1
$\begingroup$

This question already has an answer here:

Question

Please note: "She never had to cut an orange".

$\endgroup$

marked as duplicate by Oray, Kate Gregory, Deusovi Apr 27 at 14:40

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

4
$\begingroup$

She started with

7 oranges

so she sold to the first customer

7/2 + 1/2 = 4 oranges, leaving her with 3 oranges.

She sold to the second customer

3/2 + 1/2 = 2 oranges, leaving her with 1 orange.

She sold to the last customer

1/2 + 1/2 = 1 orange,

leaving her with none, and no orange had to be cut.

$\endgroup$
2
$\begingroup$

If she started with $y$ oranges, then this is true:
$$\frac{\frac{\frac{y-1}{2}-1}{2}-1}{2}=0$$
Then,
$$\frac{\frac{y-1}{2}-1}{2}-1=0$$$$\frac{\frac{y-1}{2}-1}{2}=1$$$$\frac{y-1}{2}-1=2$$$$\frac{y-1}{2}=3$$$$y-1=6$$$$y=7$$
So she started with $7$ oranges.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.