Question

The luncheon special at the Diner costs $4.89. The math club has $23.50 in its account. What is the greatest number of luncheon specials they can buy with the money in iheir luncheon account?

Answer 1

The Math Club can buy `four` luncheon specials at the most.

Explanation:

You can solve this problem just by using logic.

The Math Club has about $20 and a special costs about $5.

That means that the Math Club can buy 4 specials.

`color(white)(mmmmmmmm)` . . . . . . . . . . . . . . .

This loose approximation works because the Math Club can't buy a part of a luncheon special.

So you need only a loosely approximate answer

They either have enough money to buy four specials, or they have enough money to afford five.

As it turns out, they don't have enough money for five specials, so the answer has to be `four  specials`.

`color(white)(mmmmmmmm)` . . . . . . . . . . . . . . .

Here's how to solve this problem with math:

`$23.50 -: $4.89`
`4.8` specials

You can't buy a fraction of a special, so you have to buy four and get change.

Answer:
`4  luncheon  specials`

`color(white)(mmmmmmmm)` . . . . . . . . . . . . . . .

Check
4 specials @ `$4.89` ea . . . . . . . `$19.56`

The Math Club doesn't have enough money to buy a fifth luncheon special.

After buying four, they have only `$3.94` left in the account.

`Check!`