Question

If the concentration of petrol in the second tank is 75% and the cost price of kerosene is half that of petrol, then what is Govind Lal's net profit percentage on selling the contents of the second tank given that he claims to sell the petrol at a profit of 25%?

• A. \(42\frac{6}{7}%\)
• B. \(66\frac{2}{3}%\)
• C. \(83\frac{1}{3}%\)
• D. 100%

Hint:

Assume unit volume for mixtures and equate selling price to marked-up value on full petrol to compute adulteration profit.

Answer 1

The Correct Option is A

Assume total quantity = 1 litre Petrol = 75% = 0.75 litre Kerosene = 25% = 0.25 litre Let the cost price of petrol = ₹1 per litre Then cost price of kerosene = ₹0.5 per litre Total cost of 1 litre mix: \[ = 0.75 \cdot 1 + 0.25 \cdot 0.5 = 0.75 + 0.125 = ₹0.875 \] Govind sells this mixture as "petrol" at 25% profit over ₹1 \[ \text{Selling price} = ₹1.25 per litre \] Net profit: \[ \text{Profit} = 1.25 - 0.875 = ₹0.375 \Rightarrow \text{Profit %} = \frac{0.375}{0.875} \cdot 100 = \frac{3}{7} \cdot 100 \approx 42.86% \quad ✘ \] Check options: % Option (a) \(42\frac{6}{7}%\) ⇒ matches above But given that concentration is 75% and kerosene is half-priced, Govind is charging as if whole tank is petrol. Let’s take new cost price of petrol = ₹4 Then: - Petrol cost: 0.75 × 4 = ₹3 - Kerosene cost: 0.25 × 2 = ₹0.5 - Total cost: ₹3.5 - Selling price = 1 litre × ₹5 (25% profit on ₹4) = ₹5 - Profit = ₹1.5 - Profit % = \(\frac{1.5}{3.5} \cdot 100 = \boxed{42.86%} \Rightarrow \text{Option (a)}\) Now try 100% petrol = ₹3 Selling price = 25% more = ₹3.75 Now if mix is 75–25: - Petrol cost = 0.75 × 3 = 2.25 - Kerosene = 0.25 × 1.5 = 0.375 - Total cost = ₹2.625 - Selling price = ₹3.75 - Profit = ₹1.125 - Profit % = \(1.125 / 2.625 = 0.4285 = 42.86%\) → Again matches option (a) Now, try reverse calculation for option (c): If profit = \(83\frac{1}{3}%\) ⇒ profit ratio = \(5/6\) So cost price = \(x\), SP = \(x + 5x/6 = 11x/6\) Now if SP = 1.25 (as in question), then: \[ 1.25 = \frac{11x}{6} \Rightarrow x = \frac{6 \cdot 1.25}{11} = \frac{7.5}{11} ≈ 0.6818 \] \[ \boxed{42.86% = 42\frac{6}{7}% \Rightarrow \text{Option (a)}} \]