Question

If a bookshelf is `21 1/8` inches long, how many `1 7/8` inch thick books will the bookshelf hold?

Answer 1

See a solution process below:

Explanation:

To make the problem easier to work with convert each mixed number into an improper fraction:

`21 1/8 = 21 + 1/8 = (8/8 xx 21) + 1/8 = 168/8 + 1/8 = (168 + 1)/8 = 169/8`

`1 7/8 = 1 + 7/8 = (8/8 xx 1) + 7/8 = 8/8 + 7/8 = (8 + 7)/8 = 15/8`

Now, to find how many `15/8` books can fit into a `169/8` bookshelf we divide the first number into the second number and use this rule for dividing fractions:

`(color(red)(a)/color(blue)(b))/(color(green)(c)/color(purple)(d)) = (color(red)(a) xx color(purple)(d))/(color(blue)(b) xx color(green)(c))`

`(color(red)(169)/color(blue)(8))/(color(green)(15)/color(purple)(8)) => (color(red)(169) xx color(purple)(8))/(color(blue)(8) xx color(green)(15)) => (color(red)(169) xx cancel(color(purple)(8)))/(cancel(color(blue)(8)) xx color(green)(15)) => 169/15 = 11.27`

However, because you cannot fit a partial book onto the bookshelf we need to round down to `11`

You can fit `color(red)(11)`, `1 7/8` inch books on a `21 1/8` inch long bookshelf.

Answer 2

A maximum of `11` books

Explanation:

Let's define a new measurement term oct
where `1` oct `= 1/8` inch.

The shelf is `21 1/8` inches
or `21xx8+1=169` octs

Each book is `1 7/8` inches thick
or `1xx8+7=15` octs thick

`"Number of books" = darr ("shelf length")/("book thickness")darr`
`color(white)("XXXXXXX")`where `darr` num `darr` indicates the largest integer not greater than num

`"Number of books"= darr (169 cancel(" oct"))/(15cancel(" oct"))darr=11`

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This could also be done as
`color(white)("XXX")(21 1/8)/(1 7/8)`

`color(white)("XXX")=(169/8)/(15/8)`

`color(white)("XXX")=169/15`

`color(white)("XXX")=11 4/15`

...and assuming we can only have complete books, the `4/15` will need to be discarded.