Question

A grouped bar chart displaying Number of units sold and Net Profit in Rupees across three years: Year 1 has 100 units sold and 240 net profit, Year 2 has 200 units sold and 296 net profit, and Year 3 has 300 units sold and 210 net profit.
Units sold and net profit over three years

The number of units of a product sold in three different years and the respective net profits are presented in the figure above. The cost/unit in Year 3 was ₹1, which was half the cost/unit in Year 2. The cost/unit in Year 3 was one-third of the cost/unit in Year 1. Taxes were paid on the selling price at 10%, 13% and 15% respectively for the three years. Net profit is calculated as the difference between the selling price and the sum of cost and taxes paid in that year.

The ratio of the selling price in Year 2 to the selling price in Year 3 is

Choose one option
Official answer and solution
Official answer
A
From
Official GATE 2021 answer key

Solution

Use unit costs, tax rates, and net profits to calculate selling prices for Year 2 and Year 3.

From the problem statement:

  • Cost per unit in Year 3: c3=1c_3 = 1.
  • Cost per unit in Year 2: c2=2×1=2c_2 = 2 \times 1 = 2.
  • Cost per unit in Year 1: c1=3×1=3c_1 = 3 \times 1 = 3.

From the bar chart:

  • Year 1: Units sold N1=100N_1 = 100, Net Profit P1=240P_1 = 240.
  • Year 2: Units sold N2=200N_2 = 200, Net Profit P2=296P_2 = 296.
  • Year 3: Units sold N3=300N_3 = 300, Net Profit P3=210P_3 = 210.

Net profit is defined as:

P=S−(Cost+Taxes)=S(1−t)−c⋅NP = S - (\text{Cost} + \text{Taxes}) = S(1 - t) - c \cdot N

where SS is the selling price, tt is the tax rate on selling price, and c⋅Nc \cdot N is the total cost.

For Year 2 (t2=13%t_2 = 13\%):

296=S2(1−0.13)−2×200  ⟹  0.87S2=296+400=696  ⟹  S2=6960.87=800296 = S_2(1 - 0.13) - 2 \times 200 \implies 0.87 S_2 = 296 + 400 = 696 \implies S_2 = \frac{696}{0.87} = 800

For Year 3 (t3=15%t_3 = 15\%):

210=S3(1−0.15)−1×300  ⟹  0.85S3=210+300=510  ⟹  S3=5100.85=600210 = S_3(1 - 0.15) - 1 \times 300 \implies 0.85 S_3 = 210 + 300 = 510 \implies S_3 = \frac{510}{0.85} = 600

The ratio of the selling price in Year 2 to that in Year 3 is:

S2S3=800600=43\frac{S_2}{S_3} = \frac{800}{600} = \frac{4}{3}
Final answer

A

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper
GATE 2021 Computer Science and Information Technology - Session 2 (CS2)
Question
9 of 65, printed as GA Q.9
Section
General Aptitude

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