IPMAT Indore 2023 · Quantitative Ability · Question 32

IPMAT Indore 2023 previous year question on Profit Loss and Discount, with the answer and a step-by-step solution.

Profit Loss and DiscountArithmeticMultiple choiceMedium
A goldsmith bought a large solid golden ball at ₹1000000 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 of each golden bead, in ₹, was
  1. A
    12000
  2. B
    48000
  3. C
    9600
  4. D
    24000
Pick an option, then check.
Answer & solution

Answer: Option A

Step 1 — count the beads. Volume scales as the cube of the radius. Each bead has 1/5 the radius, so 1/125 the volume:

\text{number of beads}={5}^{3}=125

(The gold is conserved — melting changes shape, not volume.)

Step 2 — find the total selling price. A 20% profit on a cost of ₹10,00,000:

S{P}_{\text{total}}=1.2\times 1000000=1200000

Step 3 — selling price per bead.

S{P}_{\text{bead}}=\frac{1200000}{125}=9600

Step 4 — undo the discount. The beads sold at 20% off the listed price, so SP=0.8\times LP:

LP=\frac{9600}{0.8}=12000

Answer: Option 1 — ₹12,000

Shortcut. Three steps, each one multiplication:

LP=\frac{1000000\times 1.2}{125\times 0.8}=\frac{1200000}{100}=12000

Notice 125\times 0.8=100 exactly — the numbers were chosen to collapse.

Option 3 (9600) is the planted trap: that is the selling price per bead, the value you reach one step before the end. Discount is always applied to the listed price, so you divide by 0.8 rather than multiplying. Options 2 and 4 come from using {5}^{2} or {5}^{1} beads instead of {5}^{3}.

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