Competitive Exams : Quantitative Aptitude - Profit/Loss (Set 6) | Concept of Maths

Question 1

1. A dishonest dealer marks up his goods by $20\%$ and gives a discount of $10\%$ to the customer. Beside this, he also cheats his supplier by $20\%$ and his customer by $10\%$ through faulty balances. Find his overall profit percentage.

A) $35\%$

B) $44\%$

C) $50\%$

D) $40\%$

Answer

Correct Answer: B

Let the standard weight be $1000$ g and Cost Price ($CP$) be $₹1$ per gram. 1. Cheating supplier by $20\%$: He gets $1200$ g for the price of $1000$ g. So, $CP$ of $1200$ g = $₹1000$. 2. Markup and Discount: $MP = 1.20 \times CP$, $SP = 0.90 \times MP = 0.90 \times 1.20 = 1.08$ per gram. 3. Cheating customer by $10\%$: He sells $900$ g but charges for $1000$ g. Total $SP$ for $1000$ units charged = $1000 \times 1.08 = ₹1080$. But he only gave $900$ g. In the first step, we found $CP$ of $1200$ g is $₹1000$. So, $CP$ of $900$ g = $\left(\frac{1000}{1200}\right) \times 900 = ₹750$. Profit Percentage = $\left(\frac{1080 - 750}{750}\right) \times 100 = \left(\frac{330}{750}\right) \times 100 = 44\%$.

Question 2

2. If the cost price of $18$ articles is equal to the selling price of $12$ articles, what is the profit percentage?

A) $33.33\%$

B) $25\%$

C) $50\%$

D) $66.67\%$

Answer

Correct Answer: C

Let $CP$ of $1$ article be $x$ and $SP$ of $1$ article be $y$. Given: $18x = 12y \implies \frac{y}{x} = \frac{18}{12} = \frac{3}{2}$. Profit = $SP - CP = 3 - 2 = 1$ unit. Profit Percentage = $\left(\frac{1}{2}\right) \times 100 = 50\%$.

Question 3

3. A man sells two chairs at $₹1200$ each. On one he gains $20\%$ and on the other he loses $20\%$. Find his net gain or loss in the whole transaction.

Question 4

4. A trader marks his goods such that even after allowing a discount of $12.5\%$, he makes a profit of $15\%$. If the cost price of the article is $₹420$, find the marked price.

A) $₹520$

B) $₹540$

C) $₹560$

D) $₹552$

Answer

Correct Answer: D

We know that $\frac{MP}{CP} = \frac{100 + \text{Profit}\%}{100 - \text{Discount}\%}$. $\frac{MP}{420} = \frac{100 + 15}{100 - 12.5} = \frac{115}{87.5}$. $MP = \frac{115 \times 420}{87.5} = \frac{115 \times 4200}{875}$. Dividing by $25$: $\frac{115 \times 168}{35}$. Dividing by $7$: $\frac{115 \times 24}{5} = 23 \times 24 = ₹552$.

Question 5

5. By selling $33$ meters of cloth, a person gains the cost price of $11$ meters. Find the gain percentage.

A) $25\%$

B) $50\%$

C) $33\frac{1}{3}\\%$

D) $20\%$

Answer

Correct Answer: C

Gain = $SP$ of $33$m - $CP$ of $33$m. Given: Gain = $CP$ of $11$m. $CP \text{ of } 11 = SP \text{ of } 33 - CP \text{ of } 33$. $CP \text{ of } 44 = SP \text{ of } 33$. Gain Percentage = $\frac{44-33}{33} \times 100 = \frac{11}{33} \times 100 = 33\frac{1}{3}\%$.

Question 6

6. A reduction of $20\%$ in the price of sugar enables a housewife to buy $5$ kg more for $₹600$. Find the reduced price per kg.

A) $₹30$

B) $₹24$

C) $₹20$

D) $₹36$

Answer

Correct Answer: B

Reduction in price = $20\%$ of $600 = ₹120$. This $₹120$ allows buying $5$ kg more. So, Reduced Price = $\frac{120}{5} = ₹24$ per kg.

Question 7

7. A manufacturer sells an article to a wholesale dealer at a profit of $10\%$. The wholesaler sells it to a shopkeeper at $20\%$ profit. The shopkeeper sells it to a customer for $₹56,100$ at a loss of $15\%$. Find the cost price for the manufacturer.

A) $₹45,000$

B) $₹48,000$

C) $₹52,000$

D) $₹50,000$

Answer

Correct Answer: D

Let the initial $CP$ be $x$. $x \times (1.10) \times (1.20) \times (0.85) = 56100$. $x \times 1.1 \times 1.2 \times 0.85 = 56100$. $x \times (1.122) = 56100$. $x = \frac{56100}{1.122} = 50,000$. Therefore, $CP = ₹50,000$.

Question 8

8. A person bought some oranges at the rate of $5$ for $₹1$ and the same number at the rate of $4$ for $₹1$. He sold them at the rate of $9$ for $₹2$. During the whole transaction, he incurred a loss of $₹3$. Find the total number of oranges he bought.

A) $1080$

B) $540$

C) $2160$

D) $480$

Answer

Correct Answer: A

Let the number of oranges bought at each rate be $\text{LCM}(5, 4, 9) = 180$. $CP$ of $180$ oranges at $5$ for $₹1 = ₹36$. $CP$ of $180$ oranges at $4$ for $₹1 = ₹45$. Total $CP$ for $360$ oranges $= 36 + 45 = ₹81$. $SP$ of $360$ oranges at $9$ for $₹2 = (360/9) \times 2 = ₹80$. Loss = $81 - 80 = ₹1$ for $360$ oranges. Actual loss is $₹3$, so total oranges $= 360 \times 3 = 1080$.

Question 9

9. If a merchant offers a discount of $40\%$ on the marked price of his goods and still manages to make a profit of $20\%$, then what is the percentage by which the marked price is more than the cost price?

A) $60\%$

B) $80\%$

C) $100\%$

D) $120\%$

Answer

Correct Answer: C

Let $CP = 100$. Profit = $20\% \implies SP = 120$. Discount = $40\%$ on $MP$. $0.60 \times MP = 120 \implies MP = \frac{120}{0.6} = 200$. Markup Amount = $MP - CP = 200 - 100 = 100$. Markup Percentage = $\left(\frac{100}{100}\right) \times 100 = 100\%$.

Question 10

10. An item is sold at a loss of $10\%$. Had it been sold for $₹90$ more, there would have been a gain of $5\%$. Find the $CP$ of the item.

A) $₹500$

B) $₹600$

C) $₹800$

D) $₹750$

Answer

Correct Answer: B

Difference in percentage $= 5\% - (-10\%) = 15\%$. This $15\%$ difference is caused by $₹90$. $15\% \text{ of } CP = 90$. $CP = \frac{90 \times 100}{15} = 6 \times 100 = ₹600$.

Accepted Answer

Correct Answer: A

When two items are sold at the same $SP$, one at $x\%$ gain and another at $x\%$ loss, there is always a loss of $\left(\frac{x^2}{100}\right)\%$. Percentage Loss = $\left(\frac{20^2}{100}\right) = 4\%$. Total $SP = 1200 + 1200 = ₹2400$. Total $CP = \frac{100}{100 - \text{Loss}\%} \times SP = \frac{100}{96} \times 2400 = ₹2500$. Net Loss = $CP - SP = 2500 - 2400 = ₹100$.

Competitive Exams : Quantitative Aptitude - Profit/Loss (Set 6) | Concept of Maths