2.7 Selling at the Same Price: One Profit%, One Loss% (Classic Trap)

A dealer sells two similar articles at the same selling price S, gaining x% on one and losing x% on the other (same numerical percentage, opposite sign). A very common wrong instinct is to say “profit and loss cancel out — no net profit, no net loss.” This is always false whenever the percentages are non-zero. There is always a net loss, given by:

Overall Loss% = (x/10)² = x²/100

Derivation: Let SP = S for both articles. CP₁ = S × 100/(100+x) [the article sold at x% profit] CP₂ = S × 100/(100−x) [the article sold at x% loss]

Total CP = 100S [1/(100+x) + 1/(100−x)] = 100S × [200/(100² − x²)] = 20000S/(10000 − x²) Total SP = 2S

Loss = Total CP − Total SP = 20000S/(10000−x²) − 2S = 2x²S/(10000−x²)

Loss% = Loss/Total CP × 100 = [2x²S/(10000−x²)] ÷ [20000S/(10000−x²)] × 100 = (2x²/20000) × 100 = x²/100

This confirms the shortcut. Note that this result is independent of S — it depends only on x.

Worked Example 16: A shopkeeper sells two watches, each for Rs 1980. On one he gains 10%, and on the other he loses 10%. Find his overall gain or loss percentage, and the amount.

Solution (by formula): Loss% = (10)²/100 = 1% loss. Verification: CP₁ = 1980 × 100/110 = 1800. CP₂ = 1980 × 100/90 = 2200. Total CP = 4000. Total SP = 3960. Loss = 40. Loss% = 40/4000 × 100 = 1% ✓. Loss amount = Rs 40.

Worked Example 17: Two articles are sold at Rs 2400 each — one at 20% profit, the other at 20% loss. Find the net loss percentage.

Solution: Loss% = 20²/100 = 4%. Verification: CP₁ = 2400/1.2 = 2000; CP₂ = 2400/0.8 = 3000; Total CP = 5000; Total SP = 4800; Loss = 200; Loss% = 200/5000 × 100 = 4% ✓.

Key takeaway for exams: Whenever you see “sold at the same price, x% profit on one and x% loss on the other,” the answer is a loss of x²/100 %, and you can compute it in one line without finding individual CPs — a huge time-saver in Tier-I/CHSL.

Paper-setters favour this exact phrasing because it reliably catches aspirants who reason intuitively (“equal and opposite, so it must cancel”) instead of algebraically. The underlying reason the result is always a loss and never a profit is that the article sold at a loss has a larger CP than the article sold at a profit (since a bigger CP is needed to still reach the same fixed SP after subtracting a loss), and this larger CP is weighted more heavily when the total loss is computed — the asymmetry always tips the outcome towards a net loss.

Worked Example 17A (reverse — given the net loss amount, find SP and individual CPs): Two washing machines are sold at the same price. On one, the dealer gains 20%, and on the other he loses 20%. If his overall loss on the two transactions together is Rs 400, find the selling price of each washing machine and the cost price of each.

Solution: By the shortcut, Loss% = 20²/100 = 4%. Let SP of each = S, so Total SP = 2S. Since Loss% is measured on Total CP: Loss = 0.04 × Total CP. Also, Total CP − Total SP = Loss ⇒ Total CP − 2S = 0.04 × Total CP ⇒ 0.96 × Total CP = 2S ⇒ Total CP = 2S/0.96 = 25S/12. Now Loss = Total CP − 2S = 25S/12 − 2S = S/12. Given Loss = 400 ⇒ S/12 = 400 ⇒ S = Rs 4800 (SP of each machine). Total CP = 25 × 4800/12 = Rs 10,000. CP of profit-machine = 4800/1.2 = Rs 4000; CP of loss-machine = 4800/0.8 = Rs 6000. Check: Total CP = 4000+6000 = 10,000; Total SP = 9600; Loss = 400; Loss% = 400/10000×100 = 4% ✓ (matches x²/100 with x=20).

Worked Example 17B (same trap, larger x, reverse for SP): A trader sells two mobile phones at the same selling price. He gains 25% on one and loses 25% on the other. If his overall loss on the transaction is Rs 750, find the selling price of each phone.

Solution: Loss% = 25²/100 = 6.25%. Let SP of each phone = S, Total SP = 2S. Total CP − 2S = 0.0625 × Total CP ⇒ 0.9375 × Total CP = 2S ⇒ Total CP = 2S/0.9375 = 32S/15. Loss = Total CP − 2S = 32S/15 − 2S = 2S/15. Given Loss = 750 ⇒ 2S/15 = 750 ⇒ S = Rs 5625. Check: Total SP = 11,250. CP₁ = 5625/1.25 = 4500; CP₂ = 5625/0.75 = 7500; Total CP = 12,000; Loss = 750; Loss% = 750/12000×100 = 6.25% ✓.