tanyajoseph wrote: ↑Sat Sep 01, 2007 10:36 am
A company wants to buy computers and printers for a new branch office, and the number of computers can be at most 3 times the number of printers. Computers cost $1500 each, and printers cost $300 each. What is the greatest number of computers that the company can buy if it has a total of $9100 to spend on computers and printers?
A) 2
B) 3
C) 4
D) 5
E) 6
The number of computers can be at most 3 times the number of printers
Let C = the number of computers purchased
Let P = the number of printers purchased
So, we can write:
C ≤ 3P
Subtract 3P from both sides of the inequality to get:
C - 3P ≤ 0
Computers cost $1,500 each, and printers cost $300 each.
So, the total cost of purchasing C computers and P printers =
1500C + 300P
What is the greatest number of computers that the company can buy if it has a total of $9,100 to spend on computers and printers?
This tells us that the total cost cannot exceed $9100
So, we can write,
1500C + 300P ≤ 9100
We now have the following system of inequalities:
C - 3P ≤ 0
1500C + 300P ≤ 9100
Take the top inequality and multiply both sides by 100 to get:
100C - 300P ≤ 0
1500C + 300P ≤ 9100
Since the inequality symbols are facing the same direction, we can add the two inequalities to get:
1600C ≤ 9100
Divide both sides of the inequality by 1600 to get:
C ≤ 9100/1600
Simplify to get:
C ≤ 5.something
Since C must be a positive integer, the greatest possible value of C is 5
Answer: D
