Past Year QuestionsIPMAT IndoreQAProfit & Loss

IPMAT Indore Profit & Loss — PYPs

5 solved Profit & Loss previous year questions (PYQs) from IPMAT Indore past year papers — attempt each and check the answer.

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Q1:ipmat indore 2022QAProfit & LossMediumSA · TITA
Aruna purchases a certain number of apples for INR 20 each and a certain number of mangoes for INR 25 each. If she sells all the apples at 10%10 \% profit and all the mangoes at 20%20 \% loss, overall she makes neither profit nor loss. Instead, if she sells all the apples at 20%20 \% loss and all the mangoes at 10%10 \% profit, overall she makes a loss of INR 150. Then the number of apples purchased by Aruna is _________.
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The Setup: We have a system of profit/loss constraints across two different scenarios for an unknown quantity of apples and mangoes with fixed unit costs. We must solve the linear equations to find the exact number of apples. Step 1: Define variables and calculate per-item cost variations. Let AA be the number of apples (Cost = 2020 each). Let MM be the number of mangoes (Cost = 2525 each). * +10%+10\% on apples = +2+2 profit per apple. * 20%-20\% on mangoes = 5-5 loss per mango. * 20%-20\% on apples = 4-4 loss per apple. * +10%+10\% on mangoes = +2.5+2.5 profit per mango. Step 2: Formulate Scenario 1 to find the quantity ratio. In the first scenario, the net profit is exactly 00: 2A5M=0    2A=5M    M=0.4A2A - 5M = 0 \implies 2A = 5M \implies M = 0.4A Step 3: Formulate Scenario 2 to solve for AA. In the second scenario, the net loss is 150150 (meaning profit is 150-150): 4A+2.5M=150-4A + 2.5M = -150 Multiply by 1-1 for clarity: 4A2.5M=1504A - 2.5M = 150 Step 4: Substitute the ratio and calculate AA. Substitute M=0.4AM = 0.4A into the second equation: 4A2.5(0.4A)=1504A - 2.5(0.4A) = 150 4A1A=1504A - 1A = 150 3A=150    A=503A = 150 \implies A = 50 Final Answer: 50
Q2:ipmat indore 2019QAProfit & LossEasySA · TITA
A shopkeeper reduces the price of a pen by 25% as a result of which the sales quantity increased by 20%. If the revenue made by the shopkeeper decreases by x% then x is
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The Setup: Revenue is Price×Quantity\text{Price}\times\text{Quantity}, so a percentage change in each multiplies rather than adds. Anyone who subtracts 25%20%=5%25\%-20\%=5\% has fallen for the trap this question exists to set. Step 1: Define the baseline. Let the original price be PP and quantity QQ, so the original revenue is R1=PQR_1=PQ. Step 2: Apply the percentage shifts as multipliers. * Price falls 25%: new price =(10.25)P=0.75P=(1-0.25)P=0.75P * Quantity rises 20%: new quantity =(1+0.20)Q=1.20Q=(1+0.20)Q=1.20Q Step 3: Calculate the new revenue. R2=(0.75P)(1.20Q)=(0.75×1.20)PQ=0.90PQR_2=(0.75P)(1.20Q)=(0.75\times 1.20)PQ=0.90\,PQ Step 4: Convert to a percentage drop. Revenue went from 1.00PQ1.00PQ to 0.90PQ0.90PQ, a fall of 0.10PQ0.10PQ: R1R2R1×100=0.10PQPQ×100=10%\frac{R_1-R_2}{R_1}\times 100=\frac{0.10PQ}{PQ}\times 100=10\% So x=10x=10. Sanity check with real numbers: at P=100P=100, Q=100Q=100, revenue goes from 10,000 to 75×120=9,00075\times 120=9{,}000 - down 1,000, which is 10%. Final Answer: 10
Q3:ipmat indore 2024QAProfit & LossEasySA · TITA
The price of a chocolate is increased by x% and then reduced by x%. The new price is 96.76% of the original price. Then x is:
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The Setup: A chocolate's price undergoes a sequential x%x\% increase followed by an x%x\% decrease. The resulting price is 96.76%96.76\% of the original price, and we must determine the value of xx. Step 1: Construct the sequential price multiplier function. Let PP represent the baseline price. An x%x\% increase applies a multiplier of (1+x100)(1 + \frac{x}{100}). An x%x\% decrease applies a subsequent multiplier of (1x100)(1 - \frac{x}{100}). New Price=P×(1+x100)×(1x100)\text{New Price} = P \times \left(1 + \frac{x}{100}\right) \times \left(1 - \frac{x}{100}\right) Step 2: Simplify via the difference of squares identity. New Price=P(1x210000)\text{New Price} = P \left(1 - \frac{x^2}{10000}\right) Step 3: Equate to the provided net proportional change. The final state is 96.76%96.76\% (or 0.96760.9676) of PP. P(1x210000)=0.9676PP \left(1 - \frac{x^2}{10000}\right) = 0.9676 P Divide out PP from both sides since initial price is arbitrary: 1x210000=0.96761 - \frac{x^2}{10000} = 0.9676 Step 4: Solve for the absolute rate xx. x210000=10.9676=0.0324\frac{x^2}{10000} = 1 - 0.9676 = 0.0324 x2=324x^2 = 324 Because a percentage rate scaling magnitude must be positive, take the principal square root: x=18x = 18 Final Answer: 18
Q4:ipmat indore 2023QAProfit & LossEasyMCQ · MCQ
A goldsmith bought a large solid golden ball at INR 1,000,000 and melted it to make a certain number of solid spherical beads such that the radius of each bead was one-fifth of the radius of the original ball. Assume that the cost of making golden beads is negligible. If the goldsmith sold all the beads at 20% discount on the listed price and made a total profit of 20%, then the listed price or each golden bead, in INR, was
  • A48000
  • B12000
  • C9600
  • D24000
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The Setup: A solid golden ball costing INR 1,000,0001,000,000 is melted into smaller spherical beads. The radius of each bead is one-fifth of the original ball's radius. The beads are sold at a 20%20\% discount for a total overall profit of 20%20\%. We must find the listed price per bead. Step 1: Calculate the number of beads produced. Let the radius of the original ball be RR and the radius of a bead be rr. We are given r=R5r = \frac{R}{5}. The volume of a sphere is proportional to the cube of its radius. Number of beads=Volume of original ballVolume of one bead=43πR343πr3=(Rr)3=(5)3=125\text{Number of beads} = \frac{\text{Volume of original ball}}{\text{Volume of one bead}} = \frac{\frac{4}{3}\pi R^3}{\frac{4}{3}\pi r^3} = \left(\frac{R}{r}\right)^3 = (5)^3 = 125 So, 125125 identical beads are produced. Step 2: Calculate the required revenue per bead. The total cost is INR 1,000,0001,000,000. The total profit made is 20%20\%, so the total revenue generated is 120%120\% of the cost: Total Revenue=1,000,000×1.20=1,200,000\text{Total Revenue} = 1,000,000 \times 1.20 = 1,200,000 Since this revenue comes from selling 125125 beads, the selling price (SPSP) per bead is: SP=1,200,000125=9600SP = \frac{1,200,000}{125} = 9600 Step 3: Calculate the listed price. The selling price is derived after a 20%20\% discount is applied to the listed price (LL). 0.80×L=96000.80 \times L = 9600 L=96000.80=12000L = \frac{9600}{0.80} = 12000 Final Answer: 12000
Q5:ipmat indore 2026QAProfit & LossMediumSA · TITA
Ram purchased 3 oranges at Rs. 20 each and 5 mangoes at Rs. 38 each and then sold them by offering discounts of 25% and 20% on the fixed marked prices of oranges and mangoes, respectively. If he earned a profit of 50% on selling the oranges alone, and a total profit of 32% on selling all the fruits, then the marked price of each mango was ___
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The Setup: This is a classic Profit & Loss side quest. Ram is basically running a whole fruit cartel. We are given the final profit margins and need to reverse-engineer the original marked price (the bait) of the mangoes. We'll break this down by isolating the cost and revenue for each fruit type. Step 1: Calculate the base stats (Total Cost Price). Ram drops some cash upfront to acquire the inventory. We calculate the total cost price by multiplying the quantity by the price per unit. Total CP=3×20+5×38=60+190=250 rupeesTotal\ CP = 3 \times 20 + 5 \times 38 = 60 + 190 = 250\ rupees Step 2: Calculate the total loot (Total Selling Price). The problem states he secures a massive 32% overall profit on the entire stash. We apply a 1.321.32 multiplier to the total cost. Total SP=250×1.32=330 rupeesTotal\ SP = 250 \times 1.32 = 330\ rupees Step 3: Isolate the orange revenue. The oranges alone gave a cracked 50% profit buff. The cost price for just the oranges was 60 rupees. Oranges SP=60×1.5=90 rupeesOranges\ SP = 60 \times 1.5 = 90\ rupees Step 4: Find the mango selling price. By subtracting the orange revenue from the total revenue, we find exactly how much the mango bag sold for. Mangoes SP (for 5)=33090=240 rupeesMangoes\ SP\ (for\ 5) = 330 - 90 = 240\ rupees To find the selling price of just one mango, divide by the squad size of 5. SP of one mango=2405=48 rupeesSP\ of\ one\ mango = \frac{240}{5} = 48\ rupees Step 5: Reverse-engineer the Marked Price. Ram offered a 20% discount on mangoes to pass the vibe check for his customers. This means the final selling price is exactly 80% of the marked price (SP=0.80×MPSP = 0.80 \times MP). We divide the single mango SP by 0.800.80 to find the original inflated tag. MP=480.80=60 rupeesMP = \frac{48}{0.80} = 60\ rupees Final Answer: 60

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