The problem states that three successive discounts of 20%, 50%, and 10% are applied to an item. We need to find the single equivalent discount.

ArithmeticPercentageDiscountsFinancial Mathematics
2025/4/27

1. Problem Description

The problem states that three successive discounts of 20%, 50%, and 10% are applied to an item. We need to find the single equivalent discount.

2. Solution Steps

Let the original price of the item be PP.
After the first discount of 20%, the price becomes P1=P0.20P=0.80PP_1 = P - 0.20P = 0.80P.
After the second discount of 50%, the price becomes P2=P10.50P1=0.50P1=0.50(0.80P)=0.40PP_2 = P_1 - 0.50P_1 = 0.50P_1 = 0.50(0.80P) = 0.40P.
After the third discount of 10%, the price becomes P3=P20.10P2=0.90P2=0.90(0.40P)=0.36PP_3 = P_2 - 0.10P_2 = 0.90P_2 = 0.90(0.40P) = 0.36P.
The final price after all three discounts is 0.36P0.36P. The total discount is P0.36P=0.64PP - 0.36P = 0.64P.
The equivalent single discount percentage is 0.64PP×100%=0.64×100%=64%\frac{0.64P}{P} \times 100\% = 0.64 \times 100\% = 64\%.

3. Final Answer

The equivalent single discount is 64%.
Therefore, the answer is D) 64%.

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