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Question

A woman sells to the first customer half of her stock of apples and half an apple, to the second customer she sells half her remaining stock and half an apple, and so on to the third and to the fourth customer. She finds that she has now $$15$$ apples left. How many apples did she have before she started selling?


A
422
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B
375
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C
255
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D
182
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Solution

The correct option is D $$255$$
Suppose she had $$x$$ apples in the beginning. 
Sold to the first customer $$\displaystyle =\frac{x}{2}+\frac{1}{2}=\frac{x+1}{2}$$
Remaining stock $$\displaystyle =x-\frac{x+1}{2}=\frac{2x-x-1}{2}=\frac{x-1}{2}$$
Sold to the second customer $$\displaystyle =\frac{1}{2}\times\frac{x-1}{2}+\frac{1}{2}=\frac{x-1}{4}+\frac{1}{2}=\frac{x-1+2}{4}=\frac{x+1}{4}$$
Remaining stock $$\displaystyle =\left ( \frac{x-1}{2} \right )-\left ( \frac{x+1}{4} \right )=\frac{2x-2-x-1}{4}=\frac{x-3}{4}$$
Sold to the third customer $$\displaystyle \frac{1}{2}\times\frac{x-3}{4}+\frac{1}{2}=\frac{x-3}{8}+\frac{1}{2}=\frac{x-3+4}{8}=\frac{x+1}{8}$$
Remaining sock $$\displaystyle =\left ( \frac{x-3}{4} \right )-\left ( \frac{x+1}{8} \right )=\frac{2x-6-x-1}{8}=\frac{x-7}{8}$$
Sold to the fourth cutomer $$\displaystyle =\frac{1}{2}\times\frac{x-7}{8}+\frac{1}{2}$$$$=\dfrac{x-7}{16}+\dfrac{1}{2}$$$$=\dfrac{x-7+8}{16}=\dfrac{x+1}{16}$$
Therefore, $$ x-\left [ \dfrac{x+1}{2}+\dfrac{x+1}{4}+\dfrac{x+1}{8}+\dfrac{x+1}{16} \right ]=15$$
$$\Rightarrow x-\left [ \dfrac{8x+8+4x+4+2x+2+x+1}{16} \right ]=15$$
$$\Rightarrow x-\left [ \dfrac{15x+15}{16} \right ]=15$$
$$\Rightarrow \dfrac{16x-15x-15}{16}=15\Rightarrow x-15 =16\times 15 =240$$
$$\Rightarrow x= 240+15 = 255$$ 
Therefore, she had $$255$$ apples before she stared selling.

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