This problem involves two transactions, one profit and one loss, with same percentage. Hence the whole will definitely result in a loss with a percentage of $x^2/100$, $$=10^2/100$$$$=1\%$$. We can say, the value of the watch gets reduced by $1\%$ of original value. Hence Ranjan has to pay $1 Rs$ less than the cost price

Dishonest dealers and Faulty weights:

A dishonest dealer is someone who gives us the commodity with less weight, thereby he gains profit. 

Weight gain only: If the dishonest dealer sells you the commodity in its cost price itself means, he gains a profit only by means of faulty weight.

Error = True weight – False weight

% of Gain by means of faulty weight is $$W\%=\frac{True\;weight-False\;weight}{False\;weight}$$

Weight gain $+$ Monetary gain: If the dishonest dealer sells you the commodity with some profit percentage, then it is said to be two successive profits (weight gain $W\%$ and monetary gain $M\%$). $$W+M+\frac{WM}{100}\%$$

Practice Question

🌾John bought a watch for $Rs.110$ with a loss of $10\%$ and sells it to Ranjan with a loss of $10\%$. How much amount did Ranjan pay?

Solution:

Read this question very carefully. Eventhough it says loss and loss, this question is a “Profit and a loss” problem. Don’t start this problem from John, start it from the shopkeeper. For John it’s a loss but for shopkeeper it’s a profit. $$Shop\xrightarrow{\text{P%}} John \xrightarrow{\text{L%}}Ranjan$$John bought the watch for $Rs.110$ with a loss of $10\%$ (or) shopkeeper sold the watch to John with $10\%$ profit, then the original price of the watch is$$CP=\frac{100}{100+10\%}×110$$ $$=100Rs$$

John sells the watch with a loss of $10%$, then the selling price will be, $$SP=\frac{100-L\%}{100}×CP$$ $$=\frac{100-10}{100}×110$$ $$=\frac{90}{100}110$$$$=99Rs$$

Hence Ranjan paid $99$ rupees.

Shortcut: This problem involves two transactions, one profit and one loss, with same percentage. Hence the whole will definitely result in a loss with a percentage of $x^2/100$, $$=10^2/100$$$$=1\%$$. We can say, the value of the watch gets reduced by $1\%$ of original value. Hence Ranjan has to pay $100-1=99\;Rs$.

Discount:

Marked price (MP): Marked price is the maximum retail price (MRP) or tag price which is always greater then the cost price.

Marked price includes cost price, amount spent on advertising, transportation, profit for the retailer, etc.,

A retailer can give us discount from the marked price to attract customers.

Marked price is always greater than the Cost price.$$MP\gt CP$$

Discount (D): A retailer can give us discount from the marked price to attract customers.$$MP-Discount\;=\;SP$$ $$CP=MP-D-Profit$$

Note: Remember, Profit has to be calculated w.r.t CP not MP.

Discount percentage (D%): How much percentage of amount to be reduced from MP in order to sell an item. $$D\%=\frac{MP-SP}{MP}×100$$

Two successive discounts: A retailer can give you store discount $(D_1\%)$ and from the new price they can give you credit card discount $(D_2\%)$ and so on. The overall discount can be found using, $$D_1+D_2-\frac{D_1D_2}{100}\%$$

Practice Question

A shopkeeper makes a net profit of $44\%$ on selling an article at successive discounts of $10\%$ and $20\%$. Find the net profit percentage, if the shopkeeper sells the same article at a discount of $15\%$.

a) $50\%$

b) $70\%$

c) $30\%$

d) $40\%$

Solution:

Correct Answer: Option B

First problem:

  1. Two successive discounts $10\%$ and $20\%$, overall discount is, $$D_1+D_2-\frac{D_1D_2}{100}$$$$=10+20-200/100$$$$=10+20-2$$$$=28\%$$Hence,$$SP=MP-28\%\;of\;MP$$ Assume Mp=$100$ $$SP=100-28=72$$
  2. Profit =$44\%$, we have SP in hand, $$CP=\frac{100}{100+P}SP$$$$CP=\frac{100}{144}72$$$$=50$$

Second problem:

  1. Same article with different discount $15\%$,$$SP=100-15=85$$
  2. Profit\% will be $$P\%=\frac{SP-CP}{CP}×100$$
  3. $$P\%=\frac{85-50}{50}×100$$$$=\frac{35}{50}100$4$$=70\%$$