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# With 4 identical servers working at a constant rate, a new I

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With 4 identical servers working at a constant rate, a new I [#permalink]  18 Jun 2018, 14:56
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Question Stats:

88% (01:43) correct 11% (00:00) wrong based on 9 sessions
With 4 identical servers working at a constant rate, a new Internet search provider processes 9,600 search requests per hour. If the search provider adds 2 more identical servers, and server work rate never varies, the search provider can process 216,000 search requests in how many hours?

(A) 15
(B) 16
(C) 18
(D) 20
(E) 24
[Reveal] Spoiler: OA

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Sandy
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Re: With 4 identical servers working at a constant rate, a new I [#permalink]  23 Jun 2018, 11:28
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sandy wrote:
With 4 identical servers working at a constant rate, a new Internet search provider processes 9,600 search requests per hour. If the search provider adds 2 more identical servers, and server work rate never varies, the search provider can process 216,000 search requests in how many hours?

(A) 15
(B) 16
(C) 18
(D) 20
(E) 24

Here,
4 identical servers work at a constant rate, a new internet search provider processes 9,600 search requests per hour.

1 server work at a rate = $$\frac{9600}{4}$$ = 2400 search request per hour.

Now we know

Work = TIme * Rate * No. of units

i.e 216000 = Time * 2400 * 6

Time = $$\frac{(216,000)}{(2400*6)}$$ = 15hours
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Re: With 4 identical servers working at a constant rate, a new I [#permalink]  24 Jun 2018, 17:00
Expert's post
sandy wrote:
With 4 identical servers working at a constant rate, a new Internet search provider processes 9,600 search requests per hour. If the search provider adds 2 more identical servers, and server work rate never varies, the search provider can process 216,000 search requests in how many hours?

(A) 15
(B) 16
(C) 18
(D) 20
(E) 24

Let’s determine the rate of 6 servers with the following proportion:

4/9600 = 6/x

1/2400 = 6/x

x = 14,400

We see that 6 servers can process a total of 14,400 requests in 1 hour.

So it will take 216,000/14,400 = 15 hours to process 216,000 search requests.

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Re: With 4 identical servers working at a constant rate, a new I [#permalink]  25 Jun 2018, 07:37
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4 identical servers work at a constant rate can process 9,600 search requests per hour.

1 server work at a constant rate can process 2400 search request per hour.

so, 6 server work at a constant rate can process 2400*6 = 14400 search request per hour.

so it will take 216,000/14,400 = 15 hours to process 216,000 search requests.

GMAT Club Legend
Joined: 07 Jun 2014
Posts: 4749
GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
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Re: With 4 identical servers working at a constant rate, a new I [#permalink]  10 Jul 2018, 05:30
Expert's post
Explanation

If the search provider adds 2 identical servers to the original 4, there are now 6 servers.

Because 6 ÷ 4 = 1.5, the rate at which all 6 servers work is 1.5 times the rate at which 4 servers work:

9,600 searches per hour × 1.5 = 14,400 searches per hour

Now apply this rate to the given amount of work (216,000 searches), using the W = RT formula:

216,000 = 14,400 × T
216,000 ÷ 14,400 = 15 hours
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Re: With 4 identical servers working at a constant rate, a new I [#permalink]  18 Oct 2018, 08:42
As 4 servers can process 9600 requests in 1 hour,
• 1 server can process 96004=240096004=2400 requests per hour

If 2 more, servers are added,
• Total number of servers = 4 + 2 = 6
• Total requests processed in 1 hour = 2400 * 6 = 14400

Therefore, to process 216000 search requests,
• The servers will take 21600014400=1521600014400=15 hours

Hence, the correct answer is option A.
Re: With 4 identical servers working at a constant rate, a new I   [#permalink] 18 Oct 2018, 08:42
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