How exactly does the response "infintely many" answer the question of "how many"?

Question

I admit that the level of this question is roughly about middle school, but this is what the question asks:

The ratio of nickels to dimes to quarters is 3:8:1. If all the coins were dimes, the amount of money would be the same. Show that there are infinitely many solutions to this problem.

Answer 1

let $n,d,q$ be the quantity of nickels, dimes and quarters respectively.

The fact they are in ratio 3:8:1 means for every 12 coins: 3 are nickels , 8 are dimes and 1 is a quarter.

So if we have 12 coins exactly there are 3 nickels, 8 dimes and 1 quarter (1.20 dollars total). On the other hand if we had 12 dimes it would be the same.

So the case with 12 works. What about the case with 24? Well we would just end up having twice of each coin and the cash would add up to 2.40 dollars. And 24 dimes also give 2.40 dollars.

We can use this method for any multiple of 12. And there are infinite of those.

Hope this helps, Regards.

Answer 2

There are 2 statements here

1. The ratio of nickels to dimes to quarters is 3:8:1.
2. If all the coins were dimes, the amount of money would be the same.

Believe it or not (do the math and you'll see), the second statement is another way of saying that there are 3 times as many nickels as there are quarters. But we already have that information in the first statement. So you could restate the problem as follows

The ratio of nickels to dimes to quarters is 3:8:1. Show that there are infinitely many solutions to this problem.

for any $q\in\mathbb{N^+}, (n,d,q)=(3q,8q,q)$ is a solution. Where $\mathbb{N^+}$={1,2,3,4,...}. Because $\mathbb{N^+}$ has an infinite number of elements, there are an infinite number of solutions.

Answer 3

Three nickels and a quarter make up $40$ cents, as do four dimes. As a consequence, the second sentence in your problem does not amount to an additional condition. It follows that any multiple of the package "$3$ nickels, $8$ dimes, and $1$ quarter" solves the problem.