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A computer chip manufacturer expects the ratio of the number

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A computer chip manufacturer expects the ratio of the number [#permalink]

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Shipment --- No. of Defective Chips/shipment --- Total Chips in shipment
S1 ---------------------- 2 ------------------------------------------ 5,000
S2 ---------------------- 5 ------------------- ---------------------- 12,000
S3 ---------------------- 6 ------------------------------------------ 18,000
S4 ---------------------- 4 ------------------------------------------ 16,000

A computer chip manufacturer expects the ratio of the number of defective chips to the total number of chips in all future shipments to equal the corresponding ratio for shipments S1, S2, S3, and S4 combined, as shown in the table above. What’s the expected number of defective chips in a shipment of 60,000 chips?

A. 14
B. 20
C. 22
D. 24
E. 25
[Reveal] Spoiler: OA
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Re: Weighted Average [#permalink]

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Shipment --- No. of Defective Chips/shipment --- Total Chips in shipment
S1 ---------------------- 2 ------------------------------------------ 5,000
S2 ---------------------- 5 ------------------- ---------------------- 12,000
S3 ---------------------- 6 ------------------------------------------ 18,000
S4 ---------------------- 4 ------------------------------------------ 16,000

A computer chip manufacturer expects the ratio of the number of defective chips to the total number of chips in all future shipments to equal the corresponding ratio for shipments S1, S2, S3, and S4 combined, as shown in the table above. What’s the expected number of defective chips in a shipment of 60,000 chips?


A. 14
B. 20
C. 22
D. 24
E. 25

Set up equation: \(\frac{x}{60,000}=\frac{2+5+6+4}{5,000+12,000+18,000+16,000}\) --> \(x=20\);

Or: \(2+5+6+4=17\) defective chips in \(5,000+12,000+18,000+16,000=51,000\) chips, so \(\frac{17}{51,000}=\frac{1}{3,000}\): 1 in 3,000. So, expected number of defective chips in a shipment of 60,000 chips is \(\frac{60,000}{3,000}=20\).

Answer: B.
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Re: Weighted Average [#permalink]

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New post 25 Oct 2010, 08:16
Thanks guys! I don't why was i trying to take weighted average for both quantities!

By the way Bunuel, I have my GMAT on 27th (Wed).. Can you please suggest any last moment preparations? I was planning to go through the basic principles/formulas/tips & tricks, etc once that's it..
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Re: ROOTS - Gmatprep [#permalink]

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Janealams wrote:
I don't have much time left in the exam I would really appreciate explanation to following question!


for a total of 51000 chips (adding S1,S2,S3,S4) total number of defective chips is 17 ((adding defective chips of S1,S2,S3,S4) so ratio is 17/51000 or 1 every 3000 chips.

Keeping this ratio constant for 60000 chips number of defective chips will be (1/3000) * 60000 = 20
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New post 04 Jan 2012, 07:01
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let X be the # of defective chips.
17/51,000=X/60,000
X= 20
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Re: ROOTS - Gmatprep [#permalink]

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If we sum defective chips and total chips, we get 17 : 51000. For additional 9000 total chips, which shipment makes sense to split such that we can find an easy 9000 number. S3 makes sense here since 18000/2 = 9000. Therefore, number of additional defective chips = 6/3 = 3.

Total = 17 + 3 = 20 -> B
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Re: A computer chip manufacturer expects the ratio of the number [#permalink]

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New post 10 Aug 2012, 23:38
Hi just require little clarification---
In the original question it says " ratio s1, s2, s3 and s4 combined"
lets disregard the question for a while
suppose there are four ratios s1=a/b , s2=c/d ,s3=e/f and s4= g/h
doesn't combined ratio mean s1+s2+s3+s4= a/b+c/d + e/f + g/h
if yes why are we using a+c+e+g/ b+d+f+h
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Re: A computer chip manufacturer expects the ratio of the number [#permalink]

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New post 11 Aug 2012, 00:45
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ASHISH3344 wrote:
Hi just require little clarification---
In the original question it says " ratio s1, s2, s3 and s4 combined"
lets disregard the question for a while
suppose there are four ratios s1=a/b , s2=c/d ,s3=e/f and s4= g/h
doesn't combined ratio mean s1+s2+s3+s4= a/b+c/d + e/f + g/h
if yes why are we using a+c+e+g/ b+d+f+h


The question says: "...ratio for shipments S1, S2, S3, and S4 combined..."

The ratio of the number of defective chips to the total number of chips for these 4 shipments is \(\frac{2+5+6+4}{5,000+12,000+18,000+16,000}\).

Hope it's clear.
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Re: A computer chip manufacturer expects the ratio of the number [#permalink]

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the ratio of the number of defective chips to the total number of chips

i.e. (2+5+6+4) = 17 : (5,000+12,000+18,000+16,000) = 51,000

i.e. 17 : 51,000 = 1 : 3,000

which means that 1 out of every 3,000 chips is defective, so out of a total 60,000 how many are defective?

Cross multiply : (60,000 *1)/3,000 = 20 chips are defective
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Re: A computer chip manufacturer expects the ratio of the number [#permalink]

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Re: A computer chip manufacturer expects the ratio of the number [#permalink]

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New post 24 Sep 2014, 23:40
Total defective chips expected\(= 60000 * \frac{(2+5+6+4)}{(5000+12000+18000+16000)} = 20\)
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Re: A computer chip manufacturer expects the ratio of the number [#permalink]

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New post 28 Apr 2015, 17:35
shinbhu wrote:
If we sum defective chips and total chips, we get 17 : 51000. For additional 9000 total chips, which shipment makes sense to split such that we can find an easy 9000 number. S3 makes sense here since 18000/2 = 9000. Therefore, number of additional defective chips = 6/3 = 3.

Total = 17 + 3 = 20 -> B



I think you got the division wrong 6/3 = 2 Right?
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Re: A computer chip manufacturer expects the ratio of the number   [#permalink] 05 May 2016, 13:52
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