Question

Each of 435 bags contains at least one of the following three items: raisins, almonds, and peanuts. The number of bags that contain only raisins is 10 times the number of bags that contain only peanuts. The number of bags that contain only almonds is 20 times the number of bags that contain only raisins and peanuts. The number of bags that contain only peanuts is one-fifth the number of bags that contain only almonds. 210 bags contain almonds. How many bags contain only one kind of item?

Answer 1

320 bags

Explanation:

Introduction

Let's first assign some literals, to simplify the problem. The goal is to set everything up, in order to only use one symbol.

`p`: number of bags with only peanuts.

`a`: number of bags with only almonds.

`r`: number of bags with only raisins.

`x`: number of bags with only raisins and peanuts.

Now, the problem establish 3 useful equations. We can find equations equivalences for the next sentences.

"The number of bags that contain only raisins is 10 times the number of bags that contain only peanuts" is equivalent to
`r = 10p`.

"The number of bags that contain only almonds is 20 times the number of bags that contain only raisins and peanuts" is equivalent to
`a= 20x`.

"The number of bags that contain only peanuts is one-fifth the number of bags that contain only almonds" is equivalent to
`p = (1)/(5) a`.

Now, let's set every variable in function of
`x`

We already know that
`a = 20x`. And because of that, we also know that

`p = (1)/(5)a = (1)/(5)(20x) = 4x`

and to conclude this stage of the problem, we also know that
`r = 10p =10(4x) = 40x`

¡Let's draw it!

As there are only 3 items, it is possible to use a Venn diagram. As we can see in the diagram, the entire quantity of bags is going to be

`210 + 4x + x + 40x = 210 + 45x`

But, we also know that there are 435 bags, then we only have to solve the equation:

`210 + 45x = 435`

`45x = 435 - 210`

`45x = 225`

`x = 225/45`

`x = 5`

Conclude

Substituting
`x = 5` we get

`a = 20 x = 20(5) = 100`

`p = 4 x = 4(5) = 20`

`r = 40 x = 40(5) = 200`

Finally
`ans = 100 + 20 + 200 = 320`