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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 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 \(\frac{60,000}{3,000}=20\).

Re: Weighted Average [#permalink]
25 Oct 2010, 07:16

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..

Re: ROOTS - Gmatprep [#permalink]
04 Jan 2012, 02:50

1

This post received KUDOS

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

Re: ROOTS - Gmatprep [#permalink]
04 Jan 2012, 10:52

1

This post received KUDOS

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]
10 Aug 2012, 22:38

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

Re: A computer chip manufacturer expects the ratio of the number [#permalink]
10 Aug 2012, 23:45

Expert's post

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 \(\frac{2+5+6+4}{5,000+12,000+18,000+16,000}\).

Re: A computer chip manufacturer expects the ratio of the number [#permalink]
24 Sep 2014, 10:05

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Re: A computer chip manufacturer expects the ratio of the number [#permalink]
28 Apr 2015, 16:35

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?

gmatclubot

Re: A computer chip manufacturer expects the ratio of the number
[#permalink]
28 Apr 2015, 16:35

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