Question

Suppose a certain company sells regular keyboards for $83 and wireless keyboards for $110. Last week the store sold three times as many regular keyboards as wireless. If total keyboard sales were $5,385, how many of each type were sold?

regular keyboards :

wireless keyboards:

Answer 1

Let's solve the problem step by step.

Step 1: Let's define our variables:
Let \( x \) be the number of wireless keyboards sold.
Since the store sold three times as many regular keyboards as wireless ones, the number of regular keyboards sold is \( 3x \).

Step 2: Create the equation based on the total sales:
We are given that the price of a regular keyboard is $83 and the price of a wireless keyboard is $110.
The total sales from the regular keyboards is \( 83 \times 3x \) because the store sold three times as many regular keyboards.
The total sales from the wireless keyboards is \( 110 \times x \).
The total sales for both types of keyboards last week were $5,385.

Step 3: Formulate the equation:
\( 83 \times 3x + 110 \times x = 5,385 \)

Step 4: Simplify the equation:
\( 249x + 110x = 5,385 \)
\( 359x = 5,385 \)

Step 5: Solve for \( x \):
\( x = \frac{5,385}{359} \)
\( x = 15 \)

So the store sold \( 15 \) wireless keyboards.

Step 6: Find the number of regular keyboards sold:
Since the store sold \( 3 \) times as many regular keyboards, we multiply \( 15 \) by \( 3 \):
\( 3 \times 15 = 45 \)

The store sold \( 45 \) regular keyboards.

Therefore, the answer is:
Regular keyboards: 45
Wireless keyboards: 15

Answer 2

• 45 regular keyboards
• 15 wireless keyboards

Explanation:

A set of 3 regular and 1 wireless keyboard would sell for ...

3×$83 +110 = $359

For the given sales, the number of sets sold was ...

$5385/($359/set) = 15 sets

Since there are 3 regular keyboards in each set, there were 3×15 = 45 regular keyboards sold. The number of each type of keyboards sold is ...

45 regular keyboards and 15 wireless keyboards

_____

Comment on this solution

When a problem statement tells you the ratio of one kind of item to another, it is often convenient to group the items in that ratio and deal with the groups. Sometimes, there will be a few missing or left over, for example "10 more than 3 times as many." In those cases, you can make an adjustment to the total and still deal with the groups. (Any equations you might write will effectively do this same thing.)