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No of computers - 5

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by tanyajoseph » 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
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by agps » Sat Sep 01, 2007 10:52 am
if they buy 3 computer and 1 printer, they spend

3*1500+300 = 4800
add 1 computer and 1 printer 4800+1500+300 = 6600.
now, you can still buy 2 more computers before needing to buy another printer, but 2 computers coste 3000, which would bring us out of budget, so the answer is 1 more computer, 5 computers and 2 printers = 8100
there is still money left, but you can't buy any more computers adn you need those 2 printers to satisfy the requirements.
therefore the answer is 5 or D.
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by anniev2 » Wed Feb 18, 2009 8:21 pm
I got this same question incorrect during the prep and it ended up just being a careless error. I did not multiply the # of computers by the cost to determine the money spent. Therefore I was getting 15 as my answer. :oops:
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Re: No of computers - 5

by Brent@GMATPrepNow » Thu Feb 10, 2022 12:52 pm
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
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