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A company wants to buy computers and printers for a new branch office

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New post 17 Jan 2016, 02:37
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A
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D
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Difficulty:

  45% (medium)

Question Stats:

67% (02:11) correct 33% (02:31) wrong based on 303 sessions

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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 $1,500 each, and printers cost $300 each. 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?

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6

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A company wants to buy computers and printers for a new branch office  [#permalink]

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New post Updated on: 17 Jan 2016, 13:54
1
BrainLab wrote:
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 $1,500 each, and printers cost $300 each. 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?

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6


\(1500c+300p=9100\) (max) -> \(15c+3p=91\) -> \(c=\frac{91-3p}{15}\)
Max # of computers that one can buy is 6*15=90, BUT we have a restriction C ≤ 3P
If we buy 1 printer we can buy max. 5 computers - but the restriction is not satisfied, 2 printers -> 5 computers OK
Answer D
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Originally posted by BrainLab on 17 Jan 2016, 02:41.
Last edited by BrainLab on 17 Jan 2016, 13:54, edited 2 times in total.
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Re: A company wants to buy computers and printers for a new branch office  [#permalink]

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New post 17 Jan 2016, 07:59
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BrainLab wrote:
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 $1,500 each, and printers cost $300 each. 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?

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6


Tricky question in it that the total money of 9100$ is not the exact total required to buy the maximum number of computers at hand.

You are given c \(\leq\) 3p ---> c/3 \(\leq\) p ...(1)

Additionally, 1500c+300p=9100 ---> p = (91-15c)/3 and from 1 we get, (91-15c)/3 \(\geq\) c/3

Thus we get, 16c \(\leq\) 91 ---> c \(\leq\) 5.XXX and as number of computers MUST be integers, the maximum value of 'c' can thus be = 5.

D is the correct answer.

Hope this helps.
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New post 19 Jan 2016, 14:00
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1
if C<= 3P, then C/3 <= P

1500C + 300P = 9100 can simplify to 15C + 3P = 91

The maximum C can equal is 3P so let's plug in C/3 for P in the above equation.

15C + (3)*(C/3) = 15C + C = 91 --> 16C = 91

So we need to figure out the max number of C that fits into 91. The answer is 5 (6 is too much as 16*6 = 96) and thus D.
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Re: A company wants to buy computers and printers for a new branch office  [#permalink]

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New post 08 Aug 2018, 22:36
1
BrainLab wrote:
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 $1,500 each, and printers cost $300 each. 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?

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6

Simple calculation is as below.

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Re: A company wants to buy computers and printers for a new branch office  [#permalink]

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New post 08 Aug 2018, 22:44
1
BrainLab wrote:
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 $1,500 each, and printers cost $300 each. 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?

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6



15c + 3p = 91

put the value of c as 6 but the condition is the c=3p (max)

so c =5 satisfies
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A company wants to buy computers and printers for a new branch office  [#permalink]

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New post 17 Jan 2019, 08:13
Using single variable :

To get maximum number of computers -
Take the "atmost" ratio of computers : printers = \(3 : 1\)
Let number of computers = \(x\) ------ then number of printers =\(x/3\)

\(1500x + 300 * \frac{x}{3}= 9100\)

\(15x + x = 91\) -------- (just dividing each side by 100)

Now just plug in values of \(x\) from AC and we see 5 works fine!
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A company wants to buy computers and printers for a new branch office   [#permalink] 17 Jan 2019, 08:13
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