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Six computers, each working at the same constant rate, together can [#permalink]
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13 Mar 2011, 10:10
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Six computers, each working at the same constant rate, together can process a certain amount of data in 15 days. How many additional computers, each working at the same constant rate, will be needed to process the same amount of data in 10 days? (A) 3 (B) 5 (C) 6 (D) 9 (E) 12
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Last edited by Bunuel on 21 May 2016, 00:16, edited 1 time in total.
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Re: Six computers, each working at the same constant rate, together can [#permalink]
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13 Mar 2011, 10:11
4. A Explanation: If six computers require 15 days to process the data, thats a total of 90 computerdays the product of 6 and 15. If you change the number of computers or the number of days, 90 will have to remain the product, whether that means 90 days of one computer or one day with 90 computers. In 10 days, the number of computers is: 10c = 90 c = 9 9 computers is 3 more than the 6 that it took to do the job in 15 days, so the correct choice is (A).
...WHAT I DON'T GET, is this business about "90 computer days"... is there another way to solve it? I wouldn't have known at all to solve it the way they solved it



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Re: Six computers, each working at the same constant rate, together can [#permalink]
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13 Mar 2011, 11:45
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n2739178 wrote: ...WHAT I DON'T GET, is this business about "90 computer days"... is there another way to solve it? I wouldn't have known at all to solve it the way they solved it
Number of computers and days taken have an inverse relation. i.e. if number of computers increase, days taken will reduce... Agreed? If to complete the work in 15 days, you need 6 computers, then to complete the work in 1 day, how many computers will you need? You will need more computers now so you multiply 15 by 6 i.e. 15*6 = 90 computers. To complete the work in 1 day, you need 90 computers. To complete the work in 10 days, how many computers will you need? You will need fewer computers now, right? So you divide 90 by 10 to get 90/10 = 9 computers. You need 3 extra computers. Another way: 15 days .................. 6 computers 10 days ...................? computers Unknown is the number of computers. You need to change the previous number of computers to get the new number. You will need to multiply either by 15/10 or 10/15. If you have fewer number of days, you will need more computers so you multiply by 15/10 (i.e. greater than 1) to give a bigger number 6*(15/10) = 9 Another way: Days (d) and computers (c) are inversely proportional. dc = k (a constant) Say x is the new number of computers. 15*6 = k 10*x = k 15*6 = 10*x x = 9 Finally, the given explanation way: 6 computers need 15 days. This means each computer is working for 15 days. How much total work is there? Work which needs 6 computers working simultaneously for 15 days so 15+15+15+15+15+15 = 90 days will be needed if only 1 computer were working. Now, if we need to finish the total work in 10 days, how many computers do we need? 10 + 10 + 10 ..... = 90 We will need 9 such computers to finish the given work.
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Re: Six computers, each working at the same constant rate, together can [#permalink]
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13 Mar 2011, 12:33
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Re: Six computers, each working at the same constant rate, together can [#permalink]
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Re: Six computers, each working at the same constant rate, together can [#permalink]
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21 May 2016, 11:08
n2739178 wrote: Six computers, each working at the same constant rate, together can process a certain amount of data in 15 days. How many additional computers, each working at the same constant rate, will be needed to process the same amount of data in 10 days?
(A) 3 (B) 5 (C) 6 (D) 9 (E) 12 Total work = Computer * Days Total work = 15 * 6 => 90 Quote: How many additional computers, each working at the same constant rate, will be needed to process the same amount of data in 10 days? Total work = Computer * Days 90 = Computer * 10 So, Total computers required is 9 So, additional computers required is 3 ( 9  6 ) So, Correct answer will be (A)
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Six computers, each working at the same constant rate, together can [#permalink]
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31 May 2016, 03:25
6 computers do work in 15 days, or they do \(\frac{1}{15}\) of the job per day together Thus one computer does \(\frac{1}{15} * \frac{1}{6}\) = \(\frac{1}{90}\) per day How many computers would we need in total to complete the job in 10 days? 1/ \(\frac{1}{90}\) * X = 10 days \(\frac{90}{X}\) = 10 days 90 = 10X X = 9 In total we need 9 computers to complete the work. Since we already have 6, we need extra 3.
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Re: Six computers, each working at the same constant rate, together can [#permalink]
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31 May 2016, 05:39
let the total work be 1
now, 1 work is done in 6*15 computer days = 90 computer days same work has to be done in x*10 computer days. Therefore,
6*15=x*10 x=9
additional computers = 3



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Six computers, each working at the same constant rate, together can [#permalink]
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02 Aug 2016, 21:21
Once you see enough of these problems you realize perhaps the most efficient way to solve them is to set up a simple multiplication equation.
\(6(15)=10(x)\).
\(90=10x\)
\(9=x.\)
Be careful, to answer the question we must attend to additional hours. The answer is \(96=3\).
This issue could be bypassed by setting up this equation. \(6(15)=(6+x)(10)\)
Dividing both sides by 2(5) we obtain \(9=6+x\)
\(Hence, x=3\)




Six computers, each working at the same constant rate, together can
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