Orange Juice Puzzle

You drink $\frac{1}{2}$ of it. Your friend drinks $\frac{1}{4}$ of it. You drink $\frac{1}{8}$ of it, and this pattern will continue.

Whenever you drink $X$ amount of it, your friend will then drink $\frac{X}{2}$ of it. You drink twice the amount your friend does, so the answer is the solution to the equation $x+\frac{x}{2}=1\text{,}$ which has the solution $x=\boxed{\frac{2}{3}}$

Alternate solution: The amount you drink is an infinite geometric sequence with first term $\frac{1}{2}$ and ratio $\frac{1}{4}.$ The total amount you drink is equal to the sum of this geometric sequence. $\dfrac{\frac{1}{2}}{1-\frac{1}{4}}=\dfrac{\frac{1}{2}}{\frac{3}{4}}=\boxed{\frac{2}{3}}$

Let $a$ be the amount that "I" drink. Therefore $a= \frac{1}{2}+\frac{1}{8}+\frac{1}{32}+\frac{1}{128}+...$

If we multiply $a$ by $4$ , we get $4a=2+\frac{1}{2}+\frac{1}{8}+\frac{1}{32}+\frac{1}{128}+...$

Subtracting the two equations, $3a = 2 \Rightarrow a= \boxed{\frac{2}{3}}$

The second person always drinks half of what first one drank. so 2:1 => 2/(2+1) = 2/3

This is simple question. First of all, you should modelling this problem into mathematics form. You have to focus on that question, the subject is "I". So, We count the total of pint that "I" drank.

First time, "I" drink a half of orange juice, so it was $\frac{1}{2}$ , then "I" give back the juice to you. I drink it again, but it's a half of you drink, so it was $\frac{1}{4} * \frac{1}{2} = \frac{1}{8}$ . This event happen again and next "I" drink $\frac{1}{32}$ , approximate to infinite. The solution is Infinite Geometric Series , you can check at this .

Finally, the $a$ is $\frac{1}{2}$ and the rasio, $r$ is $\frac{1}{4}$ . Total of pints that "I" drink is $S_{\infty} = \frac{a}{1-r} = \frac{\frac{1}{2}}{1-\frac{1}{4}} = \frac{2}{3}$ . The answer is $\boxed{\frac{2}{3}}$ . :)

A more detailed solution (almost the same with Trevor B's solution, but with more details).

The problem says that I drink half of the juice, give it to you, then you drink half of what's remaining. Then you pass it back to me and I drink half of what you've left, and then I give it back to you again. Then the process goes on forever.

Let $S_{1}$ be the amount of juice I drank in pints, and $S_{2}$ be the amount of juice you drank in pints. Given these, it then follows that: $S_{1} = \frac{1}{2} + \frac{1}{8} + \frac{1}{32} + ...$ $S_{2} = \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + ...$

We can see from the equations above that $S_{2}$ is just half of $S_{1}$ , so factoring $\frac{1}{2}$ out from $S_{2}$ , we will get: $S_{2} = \frac{1}{2} ( \frac{1}{2} + \frac{1}{8} + \frac{1}{32} + ... )$

It now somehow looks like the first equation $S_{1}$ , therefore: $S_{2} = \frac{1}{2} S_{1}$

We can now combine these into a single equation: $S_{1} + \frac{1}{2} S_{1} = 1$ $\frac{3}{2} S_{1} = 1$ $3 S_{1} = 2$ $S_{1} = \frac{2}{3}$

And there you have it, I have drunk $\frac{2}{3}$ pint of juice.

There are two ways to approach this problem. If you note that Person 1 (the "I" in the problem) drinks 1/2 pint, then 1/8 pint, then 1/32 pint, etc., then you can apply the formula for the sum of an infinite geometric series S that has first term a and common ratio r. In this problem a = 1/2 and r = 1/4, since Person 1 drinks 1/2 pint the first time and 1/4 of the previous amount each successive time: $S = \frac{a}{1-r} = \frac{1/2}{1-(3/4)} = \boxed{\frac{2}{3}}$

You can also approach this problem by noting that Person 1 always drinks twice as much as Person 2 (the "you" in the problem). For example, Person 1 drinks 1/2 pint, then Person 2 drinks 1/4 pint. When Person 1 drinks 1/8 pint, then Person 2 drinks 1/16 pint. As the process continues forever, the total amount of orange juice drank by the two people will approach 1 pint. If Person 1 always drinks twice as much as Person 2, then Person 1 should drink 2/3 pints.

The amount the person drinks can be represented as follows. Since (1/4)<1, the amount he drinks is:

This is a problem for which we should first know about the infinite GP series. For an infinite GP, sum $(S)=\frac{a}{1-r}$ where $a=$ First term and $r=$ Common ratio and $r\lt 1$ .

I drink at first $\frac{1}{2}$ of a pint, then you drink $\frac{1}{4}$ of a pint, then again I drink $\frac{1}{8}$ of a pint and this goes on. This is an infinite GP with $a=\frac{1}{2}$ and $r=\frac{1}{4}$ for my drinking parts.

Total pints I drank $=\frac{1}{2}+\frac{1}{8}+\frac{1}{16}+....\infty$

$=\frac{\frac{1}{2}}{1-\frac{1}{4}}=\frac{\frac{1}{2}}{\frac{3}{4}}=\frac{1}{2}\times \frac{4}{3}=\boxed{\frac{2}{3}}$

Let you drink $x$ much of the pint, we have:

$\begin{aligned} x & = & \frac { 1 }{ 2 } +{ (\frac { 1 }{ { 2 } } ) }^{ 3 }+{ (\frac { 1 }{ { 2 } } ) }^{ 5 }+\cdots \\ 2x & = & 1+{ (\frac { 1 }{ 2 } ) }^{ 2 }+{ (\frac { 1 }{ { 2 } } ) }^{ 4 }+{ (\frac { 1 }{ { 2 } } ) }^{ 6 }+\cdots \\ 2x & = & 1+{ (\frac { 1 }{ 2 } ) }^{ 2 }(1+{ (\frac { 1 }{ 2 } ) }^{ 2 }+{ (\frac { 1 }{ { 2 } } ) }^{ 4 }+{ (\frac { 1 }{ { 2 } } ) }^{ 6 }+\cdots ) \\ 2x & = & 1+\frac { 1 }{ 4 } (2x) \\ 4x & = & 2+x \\ 3x & = & 2 \\ x & = & \boxed { \frac { 2 }{ 3 } } \end{aligned}$

$\therefore$ You drink two-thirds of the pint.

Trevor B's solution is terrific. I took advantage of the multiple choice aspect of the problem. I start with 1/2 and go up from there so it's not D. You start with 1/4 and go up from there so it's not C. After the second sip, I've had 5/8 and go up from there so it's not B. Thus A.

I drink at first 1/2 of juice next turn i drink,1/8 So it is in a GP 1/2+1/8+1/32+............. Sum =1/2(1/(1-1/4))) =2/3 $LaTeX$

First, he drinks 1/2 of the juice then 1/8th, then, drinks 1/32. This forms a GP continuing till infinity where a=1/2, r=1/4 Therfore , S(n)=a/(1-r)= (1/2)/(1-1/4)= 2/3

It's simple. The answer can be obtained by analysing the given choices.

$\sum_{i=2}^(infinity)$ (1/2)^(n-1) - (1/4)^(n-1)