Question

1. John picked a bad of oranges to give to his friends. To the first of his friends, he gave half of the oranges he had and another one beside. to his second friend he gave half of the remaining oranges and another one beside. By this time, John had one orange left. How many did he start with?

Answer 1

This is really long. Continue, Kreshnik

Answer 2

x/2 +(x/2)/2+1=x
solving for X
we get:
\[\LARGE \frac x2+\frac x4+1=x\]
\[\LARGE \frac {2x}{4}+\frac x4+\frac44 =x\]
\[\LARGE \frac {3x+4}{4} =x\]
\[\LARGE 4x=3x+4\]
\[\LARGE 4x-3x=4\]
\[\LARGE x=4\]

let's pfove it:
\[\LARGE \frac42 +\frac44+1=4\]
\[\LARGE 2+1+1=4\]
\[\LARGE 4=4\]
TRUE ... So he had 4 oranges :)

Answer 3

sorry prove ... lol

Answer 4

@Kreshnik: Wait, the first time, he game half oranges plus one more. Second time, he gave half of that AND THEN ONE MORE.(where is that gone in your equation)?

Answer 5

huh?

\[\LARGE \left(\frac x2+1\right)+\frac{\frac x2+1}{2}+1=x\]
sorry I guess I didn't understand it well enough... Is this what we are talking about? :) @ParthKohli