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With 4 identical servers working at a constant rate, a new I
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18 Jun 2018, 14:56
18 Jun 2018, 14:56
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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
(A) 15
(B) 16
(C) 18
(D) 20
(E) 24
Re: With 4 identical servers working at a constant rate, a new I
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23 Jun 2018, 11:28
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
(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
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24 Jun 2018, 17:00
24 Jun 2018, 17:00
Expert Reply
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
(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.
Answer: A
