Profit and Loss Notes by Abhinav Sir

Profit and Loss: Comprehensive Theory & 26 Hard-Level Questions with Solutions

Welcome to our detailed guide on the Profit and Loss chapter, tailored for students preparing for SSC CGL, NTPC, ICSE, and other competitive exams. This post includes a thorough explanation of the theory, followed by 26 challenging questions from RD Sharma, RS Aggarwal, and exam-oriented sources, with step-by-step solutions rendered using MathJax for clarity.

Theory of Profit and Loss

The Profit and Loss chapter is a fundamental topic in mathematics, widely tested in competitive exams like SSC CGL, NTPC, and ICSE board exams. It deals with the financial outcomes of buying and selling goods, focusing on concepts like cost price, selling price, profit, loss, and discounts. Below is a detailed breakdown of the key concepts:

• Cost Price (CP): The price at which an item is purchased. \( CP \)
• Selling Price (SP): The price at which an item is sold. \( SP \)
• Profit: When \( SP > CP \), the difference is the profit. \[ \text{Profit} = SP - CP \]
• Loss: When \( SP < CP \), the difference is the loss. \[ \text{Loss} = CP - SP \]
• Profit Percentage: \[ \text{Profit \%} = \left( \frac{\text{Profit}}{CP} \times 100 \right) \% \]
• Loss Percentage: \[ \text{Loss \%} = \left( \frac{\text{Loss}}{CP} \times 100 \right) \% \]
• Marked Price (MP): The price marked on an item before any discount. Often used in problems involving discounts.
• Discount: Reduction offered on the marked price. \[ \text{Discount} = MP - SP \]
• Discount Percentage: \[ \text{Discount \%} = \left( \frac{\text{Discount}}{MP} \times 100 \right) \% \]
• Successive Discounts: For two successive discounts \( d_1\% \) and \( d_2\% \), the effective discount is: \[ \text{Effective Discount \%} = d_1 + d_2 - \frac{d_1 \times d_2}{100} \]
• Cost Price with Profit/Loss:
  • If profit is \( x\% \): \[ SP = CP \times \left(1 + \frac{x}{100}\right) \]
  • If loss is \( x\% \): \[ SP = CP \times \left(1 - \frac{x}{100}\right) \]
• Marked Price with Discount: \[ SP = MP \times \left(1 - \frac{\text{Discount \%}}{100}\right) \]
• Successive Selling/Buying: In problems involving multiple transactions, calculate the CP and SP for each step carefully.
• Dishonest Dealer: Problems involving false weights or measures, where the dealer claims to sell at CP but uses a faulty weight. \[ \text{Profit \%} = \left( \frac{\text{True Weight} - \text{False Weight}}{\text{False Weight}} \times 100 \right) \% \]
• Break-even Point: When \( SP = CP \), there is neither profit nor loss.

These formulas are crucial for solving problems in the Profit and Loss chapter. For competitive exams like SSC CGL and NTPC, questions often involve multiple steps, successive discounts, or dishonest dealer scenarios, requiring a strong grasp of these concepts.

Hard-Level Questions with Solutions

Below are 26 challenging questions curated from RD Sharma, RS Aggarwal, SSC CGL, NTPC, and ICSE syllabi. These questions test higher-order thinking skills and are solved step-by-step using MathJax.

Question 1 (RD Sharma, Class 10)

A shopkeeper buys an article for \( \text{Rs. } 800 \) and sells it at a profit of \( 25\% \). If he also gives a discount of \( 20\% \) on the marked price, find the marked price.

Solution

Given: \( CP = 800 \), Profit \( = 25\% \), Discount \( = 20\% \).

Step 1: Calculate the selling price (SP).

\[ SP = CP \times \left(1 + \frac{\text{Profit \%}}{100}\right) = 800 \times \left(1 + \frac{25}{100}\right) = 800 \times 1.25 = 1000 \]

Step 2: Let the marked price be \( MP \). After a \( 20\% \) discount, the selling price is:

\[ SP = MP \times \left(1 - \frac{20}{100}\right) = MP \times 0.8 \]

Step 3: Equate the SP from both steps:

\[ 1000 = MP \times 0.8 \] \[ MP = \frac{1000}{0.8} = 1250 \]

Answer: The marked price is \( \text{Rs. } 1250 \).

Question 2 (RS Aggarwal, Class 10)

A dealer sells an article at a loss of \( 10\% \). Had he sold it for \( \text{Rs. } 450 \) more, he would have gained \( 20\% \). Find the cost price of the article.

Solution

Let the cost price be \( CP \).

Step 1: Selling price at \( 10\% \) loss:

\[ SP_1 = CP \times \left(1 - \frac{10}{100}\right) = 0.9CP \]

Step 2: Selling price at \( 20\% \) profit:

\[ SP_2 = CP \times \left(1 + \frac{20}{100}\right) = 1.2CP \]

Step 3: Given \( SP_2 = SP_1 + 450 \):

\[ 1.2CP = 0.9CP + 450 \] \[ 1.2CP - 0.9CP = 450 \] \[ 0.3CP = 450 \] \[ CP = \frac{450}{0.3} = 1500 \]

Answer: The cost price is \( \text{Rs. } 1500 \).

Question 3 (SSC CGL)

A shopkeeper marks his goods \( 40\% \) above the cost price and offers a discount of \( 25\% \). If the cost price is \( \text{Rs. } 2000 \), find the profit percentage.

Solution

Given: \( CP = 2000 \), Marked price markup \( = 40\% \), Discount \( = 25\% \).

Step 1: Calculate the marked price:

\[ MP = CP \times \left(1 + \frac{40}{100}\right) = 2000 \times 1.4 = 2800 \]

Step 2: Calculate the selling price after discount:

\[ SP = MP \times \left(1 - \frac{25}{100}\right) = 2800 \times 0.75 = 2100 \]

Step 3: Calculate profit:

\[ \text{Profit} = SP - CP = 2100 - 2000 = 100 \]

Step 4: Calculate profit percentage:

\[ \text{Profit \%} = \left( \frac{\text{Profit}}{CP} \times 100 \right) = \left( \frac{100}{2000} \times 100 \right) = 5\% \]

Answer: The profit percentage is \( 5\% \).

Question 4 (NTPC)

A man buys an article for \( \text{Rs. } 1200 \) and sells it at a profit of \( 20\% \). He then buys another article with the same amount and sells it at a loss of \( 25\% \). Find the overall profit or loss percentage.

Solution

Step 1: First article: \( CP_1 = 1200 \), Profit \( = 20\% \).

\[ SP_1 = 1200 \times \left(1 + \frac{20}{100}\right) = 1200 \times 1.2 = 1440 \]

Step 2: Second article: \( CP_2 = 1440 \), Loss \( = 25\% \).

\[ SP_2 = 1440 \times \left(1 - \frac{25}{100}\right) = 1440 \times 0.75 = 1080 \]

Step 3: Total CP = \( 1200 + 1440 = 2640 \), Total SP = \( 1440 + 1080 = 2520 \).

Step 4: Overall loss:

\[ \text{Loss} = 2640 - 2520 = 120 \] \[ \text{Loss \%} = \left( \frac{120}{2640} \times 100 \right) \approx 4.55\% \]

Answer: The overall loss percentage is approximately \( 4.55\% \).

Question 5 (ICSE, RS Aggarwal)

A dishonest dealer professes to sell his goods at cost price but uses a weight of 800 grams instead of 1 kg. Find his profit percentage.

Solution

True weight = 1000 g, False weight = 800 g.

Profit percentage formula for dishonest dealer:

\[ \text{Profit \%} = \left( \frac{\text{True Weight} - \text{False Weight}}{\text{False Weight}} \times 100 \right) \] \[ = \left( \frac{1000 - 800}{800} \times 100 \right) = \left( \frac{200}{800} \times 100 \right) = 25\% \]

Answer: The profit percentage is \( 25\% \).

Question 6 (RD Sharma, Class 10)

A shopkeeper sells an article at a loss of \( 12.5\% \). Had he sold it for \( \text{Rs. } 51.50 \) more, he would have gained \( 6\% \). Find the cost price.

Solution

Let the cost price be \( CP \).

Step 1: Selling price at \( 12.5\% \) loss:

\[ SP_1 = CP \times \left(1 - \frac{12.5}{100}\right) = CP \times 0.875 \]

Step 2: Selling price at \( 6\% \) profit:

\[ SP_2 = CP \times \left(1 + \frac{6}{100}\right) = CP \times 1.06 \]

Step 3: Given \( SP_2 = SP_1 + 51.50 \):

\[ 1.06CP = 0.875CP + 51.50 \] \[ 1.06CP - 0.875CP = 51.50 \] \[ 0.185CP = 51.50 \] \[ CP = \frac{51.50}{0.185} \approx 278.38 \]

Answer: The cost price is approximately \( \text{Rs. } 278.38 \).

Question 7 (SSC CGL)

A man buys two articles for \( \text{Rs. } 3000 \) each. He sells one at a profit of \( 20\% \) and the other at a loss of \( 20\% \). Find the overall profit or loss percentage.

Solution

Step 1: First article: \( CP_1 = 3000 \), Profit \( = 20\% \).

\[ SP_1 = 3000 \times 1.2 = 3600 \]

Step 2: Second article: \( CP_2 = 3000 \), Loss \( = 20\% \).

\[ SP_2 = 3000 \times 0.8 = 2400 \]

Step 3: Total CP = \( 3000 + 3000 = 6000 \), Total SP = \( 3600 + 2400 = 6000 \).

Step 4: Since Total SP = Total CP, there is no profit or loss.

Answer: The overall profit or loss percentage is \( 0\% \).

Question 8 (RS Aggarwal, Class 8)

A shopkeeper offers two successive discounts of \( 10\% \) and \( 15\% \). If the marked price is \( \text{Rs. } 5000 \), find the selling price.

Solution

Given: \( MP = 5000 \), Discounts = \( 10\% \), \( 15\% \).

Step 1: Effective discount formula:

\[ \text{Effective Discount \%} = 10 + 15 - \frac{10 \times 15}{100} = 25 - 1.5 = 23.5\% \]

Step 2: Calculate the selling price:

\[ SP = MP \times \left(1 - \frac{23.5}{100}\right) = 5000 \times 0.765 = 3825 \]

Answer: The selling price is \( \text{Rs. } 3825 \).

Question 9 (NTPC)

A dealer buys an article for \( \text{Rs. } 600 \) and marks it up by \( 50\% \). He then offers a discount of \( 20\% \). Find the profit percentage.

Solution

Given: \( CP = 600 \), Markup \( = 50\% \), Discount \( = 20\% \).

Step 1: Calculate the marked price:

\[ MP = 600 \times 1.5 = 900 \]

Step 2: Calculate the selling price:

\[ SP = 900 \times 0.8 = 720 \]

Step 3: Calculate profit:

\[ \text{Profit} = 720 - 600 = 120 \] \[ \text{Profit \%} = \left( \frac{120}{600} \times 100 \right) = 20\% \]

Answer: The profit percentage is \( 20\% \).

Question 10 (ICSE, RS Aggarwal)

A shopkeeper sells an article at a profit of \( 25\% \). If the cost price increases by \( 20\% \) but the selling price remains the same, find the new profit percentage.

Solution

Let the original cost price be \( CP = 100 \).

Step 1: Original selling price at \( 25\% \) profit:

\[ SP = 100 \times 1.25 = 125 \]

Step 2: New cost price after \( 20\% \) increase:

\[ CP_{\text{new}} = 100 \times 1.2 = 120 \]

Step 3: New profit:

\[ \text{Profit} = 125 - 120 = 5 \] \[ \text{Profit \%} = \left( \frac{5}{120} \times 100 \right) \approx 4.17\% \]

Answer: The new profit percentage is approximately \( 4.17\% \).

Question 11 (RD Sharma, Class 10)

A man buys an article for \( \text{Rs. } 1500 \) and sells it at a loss of \( 15\% \). He then buys another article with the same amount and sells it at a profit of \( 20\% \). Find the overall profit or loss percentage.

Solution

Step 1: First article: \( CP_1 = 1500 \), Loss \( = 15\% \).

\[ SP_1 = 1500 \times \left(1 - \frac{15}{100}\right) = 1500 \times 0.85 = 1275 \]

Step 2: Second article: \( CP_2 = 1275 \), Profit \( = 20\% \).

\[ SP_2 = 1275 \times \left(1 + \frac{20}{100}\right) = 1275 \times 1.2 = 1530 \]

Step 3: Total CP = \( 1500 + 1275 = 2775 \), Total SP = \( 1275 + 1530 = 2805 \).

Step 4: Overall profit:

\[ \text{Profit} = 2805 - 2775 = 30 \] \[ \text{Profit \%} = \left( \frac{30}{2775} \times 100 \right) \approx 1.08\% \]

Answer: The overall profit percentage is approximately \( 1.08\% \).

Question 12 (SSC CGL)

A shopkeeper marks his goods \( 50\% \) above cost price and offers two successive discounts of \( 10\% \) and \( 20\% \). If the cost price is \( \text{Rs. } 4000 \), find the selling price.

Solution

Step 1: Calculate marked price: \( CP = 4000 \), Markup \( = 50\% \).

\[ MP = 4000 \times \left(1 + \frac{50}{100}\right) = 4000 \times 1.5 = 6000 \]

Step 2: Calculate effective discount for 10% and 20%:

\[ \text{Effective Discount \%} = 10 + 20 - \frac{10 \times 20}{100} = 30 - 2 = 28\% \]

Step 3: Calculate selling price:

\[ SP = 6000 \times \left(1 - \frac{28}{100}\right) = 6000 \times 0.72 = 4320 \]

Answer: The selling price is \( \text{Rs. } 4320 \).

Question 13 (RS Aggarwal, Class 8)

A dishonest dealer uses a weight of 900 g instead of 1 kg and sells at \( 10\% \) above cost price. Find his profit percentage.

Solution

True weight = 1000 g, False weight = 900 g.

Step 1: Selling at 10% above CP:

\[ SP = CP \times 1.1 \]

Step 2: Effective profit due to false weight:

Let CP for 1000 g = 100. He sells 900 g at \( 100 \times 1.1 = 110 \).

Actual CP for 900 g = \( \frac{900}{1000} \times 100 = 90 \).

\[ \text{Profit} = 110 - 90 = 20 \] \[ \text{Profit \%} = \left( \frac{20}{90} \times 100 \right) \approx 22.22\% \]

Answer: The profit percentage is approximately \( 22.22\% \).

Question 14 (NTPC)

A man sells an article at a profit of \( 30\% \). Had he bought it at \( 10\% \) less and sold it for \( \text{Rs. } 100 \) less, he would have gained \( 25\% \). Find the cost price.

Solution

Step 1: Let the original cost price be \( CP \).

Selling price at 30% profit:

\[ SP = CP \times 1.3 \]

Step 2: New CP = \( CP \times 0.9 \), New SP = \( CP \times 1.3 - 100 \), New profit = 25%.

\[ SP_{\text{new}} = (CP \times 0.9) \times 1.25 \] \[ CP \times 1.3 - 100 = (CP \times 0.9) \times 1.25 \] \[ 1.3CP - 100 = 1.125CP \]

Step 3: Solve:

\[ 1.3CP - 1.125CP = 100 \] \[ 0.175CP = 100 \] \[ CP = \frac{100}{0.175} \approx 571.43 \]

Answer: The cost price is approximately \( \text{Rs. } 571.43 \).

Question 15 (ICSE, RS Aggarwal)

A shopkeeper buys 100 pens for \( \text{Rs. } 500 \) and sells 80 pens at a profit of \( 20\% \). The remaining pens are sold at a loss of \( 10\% \). Find the overall profit or loss percentage.

Solution

Step 1: CP of 100 pens = 500, so CP per pen = \( \frac{500}{100} = 5 \).

Step 2: For 80 pens, Profit = 20%:

\[ SP \text{ per pen} = 5 \times 1.2 = 6 \] \[ \text{Total SP for 80 pens} = 80 \times 6 = 480 \]

Step 3: For 20 pens, Loss = 10%:

\[ SP \text{ per pen} = 5 \times 0.9 = 4.5 \] \[ \text{Total SP for 20 pens} = 20 \times 4.5 = 90 \]

Step 4: Total SP = \( 480 + 90 = 570 \), Total CP = 500.

\[ \text{Profit} = 570 - 500 = 70 \] \[ \text{Profit \%} = \left( \frac{70}{500} \times 100 \right) = 14\% \]

Answer: The overall profit percentage is \( 14\% \).

Question 16 (RD Sharma, Class 10)

A dealer sells an article at a loss of \( 5\% \). Had he sold it for \( \text{Rs. } 75 \) more, he would have gained \( 10\% \). Find the cost price.

Solution

Step 1: Let the cost price be \( CP \).

Selling price at 5% loss:

\[ SP_1 = CP \times 0.95 \]

Step 2: Selling price at 10% profit:

\[ SP_2 = CP \times 1.1 \]

Step 3: Given \( SP_2 = SP_1 + 75 \):

\[ 1.1CP = 0.95CP + 75 \] \[ 1.1CP - 0.95CP = 75 \] \[ 0.15CP = 75 \] \[ CP = \frac{75}{0.15} = 500 \]

Answer: The cost price is \( \text{Rs. } 500 \).

Question 17 (SSC CGL)

A shopkeeper offers a discount of \( 15\% \) on the marked price. If the marked price is \( \text{Rs. } 2300 \) and the cost price is \( \text{Rs. } 1700 \), find the profit percentage.

Solution

Step 1: Calculate selling price: \( MP = 2300 \), Discount = 15%.

\[ SP = 2300 \times \left(1 - \frac{15}{100}\right) = 2300 \times 0.85 = 1955 \]

Step 2: Profit: \( CP = 1700 \).

\[ \text{Profit} = 1955 - 1700 = 255 \] \[ \text{Profit \%} = \left( \frac{255}{1700} \times 100 \right) \approx 15\% \]

Answer: The profit percentage is \( 15\% \).

Question 18 (RS Aggarwal, Class 10)

A man buys an article for \( \text{Rs. } 800 \) and sells it at a profit of \( 25\% \). He then buys another article with the same amount and sells it at a loss of \( 20\% \). Find the overall profit or loss percentage.

Solution

Step 1: First article: \( CP_1 = 800 \), Profit = 25%.

\[ SP_1 = 800 \times 1.25 = 1000 \]

Step 2: Second article: \( CP_2 = 1000 \), Loss = 20%.

\[ SP_2 = 1000 \times 0.8 = 800 \]

Step 3: Total CP = \( 800 + 1000 = 1800 \), Total SP = \( 1000 + 800 = 1800 \).

Step 4: Since Total CP = Total SP, no profit or loss.

Answer: The overall profit or loss percentage is \( 0\% \).

Question 19 (NTPC)

A shopkeeper marks his goods at \( 60\% \) above cost price and offers a discount of \( 25\% \). If the selling price is \( \text{Rs. } 4500 \), find the cost price.

Solution

Step 1: Let the cost price be \( CP \).

Marked price: \( MP = CP \times 1.6 \).

Step 2: Selling price after 25% discount:

\[ SP = MP \times 0.75 = (CP \times 1.6) \times 0.75 = 1.2CP \]

Step 3: Given \( SP = 4500 \):

\[ 1.2CP = 4500 \] \[ CP = \frac{4500}{1.2} = 3750 \]

Answer: The cost price is \( \text{Rs. } 3750 \).

Question 20 (ICSE, RS Aggarwal)

A dealer sells an article at a profit of \( 15\% \). If the cost price and selling price are both increased by \( \text{Rs. } 200 \), the profit percentage becomes \( 10\% \). Find the original cost price.

Solution

Step 1: Let the original cost price be \( CP \).

Selling price: \( SP = CP \times 1.15 \).

Step 2: New CP = \( CP + 200 \), New SP = \( CP \times 1.15 + 200 \), New profit = 10%.

\[ \frac{(CP \times 1.15 + 200) - (CP + 200)}{CP + 200} \times 100 = 10 \] \[ \frac{1.15CP + 200 - CP - 200}{CP + 200} = 0.1 \] \[ \frac{0.15CP}{CP + 200} = 0.1 \]

Step 3: Solve:

\[ 0.15CP = 0.1(CP + 200) \] \[ 0.15CP = 0.1CP + 20 \] \[ 0.05CP = 20 \] \[ CP = \frac{20}{0.05} = 400 \]

Answer: The original cost price is \( \text{Rs. } 400 \).

Question 21 (RD Sharma, Class 10)

A shopkeeper sells two articles for \( \text{Rs. } 1500 \) each, one at a profit of \( 25\% \) and the other at a loss of \( 25\% \). Find the overall profit or loss percentage.

Solution

Step 1: First article: \( SP_1 = 1500 \), Profit = 25%.

\[ CP_1 = \frac{SP_1}{1.25} = \frac{1500}{1.25} = 1200 \]

Step 2: Second article: \( SP_2 = 1500 \), Loss = 25%.

\[ CP_2 = \frac{SP_2}{0.75} = \frac{1500}{0.75} = 2000 \]

Step 3: Total CP = \( 1200 + 2000 = 3200 \), Total SP = \( 1500 + 1500 = 3000 \).

Step 4: Loss:

\[ \text{Loss} = 3200 - 3000 = 200 \] \[ \text{Loss \%} = \left( \frac{200}{3200} \times 100 \right) = 6.25\% \]

Answer: The overall loss percentage is \( 6.25\% \).

Question 22 (SSC CGL)

A dealer buys an article for \( \text{Rs. } 2400 \) and marks it up by \( 50\% \). He offers two successive discounts of \( 10\% \) and \( 15\% \). Find the profit percentage.

Solution

Step 1: \( CP = 2400 \), Marked price:

\[ MP = 2400 \times 1.5 = 3600 \]

Step 2: Effective discount for 10% and 15%:

\[ \text{Effective Discount \%} = 10 + 15 - \frac{10 \times 15}{100} = 25 - 1.5 = 23.5\% \] \[ SP = 3600 \times \left(1 - \frac{23.5}{100}\right) = 3600 \times 0.765 = 2754 \]

Step 3: Profit:

\[ \text{Profit} = 2754 - 2400 = 354 \] \[ \text{Profit \%} = \left( \frac{354}{2400} \times 100 \right) \approx 14.75\% \]

Answer: The profit percentage is approximately \( 14.75\% \).

Question 23 (RS Aggarwal, Class 8)

A dishonest dealer sells goods at a loss of \( 20\% \) but uses a weight of 750 g instead of 1 kg. Find the actual profit or loss percentage.

Solution

True weight = 1000 g, False weight = 750 g.

Step 1: Sells at 20% loss:

\[ SP = CP \times 0.8 \]

Step 2: Profit due to false weight:

Let CP for 1000 g = 100. He sells 750 g at \( 100 \times 0.8 = 80 \).

Actual CP for 750 g = \( \frac{750}{1000} \times 100 = 75 \).

\[ \text{Profit} = 80 - 75 = 5 \] \[ \text{Profit \%} = \left( \frac{5}{75} \times 100 \right) \approx 6.67\% \]

Answer: The actual profit percentage is approximately \( 6.67\% \).

Question 24 (NTPC)

A man buys an article for \( \text{Rs. } 5000 \) and sells it at a profit of \( 20\% \). He then buys another article with the same amount and sells it at a loss of \( 10\% \). Find the overall profit or loss percentage.

Solution

Step 1: First article: \( CP_1 = 5000 \), Profit = 20%.

\[ SP_1 = 5000 \times 1.2 = 6000 \]

Step 2: Second article: \( CP_2 = 6000 \), Loss = 10%.

\[ SP_2 = 6000 \times 0.9 = 5400 \]

Step 3: Total CP = \( 5000 + 6000 = 11000 \), Total SP = \( 6000 + 5400 = 11400 \).

Step 4: Profit:

\[ \text{Profit} = 11400 - 11000 = 400 \] \[ \text{Profit \%} = \left( \frac{400}{11000} \times 100 \right) \approx 3.64\% \]

Answer: The overall profit percentage is approximately \( 3.64\% \).

Question 25 (ICSE, RS Aggarwal)

A shopkeeper buys 50 articles for \( \text{Rs. } 10000 \) and sells them at a profit of \( 25\% \). If 10 articles are damaged and sold at a loss of \( 50\% \), find the overall profit or loss percentage.

Solution

Step 1: CP of 50 articles = 10000, CP per article = \( \frac{10000}{50} = 200 \).

Step 2: For 40 articles, Profit = 25%:

\[ SP \text{ per article} = 200 \times 1.25 = 250 \] \[ \text{Total SP for 40 articles} = 40 \times 250 = 10000 \]

Step 3: For 10 damaged articles, Loss = 50%:

\[ SP \text{ per article} = 200 \times 0.5 = 100 \] \[ \text{Total SP for 10 articles} = 10 \times 100 = 1000 \]

Step 4: Total SP = \( 10000 + 1000 = 11000 \), Total CP = 10000.

\[ \text{Profit} = 11000 - 10000 = 1000 \] \[ \text{Profit \%} = \left( \frac{1000}{10000} \times 100 \right) = 10\% \]

Answer: The overall profit percentage is \( 10\% \).

Question 26 (SSC CGL)

A shopkeeper marks his goods at \( 50\% \) above cost price and offers a discount of \( 30\% \). If he still makes a profit of \( 5\% \), find the cost price.

Solution

Step 1: Let the cost price be \( CP \).

Marked price: \( MP = CP \times 1.5 \).

Step 2: Selling price after 30% discount:

\[ SP = MP \times 0.7 = (CP \times 1.5) \times 0.7 = 1.05CP \]

Step 3: Given profit = 5%:

\[ SP = CP \times 1.05 \]

Since \( SP = 1.05CP \), the equation holds true for any CP, confirming consistency.

Step 4: The problem is an identity, so CP can be any positive value. For example, let \( CP = 100 \):

\[ MP = 100 \times 1.5 = 150 \] \[ SP = 150 \times 0.7 = 105 \] \[ \text{Profit \%} = \left( \frac{105 - 100}{100} \times 100 \right) = 5\% \]

Answer: The cost price can be any positive value (e.g., \( \text{Rs. } 100 \)).

Conclusion

The Profit and Loss chapter is crucial for mastering quantitative aptitude in competitive exams like SSC CGL, NTPC, and ICSE boards. By understanding the theory and practicing these 26 hard-level questions, you can build confidence and improve your problem-solving skills.