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

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.

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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\(1500c+300p=9100\) (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

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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Simple calculation is as below.

Posted from my mobile device

15c + 3p = 91

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

so c =5 satisfies

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!

We can create the following equation and inequality:

C ≤ 3P

and

1500C + 300P ≤9100

15C + 3P ≤ 91

3P ≤ 91 - 15C

P ≤ (91 - 15C)/3

Substituting, we have:

C ≤ 3(91 - 15C)/3

C ≤ 91 - 15C

16C ≤ 91

C ≤ 91/16 = 5 11/16

So, the greatest number of computers is 5.

Answer: D
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Hey VeritasKarishma @scottttargettestprep mikemcgarry having a little difficulty in navigating these particular kind of inequality problems where 2 seperate "less than" equations are given and we are supposed to take the transition points and make an equation. I will illustrate:

My understanding was that C<=3P
15C+3P<=91

(NOW TRYING TO COMBINE THE TWO INEQUALITIES I GOT)

15(3P)+3P<=91
P<=91/48 i.e. 1 printer which is obviously incorrect

I am definitely missing something but am unable to put my finger on it.Please help. Your help is much appreciated.

Best

Sidharth003

C is not equal to 3P so you cannot substitute 3P in place of C. You can add equations that have the same inequality sign (< or > etc)

C <= 3P which is same as
C - 3P <= 0 ... (I)

15C+3P<=91 ... (II)

Now, adding equation I and II, we get

16C <= 91
C <= 91/16
C <= 5.7

Check: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2015/1 ... ties-gmat/
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Since, Computers and printers are in ratio of 3k:k (at most)

Total max price paid for both would be :
$1500 x 3k + $300 x k = 9100
4800k = 9100
k= 91/48

Since we need value of 3k, multiply the value of k by 3:
3K= (91*3)/48
3K= 91/16
91= 16 x 5 + 11 (remainder is 11), so max value '3K' can take is 5 only.

Please correct me, if my process is wrong?

Logic is your best friend here. The question says 'maximise' the no.of computers.

Looking at the options,

(E) 6 : 6*1500 (cost of each computer)= 9000. Remaining= 100, but each printer is 300. So this holds no good.
(D) 5: 5*1500= 7500; leaving 1600 for printers. Now this maximises the Computer number.

2,3 and 4 could also be working but since we are maximising computers number, 5 could be the best option.

Lots of algebra…Grrrrrr..

Think pretty simple and you will get the answer in 30secs.

We want to maximise number of computers and also satisfy the constraint that number of computers should not exceed number of printers by a factor of 3.
So, why not take the combination of 3 Computers and 1 printer?
Cost = 3*1500 + 1*300 = 4800

Let’s take one more such combination
So, 6 computers and 2 printers now cost 9600 but we just have budget of 9100.

Fine, give back one computer and we satisfy the budget.

Answer is 5 computers max (definitely less than 3 times of number of printers (2))

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