Profit & Loss, Discount & Tax

Q1: Find the selling price when:

i. C.P. = ₹7640, Gain = 15%

Step 1: Selling Price = C.P. × \(\left(1 + \frac{\text{Gain}}{100}\right)\)
= ₹7640 × \(\left(1 + \frac{15}{100}\right) = 7640 × \frac{115}{100} = ₹8786\)
Answer: ₹8786

ii. C.P. = ₹4850, Loss = 12%

Step 1: S.P. = C.P. × \(\left(1 – \frac{12}{100}\right) = 4850 × \frac{88}{100} = ₹4268\)
Answer: ₹4268

iii. C.P. = ₹720, Loss = \(8\frac{3}{4}\)%

Step 1: Convert mixed fraction: \(8\frac{3}{4} = \frac{35}{4}\)%
Step 2: S.P. = ₹720 × \(\left(1 – \frac{35}{400}\right) = 720 × \frac{365}{400} = ₹657\)
Answer: ₹657

iv. C.P. = ₹2652, Gain = \(16\frac{2}{3}\)%

Step 1: Convert mixed fraction: \(16\frac{2}{3} = \frac{50}{3}\)%
Step 2: S.P. = ₹2652 × \(\left(1 + \frac{50}{300}\right) = 2652 × \frac{350}{300} = ₹3094\)
Answer: ₹3094

Q2: Find the cost price when:

i. S.P. = ₹207, Gain = 15%

Step 1: C.P. = S.P. ÷ \(\left(1 + \frac{\text{Gain}}{100}\right)\)
= ₹207 ÷ \(\left(1 + \frac{15}{100}\right) = 207 ÷ \frac{115}{100} = 207 × \frac{100}{115}\)
= ₹180
Answer: ₹180

ii. S.P. = ₹448.20, Loss = 17%

Step 1: C.P. = ₹448.20 ÷ \(\left(1 – \frac{17}{100}\right) = 448.20 ÷ \frac{83}{100} = 448.20 × \frac{100}{83}\)
= ₹540
Answer: ₹540

iii. S.P. = ₹1479, Gain = \(6\frac{1}{4}\)%

Step 1: Convert mixed fraction: \(6\frac{1}{4} = \frac{25}{4}\)%
Step 2: C.P. = ₹1479 ÷ \(\left(1 + \frac{25}{400}\right) = 1479 ÷ \frac{425}{400} = 1479 × \frac{400}{425}\)
= ₹1392
Answer: ₹1392

iv. S.P. = ₹611.80, Loss = 8%

Step 1: C.P. = ₹611.80 ÷ \(\left(1 – \frac{8}{100}\right) = 611.80 ÷ \frac{92}{100} = 611.80 × \frac{100}{92}\)
= ₹665
Answer: ₹665

Q3: A sells a bicycle to B at a profit of 20% and B sells it to C at a profit of 5%. If C pays ₹3780, what did A pay for it?

Step 1: Let the Cost Price for A = ₹x
Step 2: A sells to B at 20% profit → B’s C.P. = \(x + \frac{20}{100}x = \frac{120}{100}x = \frac{6x}{5}\)
Step 3: B sells to C at 5% profit → C’s Price = \(\frac{6x}{5} + \frac{5}{100} \cdot \frac{6x}{5} = \frac{6x}{5} \cdot \frac{105}{100} = \frac{6x \cdot 105}{5 \cdot 100} = \frac{630x}{500}\)
Step 4: Given: C pays ₹3780
So, \(\frac{630x}{500} = 3780\)
Step 5: Multiply both sides by 500:
630x = 3780 × 500 = 1890000
Step 6: Divide both sides by 630:
x = \(\frac{1890000}{630} = ₹3000\)
Answer: A paid ₹3000 for the bicycle.

Q4: Raju sold a watch to Sonu at 12% gain and Sonu had to sell it to Manu at a loss of 5%. If Manu paid ₹5320, how much did Raju pay for it?

Step 1: Let Raju’s cost price = ₹x
Step 2: Raju sold to Sonu at 12% gain
⇒ Sonu’s C.P. = ₹x × \(\frac{112}{100} = \frac{28x}{25}\)
Step 3: Sonu sold to Manu at 5% loss
⇒ Manu’s Price = \(\frac{28x}{25} × \frac{95}{100} = \frac{2660x}{2500}\)
Step 4: Given: Manu paid ₹5320
\(\frac{2660x}{2500} = 5320\)
Step 5: Multiply both sides by 2500:
2660x = 5320 × 2500 = 13300000
Step 6: Divide both sides by 2660:
x = \(\frac{13300000}{2660} = ₹5000\)
Answer: Raju paid ₹5000 for the watch.

Q5: A grocer purchases 80 kg of rice at ₹27 per kg and mixes it with 120 kg of rice purchased at ₹32 per kg. At what rate per kg should he sell the mixture to gain 16%?

Step 1: Quantity of cheaper rice = 80 kg at ₹27/kg
Total cost = 80 × 27 = ₹2160
Step 2: Quantity of costlier rice = 120 kg at ₹32/kg
Total cost = 120 × 32 = ₹3840
Step 3: Total rice = 80 + 120 = 200 kg
Total cost = ₹2160 + ₹3840 = ₹6000
Step 4: Cost Price per kg of mixture = ₹6000 ÷ 200 = ₹30
Step 5: To gain 16%, Selling Price per kg = ₹30 × \(\left(1 + \frac{16}{100}\right)\)
= ₹30 × \(\frac{116}{100} = ₹34.80\)
Answer: ₹34.80 per kg

Q6: Mrs. Harjeet two bags for ₹1150 each. She sold of them at a gain of 6% and the other at a loss of 2%. How much did she gain?

Step 1: Cost Price of each bag = ₹1150
Total Cost Price = ₹1150 × 2 = ₹2300
Step 2: Selling Price of 1st bag at 6% gain = ₹1150 × \(\frac{106}{100} = ₹1219\)
Step 3: Selling Price of 2nd bag at 2% loss = ₹1150 × \(\frac{98}{100} = ₹1127\)
Step 4: Total Selling Price = ₹1219 + ₹1127 = ₹2346
Step 5: Gain = Total S.P. − Total C.P. = ₹2346 − ₹2300 = ₹46
Answer: Mrs. Harjeet gained ₹46.

Q7: A trader purchased a wall clock and a watch for a sum of ₹5070. He sold them making a profit of 10% on the wall clock and 15% on the watch. He earns a profit of ₹669.50. Find the cost price of the wall clock and that of the watch.

Step 1: Let the cost price of the wall clock be ₹x.
Then, the cost price of the watch = ₹(5070 − x)
Step 2: Profit on wall clock = 10% of x = \(\frac{10}{100}x = \frac{x}{10}\)
Profit on watch = 15% of (5070 − x) = \(\frac{15}{100}(5070 − x) = \frac{3}{20}(5070 − x)\)
Step 3: Total profit = ₹669.50 \[ \frac{x}{10} + \frac{3}{20}(5070 − x) = 669.50 \]Step 4: Multiply entire equation by 20 to eliminate denominators: \[ 2x + 3(5070 − x) = 13390 \\ 2x + 15210 − 3x = 13390 \\ −x + 15210 = 13390 \\ x = 15210 − 13390 = ₹1820 \]Step 5: Cost Price of wall clock = ₹1820
Cost Price of watch = ₹(5070 − 1820) = ₹3250
Answer: Wall clock = ₹1820, Watch = ₹3250

Q8: Toffees are bought at 15 for ₹20. How many toffees would be sold for ₹20 so as to gain 25%?

Step 1: Cost Price of 15 toffees = ₹20
⇒ Cost Price of 1 toffee = ₹\(\frac{20}{15} = ₹\frac{4}{3}\)
Step 2: To gain 25%, Selling Price of 1 toffee = ₹\(\frac{4}{3} × \frac{125}{100} = \frac{4}{3} × \frac{5}{4} = ₹\frac{5}{3}\)
Step 3: Now we want to sell **toffees for ₹20**, so:
Let number of toffees sold = x
Then: x × Selling Price per toffee = ₹20 \[ x × \frac{5}{3} = 20 \]Step 4: Solve for x: \[ x = \frac{20 × 3}{5} = 12 \]Answer: 12 toffees should be sold for ₹20 to gain 25%.

Q9: Two-thirds of a consignment was sold at a profit of 5% and the remainder at a loss of 2%. If the total profit was ₹4000, find the value at which the consignment was purchased.

Step 1: Let the total cost price of the consignment be ₹x.
Step 2: Two-thirds of the consignment sold at 5% profit:
⇒ C.P. of this part = \(\frac{2}{3}x\) ⇒ Profit = 5% of \(\frac{2}{3}x = \frac{5}{100} × \frac{2x}{3} = \frac{10x}{300} = \frac{x}{30}\)
Step 3: Remaining one-third sold at 2% loss:
⇒ C.P. = \(\frac{1}{3}x\) ⇒ Loss = 2% of \(\frac{1}{3}x = \frac{2}{100} × \frac{x}{3} = \frac{2x}{300} = \frac{x}{150}\)
Step 4: Net Profit = Profit − Loss = ₹4000 \[ \frac{x}{30} − \frac{x}{150} = 4000 \]Step 5: Take LCM of 30 and 150 = 150 \[ \frac{5x − x}{150} = 4000 ⇒ \frac{4x}{150} = 4000 \]Step 6: Multiply both sides by 150: \[ 4x = 4000 × 150 = 600000 \\ x = \frac{600000}{4} = ₹150000 \]Answer: The consignment was purchased for ₹1,50,000.

Q10: A grocer bought sugar worth of ₹4500. He sold one-third of it at 10% gain. At what gain per cent, the remaining sugar be sold to have a 12% gain on the whole?

Step 1: Total C.P. = ₹4500
⇒ C.P. of 1/3 of sugar = \(\frac{1}{3} × 4500 = ₹1500\)
⇒ C.P. of remaining 2/3 sugar = ₹4500 − ₹1500 = ₹3000
Step 2: Profit on 1st part = 10% of ₹1500 = ₹150
Step 3: Let required gain % on remaining sugar = x%
⇒ Profit on second part = \(\frac{x}{100} × 3000 = \frac{30x}{1}\)
Step 4: Total profit = 12% of ₹4500 = \(\frac{12}{100} × 4500 = ₹540\)
Now, \[ 150 + 30x = 540 \]Step 5: Solve for x: \[ 30x = 540 − 150 = 390 ⇒ x = \frac{390}{30} = 13 \]Answer: The remaining sugar should be sold at 13% gain.

Q11: A man buys a piece of land for ₹ 384000.He sells two-fifths of it at a loss of 6%. At what gain per cent should he sell the remaining piece of land to 10% gain on the whole?

Step 1: Total cost price of land = ₹384000
Step 2: C.P. of two-fifths = \(\frac{2}{5} × 384000 = ₹153600\)
C.P. of remaining = ₹384000 − ₹153600 = ₹230400
Step 3: Loss on the first part = 6% of ₹153600 = \(\frac{6}{100} × 153600 = ₹9216\)
Step 4: Let required gain % on remaining = x%
Gain = \(\frac{x}{100} × 230400 = \frac{2304x}{1}\)
Step 5: Total gain = 10% of ₹384000 = \(\frac{10}{100} × 384000 = ₹38400\)
Now use: Total gain = Gain on 2nd part − Loss on 1st part \[ 2304x − 9216 = 38400 \]Step 6: Solve for x: \[ 2304x = 38400 + 9216 = 47616 \\ x = \frac{47616}{2304} = \frac{62}{3} = 20 \frac{2}{3} \]Answer: The remaining land must be sold at a gain of \(20\frac{2}{3}\)%).

Q12: By selling an almirah for ₹10416, a man gains 12%. What will be his gain or loss per cent if it is sold ₹9114?

Step 1: Let the Cost Price (C.P.) of almirah be ₹x.
Given, S.P. = ₹10416, Gain = 12%
So, \[ 10416 = x × \frac{112}{100} = \frac{112x}{100} \]Step 2: Calculate cost price x: \[ x = \frac{10416 × 100}{112} = ₹9300 \]Step 3: Now, new S.P. = ₹9114
Difference = ₹9114 − ₹9300 = −₹186 (loss)
Step 4: Calculate loss percentage: \[ \text{Loss \%} = \frac{186}{9300} × 100 = 2\% \]Answer: There is a loss of 2% if sold at ₹9114.

Q13: A chair was sold ₹ 2142 at a gain of 5%. At what it should have been sold to gain 10%?

Step 1: Let the cost price (C.P.) of the chair be ₹x.
Given: S.P. = ₹2142, Gain = 5% \[ 2142 = x × \frac{105}{100} = \frac{21x}{20} \]Step 2: Calculate the cost price: \[ x = \frac{2142 × 20}{21} = ₹2040 \]Step 3: Find the selling price for 10% gain: \[ S.P. = x × \frac{110}{100} = 2040 × 1.10 = ₹2244 \]Answer: The chair should be sold for ₹2244 to gain 10%.

Q14: A television is sold for ₹9360 at a loss of 4%. For how much it should have been sold to gain 4%?

Step 1: Let the cost price (C.P.) be ₹x.
Given, S.P. = ₹9360, Loss = 4% \[ 9360 = x × \frac{96}{100} = \frac{24x}{25} \]Step 2: Calculate cost price: \[ x = \frac{9360 × 25}{24} = ₹9750 \]Step 3: Selling price for 4% gain: \[ S.P. = x × \frac{104}{100} = 9750 × 1.04 = ₹10140 \]Answer: The television should be sold for ₹10140 to gain 4%.

Q15: A shopkeeper sold two fans at ₹ 1980 each. On one he gained 10% while on the other he lost 10%. Calculate the gain or loss per cent on the whole transaction.

Step 1: Selling Price (S.P.) of each fan = ₹1980
Step 2: For fan with 10% gain:
Let C.P. = ₹x \[ 1980 = x × \frac{110}{100} = \frac{11x}{10} \\ ⇒ x = \frac{1980 × 10}{11} = ₹1800 \]Step 3: For fan with 10% loss:
Let C.P. = ₹y \[ 1980 = y × \frac{90}{100} = \frac{9y}{10} \\ ⇒ y = \frac{1980 × 10}{9} = ₹2200 \]Step 4: Total Cost Price = ₹1800 + ₹2200 = ₹4000
Total Selling Price = ₹1980 + ₹1980 = ₹3960
Step 5: Net Loss = ₹4000 − ₹3960 = ₹40
Step 6: Loss percent: \[ \frac{40}{4000} × 100 = 1\% \]Answer: There is a loss of 1% on the whole transaction.

Q16: Shanti sold two cameras for ₹ 6555 each. On one she lost 5%, while on the other she gained 15%. Find the gain or per cent in the whole transaction.

Step 1: Selling Price (S.P.) of each camera = ₹6555
Step 2: For camera sold at 5% loss:
Let cost price = ₹x \[ 6555 = x × \frac{95}{100} = \frac{95x}{100} \\ ⇒ x = \frac{6555 × 100}{95} = ₹6900 \]Step 3: For camera sold at 15% gain:
Let cost price = ₹y \[ 6555 = y × \frac{115}{100} = \frac{115y}{100} \\ ⇒ y = \frac{6555 × 100}{115} = ₹5700 \]Step 4: Total Cost Price = ₹6900 + ₹5700 = ₹12600
Total Selling Price = ₹6555 + ₹6555 = ₹13110
Step 5: Net Gain = ₹13110 − ₹12600 = ₹510
Step 6: Gain percent: \[ \frac{510}{12600} × 100 = \frac{85}{21} = 4 \frac{1}{21}\% \]Answer: The overall gain is \(4 \frac{1}{21}\)% on the whole transaction.

Q17: By selling 45 lemons for ₹40, a man loses 20%. How many should he sell for ₹24 to gain 20% on the transaction?

Step 1: Selling price (S.P.) of 45 lemons = ₹40
Loss = 20%
Step 2: Let the cost price (C.P.) of 45 lemons = ₹x
Since loss = 20%, \[ S.P. = 80\% \text{ of } C.P. = \frac{80}{100} × x = \frac{4}{5}x \]Given, \[ \frac{4}{5}x = 40 \\ ⇒ x = \frac{40 × 5}{4} = ₹50 \]Step 3: Cost price of 45 lemons = ₹50
⇒ C.P. per lemon = \(\frac{50}{45} = \frac{10}{9}\) ₹
Step 4: To gain 20%, S.P. per lemon should be: \[ \text{S.P.} = 120\% \text{ of } C.P. = \frac{120}{100} × \frac{10}{9} = \frac{12}{10} × \frac{10}{9} = \frac{12}{9} = \frac{4}{3} \text{ ₹ per lemon} \]Step 5: Let the number of lemons to be sold for ₹24 = y
Then, \[ y × \frac{4}{3} = 24 \\ ⇒ y = \frac{24 × 3}{4} = 18 \]Answer: He should sell 18 lemons for ₹24 to gain 20%.

Q18: Rajni sold a pressure cooker at a loss of 8%. Had she bought it at 10% less and sold for ₹176 more, She would have gained 20%. Find the cost price of the pressure cooker.

Step 1: Let the original cost price be ₹x.
Step 2: Selling price when sold at 8% loss: \[ S.P._1 = x × \frac{92}{100} = \frac{92x}{100} \]Step 3: New cost price if bought at 10% less: \[ C.P._2 = x × \frac{90}{100} = \frac{90x}{100} = \frac{9x}{10} \]Selling price in second scenario (₹176 more than \(S.P._1\)): \[ S.P._2 = \frac{92x}{100} + 176 \]Step 4: Given gain of 20% on new cost price: \[ S.P._2 = C.P._2 × \frac{120}{100} = \frac{9x}{10} × \frac{6}{5} = \frac{54x}{50} = \frac{27x}{25} \]Step 5: Equate the two expressions for \(S.P._2\): \[ \frac{92x}{100} + 176 = \frac{27x}{25} \]Multiply both sides by 100 to clear denominators: \[ 92x + 17600 = 4 × 27x = 108x \]Step 6: Simplify: \[ 108x – 92x = 17600 \\ 16x = 17600 \\ x = \frac{17600}{16} = ₹1100 \]Answer: The cost price of the pressure cooker is ₹1100.

Q19: A man sold a toaster at a profit of 10%. Had he purchased it for less and sold it for ₹56 more, he could have gained 25%. For how much did he buy it?

Step 1: Let the original cost price be ₹x.
Step 2: Selling price at 10% profit: \[ S.P._1 = x × \frac{110}{100} = \frac{11x}{10} \]Step 3: New purchase price (5% less): \[ C.P._2 = x × \frac{95}{100} = \frac{19x}{20} \]Step 4: New selling price (₹56 more): \[ S.P._2 = S.P._1 + 56 = \frac{11x}{10} + 56 \]Step 5: New gain = 25% on new cost price: \[ S.P._2 = C.P._2 × \frac{125}{100} = \frac{19x}{20} × \frac{5}{4} = \frac{95x}{80} = \frac{19x}{16} \]Step 6: Equate the two expressions for \(S.P._2\): \[ \frac{11x}{10} + 56 = \frac{19x}{16} \]Multiply both sides by 80 (LCM of 10 and 16) to clear denominators: \[ 80 × \frac{11x}{10} + 80 × 56 = 80 × \frac{19x}{16} \\ 8 × 11x + 4480 = 5 × 19x \\ 88x + 4480 = 95x \]Step 7: Rearrange: \[ 95x − 88x = 4480 \\ 7x = 4480 \\ x = \frac{4480}{7} = ₹640 \]Answer: The original cost price of the toaster is ₹640.

Q20: A shopkeeper sells each of his goods at a gain of \(22\frac{1}{2}\)%. If on any day, his total sale was of ₹9408, what was

i. The total cost of all the goods sold on that day.

Step 1: Given gain percent = \(22\frac{1}{2}\)% = \(\frac{45}{2}\)% = 22.5%
Step 2: Let total cost price = ₹x
Selling Price (S.P.) = Cost Price (C.P.) + Profit = \(x + \frac{45}{200}x = \frac{245}{200}x\)
Or, \[ S.P. = \frac{245}{200} x \]Step 3: Given total sale (S.P.) = ₹9408
So, \[ 9408 = \frac{245}{200} x \]Step 4: Find \(x\): \[ x = \frac{9408 × 200}{245} = \frac{1,881,600}{245} = ₹7680 \]Answer: Total cost of all goods sold = ₹7680.

ii. His profit of that day.

Step 5: Profit = S.P. − C.P. = ₹9408 − ₹7680 = ₹1728
Answer: Profit earned = ₹1728.