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
B. 375
C. 255
D. 182

Answer 1

Hint: Here, we will first assume the apples the woman had at the beginning to be ‘x’. Then, as we have given that every time she sells half of her stock along with half an apple, we will find out the remaining apples using this after the first customer and then we will do this for the next three customers too. As a result we will get the number of remaining apples after she has sold to 4 customers. Then we will keep that equal to 15 and hence obtain an equation in ‘x’. Then we will solve that equation and obtain the value of x and hence we will get the required answer.

Complete step by step answer:
Now, let us first assume that the woman had x apples before she started selling them. Hence, we have to find the value of ‘x’.
Now, we have been given that, every time she sells, she sells half her stock with another half of an apple. Thus, she is always left with half of her stock minus half an apple.
Now, at first she has ‘x’ apples. Thus, after the first customer, she has half of x minus half apples left.
Hence, apples left with the woman after the first customer are given as:
$\dfrac{x}{2}-\dfrac{1}{2}$
Now, she sells half of the remaining apples and another half of an apple to next customer.
Hence, the number of apples left with the woman after she sells to the second customer are given as:
$\dfrac{\left( \dfrac{x}{2}-\dfrac{1}{2} \right)}{2}-\dfrac{1}{2}$
Similarly, the apples left with this woman after she sells to the third customer are:
$\dfrac{\left( \dfrac{\left( \dfrac{x}{2}-\dfrac{1}{2} \right)}{2}-\dfrac{1}{2} \right)}{2}-\dfrac{1}{2}$
And in the same way, the number of apples left with the woman after she sells to the fourth customer are:
$\dfrac{\dfrac{\left( \dfrac{\left( \dfrac{x}{2}-\dfrac{1}{2} \right)}{2}-\dfrac{1}{2} \right)}{2}-\dfrac{1}{2}}{2}-\dfrac{1}{2}$
Now, we have also been given that after selling to the fourth customer, the number of apples left with the woman are 15.
Hence, we can say that:
$\dfrac{\dfrac{\left( \dfrac{\left( \dfrac{x}{2}-\dfrac{1}{2} \right)}{2}-\dfrac{1}{2} \right)}{2}-\dfrac{1}{2}}{2}-\dfrac{1}{2}=15$
Now, we will solve this equation. Hence, we will get:
$\begin{align}
  & \dfrac{\dfrac{\left( \dfrac{\left( \dfrac{x}{2}-\dfrac{1}{2} \right)}{2}-\dfrac{1}{2} \right)}{2}-\dfrac{1}{2}}{2}-\dfrac{1}{2}=15 \\
 & \Rightarrow \dfrac{\dfrac{\left( \dfrac{\left( \dfrac{x}{2}-\dfrac{1}{2} \right)}{2}-\dfrac{1}{2} \right)}{2}-\dfrac{1}{2}}{2}=15+\dfrac{1}{2} \\
 & \Rightarrow \dfrac{\dfrac{\left( \dfrac{\left( \dfrac{x}{2}-\dfrac{1}{2} \right)}{2}-\dfrac{1}{2} \right)}{2}-\dfrac{1}{2}}{2}=\dfrac{31}{2} \\
 & \Rightarrow \dfrac{\left( \dfrac{\left( \dfrac{x}{2}-\dfrac{1}{2} \right)}{2}-\dfrac{1}{2} \right)}{2}-\dfrac{1}{2}=31 \\
 & \Rightarrow \dfrac{\left( \dfrac{\left( \dfrac{x}{2}-\dfrac{1}{2} \right)}{2}-\dfrac{1}{2} \right)}{2}=31+\dfrac{1}{2} \\
 & \Rightarrow \dfrac{\left( \dfrac{\left( \dfrac{x}{2}-\dfrac{1}{2} \right)}{2}-\dfrac{1}{2} \right)}{2}=\dfrac{63}{2} \\
 & \\
\end{align}$
Now, further solving it we get:
$\begin{align}
  & \Rightarrow \dfrac{\left( \dfrac{x}{2}-\dfrac{1}{2} \right)}{2}-\dfrac{1}{2}=63 \\
 & \Rightarrow \dfrac{\left( \dfrac{x}{2}-\dfrac{1}{2} \right)}{2}=63+\dfrac{1}{2} \\
 & \Rightarrow \dfrac{\left( \dfrac{x}{2}-\dfrac{1}{2} \right)}{2}=\dfrac{127}{2} \\
 & \Rightarrow \dfrac{x}{2}-\dfrac{1}{2}=127 \\
 & \Rightarrow \dfrac{x}{2}=127+\dfrac{1}{2} \\
 & \Rightarrow \dfrac{x}{2}=\dfrac{255}{2} \\
 & \therefore x=255 \\
\end{align}$
Hence, at the beginning, the woman had 255 apples.

So, the correct answer is “Option C”.

Note: Be very careful while doing these kinds of equations as there are multiple fractions used. If it gets hard or confusing, it’s better to assign a variable to each fraction and then put the value of all the variables one by one so that there will be no chance of confusion and mistakes.