Frank Brothers Solutions for ICSE Class 9 Mathematics Chapter 2 Profit Loss And Discount

Exercise 2.1

Question 1.
Answer:
C.P of the watch = Rs. 1750
S.P of the watch = Rs. 1610
Loss = C.P. - S.P
= Rs. (1750 - 1610) = Rs. 140
Loss% = \( \frac{\text{Loss}}{\text{C.P.}} \times 100 \)
= \( \frac{140}{1750} \times 100 = 8\% \)
In simple words: When an item is sold for less than what it cost to buy, you have a loss. In this case, the watch lost 8% of its value during the sale.

📝 Teacher's Note: Help students understand that Loss% is a way of seeing how much money was lost for every Rs. 100 spent. Always emphasize that the base for this calculation is the Cost Price.

🎯 Exam Tip: Ensure you clearly show the subtraction step for Loss and the division step for Percentage to get full marks for the procedure.

Question 2.
Answer:
C.P of the camera = Rs. 4600
Profit = 15%
\( \frac{\text{S.P.}}{\text{C.P.}} = 1 + \frac{\text{Profit}}{100} \)

\( \implies \frac{\text{S.P.}}{4600} = 1 + \frac{15}{100} \)

\( \implies \frac{\text{S.P.}}{4600} = \frac{100 + 15}{100} \)

\( \implies \text{S.P.} = \frac{115}{100} \times 4600 = \text{Rs. } 5290 \)
In simple words: To make a 15% profit, you need to sell the camera for its original price plus an extra 15% of that price. This brings the total selling price to Rs. 5290.

📝 Teacher's Note: You can also explain this by finding 15% of 4600 first (which is 690) and then adding it to the C.P. (4600 + 690). Both methods are mathematically identical.

🎯 Exam Tip: Using the formula method shown in the answer is generally faster and less prone to errors in multi-step problems.

Question 3.
Answer:
C.P of the watch = Rs. 4050
Loss = 14%
\( \frac{\text{S.P.}}{\text{C.P.}} = 1 - \frac{\text{Loss}}{100} \)

\( \implies \frac{\text{S.P.}}{4050} = 1 - \frac{14}{100} \)

\( \implies \frac{\text{S.P.}}{4050} = \frac{100 - 14}{100} \)

\( \implies \text{S.P.} = \frac{86}{100} \times 4050 = \text{Rs. } 3483 \)
In simple words: A 14% loss means the selling price is only 86% of the original cost. Multiplying the cost by 0.86 gives us the final sale price.

📝 Teacher's Note: Contrast this with Question 2. In profit we add to 1, and in loss we subtract from 1. This helps students remember the sign change.

🎯 Exam Tip: Always double-check the subtraction \( 100 - 14 \). A small calculation error here will lead to the wrong final answer.

Question 4.
Answer:
C.P. of the car = Rs. 75000
Amount spent on repairing = Rs. 15000
\( \therefore \) Total C.P. = Rs. 75000 + Rs. 15000 = Rs. 90000
S.P. of the car = Rs. 114000
\( \therefore \) Gain = S.P - C.P.
= Rs. (114000 - 90000)
= Rs. 24000
Gain% = \( \frac{\text{Gain}}{\text{C.P.}} \times 100 \)
= \( \frac{24000}{90000} \times 100 = 26.6\% \)
In simple words: When you buy a used car and fix it up, the repair costs must be added to what you paid to find your true "Total Cost." Here, the total cost was Rs. 90,000, and selling it for more led to a 26.6% profit.

📝 Teacher's Note: This is a crucial concept. "Overhead expenses" like repairs, transport, or taxes are always added to the initial C.P. to find the effective C.P. before calculating profit or loss.

🎯 Exam Tip: Do not calculate profit based on the initial Rs. 75,000 purchase price. Using the wrong C.P. is a very common trap in this chapter.

Question 5.
Answer:
C.P. of the furniture set = Rs. 21000
Amount spent on transportation = Rs. 500
Amount spent on repairing = Rs. 4500
\( \therefore \) Total C.P = Rs. 21000 + Rs. 500 + Rs. 4500 = Rs. 26000
Profit% = 20%
Now,
\( \frac{\text{S.P.}}{\text{C.P.}} = 1 + \frac{\text{Profit}}{100} \)

\( \implies \frac{\text{S.P.}}{26000} = 1 + \frac{20}{100} \)

\( \implies \frac{\text{S.P.}}{26000} = \frac{100 + 20}{100} \)

\( \implies \text{S.P.} = \frac{120}{100} \times 26000 = \text{Rs. } 31200 \)
\( \therefore \) He must sell the furniture set at Rs. 31200 to make a profit of 20%.
In simple words: After buying the furniture and paying for moving and fixing it, the owner had spent Rs. 26,000. To earn 20% more than that total cost, the final price needs to be Rs. 31,200.

📝 Teacher's Note: Reinforce the idea of Total C.P. by asking students if they would count the cost of fuel if they went to pick up the furniture themselves.

🎯 Exam Tip: When multiple expenses are given, list them clearly and sum them up as "Total C.P." before proceeding with any other formulas.

Question 6.
Answer:
One score = 20 notebooks
C.P of 20 notebooks = Rs. 240
C.P. of 1 notebook = Rs. \( 240 / 20 = \text{Rs. } 12 \)
\( \therefore \) C.P. of 1000 notebooks = Rs. \( 12 \times 1000 = \text{Rs. } 12000 \)
S.P of 1 notebook = Rs. 15
\( \therefore \) S.P of 1000 notebooks = Rs. \( 15 \times 1000 = \text{Rs. } 15000 \)
\( \because \text{ S.P. } > \text{ C.P.} \)
Profit = S.P. - C.P. = Rs. (15000 - 12000) = Rs. 3000
Profit% = \( \frac{\text{Profit}}{\text{C.P.}} \times 100 \)
= \( \frac{3000}{12000} \times 100 = 25\% \)
In simple words: By finding the cost of just one notebook (Rs. 12) and comparing it to its selling price (Rs. 15), we see a profit of Rs. 3 per book. For 1000 books, that's a total profit of 25%.

📝 Teacher's Note: Teach students that they can calculate Profit% using just the single unit price (Rs. 3 profit on Rs. 12 cost) to get the same 25% result faster. It avoids large numbers like 15000.

🎯 Exam Tip: Know your units! A "score" is 20, a "dozen" is 12. These terms are frequently used to test general mathematical literacy.

Question 7.
Answer:
Cost of one box = Rs. 180
\( \therefore \) C.P of 25 boxes = Rs. \( 180 \times 25 = \text{Rs. } 4500 \)
One box contains = 12 bars
\( \therefore \) 25 boxes contain = \( 12 \times 25 = 300 \text{ bars} \)
\( \therefore \) S.P of 25 boxes = Rs. \( 18 \times 300 = \text{Rs. } 5400 \)
\( \because \text{ S.P. } > \text{ C.P.} \)
Profit = S.P. - C.P. = Rs. (5400 - 4500) = Rs. 900
Profit% = \( \frac{\text{Profit}}{\text{C.P.}} \times 100 \)
= \( \frac{900}{4500} \times 100 = 20\% \)
In simple words: The seller bought 25 boxes of chocolate bars and sold each bar individually. By calculating the total money spent and total money earned, we find a 20% gain.

📝 Teacher's Note: This problem requires keeping track of two different "quantities": boxes and individual bars. Suggest students convert everything to "individual bars" to stay organized.

🎯 Exam Tip: Be careful when calculating the total number of items (bars) and ensure you use that number for the Selling Price, not the number of boxes.

Question 8.
Answer:
S.P of 1 kg of coffee = Rs. 135
Loss% = 10%
Now,
\( \frac{\text{S.P.}}{\text{C.P.}} = 1 - \frac{\text{Loss}}{100} \)

\( \implies \frac{135}{\text{C.P.}} = 1 - \frac{10}{100} \)

\( \implies \frac{135}{\text{C.P.}} = \frac{100 - 10}{100} \)

\( \implies \text{C.P.} = \frac{100}{90} \times 135 = \text{Rs. } 150 \)
\( \therefore \) C.P. of 1 kg of coffee = Rs. 150
\( \therefore \) Loss per 1kg of coffee = C.P. - S.P
= Rs. 150 - Rs. 135 = Rs. 15
Total loss incurred = Rs. 180
\( \therefore \) Amount of coffee sold = \( \frac{\text{Total loss}}{\text{loss per 1 kg of coffee}} \)
= \( \frac{180}{15} = 12 \text{ kg} \).
In simple words: The seller loses Rs. 15 on every kilogram sold. Since their total loss was Rs. 180, we divide that by 15 to see they must have sold 12 kilograms.

📝 Teacher's Note: This is a reverse-logic problem. First find the unit cost and unit loss, then use the total loss to find the quantity. It's a great exercise in multi-step problem-solving.

🎯 Exam Tip: First step must be finding the C.P. per unit. You cannot find the quantity sold without knowing how much was lost on a single unit.

Question 9.
Answer:
Cost of 8 locks = Rs. 520
\( \therefore \) C.P. of 1 lock = Rs. \( \frac{520}{8} = \text{Rs. } 65 \)
Selling price of 12 locks = Rs. 936
\( \therefore \) S.P. of 1 lock = Rs. \( \frac{936}{12} = \text{Rs. } 78 \)
\( \because \text{ S.P. } > \text{ C.P.} \)
Profit = S.P. - C.P. = Rs. (78 - 65) = Rs. 13
Profit% = \( \frac{\text{Profit}}{\text{C.P.}} \times 100 \)
= \( \frac{13}{65} \times 100 = 20\% \)
Net profit = Rs. 520
\( \therefore \) Number of locks sold = \( \frac{\text{Total profit}}{\text{profit per lock}} \)
= \( \frac{520}{13} = 40 \)
In simple words: The shopkeeper earns Rs. 13 on each lock. To reach a total profit of Rs. 520, they need to sell 40 locks.

📝 Teacher's Note: This problem uses different quantities (8 and 12) for buying and selling to confuse students. Finding the price of 1 unit is the most reliable way to solve it.

🎯 Exam Tip: Keep the Unit Profit (Rs. 13) and Unit Cost (Rs. 65) distinct. To find quantity, use the formula: \( \text{Total Profit} / \text{Profit per unit} \).

Question 10.
Answer:
\( \text{SP} = \left( \frac{100 + \text{Profit\%}}{100} \right) \times \text{CP} \)

\( \therefore \text{SP} = \left( \frac{100 + 20}{100} \right) \times \text{Rs. } 3000 = \text{Rs. } 3600 \)
\( \therefore \text{profit} = \text{SP} - \text{CP} = \text{Rs. } 600 \)
This profit includes tax = Rs. 360
\( \therefore \) net profit = Rs. 600 - Rs. 360 = Rs. 240
Profit% = \( \frac{\text{Profit}}{\text{CP}} \times 100 = \frac{240}{3000} \times 100 = 8 \)
So, the net profit is Rs. 240 and the profit percentage is 8.
In simple words: Even though the total gain was Rs. 600, some of that money had to be paid as tax (Rs. 360). The "Net Profit" is the money actually kept in the pocket, which is Rs. 240, or 8% of the cost.

📝 Teacher's Note: Distinguish between Gross Profit and Net Profit. Net Profit is what remains after all other deductions (like taxes) are taken out of the gross profit.

🎯 Exam Tip: Read carefully to see if the question asks for "Profit%" or "Net Profit%". Deduct taxes *only* after calculating the total profit if they are taken from the gain.

Question 11.
Answer:
CP of 800 straw at the rate of 50 paise per straw
= Rs. \( \left( \frac{50}{100} \times 800 \right) = \text{Rs. } 400 \)
Since profit is 50% of his outlay when only 640 articles are sold,
\( \therefore \) SP of 640 straws = \( \left( 1 + \frac{50}{100} \right) \) of Rs. 400 = \( \left( \frac{150}{100} \right) \times \text{Rs. } 400 = \text{Rs. } 600 \)
\( \therefore \) SP of each article = Rs. \( \frac{600}{640} = \text{Rs. } \frac{15}{16} \)
\( \therefore \) SP of 720 straws = Rs. \( \left( 720 \times \frac{15}{16} \right) = \text{Rs. } 675 \)
\( \therefore \) Actual profit = Rs. 675 - Rs. 400 = Rs. 275
\( \therefore \) Actual profit% = \( \left( \frac{\text{Profit}}{\text{CP}} \times 100 \right)\% = \left( \frac{275}{400} \times 100 \right)\% = 68.75\% \)
In simple words: The seller bought 800 straws but reached his target profit much earlier after selling only 640. After selling more of them (720 in total), the final profit rose significantly to 68.75%.

📝 Teacher's Note: This problem introduces the concept of "outlay" which just means the total Cost Price. It's a slightly more advanced problem because the S.P. is determined by a specific profit goal at a partial sales volume.

🎯 Exam Tip: Treat the "SP of 640 straws" calculation as a sub-problem to find the unit S.P. before applying it to the final 720 count.

Question 12.
Answer:
SP of the first mobile = Rs. 15000, profit = 25%
\( \therefore \text{Rs. } 15000 = \left( 1 + \frac{25}{100} \right) \text{ of CP} = \frac{5}{4} \text{ of CP} \)

\( \implies \text{CP} = \text{Rs. } \left( 15000 \times \frac{4}{5} \right) = \text{Rs. } 12000 \)
SP of the second mobile = Rs. 9945, profit = \( 10 \frac{1}{2}\% = \frac{21}{2}\% \)
Rs. 9945 = \( \left( 1 + \frac{\frac{21}{2}}{100} \right) \text{ of CP} = \frac{221}{200} \text{ of CP} \)

\( \implies \text{CP} = \text{Rs. } \left( 9945 \times \frac{200}{221} \right) = \text{Rs. } 9000 \)
Let the CP of the third article be Rs. x.
\( \therefore \) CP of all the three articles = Rs. 12000 + Rs. 9000 + Rs. x = Rs. (21000 + x)
\( \therefore \) SP of all the three articles = Rs. 15000 + Rs. 9945 + Rs. 5392 = Rs. 30337
As the loss incurred on the whole transaction = \( 8 \frac{1}{3}\% = \frac{25}{3}\% \)
So, Rs. 30337 = \( \left( 1 - \frac{\frac{25}{3}}{100} \right) \text{ of Rs. } (21000 + x) \)

\( \implies 30337 = \left( 1 - \frac{1}{12} \right) \times (21000 + x) \)

\( \implies 30337 = \left( \frac{11}{12} \right) \times (21000 + x) \)

\( \implies \frac{364044}{11} = 21000 + x \)

\( \implies \text{Rs. } 33094.90 \approx 21000 + x \)

\( \implies x = \text{Rs. } 12095 \text{ approximately} \)
In simple words: We find the individual costs of the first two phones using their selling prices and profits. Then, we use the total loss on all three items combined to work backward and find the missing cost of the third item.

📝 Teacher's Note: This is a comprehensive problem. It tests the ability to work backwards from S.P. to C.P. multiple times and then handle a combined transaction. Remind students to convert mixed fractions to improper fractions immediately.

🎯 Exam Tip: Maintain precision during division (\( 364044 / 11 \)). Carry the decimals until the final step and then round off if necessary.

Question 13.
Answer:
Let CP of the car at Kolkata be Rs. x.
As the car is available at 12% less price at Chennai,
CP of the car at Chennai = \( \left( 1 - \frac{12}{100} \right) \text{ of Rs. x} = \text{Rs. } \frac{22}{25}x \)
Since he incurs Rs. 9000 as overhead expenses,
total CP of the car = Rs. \( \left( \frac{22}{25}x + 9000 \right) \)
By selling the car at Kolkata for Rs. x, he makes a profit of 10%
\( \therefore \text{Rs. x} = \left( 1 + \frac{10}{100} \right) \text{ of Rs. } \left( \frac{22}{25}x + 9000 \right) \)

\( \implies x = \frac{11}{10} \left( \frac{22}{25}x + 9000 \right) \)

\( \implies \frac{10}{11}x = \frac{22}{25}x + 9000 \)

\( \implies \frac{10}{11}x - \frac{22}{25}x = 9000 \)

\( \implies \frac{250x - 242x}{275} = 9000 \)

\( \implies \frac{8x}{275} = 9000 \implies x = \frac{275 \times 9000}{8} \implies x = \text{Rs. } 309375 \)
In simple words: The car is cheaper in another city. Even after paying for the trip (Rs. 9,000), the buyer still makes a 10% profit by selling it back in his home city. The original price of the car must be Rs. 3,09,375.

📝 Teacher's Note: This is a sophisticated algebraic problem. Define the unknown variable 'x' early and build the equation based on the "Total C.P. + Profit = S.P." logic.

🎯 Exam Tip: The trick here is that the Selling Price (S.P.) is the same as the original Cost Price in Kolkata (x). Identifying this equality is the key to setting up the equation correctly.

Exercise 2.2

Question 1.
Answer:
Let the cost price be Rs. 100
So, the profit will be Rs. \( \left( \frac{25}{100} \times 100 \right) = \text{Rs. } 25 \)

\( \implies \text{SP} = \text{CP} + \text{Profit} = \text{Rs. } (100 + 25) = \text{Rs. } 125 \)
When the profit is Rs. 25, the sale is Rs. 125
So, let x be the profit when the sale is Rs. 5000

\( \implies x = \frac{25 \times 5000}{125} = \text{Rs. } 1000 \)
Hence, the profit is Rs. 1000.
In simple words: If for every Rs. 100 cost, the sale price is Rs. 125, we use that ratio to find out how much profit exists in a total sale of Rs. 5000.

📝 Teacher's Note: Using "Rs. 100" as a starting point is a great mental tool for percentage problems. It converts percentages into tangible amounts.

🎯 Exam Tip: Alternatively, you can use the formula \( \text{Profit} = \text{S.P.} - \left( \frac{\text{S.P.}}{1 + \frac{r}{100}} \right) \), but the unitary method shown above is less abstract.

Question 2.
Answer:
Let the CP of 3 watches be Rs. x.
\( \therefore \) CP of 1 watch = Rs. \( \frac{x}{3} \)
\( \implies \text{CP of 10 watches} = \text{Rs. } \frac{10x}{3} \)
Loss on selling 10 watches = CP of 3 watches = Rs. x
SP of 10 watches is Rs. 1400
Loss incurred on selling 10 watches = CP of 3 watches = Rs. x
Since CP - SP = Loss

\( \implies \text{Rs. } \frac{10x}{3} - \text{Rs. } 1400 = \text{Rs. x} \)

\( \implies \frac{10x - 4200}{3} = x \)

\( \implies 10x - 4200 = 3x \)

\( \implies 7x = 4200 \)

\( \implies x = 600 \)
Hence, CP of a watch = Rs. \( \frac{x}{3} = \text{Rs. } \frac{600}{3} = \text{Rs. } 200 \).
In simple words: The seller lost the value of 3 watches while selling 10 of them. This means the money he got back (Rs. 1400) was equal to the cost of only 7 watches. So, each watch must cost Rs. 200.

📝 Teacher's Note: This is a logic-based problem. The "Loss" is given in terms of the cost of articles rather than a currency amount. Help students write the equation in terms of the number of items.

🎯 Exam Tip: Clearly define 'x' as the C.P. of the number of items specified in the loss statement to avoid mixing up the variables.

Question 3.
Answer:
CP of 5 toffees = Re. 1
SP of 5 toffees = \( \left( \frac{100 + \text{Profit\%}}{100} \right) \text{ of CP} \)
= \( \left( \frac{100 + 25}{100} \right) \times \text{Re. } 1 \)
= \( \text{Rs. } \frac{125}{100} \times 1 \)
= Rs. \( \frac{5}{4} \)
For Rs. \( \frac{5}{4} \), toffees sold = 5
For Re. 1, toffees sold = \( \left( 5 \times \frac{4}{5} \right) = 4 \)
Hence, 4 toffees were sold to gain 25%.
In simple words: If you buy 5 toffees for a rupee and want to make a 25% profit while still only charging a rupee, you have to give the customer fewer toffees—just 4 instead of 5.

📝 Teacher's Note: This is a "Value per Rupee" problem. It's an inverse relationship: to increase profit at the same price, the quantity given must decrease.

🎯 Exam Tip: Keep the calculation in fractions (\( 5/4 \)) rather than decimals (1.25) to make the final unitary flip easier.

Question 4.
Answer:
S.P of a tie = Rs. 648
Gain = 8%
\( \frac{\text{S.P.}}{\text{C.P.}} = 1 + \frac{\text{Profit}}{100} \)

\( \implies \frac{648}{\text{C.P.}} = 1 + \frac{8}{100} \)

\( \implies \frac{648}{\text{C.P.}} = \frac{100 + 8}{100} \)

\( \implies \text{C.P.} = \frac{100}{108} \times 648 = \text{Rs. } 600 \)
Now, C.P. of the tie = Rs. 600
Gain = 10%
\( \therefore \) Gain = \( \frac{10}{100} \times \text{C.P.} \)
= \( \frac{10}{100} \times 600 = \text{Rs. } 60 \)
\( \therefore \) S.P. = Rs. (600 + 60) = Rs. 660
\( \therefore \) He must sell the tie at Rs. 660 to make a gain of 10%
In simple words: First we find that the tie originally cost Rs. 600. Then we calculate a new selling price by adding a 10% profit (Rs. 60) to that cost.

📝 Teacher's Note: This is a very common two-stage problem: first find the C.P., then find the new S.P. based on a different profit/loss requirement.

🎯 Exam Tip: Do not use the initial S.P. (648) to calculate the 10% gain. Profit percentages are always on the C.P. (600).

Question 5.
Answer:
S.P. of the cupboard = Rs. 6480
Loss = 10%
Now,
\( \frac{\text{S.P.}}{\text{C.P.}} = 1 - \frac{\text{Loss}}{100} \)

\( \implies \frac{6480}{\text{C.P.}} = 1 - \frac{10}{100} \)

\( \implies \frac{6480}{\text{C.P.}} = \frac{100 - 10}{100} \)

\( \implies \text{C.P.} = \frac{100}{90} \times 6480 = \text{Rs. } 7200 \)
Now, C.P. of the cupboard = Rs. 7200
S.P. of the cupboard = Rs. 7560
\( \because \text{ S.P. } > \text{ C.P.} \)
\( \therefore \text{ Gain} = \text{ S.P. } - \text{ C.P. } \)
= Rs. (7560 - 7200)
= Rs. 360
\( \therefore \text{ Gain\% } = \frac{\text{gain}}{\text{C.P.}} \times 100 \)
= \( \frac{360}{7200} \times 100 = 5\% \)
In simple words: The cupboard originally cost Rs. 7200. If it is sold for Rs. 7560 instead of the previous low price, the seller makes a 5% profit instead of a loss.

📝 Teacher's Note: This problem shows how knowing the Cost Price allows you to evaluate any future selling price for its profitability.

🎯 Exam Tip: Once you find the C.P. (7200), compare it to the *new* S.P. (7560). The gain is the simple difference between these two values.

Question 6.
Answer:
Let the S.P. of 4 pens = Rs. x
\( \therefore \) S.P. of 1 pen = Rs. \( \frac{x}{4} \)
C.P. of 5 pens will also be Rs. x
\( \therefore \) C.P. of 1 pen = Rs. \( \frac{x}{5} \)
As S.P. > C.P.
\( \therefore \text{ Profit} = \text{ S.P. } - \text{ C.P. } \)
= Rs. \( \left( \frac{x}{4} - \frac{x}{5} \right) = \text{Rs. } \left( \frac{5x - 4x}{20} \right) = \text{Rs. } \frac{x}{20} \)
Now, Profit% = \( \frac{\text{Profit}}{\text{C.P.}} \times 100 \)
= \( \frac{\frac{x}{20}}{\frac{x}{5}} \times 100 \)
= \( \frac{x}{20} \times \frac{5}{x} \times 100 \)
= 25%
In simple words: You bought 5 pens for the same amount of money you got back by selling only 4 of them. That means the 5th pen is essentially "free" profit, which works out to 25% gain.

📝 Teacher's Note: This is a classic "Items-based Profit" problem. The number of items purchased for a price is compared to the number of items sold for the same price.

🎯 Exam Tip: The algebraic 'x' will always cancel out in the percentage formula. Don't worry if the problem doesn't give you a dollar or rupee amount.

Question 7.
Answer:
Initial S.P. of a computer = Rs. 32200
Profit = 15%
\( \frac{\text{S.P.}}{\text{C.P.}} = 1 + \frac{\text{Profit}}{100} \)

\( \implies \frac{32200}{\text{C.P.}} = 1 + \frac{15}{100} \)

\( \implies \frac{32200}{\text{C.P.}} = \frac{100 + 15}{100} \)

\( \implies \text{C.P.} = \frac{100}{115} \times 32200 = \text{Rs. } 28000 \)
\( \therefore \) C.P. of the computer = Rs. 28000
If the S.P. of the computer is Rs. 29960,
S.P. > C.P
\( \therefore \) There would be a profit of = S.P. - C.P.
= Rs. (29960 - 28000) = Rs. 1960
In simple words: The computer was bought for Rs. 28,000. If we sell it for Rs. 29,960, we make a clear profit of Rs. 1,960.

📝 Teacher's Note: This mirrors the logic of Question 5 but asks for the monetary profit instead of the percentage.

🎯 Exam Tip: Calculate the base C.P. accurately; every subsequent part of the question depends on that number.

Question 8.
Answer:
For the first refrigerator,
S.P. = Rs. 37500
Profit = 25%
\( \frac{\text{S.P.}}{\text{C.P.}} = 1 + \frac{\text{Profit}}{100} \)

\( \implies \frac{37500}{\text{C.P.}} = 1 + \frac{25}{100} \)

\( \implies \frac{37500}{\text{C.P.}} = \frac{100 + 25}{100} \)

\( \implies \text{C.P.} = \frac{100}{125} \times 37500 = \text{Rs. } 30000 \)
For the second refrigerator,
S.P. = Rs. 37500
Loss = 25%
\( \frac{\text{S.P.}}{\text{C.P.}} = 1 - \frac{\text{Loss}}{100} \)

\( \implies \frac{37500}{\text{C.P.}} = 1 - \frac{25}{100} \)

\( \implies \text{C.P.} = \frac{100}{75} \times 37500 = \text{Rs. } 50000 \)
Total C.P. of both the refrigerators = Rs. 30000 + Rs. 50000 = Rs. 80000
Total S.P. of both the refrigerators = Rs. 37500 \( \times \) 2 = Rs. 75000
Since C.P. > S.P., so there is a loss
Loss = C.P. - S.P. = Rs. (80000 - 75000) = Rs. 5000
Loss % = \( \frac{\text{Loss}}{\text{C.P.}} \times 100 \)
= \( \frac{5000}{80000} \times 100 = 6.25\% \)
In simple words: When you sell two things for the same price, but one at a profit and one at a loss, you don't break even. In this case, the total loss was 6.25%.

📝 Teacher's Note: This is a very important "Competitive Exam" style problem. When profit and loss percentages are the same (\( x\% \)) on the same S.P., there is always a loss equal to \( (x^2 / 100)\% \). For \( 25\% \), that's \( 625 / 100 = 6.25\% \).

🎯 Exam Tip: You must calculate total C.P. and total S.P. separately before finding the overall profit or loss percentage for the whole transaction.

Question 9.
Answer:
Let the C.P. of briefcase be Rs. 100
Profit = 10%
\( \frac{\text{S.P.}}{\text{C.P.}} = 1 + \frac{\text{Profit}}{100} \)

\( \implies \frac{\text{S.P.}}{100} = 1 + \frac{10}{100} \)

\( \implies \text{S.P.} = \frac{110}{100} \times 100 = \text{Rs. } 110 \)
When buying at 5% less,
C.P. of the briefcase = Rs. 100 - 5% of Rs. 100 = Rs. (100 - 5) = Rs. 95
Gain % = 20%
Gain = \( \frac{20}{100} \times \text{Rs. } 95 = \text{Rs. } 19 \)
\( \therefore \) S.P. of the briefcase = Rs. 95 + Rs. 19 = Rs. 114
\( \therefore \) Difference between the two S.P.’s = Rs. 114 - Rs. 110 = Rs. 4
When the difference in S.P. is Rs. 4, the C.P of the briefcase is Rs. 100
\( \therefore \) When the difference in S.P. is Rs. 120, the C.P of the
briefcase is = \( \text{Rs. } \left( \frac{100 \times 120}{4} \right) = \text{Rs. } 3000 \)
In simple words: This problem explores a "what-if" scenario. If the briefcase were bought cheaper and sold for more profit, the difference in the final price helps us find its real cost, which is Rs. 3000.

📝 Teacher's Note: This problem uses the "Assumed Value" method. By starting with Rs. 100, we find a proportional difference (Rs. 4) which we then scale up to the real difference (Rs. 120).

🎯 Exam Tip: Be very careful tracking the two different S.P. values (110 and 114) generated by the different purchase prices and profit goals.

Question 10.
Answer:
S.P of the shirt = Rs. 1265 + Rs. 55 = Rs. 1320
Gain = 20%
\( \frac{\text{S.P.}}{\text{C.P.}} = 1 + \frac{\text{Profit}}{100} \)

\( \implies \frac{1320}{\text{C.P.}} = 1 + \frac{20}{100} \)

\( \implies \frac{1320}{\text{C.P.}} = \frac{120}{100} \)

\( \implies \text{C.P.} = \frac{1320 \times 100}{120} = \text{Rs. } 1100 \)
In simple words: The shirt was bought for Rs. 1100. Selling it for a total of Rs. 1320 gave the seller a 20% gain.

📝 Teacher's Note: A simple check of the formula \( \text{C.P. } = \frac{\text{S.P. } \times 100}{100 + \text{Profit\%}} \).

🎯 Exam Tip: Make sure you add any extra costs (like Rs. 55) to the sales figure if they are part of the total S.P.

Question 11.
Answer:
C.P. of 200 kg sugar = Rs. \( (18 \times 200) = \text{Rs. } 3600 \)
C.P. of 100 kg sugar = Rs. \( (22 \times 100) = \text{Rs. } 2200 \)
\( \therefore \) C.P. of 300 kg sugar = Rs. (3600 + 2200) = Rs. 5800
S.P. of 300 kg sugar = Rs. \( (20 \times 300) = \text{Rs. } 6000 \)
As S.P > C.P, so there is a profit
\( \therefore \) Profit = S.P. - C.P.
= Rs. (6000 - 5800) = Rs. 200
Profit % = \( \frac{\text{Profit}}{\text{C.P.}} \times 100 \)
= \( \frac{200}{5800} \times 100 = 3.44\% \)
In simple words: The seller mixed two batches of sugar bought at different prices. By selling the whole mixture at Rs. 20 per kg, they made a small profit of 3.44%.

📝 Teacher's Note: This is a mixture problem. Students must first find the "Average C.P." of the whole mixture before they can determine profit.

🎯 Exam Tip: Calculate the Total C.P. by summing the individual batch costs. Do not average the rates (18 and 22) unless the quantities are equal.

Question 12.
Answer:
S.P. of 12 glasses = Rs. 600
S.P. of 1 glass = Rs. \( \frac{600}{12} = \text{Rs. } 50 \)
Profit = 25%
\( \frac{\text{S.P.}}{\text{C.P.}} = 1 + \frac{\text{Profit}}{100} \)

\( \implies \frac{600}{\text{C.P.}} = 1 + \frac{25}{100} \)

\( \implies \text{C.P.} = \frac{600 \times 100}{125} = \text{Rs. } 480 \)
\( \therefore \) C.P. of 12 glasses = Rs. 480
\( \therefore \) C.P. of 1 glass = Rs. \( \frac{480}{12} = \text{Rs. } 40 \)
C.P. of 15 such glasses = Rs. \( 40 \times 15 = \text{Rs. } 600 \)
S.P. of 15 glasses = Rs. 540
\( \because \text{ C.P. } > \text{ S.P.} \)
There is a loss of C.P. - S.P = Rs. (600 - 540) = Rs. 60
Loss % = \( \frac{\text{Loss}}{\text{C.P.}} \times 100 \)
= \( \frac{60}{600} \times 100 = 10\% \)
In simple words: We find out that one glass costs the seller Rs. 40. Later, they sold 15 glasses for Rs. 540. Since the cost for those 15 glasses was Rs. 600, they ended up with a 10% loss.

📝 Teacher's Note: This problem requires switching between two different sets of numbers. Suggest students find the "C.P. per unit" as their anchor point to solve both halves of the problem.

🎯 Exam Tip: When the number of items changes (from 12 to 15), the Unit C.P. is the only thing that stays constant. Always find it first.

Question 13.
Answer:
Let C.P. of an article be Rs. 100
Profit = 8%
\( \therefore \text{ S.P. } = \text{Rs. } 100 + 8\% \text{ of Rs. } 100 = \text{Rs. } 100 + \text{Rs. } 8 = \text{Rs. } 108 \)
Again, Profit = 12%
\( \therefore \text{ S.P. } = \text{Rs. } 100 + 12\% \text{ of Rs. } 100 = \text{Rs. } 100 + \text{Rs. } 12 = \text{Rs. } 112 \)
Difference between the two S.P. s = Rs. 112 - Rs. 108 = Rs. 4
When difference is Rs. 4, then C.P = Rs. 100
When difference is Rs. 72, then C.P = \( \frac{100 \times 72}{4} = \text{Rs. } 1800 \)
\( \therefore \) The cost price of the article is Rs. 1800
First S.P
= Rs. 1800 + 8% of Rs. 1800
= \( \text{Rs. } 1800 + \frac{8}{100} \times 1800 = \text{Rs. } 1800 + \text{Rs. } 144 = \text{Rs. } 1944 \)
Second S.P
= Rs. 1800 + 12% of Rs. 1800
= \( \text{Rs. } 1800 + \frac{12}{100} \times 1800 = \text{Rs. } 1800 + \text{Rs. } 216 = \text{Rs. } 2016 \)
In simple words: Changing the profit from 8% to 12% adds Rs. 72 to the price. By using algebra, we find the item cost Rs. 1800. We then calculate what those two specific sale prices were.

📝 Teacher's Note: This is a "Differential Profit" problem. The difference in percentages (12% - 8% = 4%) is directly proportional to the difference in S.P. (Rs. 72).

🎯 Exam Tip: A faster way: \( 4\% \text{ of CP} = \text{Rs. } 72 \). This simple equation gets you the C.P. in just one step.

Question 14.
Answer:
S.P of retailer = Rs. 12474
Profit = 5%
\( \frac{\text{S.P.}}{\text{C.P.}} = 1 + \frac{\text{Profit}}{100} \)

\( \implies \frac{12474}{\text{C.P.}} = 1 + \frac{5}{100} \)

\( \implies \text{C.P.} = \frac{100}{105} \times 12474 = \text{Rs. } 11880 \)
S.P. of dealer = Rs. 11880
Profit = 8%
\( \frac{\text{S.P.}}{\text{C.P.}} = 1 + \frac{\text{Profit}}{100} \)

\( \implies \frac{11880}{\text{C.P.}} = 1 + \frac{8}{100} \)

\( \implies \text{C.P.} = \frac{100}{108} \times 11880 = \text{Rs. } 11000 \)
S.P of manufacturer = Rs. 11000
Profit = 10%
\( \frac{\text{S.P.}}{\text{C.P.}} = 1 + \frac{\text{Profit}}{100} \)

\( \implies \frac{11000}{\text{C.P.}} = 1 + \frac{10}{100} \)

\( \implies \text{C.P.} = \frac{100}{110} \times 11000 = \text{Rs. } 10000 \)
In simple words: This is a chain of sales from factory to dealer to shop to customer. Each person adds a bit of profit. By starting at the final customer's price and working backward, we find the factory cost was Rs. 10,000.

📝 Teacher's Note: This is called a "Chain Transaction." Emphasize that the C.P. for one person becomes the S.P. for the previous person in the chain.

🎯 Exam Tip: Work carefully one stage at a time. The result of the first calculation becomes the starting number for the second.

Question 15.
Answer:
CP of the painting for Akhil = Rs. 50000
Profit = 15%
\( \therefore \text{ Profit} = 15\% \text{ of Rs. } 50000 \)
= \( \frac{15}{100} \times 50000 = \text{Rs. } 7500 \)
SP = CP + Profit
= Rs. (50000 + 7500)
= Rs. 57500
CP of the painting for B = Rs. 57500
Loss = 15%
\( \therefore \text{ Loss} = 15\% \text{ of Rs. } 57500 \)
= \( \frac{15}{100} \times 57500 = \text{Rs. } 8625 \)
SP = CP - Loss
= Rs. (57500 - 8625)
= Rs. 48875
Total gain made by Akhil = Rs. [7500 + (50000 - 48875)]
= Rs. 8625
Gain% in the second transaction = \( \frac{\text{Gain}}{\text{CP}} \times 100 \)
= \( \frac{8625}{50000} \times 100 = 17.25\% \)
In simple words: Akhil sold a painting for a profit, then bought it back from the same person for a lower price. Because he kept the original profit AND paid less to get the painting back, his total gain was Rs. 8625.

📝 Teacher's Note: This is a double transaction problem. The gain for Akhil is the sum of the profit from the first sale plus the savings (Initial CP - Final buy-back price).

🎯 Exam Tip: Be careful with the "Total Gain" calculation. You must account for both the cash earned in the first sale and the cash saved in the second purchase.

Question 16.
Answer:
S.P of the T.V = Rs 15730
Profit = 30%
\( \frac{\text{S.P.}}{\text{C.P.}} = 1 + \frac{\text{Profit}}{100} \)

\( \implies \frac{15730}{\text{C.P.}} = 1 + \frac{30}{100} \)

\( \implies \text{C.P.} = \frac{100}{130} \times 15730 = \text{Rs. } 12100 \)
C.P. of the T.V = Rs 12100
Increase in C.P = 30%
New C.P = Rs 12100 + 30% of Rs 12100
= Rs 12100 + Rs 3630 = Rs 15730
S.P. of the T.V = Rs 15730
Increase in S.P = 20%
New S.P = Rs 15730 + 20% of Rs 15730
= Rs 15730 + Rs 3146 = Rs 18876
Profit = S.P - C.P
= Rs. (18876 - 15730) = Rs 3146
Profit% = \( \frac{\text{Profit}}{\text{CP}} \times 100 = \frac{3146}{15730} \times 100 = 20\% \)
In simple words: The manufacturer's costs went up, so they raised their prices too. After all the changes, they are now making a 20% profit on each TV.

📝 Teacher's Note: This problem involves "Updated Values." We first find the base CP, then apply the market changes to both CP and SP to find the new profitability.

🎯 Exam Tip: Note that the new C.P. (15730) happens to be exactly the same as the old S.P. in this specific problem. Use this to your advantage to save calculation time.

Question 17.
Answer:
Let the cost price of one of the cycles be Rs x
\( \therefore \) The cost price of the other cycle = Rs (8000 - x)
For the first cycle,
C.P = Rs x
Loss = 20%
\( \therefore \) Loss = 20% of Rs x = Rs 0.20 x
For the second cycle,
C.P = Rs (8000 - x)
Profit = 30%
\( \therefore \) Profit = 30% of Rs (8000 - x) = Rs 0.3 (8000 - x) = Rs (2400 - 0.3x)
Given, overall profit = Rs 650
\( \therefore \) (2400 - 0.3x) - 0.2x = 650

\( \implies 0.5x = 2400 - 650 = 1750 \)

\( \implies x = 1750 / 0.5 = 3500 \)
\( \therefore \) C.P of 1st cycle = Rs 3500
C.P of 2nd cycle = Rs 8000 - Rs 3500 = Rs 4500
In simple words: A man bought two cycles for Rs. 8000 total. By selling one at a loss and one at a profit, he ended up with Rs. 650 extra. We use algebra to find that the cycles cost Rs. 3500 and Rs. 4500.

📝 Teacher's Note: This is a "Simultaneous Equation" type problem. The total profit is the difference: Profit from second item - Loss from first item.

🎯 Exam Tip: Check your final answer by calculating the profit/loss on both results: \( 30\% \text{ of } 4500 (1350) - 20\% \text{ of } 3500 (700) = 650 \). It takes 10 seconds and guarantees your marks.

Question 18.
Answer:
C.P. of both the transistors = Rs 7200
Let C.P of the 1st transistor be Rs x
\( \therefore \) C.P of the 2nd transistor is Rs (7200 - x)
For the 1st transistor,
Loss = 15%
\( \therefore \) S.P = C.P - Loss
= Rs x - 15% of Rs x = Rs 0.85x
For the 2nd transistor,
Profit = 19%
\( \therefore \) S.P = C.P + Profit
= Rs (7200 - x) + 19% of Rs (7200 - x)
= Rs (8568 - 1.19x)
Given, both the S.P's are equal

\( \implies 0.85x = 8568 - 1.19x \)

\( \implies 1.19x + 0.85x = 8568 \)

\( \implies 2.04x = 8568 \)

\( \implies x = \frac{8568}{2.04} = 4200 \)
\( \therefore \) C.P of 1st transistor = Rs 4200
C.P of 2nd transistor = Rs 7200 - Rs 4200 = Rs 3000
In simple words: Two radios were sold for the exact same price. Even though one was a loss and one was a profit, the equal final price helps us find they originally cost Rs. 4200 and Rs. 3000.

📝 Teacher's Note: The condition "Selling Prices are equal" is the key. Set up the S.P. expressions for both items and set them equal to each other.

🎯 Exam Tip: Be careful with decimals like 0.85 and 1.19. It's often easier to work with percentages directly: \( 85\% \text{ of } x = 119\% \text{ of } (7200 - x) \).

Question 19.
Answer:
Let the S.P of both the cycles be Rs. x each.
For the first cycle,
Profit = 20%
\( \frac{\text{S.P.}}{\text{C.P.}} = 1 + \frac{\text{Profit}}{100} \)

\( \implies \frac{x}{\text{C.P.}} = 1 + \frac{20}{100} \)

\( \implies \frac{x}{\text{C.P.}} = \frac{100 + 20}{100} \)

\( \implies \text{CP} = \frac{100}{120}x = \frac{5}{6}x \)
Profit = S.P - C.P = \( x - \frac{5}{6}x = \frac{(6-5)}{6}x = \frac{x}{6} \)
For the second cycle,
Loss = 20%
\( \frac{\text{S.P.}}{\text{C.P.}} = 1 - \frac{\text{Loss}}{100} \)

\( \implies \frac{x}{\text{C.P.}} = 1 - \frac{20}{100} \)

\( \implies \frac{x}{\text{C.P.}} = \frac{100 - 20}{100} \)

\( \implies \text{CP} = \frac{100}{80}x = \frac{5}{4}x \)
Loss = C.P - S.P = \( \frac{5}{4}x - x = \frac{(5-4)}{4}x = \frac{x}{4} \)
Given, total loss = Rs 180

\( \implies \frac{x}{4} - \frac{x}{6} = 180 \)

\( \implies \frac{2x}{24} = 180 \)

\( \implies x = 12 \times 180 = 2160 \)
\( \therefore \) C.P. of first bicycle = \( \frac{5}{6} \times \text{Rs. } 2160 = \text{Rs. } 1800 \)
\( \therefore \) C.P. of second bicycle = \( \frac{5}{4} \times \text{Rs. } 2160 = \text{Rs. } 2700 \)
In simple words: This seller lost more money on the second bike than he made on the first. The total difference (the loss) was Rs. 180. We find the bikes cost him Rs. 1800 and Rs. 2700.

📝 Teacher's Note: This is a sophisticated problem. The "total loss" is the Net Loss across the whole transaction. Equation: Loss on bike 2 - Profit on bike 1 = 180.

🎯 Exam Tip: Keep the variable 'x' for Selling Price. Solving for x first allows you to find both C.P. values easily at the end.

Question 20.
Answer:
S.P of 12 pens = Rs 72
\( \therefore \) S.P of 1 pen = Rs \( \frac{72}{12} = \text{Rs. } 6 \)
Gain % = 20%
\( \frac{\text{S.P.}}{\text{C.P.}} = 1 + \frac{\text{Profit}}{100} \)

\( \implies \frac{6}{\text{C.P.}} = 1 + \frac{20}{100} \)

\( \implies \frac{6}{\text{C.P.}} = \frac{100 + 20}{100} \)

\( \implies \text{C.P.} = \frac{100}{120} \times 6 = \text{Rs. } 5 \)
Given, S.P = Rs 100
Gain = 25%
\( \frac{\text{S.P.}}{\text{C.P.}} = 1 + \frac{\text{Profit}}{100} \)

\( \implies \frac{100}{\text{C.P.}} = 1 + \frac{25}{100} \)

\( \implies \text{C.P.} = \frac{100}{125} \times 100 = \text{Rs. } 80 \)
\( \therefore \) Number of pens sold = \( \frac{\text{Rs. } 80}{\text{Rs. } 5} = 16 \)
In simple words: First we find that each pen cost Rs. 5 to buy. To earn a 25% profit on a total sale of Rs. 100, we calculate that the seller must have sold 16 pens.

📝 Teacher's Note: This problem asks for a change in quantity. The "anchor" is the unit C.P. (Rs. 5). Find it first and use it for the second scenario.

🎯 Exam Tip: Note that "SP = Rs 100" in the second part refers to a new sales goal, not a per-pen price. This is a common point of confusion.

Question 21.
Answer:
Let the quantity of milk purchased be x litres.
\( \therefore \) C.P of 1 litre = Rs 14
\( \therefore \) C.P of x litre = Rs 14x
Quantity of water mixed = 40% of x = \( \frac{40}{100}x = \frac{2}{5}x \text{ litres} \)
\( \therefore \) Quantity of milk now becomes = \( x + \frac{2}{5}x = \frac{7}{5}x \text{ litres} \)
S.P of 1 litre mixture = Rs 16
\( \therefore \) S.P of \( \frac{7}{5}x \) litres mixture = \( 16 \times \frac{7}{5}x = \text{Rs. } \frac{112}{5}x \)
Profit = S.P - C.P
= \( \frac{112}{5}x - 14x \)
= \( \frac{112x - 70x}{5} = \frac{42}{5}x \)
\( \therefore \) Profit % = \( \frac{\text{Profit}}{\text{CP}} \times 100 \)
= \( \frac{42x/5}{14x} \times 100 = 60\% \)
In simple words: A milkman added 40% water to his milk (which costs him nothing) and then sold the mixture for more than the original milk price. This "double trick" gave him a huge 60% profit.

📝 Teacher's Note: This is a "Dishonest Milkman" problem. Since water is free, the S.P. applies to the *total* volume (Milk + Water), while the C.P. only applies to the original milk.

🎯 Exam Tip: Ensure your "Total Quantity" used for S.P. includes the added water. Using just 'x' will lead to an incorrect profit result.

Question 22.
Answer:
Let A buy the cycle for Rs x.
For A, C.P of the cycle = Rs x + Rs 110
Profit = 20%
\( \frac{\text{S.P.}}{\text{C.P.}} = 1 + \frac{\text{Profit}}{100} \)

\( \implies \frac{\text{S.P.}}{x + 110} = 1 + \frac{20}{100} \)

\( \implies \text{S.P.} = \frac{120}{100}(x + 110) = \text{Rs. } \frac{12}{10}(x + 110) \)
For B, C.P of the cycle = Rs. \( \frac{12}{10}(x + 110) \)
Loss = 10%
\( \frac{\text{S.P.}}{\text{C.P.}} = 1 - \frac{\text{Loss}}{100} \)

\( \implies \frac{\text{S.P.}}{\frac{12}{10}(x + 110)} = 1 - \frac{10}{100} \)

\( \implies \text{S.P.} = \frac{90}{100} \times \frac{12}{10}(x + 110) = \text{Rs. } \frac{9}{10} \times \frac{12}{10}(x + 110) \)
For C, C.P of the cycle = Rs. \( \frac{9}{10} \times \frac{12}{10}(x + 110) \)
Profit = 10%
S.P of the cycle =
\( \text{Rs } \left( 1 + \frac{10}{100} \right) \frac{9}{10} \times \frac{12}{10}(x + 110) = \text{Rs } \frac{11}{10} \times \frac{9}{10} \times \frac{12}{10}(x + 110) \)
ATQ, \( \frac{11}{10} \times \frac{9}{10} \times \frac{12}{10}(x + 110) = 1188 \)

\( \implies x + 110 = 1000 \)

\( \implies x = 1000 - 110 = 890 \)
\( \therefore \) A paid Rs 890 for the cycle.
In simple words: A cycle passes through three people, with each one either making money or losing it. By setting up a long chain equation, we find that the first person, A, originally bought the cycle for Rs. 890.

📝 Teacher's Note: This is a "Successive Transaction" problem. The formula is: Final Price = Initial Cost \( \times (1 \pm p_1/100) \times (1 \pm p_2/100) \dots \) This is much faster than solving stage-by-stage.

🎯 Exam Tip: Don't forget to add the initial overhead (Rs. 110) to the purchase price 'x' at the very beginning. Overheads are part of the C.P.

Question 23.
Answer:
Cost of 40 articles = Rs 54400
\( \therefore \) Cost of each article = Rs \( \frac{54400}{40} = \text{Rs. } 1360 \)
Cost of finishing of one article = Rs 140
\( \therefore \) C.P of finished article = Rs 1360 + Rs 140 = Rs 1500
\( \therefore \) C.P of 40 finished articles = Rs \( 1500 \times 40 = \text{Rs. } 60000 \)
S.P of one-fourth articles = \( \frac{1}{4} \times 40 \times \text{Rs. } 2100 \)
S.P of rest of articles = \( \frac{3}{4} \times 40 \times \text{Rs. } 1800 \)
\( \therefore \) Total S.P = \( \frac{1}{4} \times 40 \times \text{Rs. } 2100 + \frac{3}{4} \times 40 \times \text{Rs. } 1800 \)
= Rs 21000 + Rs 54000 = Rs 75000
Profit = S.P - C.P
= Rs 75000 - Rs 60000 = Rs 15000
\( \therefore \text{ Profit \% } = \frac{\text{Profit}}{\text{CP}} \times 100 \)
= \( \frac{15000}{60000} \times 100 = 25\% \)
In simple words: The merchant bought 40 things and spent money to make them look nice. He sold some at a high price and the rest at a lower price. Overall, he made a 25% profit on all the money he spent.

📝 Teacher's Note: This is an overhead + segmented sales problem. Ensure students calculate the *weighted* S.P. for the whole lot before finding profit.

🎯 Exam Tip: Clearly show the calculation for "rest of the articles" (\( 1 - 1/4 = 3/4 \)). Breaking the total sales into parts prevents logic errors.

Question 24.
Answer:
Let the C.P. of the briefcase is Rs x
Profit = 15%
\( \therefore \) S.P = C.P + Profit
= Rs x + 15% of Rs x = Rs 1.15 x
In the later case, C.P = Rs x - 5% of Rs x = Rs 0.95x
S.P = Rs (1.15x - 35)
\( \therefore \text{ Gain } = \text{ S.P - C.P } \)
= Rs (1.15x - 35) - Rs 0.95x = Rs (0.2x - 35)
Gain % = 20%

\( \implies \frac{\text{gain}}{\text{C.P.}} \times 100 = 20 \)

\( \implies \left( \frac{0.2x - 35}{0.95x} \right) \times 100 = 20 \)

\( \implies \frac{0.2x - 35}{0.95x} = 0.20 \)

\( \implies 0.2x - 35 = 0.19x \)

\( \implies 0.01x = 35 \)

\( \implies x = 3500 \)
\( \therefore \) C.P of the briefcase is Rs 3500
In simple words: Similar to Question 9, we use an algebraic 'x' to figure out how much a bag costs based on two different sale scenarios. The bag costs Rs. 3500.

📝 Teacher's Note: This problem is a variation of the "Hypothetical Scenario" type. It requires careful setting up of the final gain percentage equation using the modified CP and modified SP.

🎯 Exam Tip: Note that "Rs 35 less" is subtracted from the *first S.P.* (1.15x), not the cost price.

Question 25.
Answer:
Let the number of eggs bought at 4 for Rs 5 be x.
\( \therefore \) The number of eggs bought at 9 for Rs 10 are x
\( \therefore \) Total number of eggs bought = x + x = 2x
When eggs are bought at 4 for Rs 5, C.P. of each egg = Rs \( \frac{5}{4} \)
C.P. of x eggs = Rs \( \frac{5}{4}x \)
When eggs are bought at 9 for Rs 10, C.P. of each egg = Rs \( \frac{10}{9} \)
C.P. of x eggs = Rs \( \frac{10}{9}x \)
\( \therefore \text{ Total C.P } = \text{Rs } \frac{5}{4}x + \text{Rs } \frac{10}{9}x = \frac{85}{36}x \)
Number of eggs broken = 15% of 2x
= \( \frac{15}{100} \times 2x = \frac{3x}{10} \)
Eggs left = \( 2x - \frac{3x}{10} = \frac{17x}{10} \)
When eggs are sold at 2 for Rs 3, S.P of each egg = Rs \( \frac{3}{2} \)
S.P of \( \frac{17}{10}x \) eggs = Rs \( \frac{3}{2} \times \frac{17}{10}x = \text{Rs } \frac{51}{20}x \)
\( \therefore \) Gain = S.P - C.P
= \( \frac{51}{20}x - \frac{85}{36}x = \left( \frac{459 - 425}{180} \right)x \)
= \( \frac{34}{180}x = \text{Rs } \frac{17}{90}x \)
\( \therefore \text{ Gain\% } = \frac{\text{gain}}{\text{C.P.}} \times 100 \)
= \( \frac{\frac{17}{90}x}{\frac{85}{36}x} \times 100 = \frac{17}{90} \times \frac{36}{85} \times 100 = 8\% \)
Also, gain = Rs 510

\( \implies \frac{17}{90}x = 510 \)

\( \implies x = \frac{90}{17} \times 510 = 2700 \)
\( \therefore \) Number of eggs of each kind bought = 2700.
In simple words: A dealer bought eggs from two different sources and some of them broke. Even though he had fewer eggs to sell, he still made an 8% profit. In total, he bought 2700 eggs of each type.

📝 Teacher's Note: This is a highly complex multi-stage problem involving two cost rates, breakage loss, and a single sales rate. Suggest students solve for the "Unit C.P." and "Unit S.P." per egg to simplify the fractions.

🎯 Exam Tip: Be very careful with the "Total Quantity" (2x). The breakage percentage applies to the *total* number of eggs bought, not just one batch.

Question 26.
Answer:
Let the C.P of the shirt be Rs x
When sold at a profit of 10%, S.P = \( \left( 1 + \frac{10}{100} \right)x = \text{Rs. } 1.1x \)
When sold at a profit of 15%, S.P = \( \left( 1 + \frac{15}{100} \right)x = \text{Rs. } 1.15x \)
Difference between the two S.P.’s = Rs (1.15x - 1.10x) = Rs 0.05 x
ATQ, \( 0.05x = 80 \)

\( \implies x = \frac{80}{0.05} = 1600 \)
\( \therefore \) C.P of each shirt = Rs 1600
\( \therefore \) S.P of 1st shirt = Rs \( 1.1 \times 1600 = \text{Rs } 1760 \)
\( \therefore \) S.P of 2nd shirt = Rs \( 1.15 \times 1600 = \text{Rs } 1840 \)
In simple words: Selling the shirt for 5% more profit adds Rs. 80 to the price. This tells us the shirt originally cost Rs. 1600.

📝 Teacher's Note: This is another example of a "Differential Profit" problem. The step from 10% to 15% is a 5% increase in total cost.

🎯 Exam Tip: To save time, you can write the equation directly as: \( 5\% \text{ of } x = 80 \).

Question 27.
Answer:
SP of 4 identical kites = Rs. 12
SP = \( \left( \frac{100 + \text{Profit\%}}{100} \right) \times \text{CP} \)

\( \implies 12 = \left( \frac{100 + 20}{100} \right) \times \text{CP} \)

\( \implies 1200 = (100 + 20) \times \text{CP} \)

\( \implies \frac{1200}{120} = \text{CP} \)

\( \implies \text{CP} = \text{Rs. } 10 \)
So, the CP of 1 kite = Rs. \( \frac{10}{4} = \text{Rs. } \frac{5}{2} \)
Now, SP of 6 kites = Rs. 24
So, SP of 1 kite = Rs. \( \frac{24}{6} = \text{Rs. } 4 \)
Profit = SP - CP = Rs. \( 4 - \text{Rs. } \frac{5}{2} = \text{Rs. } \frac{3}{2} \)
Profit% = \( \frac{\text{Profit}}{\text{CP}} \times 100 = \frac{3/2}{5/2} \times 100 = \frac{3}{5} \times 100 = 60\% \)
Hence, his gain percent is 60%.
In simple words: The seller originally bought kites for Rs. 2.50 each. By selling them at a new rate of Rs. 4 each, his profit jumped up to 60%.

📝 Teacher's Note: This problem tests the ability to find a unit C.P. from one batch and apply it to another batch with different quantities. Finding the price of 1 item is always the safest path.

🎯 Exam Tip: Do not use the S.P. of Rs. 12 in your final profit% calculation. That S.P. was for a different quantity and a different profit level.

Question 28.
Answer:
SP of 80 bananas = Rs. 240
SP = \( \left( \frac{100 - \text{Loss\%}}{100} \right) \times \text{CP} \)

\( \implies 240 = \left( \frac{100 - 25}{100} \right) \times \text{CP} \)

\( \implies 240 = \frac{75}{100} \times \text{CP} \)

\( \implies \text{CP} = \frac{240 \times 100}{75} \)

\( \implies \text{CP} = \text{Rs. } 320 \)
So, CP of 80 bananas is Rs. 320
\( \therefore \) CP of 1 banana = Rs. \( \frac{320}{80} = \text{Rs. } 4 \)
Let the number of bananas for sold for Rs. 100 be x.
So, CP of x bananas = Rs. 4x
Now, \( \text{SP} = \left( \frac{100 + \text{Profit\%}}{100} \right) \times \text{CP} \)

\( \implies 100 = \left( \frac{100 + 25}{100} \right) \times 4x \)

\( \implies 100 = \frac{125}{100} \times 4x \)

\( \implies 100 = \frac{5}{4} \times 4x \)

\( \implies 100 = 5x \)

\( \implies x = 20 \)
Hence, he should sell 20 bananas for Rs. 100 to gain 25%.
In simple words: The seller was losing money, so he found that each banana costs him Rs. 4. To make a 25% profit while selling them for Rs. 100, he can only afford to give the customer 20 bananas.

📝 Teacher's Note: This is a "Quantity for fixed price" problem. It's an inverse problem: the higher the profit desired, the fewer items can be sold for that same fixed amount (Rs. 100).

🎯 Exam Tip: Once you find the unit C.P. (Rs. 4), the second part of the problem becomes a simple equation: \( 1.25 \times (\text{Cost of x bananas}) = 100 \).

Question 29.
Answer:
For the first washing machine :
SP = Rs. 8900 and profit = 20%
\( \therefore \text{ CP} = \left( \frac{100}{100 + \text{Profit\%}} \right) \times \text{SP} \)

\( \implies \text{CP} = \left( \frac{100}{100 + 20} \right) \times 8900 \)

\( \implies \text{CP} = \frac{100}{120} \times 8900 \)

\( \implies \text{CP} = \frac{89000}{12} \)

\( \therefore \text{ CP} = \text{Rs. } \frac{22250}{3} \)
Since in the whole transaction, there is no profit and no loss,
\( \therefore \) Loss on the second washing machine = Profit on the first washing machine
\( \therefore \text{ Profit on the first washing machine} = \text{SP} - \text{CP} \)
= \( 8900 - \frac{22250}{3} = \frac{26700 - 22250}{3} = \text{Rs. } \frac{4450}{3} \)
So, Loss on the second washing machine = Rs. \( \frac{4450}{3} \)
Loss = 15% and Loss = Rs. \( \frac{4450}{3} \), so, we have to find the CP
CP of the second washing machine = Rs. \( \frac{4450}{3} \times \frac{100}{15} \approx \text{Rs. } 9888.87 \)
Hence, CP of the second washing machine is Rs. 9888.87 approximately.
In simple words: The seller broke even on two machines. He made some money on the first and lost that exact same amount of cash on the second. Using that cash amount, we calculated the second machine's cost.

📝 Teacher's Note: "No profit, no loss" means Net Profit = 0. This implies that the numerical value of the Gain on one item must equal the numerical value of the Loss on the other.

🎯 Exam Tip: Do not set the percentages equal! 20% gain on one machine is NOT necessarily 20% loss on the other. The *actual rupee amounts* must be equal.

Question 30.
Answer:
CP of 60 kg of apples = Rs. (90 \( \times \) 60) = Rs. 5400
Gain percent on the whole = 25% of Rs. 5400
= \( \frac{25}{100} \times \text{Rs. } 5400 = \text{Rs. } 1350 \)
CP of 40 kg of apples = Rs. (90 \( \times \) 40) = Rs. 3600
Loss = 10% of 3600
= \( \frac{10}{100} \times \text{Rs. } 3600 = \text{Rs. } 360 \)
Quantity of apples left to be sold = 60 - 40 = 20kg
CP of 20 kg apples = Rs. (90 \( \times \) 20) = Rs. 1800
Profit to be made by selling 20 kg apples = Rs. 1350 + 360 = Rs. 1710
\( \therefore \) SP of 20 kg apples = CP + Profit = Rs. 1800 + Rs. 1710 = Rs. 3510
\( \therefore \text{ SP of 1 kg apples} = \text{Rs. } \frac{3510}{20} = \text{Rs. } 175.50 \)
Hence, he should sell the remaining apples at Rs. 175.50 per kg to gain 25% on the whole.
In simple words: The seller lost money on the first big batch of apples. To still make a huge profit on the whole lot, he has to sell the last few kilos at a very high price.

📝 Teacher's Note: This is a "Target Profit" problem. The target is the profit on the *entire* original outlay. The last batch must cover both the missing target profit and the previous loss.

🎯 Exam Tip: First calculate the "Total Required S.P." for the whole 60kg. Then subtract the S.P. already received from the 40kg. The remaining amount is what you must earn from the final 20kg.

Question 31.
Answer:
CP of the TV = Rs. 15000
Profit made on the TV = 20% of CP = \( \frac{20}{100} \times 15000 = \text{Rs. } 3000 \)
So, SP = CP + Profit = Rs. 15000 + Rs. 3000 = Rs. 18000
Since the SP includes tax = Rs. 1000 as tax,
So, the actual SP (including tax) = SP(without tax) - tax
= Rs. 18000 - Rs. 1000
= Rs. 17000
So, the net profit = SP (including tax) - CP of the TV
= Rs. 17000 - Rs. 15000
= Rs. 2000
Profit % = \( \frac{\text{Profit}}{\text{CP}} \times 100 = \frac{2000}{15000} \times 100 = 13 \frac{1}{3}\% \)
Hence, the net profit is Rs. 2000 and the profit percent is \( 13 \frac{1}{3} \).
In simple words: After making a 20% profit, the seller had to pay Rs. 1000 in tax. This left him with a final "Net Profit" of only Rs. 2000, which is about 13.33%.

📝 Teacher's Note: This is a variation of Question 10. The key is to correctly identify when to subtract the tax to find the "Net" gain.

🎯 Exam Tip: Note that the tax is subtracted from the *gain*, not the cost price. Always calculate the gross SP first.

Question 32.
Answer:
Let the common multiple be x.
So, 3x litres of oil A is mixed with 2x litres of oil B.
Total mixture = 3x + 2x = 5x litres
CP of oil A = Rs. 300 per litre
\( \therefore \) CP of 3x litres = Rs. (300 \( \times \) 3x) = Rs. 900x
CP of oil B = Rs. 400 per litre
\( \therefore \) CP of 2x litres = Rs. (400 \( \times \) 2x) = Rs. 800x
So, total CP of the entire mixture that is, 5x litres = Rs. 1700x
Now, one - fourth of the mixture is sold at Rs. 450 per litre
that is, \( \frac{1}{4} \text{ of } (5x) = \frac{5x}{4} \text{ litres is sold at Rs. 450 per litre} \)
So, SP of \( \frac{5x}{4} \text{ litres} = \frac{5x}{4} \times 450 = \text{Rs. } \frac{1125x}{2} \)
The remaining that is, \( \frac{3}{4} \text{ of } (5x) = \frac{3}{4} \times 5x = \frac{15x}{4} \text{ litres is sold at Rs. 500 per litre} \)
So, SP of \( \frac{15x}{4} \text{ litres} = \frac{15x}{4} \times 500 = \text{Rs. } 1875x \)
So, SP of the entire mixture = Rs. \( \frac{1125x}{2} + 1875x = \text{Rs. } \frac{4875x}{2} \)
Profit = SP - CP = Rs. \( \frac{4875x}{2} - 1700x = \text{Rs. } \frac{1475x}{2} \)
Profit% = \( \frac{\text{Profit}}{\text{CP}} \times 100 = \frac{1475x/2}{1700x} \times 100 = \frac{1475x}{3400x} \times 100 = \frac{1475}{3400} \times 100 = 43.38\% \text{ approx} \)
Hence, the profit percent on the whole is 43.38% approximately.
In simple words: The seller mixed two types of oil and sold the mixture in two parts at different prices. By combining all the costs and all the sales, we find they made about 43.4% profit.

📝 Teacher's Note: This is a complex mixture and segmented sales problem. The use of 'x' for quantity is helpful, as it eventually cancels out in the percentage calculation. This proves the profit% is independent of the total quantity mixed.

🎯 Exam Tip: Be careful summing the SP of the two parts. Use a common denominator (2) to add \( 1125x/2 \) and \( 1875x \) accurately.

Exercise 2.3

Question 1.
Answer:
a. MP = Rs. 850, Discount = 16%
\( \text{Discount\%} = \left( \frac{\text{discount}}{\text{MP}} \times 100 \right) \)

\( \implies 16 = \left( \frac{\text{discount}}{850} \times 100 \right) \)

\( \implies \frac{16 \times 850}{100} = \text{discount} \)

\( \implies \text{discount} = \text{Rs. } 136 \)
SP = MP - discount
= Rs. 850 - Rs. 136
= Rs. 714
Hence, the SP is Rs. 714.

b. MP = Rs. 5500, Discount = 30%
\( \text{Discount\%} = \left( \frac{\text{discount}}{\text{MP}} \times 100 \right) \)

\( \implies 30 = \left( \frac{\text{discount}}{5500} \times 100 \right) \)

\( \implies \frac{30 \times 5500}{100} = \text{discount} \)

\( \implies \text{discount} = \text{Rs. } 1650 \)
SP = MP - discount
= Rs. 5500 - Rs. 1650
= Rs. 3850
Hence, the SP is Rs. 3850.
In simple words: A discount is a reduction in the "Marked Price" (the price on the sticker). After taking off the discount amount, you get the final "Selling Price."

📝 Teacher's Note: This is the first question on Discount. Emphasize that Discount is always calculated on the Marked Price (MP), not the Cost Price (CP).

🎯 Exam Tip: The formula \( \text{S.P. } = \text{M.P. } \times (1 - d/100) \) is a faster way to solve these in one step.

Question 2.
Answer:
a. SP = Rs. 1892, Discount = 14%
\( \text{SP} = \left( 1 - \frac{d}{100} \right) \text{ of MP} \)

\( \implies 1892 = \left( 1 - \frac{14}{100} \right) \times \text{MP} \)

\( \implies 1892 = \frac{86}{100} \times \text{MP} \)

\( \implies \text{MP} = \frac{1892 \times 100}{86} \)

\( \implies \text{MP} = \text{Rs. } 2200 \)

b. SP = Rs. 1245, Discount = 17%
\( \text{SP} = \left( 1 - \frac{d}{100} \right) \text{ of MP} \)

\( \implies 1245 = \left( 1 - \frac{17}{100} \right) \times \text{MP} \)

\( \implies 1245 = \frac{83}{100} \times \text{MP} \)

\( \implies \text{MP} = \frac{1245 \times 100}{83} \)

\( \implies \text{MP} = \text{Rs. } 1500 \)
In simple words: If you know the final sale price and how much of a discount was given, you can work backward to find the original "sticker price."

📝 Teacher's Note: This is the reverse of Question 1. We are solving for the MP instead of the SP. The equation structure remains the same.

🎯 Exam Tip: Always verify that the MP you calculate is larger than the given SP. If it's smaller, you've made an error in the logic.

Question 3.
Answer:
a. MP = Rs. 1500, SP = Rs. 1320
Discount = MP - SP
\( \implies \text{Discount} = \text{Rs. } 1500 - \text{Rs. } 1320 \)
\( \implies \text{Discount} = \text{Rs. } 180 \)
\( \text{Discount percentage} = \left( \frac{\text{discount}}{\text{MP}} \times 100 \right)\% \)
= \( \left( \frac{180}{1500} \times 100 \right)\% \)
= 12%

b. MP = Rs. 6840, SP = Rs. 5814
Discount = MP - SP
\( \implies \text{Discount} = \text{Rs. } 6840 - \text{Rs. } 5814 \)
\( \implies \text{Discount} = \text{Rs. } 1026 \)
\( \text{Discount percentage} = \left( \frac{1026}{6840} \times 100 \right)\% \)
= 15%
In simple words: To find the discount percentage, first see how much cash was taken off. Then, see what part of the original sticker price that cash represents.

📝 Teacher's Note: This completes the trio of basic discount problems: finding SP, finding MP, and now finding Discount%.

🎯 Exam Tip: Be very careful not to divide by the SP. The denominator for discount percentage is always the Marked Price (MP).

Question 4.
Answer:
MP = Rs. 5400, discount = 12%
To find the amount paid by the customer, that is, the SP
\( \text{SP} = \left( 1 - \frac{d}{100} \right) \text{ of MP} \)

\( \implies \text{SP} = \left( 1 - \frac{12}{100} \right) \times 5400 \)

\( \implies \text{SP} = \frac{88}{100} \times 5400 \)

\( \implies \text{SP} = \text{Rs. } 4752 \)
Hence, the amount paid by the customer is Rs. 4752.
In simple words: A 12% discount on a Rs. 5400 item means the customer only has to pay Rs. 4752.

📝 Teacher's Note: This is a simple application of the direct SP formula.

🎯 Exam Tip: "Amount paid by customer" is always synonymous with Selling Price (SP) of the shopkeeper.

Question 5.
Answer:
MP = Rs. 150, discount = 8%
To find the amount paid by the customer, that is, the SP
\( \text{SP} = \left( 1 - \frac{d}{100} \right) \text{ of MP} \)

\( \implies \text{SP} = \left( 1 - \frac{8}{100} \right) \times 150 \)

\( \implies \text{SP} = \frac{92}{100} \times 150 \)

\( \implies \text{SP} = \text{Rs. } 138 \)
Hence, the amount paid by the customer is Rs. 138.
In simple words: The item was on sale for 8% off, so the final price dropped from Rs. 150 down to Rs. 138.

📝 Teacher's Note: A straightforward calculation. Encourage students to check if the result "makes sense" (it should be less than Rs. 150).

🎯 Exam Tip: In short questions, showing the fraction \( 92/100 \) is a good intermediate step.

Question 6.
Answer:
CP = Rs. 2400, discount = 10%, profit% = 12.5%
profit% = \( \frac{\text{profit}}{\text{CP}} \times 100 \)

\( \implies 12.5 = \frac{\text{profit}}{2400} \times 100 \)

\( \implies \text{profit} = \frac{12.5 \times 2400}{100} \)
\( \implies \text{profit} = \text{Rs. } 300 \)
SP = Rs. 2400 + Rs. 300
= Rs. 2700
\( \text{SP} = \left( 1 - \frac{d}{100} \right) \text{ of MP} \)

\( \implies 2700 = \left( 1 - \frac{10}{100} \right) \times \text{MP} \)

\( \implies 2700 = \frac{90}{100} \times \text{MP} \)

\( \implies \frac{2700 \times 100}{90} = \text{MP} \)
\( \implies \text{MP} = \text{Rs. } 3000 \)
Hence, the price he should mark the article at is Rs. 3000.
In simple words: The shopkeeper wants to make a profit but also wants to offer a discount to attract customers. To achieve both, they must set the sticker price higher—at Rs. 3000.

📝 Teacher's Note: This problem connects C.P., Profit, S.P., and M.P. Use a flowchart: \( \text{CP} \rightarrow \text{Profit} \rightarrow \text{SP} \leftarrow \text{Discount} \leftarrow \text{MP} \).

🎯 Exam Tip: The bridge between the two halves of this problem is the Selling Price (SP). Find SP from CP first, then use that SP to find MP.

Question 7.
Answer:
CP = Rs. 1750, discount = 20%, profit% = 20%
profit% = \( \frac{\text{profit}}{\text{CP}} \times 100 \)

\( \implies 20 = \frac{\text{profit}}{1750} \times 100 \)

\( \implies \text{profit} = \frac{20 \times 1750}{100} \)
\( \implies \text{profit} = \text{Rs. } 350 \)
SP = Rs. 1750 + Rs. 350
= Rs. 2100
\( \text{SP} = \left( 1 - \frac{d}{100} \right) \text{ of MP} \)

\( \implies 2100 = \left( 1 - \frac{20}{100} \right) \times \text{MP} \)

\( \implies 2100 = \frac{80}{100} \times \text{MP} \)

\( \implies \frac{2100 \times 100}{80} = \text{MP} \)
\( \implies \text{MP} = \text{Rs. } 2625 \)
Hence, the price he should mark the article at is Rs. 2625.
In simple words: Just like the previous problem, the shopkeeper balances a 20% discount with a 20% profit by marking up the original price to Rs. 2625.

📝 Teacher's Note: This is a classic business math problem. It helps students understand that "Sale" prices are usually calculated well in advance to ensure the shop still makes money.

🎯 Exam Tip: Don't assume that a 20% discount cancels out a 20% profit. You must calculate the C.P. to S.P. step and the S.P. to M.P. step separately.

Question 8.
Answer:
MP = Rs. 8000, discount = 15%
\( \text{SP} = \left( 1 - \frac{15}{100} \right) \times 8000 \)
\( \implies \text{SP} = \frac{85}{100} \times 8000 \)
\( \implies \text{SP} = \text{Rs. } 6800 \)
Let the cost price be Rs. x
Given that the MP = x + 25% above the CP
\( \implies 8000 = x + 25\% \text{ of CP} \)
\( \implies 8000 = x + \frac{25}{100} \times x \)
\( \implies 8000 = x + \frac{x}{4} \)
\( \implies 8000 = \frac{5x}{4} \)
\( \implies x = \frac{8000 \times 4}{5} \)
\( \implies x = \text{Rs. } 6400 \)
So, the CP is Rs. 6400.
\( \therefore \) The SP of the article is Rs. 6800 and the CP is Rs. 6400.
In simple words: The item was marked 25% above its cost, making it Rs. 8000. Even with a 15% discount, the seller still sold it for Rs. 6800, which is more than the original cost of Rs. 6400.

📝 Teacher's Note: This problem introduces the "Mark-up" concept (setting MP above CP). Help students see that Profit depends on the relationship between SP and CP, while Discount depends on MP and SP.

🎯 Exam Tip: Read the question carefully to identify what the 25% refers to. Here, it is 25% of the CP, which helps set up the equation for 'x'.

Question 9.
Answer:
CP = Rs. 4200, discount = 12.5%, profit% = 20%
profit% = \( \frac{\text{profit}}{\text{CP}} \times 100 \)

\( \implies 20 = \frac{\text{profit}}{4200} \times 100 \)

\( \implies \text{profit} = \text{Rs. } 840 \)
SP = Rs. 4200 + Rs. 840
= Rs. 5040
\( \text{SP} = \left( 1 - \frac{d}{100} \right) \text{ of MP} \)

\( \implies 5040 = \left( 1 - \frac{12.5}{100} \right) \times \text{MP} \)

\( \implies 5040 = \frac{87.5}{100} \times \text{MP} \)

\( \implies \frac{5040 \times 100}{87.5} = \text{MP} \)
\( \implies \text{MP} = \text{Rs. } 5760 \)
Hence, the price he should mark the article at is Rs. 5760.
In simple words: To make a solid 20% profit while offering a nice-looking 12.5% discount, the shopkeeper needs to set the sticker price at Rs. 5760.

📝 Teacher's Note: When dealing with decimal percentages like 12.5%, remind students that they can use fractions (\( 1/8 \)) to make the calculations easier (\( 1 - 1/8 = 7/8 \)).

🎯 Exam Tip: Always show the addition of profit to C.P. to get the S.P. before attempting to find the M.P.

Question 10.
Answer:
MP = Rs. 1200
a. \( \text{SP} = \left( 1 - \frac{d_1}{100} \right) \left( 1 - \frac{d_2}{100} \right) \text{ of MP} \)

\( \implies \text{SP} = \left( 1 - \frac{15}{100} \right) \left( 1 - \frac{10}{100} \right) \times 1200 \)

\( \implies \text{SP} = \frac{85}{100} \times \frac{90}{100} \times 1200 \)

\( \implies \text{SP} = \frac{85}{100} \times 1080 \)

\( \implies \text{SP} = \text{Rs. } 918 \)

b. \( \text{SP} = \left( 1 - \frac{d_1}{100} \right) \left( 1 - \frac{d_2}{100} \right) \left( 1 - \frac{d_3}{100} \right) \text{ of MP} \)

\( \implies \text{SP} = \left( 1 - \frac{10}{100} \right) \left( 1 - \frac{8}{100} \right) \left( 1 - \frac{5}{100} \right) \times 1200 \)

\( \implies \text{SP} = \frac{90}{100} \times \frac{92}{100} \times \frac{95}{100} \times 1200 \)

\( \implies \text{SP} \approx \text{Rs. } 944 \)
In simple words: "Successive discounts" mean applying one discount after another. You take the first percentage off the original price, then the next percentage off the new, lower price.

📝 Teacher's Note: This is a key concept. Successive discounts of 15% and 10% are NOT the same as a single 25% discount. They are always less than the sum of the percentages.

🎯 Exam Tip: Use the formula shown in part (a) to solve multi-stage discount problems quickly. Multiplying the percentages directly is the standard scientific method.

Question 11.
Answer:
MP = Rs. 4000, SP = Rs. 3060
a. \( \text{SP} = \left( 1 - \frac{d_1}{100} \right) \left( 1 - \frac{d_2}{100} \right) \text{ of MP} \)

\( \implies 3060 = \left( 1 - \frac{10}{100} \right) \left( 1 - \frac{d_2}{100} \right) \times 4000 \)

\( \implies 3060 = \frac{90}{100} \times \left( 1 - \frac{d_2}{100} \right) \times 4000 \)

\( \implies 3060 = \left( 1 - \frac{d_2}{100} \right) \times 3600 \)

\( \implies 1 - \frac{d_2}{100} = \frac{3060}{3600} \)

\( \implies 1 - \frac{d_2}{100} = 0.85 \)

\( \implies \frac{d_2}{100} = 1 - 0.85 \)

\( \implies \frac{d_2}{100} = 0.15 \)
\( \implies d_2 = 15\% \)
Hence, the second discount is 15%.
In simple words: We know the original price, the first discount, and the final price. We calculate backward to find out that the mystery second discount was 15%.

📝 Teacher's Note: Solving for a missing percentage in a chain requires good algebraic isolation skills. Remind students to simplify the "known" parts of the right side (the 90% of 4000) first.

🎯 Exam Tip: Once you reach \( 1 - d/100 = 0.85 \), you can immediately see the discount is \( 0.15 \) or \( 15\% \). This saves time on the final algebraic steps.

Question 12.
Answer:
Let the MP be Rs. x
a. First discount
\( \text{SP} = \left( 1 - \frac{d_1}{100} \right) \left( 1 - \frac{d_2}{100} \right) \left( 1 - \frac{d_3}{100} \right) \text{ of MP} \)
= \( \left( 1 - \frac{25}{100} \right) \left( 1 - \frac{20}{100} \right) \left( 1 - \frac{15}{100} \right) \times x \)
= \( \frac{75}{100} \times \frac{80}{100} \times \frac{85}{100} \times x \)
= \( \frac{75}{100} \times \frac{68}{100} \times x \)
= \( 0.510x \)
b. Second discount :
\( \text{SP} = \left( 1 - \frac{d_1}{100} \right) \left( 1 - \frac{d_2}{100} \right) \left( 1 - \frac{d_3}{100} \right) \text{ of MP} \)
= \( \left( 1 - \frac{20}{100} \right) \left( 1 - \frac{20}{100} \right) \left( 1 - \frac{20}{100} \right) \times x \)
= \( \frac{80}{100} \times \frac{80}{100} \times \frac{80}{100} \times x \)
= \( 0.512x \)
Clearly, since \( 0.512x > 0.510x \), so, the SP of the first is less than that of the second.
So, the first offer is better than the second offer.
In simple words: We compare two sets of discounts. The first set makes the final price \( 0.510 \) of the original, while the second set makes it \( 0.512 \). Since the first result is lower, it's the better deal for the buyer.

📝 Teacher's Note: This is a "Comparison of Schemes" problem. Even though the second scheme looks "fairer" (all 20s), the first scheme results in a lower final cost because of the larger 25% starting point.

🎯 Exam Tip: To compare schemes, always find the final percentage of the MP (\( 0.510 \) vs \( 0.512 \)). The lower the multiplier, the better the deal for the customer.

Question 13.
Answer:
Let the MP of the article be Rs. x and a single discount be d%
be equivalent to three given successive discounts of 20%, 10% and 5%.
Equating the two selling prices of the article we get,
\( \left( 1 - \frac{d}{100} \right) \text{ of Rs. x} = \left( 1 - \frac{20}{100} \right) \left( 1 - \frac{10}{100} \right) \left( 1 - \frac{5}{100} \right) \text{ of Rs. x} \)

\( \implies \left( 1 - \frac{d}{100} \right) \times x = \frac{80}{100} \times \frac{90}{100} \times \frac{95}{100} \times x \)

\( \implies 1 - \frac{d}{100} = \frac{684000}{1000000} \)

\( \implies 1 - \frac{d}{100} = \frac{684000}{1000000} = \frac{68.4}{100} \)

\( \implies \frac{316000}{1000000} = \frac{d}{100} \)

\( \implies d = 31.6\% \)
MP of the article = Rs. 2500
\( \text{SP} = \left( 1 - \frac{31.6}{100} \right) \times 2500 \)
\( \implies \text{SP} = \frac{68.4}{100} \times 2500 \)
\( \implies \text{SP} = \text{Rs. } 1710 \)
Hence, the equivalent discount is Rs. 31.6% and the SP is Rs. 1710.
In simple words: Giving three small discounts in a row (20, 10, and 5) is exactly the same as giving one single discount of 31.6%. We use this "single rule" to find the final price easily.

📝 Teacher's Note: This is the "Equivalent Discount" formula. It's used by businesses to simplify their pricing structures.

🎯 Exam Tip: Note that the sum of \( 20+10+5 \) is 35, but the equivalent discount is only 31.6%. Always remember the "successive discount" is less than the simple sum.

Question 14.
Answer:
List price = Rs. 4000
Case 1 :
\( \text{SP} = \left( 1 - \frac{d}{100} \right) \text{ of MP(List price)} \)
\( \implies \text{SP} = \left( 1 - \frac{25}{100} \right) \times 4000 \)
\( \implies \text{SP} = \frac{75}{100} \times 4000 \)
\( \implies \text{SP} = \text{Rs. } 3000 \)
Case 2 :
\( \text{SP} = \left( 1 - \frac{d_1}{100} \right) \left( 1 - \frac{d_2}{100} \right) \text{ of MP} \)
\( \implies \text{SP} = \left( 1 - \frac{15}{100} \right) \left( 1 - \frac{12}{100} \right) \times 4000 \)
\( \implies \text{SP} = \frac{85}{100} \times \frac{88}{100} \times 4000 \)
\( \implies \text{SP} = \text{Rs. } 2992 \)
Since in the second case the SP is lesser, so the second offer is better.
The amount paid in the second offer is Rs. 2992.
In simple words: The customer has two choices: a single 25% discount or two smaller ones (15% then 12%). After doing the math, the two-part discount saves the customer an extra Rs. 8.

📝 Teacher's Note: This reinforces the concept that multiple discounts apply to progressively smaller bases. It's a great example for demonstrating the "Value of Choice" in consumer math.

🎯 Exam Tip: Clearly state which case results in a "lesser S.P." to justify why it's the better offer for the buyer.

Question 15.
Answer:
MP of the sofa = Rs. 36000, discount at Guwahati = 20%
\( \text{SP} = \left( 1 - \frac{20}{100} \right) \times 36000 \)
\( \implies \text{SP} = \left( 1 - \frac{20}{100} \right) \times 36000 \)
\( \implies \text{SP} = \frac{80}{100} \times 36000 \)
\( \implies \text{SP} = \text{Rs. } 28800 \)
So, the SP at Guwahati is Rs. 28800.
So, total expenses
= SP + travelling expenses + transportation of the article
= Rs. 28800 + Rs. 1500 + Rs. 1200
= Rs. 31500
So, the CP at Delhi = Rs. 31500

a. SP at Delhi = marked price = Rs. 36000
So, Profit = SP - CP = Rs. 36000 - Rs. 31500 = Rs. 4500
Profit% = \( \frac{\text{Profit}}{\text{CP}} \times 100 = \frac{4500}{31500} \times 100 = 14 \frac{2}{7}\% \)

b. discount = 5% of 36000 = \( \frac{5}{100} \times 36000 = \text{Rs. } 1800 \)
\( \implies \text{SP} = 36000 - 1800 = \text{Rs. } 34200 \)
So, Profit = SP - CP = Rs. 34200 - Rs. 31500 = Rs. 2700
Profit% = \( \frac{\text{Profit}}{\text{CP}} \times 100 = \frac{2700}{31500} \times 100 = 8 \frac{4}{7}\% \)
In simple words: A dealer buys a sofa cheap in one city and brings it back to another. After paying for the truck and the travel, he still makes about 14.3% profit by selling it at the full original price.

📝 Teacher's Note: This problem combines overheads (travel/transport) with discount and profit. Remind students that the dealer's final CP is his purchase price PLUS all his costs to get the item home.

🎯 Exam Tip: For part (a) and (b), the CP (31500) remains the same. Only the SP changes based on whether he gives a discount or not.

Question 16.
Answer:
Let the cost price of each article bought = Rs. 100.
Let the number of articles bought = x
MP of the articles = Rs. 100 + 50% of Rs. 100
= \( \text{Rs. } 100 + \left( \frac{50}{100} \times 100 \right) \)
= Rs. 150
Number of articles sold at Rs. 150 = \( \frac{x}{2} \)
\( \therefore \text{ SP of } \frac{x}{2} \text{ articles} = \text{Rs. } \left( 150 \times \frac{x}{2} \right) = \text{Rs. } 75x \)
Discount = 20% on Rs. 150
= \( \frac{20}{100} \times 150 \)
= Rs. 30
\( \therefore \text{ SP} = \text{Rs. } 150 - \text{Rs. } 30 = \text{Rs. } 120 \)
Remaining number of articles sold at Rs. 120 = \( x - \frac{x}{2} - \frac{x}{4} = \frac{x}{4} \)
\( \therefore \text{ SP of } \frac{x}{4} \text{ articles} = \text{Rs. } \left( 120 \times \frac{x}{4} \right) = \text{Rs. } 30x \)
Discount = 36% on Rs. 150
= \( \frac{36}{100} \times 150 \)
= Rs. 54
\( \therefore \text{ SP} = \text{Rs. } 150 - \text{Rs. } 54 = \text{Rs. } 96 \)
Number of articles sold at Rs. 96 = \( \frac{x}{4} \)
\( \therefore \text{ SP of } \frac{x}{4} \text{ articles} = \text{Rs. } \left( 96 \times \frac{x}{4} \right) = \text{Rs. } 24x \)
Total SP of all articles = Rs. 75x + Rs. 30x + Rs. 24x = 129x
Profit = SP - CP = Rs. 129x - Rs. 100x = Rs. 29x
So, profit % = \( \frac{\text{profit}}{\text{CP}} \times 100 = \frac{29x}{100x} \times 100 = 29\% \)
Hence, the gain percent altogether is 29%.
In simple words: A merchant sold his stock in three batches with different deals for each. Even with the discounts on the later batches, he still made a strong overall profit of 29%.

📝 Teacher's Note: This is a weighted average problem. Using '100' for CP and 'x' for quantity makes the algebra very clean because 'x' cancels out in the final percentage step.

🎯 Exam Tip: Be very careful with the "Remaining number" calculation. Ensure that the fractions \( 1/2 + 1/4 + 1/4 \) add up to 1 (the whole stock).

Question 17.
Answer:
Let the cost price of each article bought = Rs. 100.
Let the number of articles bought = x
MP of the articles = Rs. 100 + 60% of Rs. 100
= \( \text{Rs. } 100 + \left( \frac{60}{100} \times 100 \right) \)
= Rs. 160
Number of articles sold at Rs. 160 = \( \frac{x}{2} \)
\( \therefore \text{ SP of } \frac{x}{2} \text{ articles} = \text{Rs. } \left( 160 \times \frac{x}{2} \right) = \text{Rs. } 80x \)
Discount = 25% on Rs. 160
= \( \frac{25}{100} \times 160 \)
= Rs. 40
\( \therefore \text{ SP} = \text{Rs. } 160 - \text{Rs. } 40 = \text{Rs. } 120 \)
Remaining number of articles sold at Rs. 120 = \( x - \frac{x}{2} - \frac{x}{4} = \frac{x}{4} \)
\( \therefore \text{ SP of } \frac{x}{4} \text{ articles} = \text{Rs. } \left( 120 \times \frac{x}{4} \right) = \text{Rs. } 30x \)
Discount = 50% on Rs. 160
= \( \frac{50}{100} \times 160 \)
= Rs. 80
\( \therefore \text{ SP} = \text{Rs. } 160 - \text{Rs. } 80 = \text{Rs. } 80 \)
Number of articles sold at Rs. 80 = \( \frac{x}{4} \)
\( \therefore \text{ SP of } \frac{x}{4} \text{ articles} = \text{Rs. } \left( 80 \times \frac{x}{4} \right) = \text{Rs. } 20x \)
Total SP of all articles = Rs. 80x + Rs. 30x + Rs. 20x = 130x
Profit = SP - CP = Rs. 130x - Rs. 100x = Rs. 30x
So, profit % = \( \frac{\text{profit}}{\text{CP}} \times 100 = \frac{30x}{100x} \times 100 = 30\% \)
Hence, the gain percent altogether is 30%.
In simple words: This is a slightly different version of Question 16. With a higher initial mark-up (60%), the final overall profit is also higher at 30%.

📝 Teacher's Note: Use this to show that a higher "Mark-up" gives the seller more "room" to offer bigger discounts (like 50% on the last batch) while still staying profitable.

🎯 Exam Tip: The logic is identical to the previous question. If you master the steps for one, you can solve any problem where stock is sold in parts.

Question 18.
Answer:
Let the CP be Rs. 100.
Given that the profit% = 21% on the CP
Profit% = \( \frac{\text{profit}}{\text{CP}} \times 100 \)

\( \implies 21\% = \frac{\text{profit}}{100} \times 100 \)

\( \implies \text{profit} = \text{Rs. } 21 \)
SP = CP + Profit
= 100 + 21
= Rs. 121
Let the marked price of the goods be Rs. x.
Discount = 12% of MP = \( \frac{12}{100} \times x = \text{Rs. } \frac{12x}{100} \)
So, SP = MP - Discount

\( \implies 121 = \text{Rs. } \left( x - \frac{12x}{100} \right) \)

\( \implies 121 = \text{Rs. } \frac{121 \times 100}{88} \)

\( \implies x = \text{Rs. } 137.5\% \)
If the goods were sold at the MP, that SP = MP
So, MP - CP = 137.7 - 100 = 37.5 = profit
Profit% = \( \frac{37.5}{100} \times 100 = 37.5\% \)
Hence, the profit percent would be 37.5%.
In simple words: The shopkeeper marks up his goods so he can give a 12% discount and still make 21%. If he stopped giving that discount, his profit would jump up to 37.5%.

📝 Teacher's Note: This is a "No-Discount" hypothetical. We find how much the item is "Marked Up" relative to the Cost Price. The Mark-up percentage IS the profit percentage if no discount is given.

🎯 Exam Tip: Note the transition from Rs. 121 (actual S.P.) to Rs. 137.5 (the hypothetical S.P. or M.P.). The calculation is much easier if you keep CP as 100 throughout.

Question 19.
Answer:
Let the CP be Rs. 100.
Given that the profit% = 36% on the CP
Profit% = \( \frac{\text{profit}}{\text{CP}} \times 100 \)

\( \implies 36\% = \frac{\text{profit}}{100} \times 100 \)

\( \implies \text{profit} = \text{Rs. } 36 \)
SP = CP + Profit
= 100 + 36
= Rs. 136
Let the marked price of the goods be Rs. x.
Discount = 15% of MP = \( \frac{15}{100} \times x = \text{Rs. } \frac{15x}{100} \)
So, SP = MP - Discount

\( \implies 136 = \text{Rs. } \left( x - \frac{15x}{100} \right) \)

\( \implies x = \text{Rs. } \frac{136 \times 100}{85} \)

\( \implies x = \text{Rs. } 160\% \)
In simple words: This seller is aggressive. He marks his items 60% above the cost so that he can offer a 15% discount and still walk away with a huge 36% profit.

📝 Teacher's Note: This is a variation of Question 18. It shows how higher mark-ups allow for higher target profits even with discounts.

🎯 Exam Tip: Identifying that \( 160\% \) MP means a \( 60\% \) mark-up is a critical conceptual link for solving these word problems.

Question 20.
Answer:
Let the CP be Rs. 100.
So, MP = CP + 45% of CP = \( 100 + \left( \frac{45}{100} \times 100 \right) = \text{Rs. } 145 \)
Discount = 20% on MP
= \( \frac{20}{100} \times 145 \)
= Rs. 29
So, SP of the goods = MP - Discount
= Rs. 145 - Rs. 29
= Rs. 116
Profit = SP - CP = Rs. 116 - Rs. 100 = Rs. 16
When the SP is Rs. 116, the profit is Rs. 16
So, when the gain is Rs. 960,
the SP = \( \frac{116 \times 960}{16} = \text{Rs. } 6960 \)
CP = SP - Profit
= Rs. 6960 - Rs. 960
= Rs. 6000
Hence, the cost price of and article on which he gains Rs. 960 is Rs. 6000.
In simple words: We start with a fake "Rs. 100" cost to find the profit ratio. We then use that ratio to find that if a seller makes Rs. 960 profit, the item must have cost them Rs. 6000.

📝 Teacher's Note: This is a "Unitary Scaling" problem. Use the Rs. 100 assumption to find the "Ratio of Profit to CP" (\( 16/100 \)) and then apply it to the actual profit given.

🎯 Exam Tip: Always subtract the discount from the MP (145) to find the SP (116) before you look at the CP (100). The profit is ONLY the part above the 100.

Question 21.
Answer:
Let the CP be Rs. 100.
So, MP = CP + 25% of CP = \( 100 + \left( \frac{25}{100} \times 100 \right) = \text{Rs. } 125 \)
Discount = 10% on MP
= \( \frac{10}{100} \times 125 \)
= Rs. 12.5
So, SP of the goods = MP - Discount
= Rs. 125 - Rs. 12.5
= Rs. 112.5
Profit = SP - CP = Rs. 112.5 - Rs. 100 = Rs. 12.5
When the SP is Rs. 112.5, the profit is Rs. 12.5
So, when the gain is Rs. 960,
the SP = \( \frac{112.5 \times 575}{12.5} = \text{Rs. } 5175 \)
CP = SP - Profit
= Rs. 5175 - Rs. 575
= Rs. 4600
Hence, the cost price of and article on which he gains Rs. 575 is Rs. 4600.
In simple words: Just like the last problem, we use the Rs. 100 model to find the math behind the deal. We find that the original cost of this item was Rs. 4600.

📝 Teacher's Note: This is another scaling problem. Note that the profit is \( 12.5/100 \) or \( 1/8\text{th} \) of the cost. So, C.P. is simply Profit \( \times 8 \). \( 575 \times 8 = 4600 \). Teaching these shortcuts builds student intuition.

🎯 Exam Tip: Ensure your final answer (4600) is the C.P. as requested, not the S.P. (5175). It's easy to stop one step too early.

Question 22.
Answer:
Let the printed price of the books be Rs. x.
Discount given by the publisher = 30% of Rs. x
= \( \frac{30}{100} \times \text{Rs. x} \)
= Rs. \( \frac{30x}{100} \)
So, the distributor bought the books at Rs. \( x - \text{Rs. } \frac{30x}{100} = \text{Rs. } \frac{70x}{100} \)
Discount given by the distributor = Rs. 23% of Rs. x
= Rs. \( \frac{23}{100} \) of Rs. x
= Rs. \( \frac{23x}{100} \)
So, the bookseller purchased the books at Rs. \( x - \text{Rs. } \frac{23x}{100} = \text{Rs. } \frac{77x}{100} \)
Profit made by the distributor = \( \text{SP} - \text{CP} = \text{Rs. } \frac{77x}{100} - \text{Rs. } \frac{70x}{100} = \text{Rs. } \frac{7x}{100} \)
Profit% = \( \frac{\text{profit}}{\text{CP}} \times 100 \)
= \( \frac{\frac{7x}{100}}{\frac{70x}{100}} \times 100 \)
= 10%
SP at which the bookseller sold the books = Rs. x
So, profit = SP - CP
= Rs. \( x - \text{Rs. } \frac{77x}{100} \)
= Rs. \( \frac{23x}{100} \)
Profit% = \( \frac{\text{profit}}{\text{CP}} \times 100 \)
= \( \frac{\frac{23x}{100}}{\frac{77x}{100}} \times 100 \)
= \( 29 \frac{67}{77}\% \)
Hence, the profit% made by the distributor is 10% and that made by the bookseller is \( 29 \frac{67}{77}\% \).
In simple words: This is a business chain. The publisher sells to the dealer at a big discount. The dealer sells to the shop at a smaller discount. The shop sells to you at full price. We find that the dealer makes 10% profit and the shop makes almost 30%.

📝 Teacher's Note: This is an excellent real-world example of a supply chain. Remind students that the "S.P." for the publisher is the "C.P." for the distributor. Each stage must be analyzed as its own separate transaction.

🎯 Exam Tip: The "Printed Price" (MP) is the reference for all discounts given in the problem. Don't use the dealer's purchase price to calculate the distributor's 23% discount.

Question 23.
Answer:
Given that the catalogue price of the laptop = Rs. 43200
SP after the discount = Rs. 43200 - Rs. \( \frac{16}{100} \times 43200 \)
= Rs. 43200 - Rs. 6912
= Rs. 36288
CP = SP - Profit = Rs. 36288 - Profit
So, \( \text{Profit\%} = \frac{\text{profit}}{\text{CP}} \times 100 \)

\( \implies 26 = \frac{\text{profit}}{36288 - \text{Profit}} \times 100 \)

\( \implies 26(36288 - \text{Profit}) = \text{profit} \times 100 \)

\( \implies 943488 - 26 \text{ profit} = 100 \text{profit} \)

\( \implies 943488 = 126 \text{ profit} \)

\( \implies \text{profit} = \frac{943488}{126} \)
\( \implies \text{profit} = \text{Rs. } 7488 \)
So, CP = Rs. 36288 - Rs. 7488 = Rs. 28800
If SP = Rs. 43200 - Rs. 9000 = Rs. 34200
Since, SP > CP, so a gain was made = Rs. 34200 - Rs. 28800 = Rs. 5400
profit% = \( \frac{5400}{28800} \times 100 = 18.75\% \)
Hence, the gain percent would be 18.75%.
In simple words: We find out that the dealer originally bought the laptop for Rs. 28,800. If he changes the discount to a flat Rs. 9000 off, his new profit margin will be 18.75%.

📝 Teacher's Note: This is a "Target Analysis" problem. First solve for the base CP using the first set of rules, then apply the second set of rules to find the new profitability.

🎯 Exam Tip: In the equation \( 26 = \text{profit} / \text{CP} \), remember that \( \text{CP} = \text{SP} - \text{Profit} \). This substitution is the only way to solve for the unknown profit when you only have the S.P. and percentage.

Question 24.
Answer:
Gurmeet gives a discount of 8% on the first Rs. 20000
So, SP on Rs. 20000
= Rs. \( 20000 - \frac{8}{100} (\text{Rs. } 20000) \)
= Rs. 20000 - Rs. 1600
= Rs. 18400
So, SP on Rs. 20000
Gurmeet gives a discount of 5% on the first Rs. 5000
= Rs. \( 5000 - \frac{5}{100} (\text{Rs. } 5000) \)
= Rs. 5000 - Rs. 250
= Rs. 4750
So, actual price at which Gurmeet sells the article
= Rs. 18400 + Rs. 4750
= Rs. 23150
In simple words: This is a "Segmented Discount." The first large portion of the price gets one discount, and the smaller remaining portion gets another. Added together, the customer pays Rs. 23,150.

📝 Teacher's Note: This mirrors how income tax or bulk billing works. The rate changes as the amount increases. Each segment must be calculated separately.

🎯 Exam Tip: Do not add the percentages (8% + 5%). They apply to different amounts of money. Calculate each part individually and then sum the final prices.

Question 25.
Answer:
Manjeet gives a discount of 6% on the first Rs. 25000
= Rs. \( 25000 - \frac{6}{100} (\text{Rs. } 25000) \)
= Rs. 25000 - Rs. 1500
= Rs. 23500
So, actual price at which Manjeet sells the article is Rs. 23150,
and that at which Gurmeet sells the article is Rs. 23150.
List price of the article = Rs. 2500
CP of the article = Rs. 2000
SP of the article at 5% discount
= Rs. 2500 - 5% of Rs. 2500
= Rs. \( 2500 - \frac{5}{100} \times \text{Rs. } 2500 \)
= Rs. 2375
Since trader gets a 5% additional discount for cash payment,
so, amount paid by the trader = Rs. 2375 - 5% of Rs. 2375
= \( \text{Rs. } 2375 - \frac{5}{100} \times \text{Rs. } 2375 \)
= Rs. 2375 - Rs. 118.75
= Rs. 2256.25
Profit made by the manufacturer
= List price - SP
= Rs. 2500 - Rs. 2256.25
= Rs. 243.75
So, \( \text{profit\%} = \frac{\text{profit}}{\text{CP}} \times 100 \)
= \( \frac{243.75}{2000} \times 100 \)
= 12.18%
Hence, the amount that the trader pays is Rs. 2256.25 and the profit % that the manufacturer makes on the sale is 12.18%.
In simple words: This problem has "Layered Discounts." First there is a standard store discount, and then a special extra discount for paying with cash. After both are taken off, the manufacturer still makes a 12.18% profit.

📝 Teacher's Note: This is a "Successive Discount" word problem. First discount is 5%, then another 5%. The second 5% is calculated on the *already reduced* price. This is a common point of error for students.

🎯 Exam Tip: Be careful with the profit calculation. Profit is based on what the buyer paid (Rs. 2256.25) minus the manufacturing cost (Rs. 2000).

Question 26.
Answer:
Let the marked price be Rs. x
CP of the computer set = Rs. 20000
\( \text{Profit\%} = \frac{\text{profit}}{\text{CP}} \times 100 \)

\( \implies 25 = \frac{\text{profit}}{20000} \times 100 \)

\( \implies \text{profit} = \text{Rs. } 5000 \)
So, SP = CP + Profit
= Rs. 20000 + Rs. 5000
= Rs. 25000
Given that a discount of 5% is given on the MP.
So, SP = Rs. x - 5% of the MP
\( \implies \text{Rs. } 25000 = x - \frac{5}{100} \times x \)
\( \implies \text{Rs. } 25000 = \frac{95x}{100} \)
\( \implies x = \text{Rs. } 26315.79 \text{ approx} \)
Hence, the price that should be marked is approximately Rs. 26315.79.
In simple words: The store wants to make a 25% profit. To do that while also giving a 5% discount, they need to put a high price tag of about Rs. 26,316 on the computer.

📝 Teacher's Note: This is the inverse of a normal discount problem. We use the profit goal to find the S.P., and then use the discount to find the M.P.

🎯 Exam Tip: When the result is a messy decimal, always include the word "approximately" or "approx" in your final answer to show you've rounded it.

Question 27.
Answer:
Total cost of production = Rs 5200
The ratio of material: labour: overheads = 5:6:2
\( \therefore \) Total of the ratio = 5 + 6 + 2 = 13
\( \therefore \text{ Cost of material} = \text{Rs } \left( \frac{5}{13} \times 5200 \right) = \text{Rs. } 2000 \)
\( \therefore \text{ Cost of labor} = \text{Rs } \left( \frac{6}{13} \times 5200 \right) = \text{Rs. } 2400 \)
\( \therefore \text{ Cost of overheads} = \text{Rs } \left( \frac{2}{13} \times 5200 \right) = \text{Rs. } 800 \)
Cost price of the video game = Rs 5200
Profit = 30%
\( \therefore \text{ Profit} = 30\% \text{ of Rs } 5200 = \text{Rs } 1560 \)
\( \therefore \text{ S.P} = \text{Rs } 5200 + \text{Rs } 1560 = \text{Rs } 6760 \)
So, marked price is Rs 6760
Cost of material = Rs 2000
Increase = 40%
\( \therefore \text{ Increase} = 40\% \text{ of Rs } 2000 = \text{Rs } 800 \)
\( \therefore \text{ New cost of material} = \text{Rs } 2000 + \text{Rs } 800 = \text{Rs } 2800 \)
Cost of labour = Rs 2400
Increase = 30%
\( \therefore \text{ Increase} = 30\% \text{ of Rs } 2400 = \text{Rs } 720 \)
\( \therefore \text{ New cost of labour} = \text{Rs } 2400 + \text{Rs } 720 = \text{Rs } 3120 \)
Cost of overheads = Rs 800
Increase = 10%
\( \therefore \text{ Increase} = 10\% \text{ of Rs } 800 = \text{Rs } 80 \)
\( \therefore \text{ New cost of overheads} = \text{Rs } 800 + \text{Rs } 80 = \text{Rs } 880 \)
Cost of manufacturing now = Rs. (2800 + 3120 + 880) = Rs 6800
Profit = 30%
\( \frac{\text{S.P.}}{\text{C.P.}} = 1 + \frac{\text{Profit}}{100} \)

\( \implies \frac{\text{S.P.}}{6800} = 1 + \frac{30}{100} \)

\( \implies \text{S.P.} = \frac{130}{100} \times 6800 = \text{Rs. } 8840 \)
The cost of manufacturing the video game now is Rs 6800, And the marked price now is Rs 8840.
In simple words: A company's costs for making a game went up because materials and workers became more expensive. To keep making their same 30% profit, they had to raise the final price from Rs. 6760 to Rs. 8840.

📝 Teacher's Note: This is a "Production Costing" problem. It combines ratios (the cost components) with percentage increases and profit calculations. It's a very thorough test of a student's ability to handle multiple variables.

🎯 Exam Tip: Step 1 is using the ratio to break down the original cost. Without the individual component costs, you cannot calculate the new higher costs correctly.