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A trader sells an article at a profit of 25%. If he had bought it at 20% less and sold it for ₹10.50 less, he would have gained 30%. Find the cost price of the article.
₹50
₹60
₹40
₹75
₹50
Assume original CP = 100x. Initial SP = 125x. New CP = 80x. New SP = 80x * 1.3 = 104x. Difference (125x - 104x) = 21x = 10.50. Thus, x = 0.5 and CP = 100 * 0.5 = 50.
Initial profit = 25%, New CP = 80% of original CP, New profit = 30%, Change in SP = ₹10.50.
CPnew×(1+profitnew)=SPnew
Assume original CP = 100x. Initial SP = 125x. New CP = 80x. New SP = 80x * 1.3 = 104x. Difference (125x - 104x) = 21x = 10.50. Thus, x = 0.5 and CP = 100 * 0.5 = 50.
Students often subtract 20% directly from 125% instead of calculating 30% of the reduced CP value (80% of original).
Define Initial State
Assume the original Cost Price (CP) is 100x. Since the profit is 25%, the initial Selling Price (SP) is 125x.
CP1=100x,SP1=125x
Define Secondary State
If the article is bought for 20% less, the new CP (CP2) is 80% of 100x. At a 30% profit, the new SP (SP2) is calculated as CP2×1.3.
CP2=80x,SP2=80x×1.3=104x
Apply Condition and Solve
The problem states the selling price decreased by ₹10.50. Therefore, the difference between SP1 and SP2 is 10.50.
125x−104x=10.50⟹21x=10.50⟹x=0.5
Calculate Final CP
Substitute x=0.5 back into our assumed CP1=100x.
CP=100×0.5=50
A is correct because the difference between initial and final selling prices equates to ₹10.50, resulting in an original cost price of ₹50.
This ratio-based approach is highly effective in time-speed-distance problems where similar percentage shifts in variables occur.
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