Number of pens and pencils sold.

Question

Pencils cost .75$ and pens cost 1$. A total of 6 dollars worth of both were sold. Assuming at least 1 of each sold, how many pencils were sold.

Answer 1

Let x = the number of pencils

Let y = the number of pens

 

Prices:

$0.75/pencil and $1/pen

For simplicity sake, let's omit the $ sign

0.75/pencil

1/pen

 

Since a total of $6 was made, then we can multiply the unit price of each item by the number of corresponding items, take the sum, and get 6:

0.75x + 1y = 6

 

If we assume at least 1 of each were sold, we can say the following:

x ≥ 1

y ≥ 1

Therefore, x + y ≥ 2

 

Let us assume, the lowest amount possible for x, which is 1

If x = 1, then 0.75 + y = 6 → y = 5.25, but we cannot have 1/4 of a pen

 

Looking at the equation 0.75x + y = 6 or (3/4)x + y = 6, we can see that x needs to be an integer that is a multiple of 4, in order to obtain an integer answer for y. Therefore, x = 4, 8, 12, etc.

 

Let us consider x = 4 → (0.75)(4) + y = 6 → y = 3

Let us consider x = 8 → (0.75)(8) + y = 6 → y = 0

Let us consider x = 12 → (0.75)(12) + y = 6 → y = -3

 

x-values of 12 and larger yield a negative y-value, which does not make sense in the context of this problem

An x-value of 8 yielded a y-value of 0, but the problem states that y ≥ 1

 

Therefore, the answer is x = 4 pencils