Question

Jane had a bottle filled with juice. At first, Jane drank 1/5 the 1/4, followed by 1/3. Jane checked how much juice was left in the bottle: there was 2/3 of a cup left. How much juice was in the bottle originally?

Answer 1

Bottle originally had `5/3` or `1 2/3` cups off juice.

Explanation:

As Jane first drank `1/5`, then `1/4` and then `1/3` and GCD of denominators `5`, `4` and `3` is `60`

Let us assume there were `60` units of juice.

Jane first drank `60/5=12` units, so `60-12=48` units were left

then she drank `48/4=12` units, and `48-12=36` were left

and then she drank `36/3=12` units,

and `36-12=24` units left

As `24` units are `2/3` cup

each unit must be `2/3xx1/24` cup and

`60` units with which Jane started are equivalent to

`2/3xx1/24xx60=2/3xx1/(2xx2xx2xx3)xx2xx2xx3xx5`

`cancel2/cancel3xx1/(cancel2xxcancel2xxcancel2xx3)xxcancel2xxcancel2xxcancel3xx5`

= `5/3`

Hence bottle originally had `5/3` or `1 2/3` cups off juice.

Answer 2

Based on the stated assumption:

`"1 bottle"= 3 1/13" cups"`

I chose the presentation to show the way of thinking when doing algebra.

Explanation:

`color(blue)("Assumption:")`

`color(blue)("The fractions are related to a full bottle each time")`

`color(blue)("Dr Cawas has opted for the different interpretation of")`

`color(blue)( 1-[1/3xx1/4xx1/5]" of a bottle left giving a different answer")`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("To determine how much of the bottle was drunk as a fraction")`

Total drank `-> 1/(color(red)(5))+1/(color(red)(4))+1/(color(red)(3))`

Consider the denominators. I chose to do it this way:

`color(red)(3xx4xx5)=60`

Convert all the denominators into `60^("ths")`

`[1/5 color(magenta)(xx1)] +[1/4color(magenta)(xx1)]+[1/3color(magenta)(xx1)]`

`[1/5 color(magenta)(xx12/12)] +[1/4color(magenta)(xx15/15)]+[1/3color(magenta)(xx20/20)]`

`" "12/60" " +" "15/60" "+" "20/60" "->" "(12+15+20)/60`

`" "color(blue)(= 47/60)`

Note that 47 is a prime number so this can not be simplified
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine the amount not drunk")`

`(1-47/60)" bottle "=" "2/3" cup"`

`color(blue)(13/60" bottle "=" "2/3" cup")`..........................Equation(1)

,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine volume of original full bottle")`

We need to change the `13/60` to 1. To do this we multiply by `60/13`

Multiply both sides of Equation(1) by `color(green)(60/13)`

`color(brown)(color(green)(60/13xx)13/60" bottle "=" "color(green)(60/13xx)2/3" cup")`

`60/60xx13/13" bottle "=" " 3 1/13" cup"`

`color(blue)("Full bottle had "3 1/13 " cups")`