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| In 1995 Division A of Company X had 4,850 customers |
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Stacey Koprince
MGMAT STAFF
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Good process, GMAT 2007.
Guest, do you want more than this for deciding sufficiency vs. insufficiency? (Note that you shouldn't really be doing this math on the test - you should be able to lay out what you need and follow through via logic, as GMAT 2007 did. Having to do the math here would put you over the 2 minute mark. But you might want to do the math in practice to make sure you understand what's going on.) If you do want the step-by-step math for proof, let us know. |
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| Yes |
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Guest
Guest
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Yes, the step by step method would be helpful to understanding the concept with more clarity.
Thanks. |
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Stacey Koprince
MGMAT STAFF
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Here you go! And just a reminder - you DO NOT want to do something like the below on the test. You want to be able to understand the below before the test so that you can see whether various statements will or won't be sufficient without doing the actual calculations!
Given: DivA had 4850 customers DivA had 86 service errors Wanted: DivB service error rate, in number of service errors per 100 customers To calculate Wanted, I need to know the total number of DivB service errors and the total number of DivB customers. I know neither of those pieces of info right now. I can figure out DivA's service error rate (though I don't really want to unless I need it): I want "per 100" and I currently have 86/4850, so I manipulate until I have 100 in the denominator and see what that makes my numerator. 86/4850 = x/100. 4850x = 8600. x=1.77. (1) DivA+DivB service error rate = 1.5 per 100 customers. You chose A here, so you thought that would be enough to find DivB rate by itself, since we already know DivA rate by itself. It isn't, though, because I don't know whether there is an equal number of customers in the two divisions. IF the number of customers is equal, then I can do a straight average calculation: (1.77 + DivBrate)/2 = 1.5. BUT if the number of customers is unequal, then I cannot do a straight average calculation. I would have to do a weighted average calculation - and that requires knowing the number of customers for both... which I don't know. So (1) is insufficient. (2) DivB customers = 9350, and no overlap between DivA and DivB. By itself, I know nothing about DivB error rate or even about combined error rate. Insufficient. (1) AND (2) I now know that there are NOT an equal number of customers and I know how many customers DivB has. Now I can do a weighted average calculation. [(1.77)(4850) + (DivBrate)(9350)] / (4850+9350) = 1.5. Solve for DivBrate. DivBrate = 1.36. (NOTE: the straight average possibility that I mentioned up in (1) would have given me a DivBrate of 1.23. So you can see that the average is not the same if the weighting - the number of customers - is different.) |
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| Re: In 1995 Division A of Company X had 4,850 customers |
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Guest
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I don't understand how A isn't sufficient. Please tell me where my logic is incorrect. The question asks us to find (Errors in B/Total # of customers in B) The error per 100 clients is the same as (Errors/# of customers) because its just the percentage. We can calculate Division A's percentage. Statement 1 says. (86/4850) + (Errors in B/Total # of customers in B)= 1.5/100 .0177 + (Errors in B/Total # of customers in B) = .015 (Errors in B/Total # of customers in B) = .015 - .0177 I don't understand why this isn't a sufficient answer. Please help me. |
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| Re: In 1995 Division A of Company X had 4,850 customers |
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Ron Purewal
MGMAT STAFF
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two things.
thing number one: BAD algebra error
nope. actually, what it says is (86 + errors in b) / (4850 + custs in b) = 1.5/100 you can't break off the 86/4850 from this fraction. if you don't realize why this is a mistake, compare similar expressions that use nothing but integers: for example, (2 + 3)/(1 + 2) vs. 2/1 + 3/2. in fact, these expressions will NEVER be equal. -- thing number two: absurd final answer
dude! 0.015 - 0.0177 is negative! obviously that's absurd, so it proves that you've done something wrong... and now, because we're awesome, you know what you did wrong. know your fractions! |
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| wtd average |
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guest612
Guest
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question,
can i potentially solve this wtd average problem as follows (i'd like to know if this is correct): 86(4850/14200)+x(9350/14200) = 1.5 Please tell me if the above equation is correct. x is the service errors for Division B and 14200 is the combined number of both Divs A + Divs B. |
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| Re: wtd average |
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Ron Purewal
MGMAT STAFF
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no, you can't do that. the left-hand side of this equation is set up as a 'weighted average' of 86 and x, weighted by the numbers of customers - an average that doesn't make sense in the first place, given that 86 and x are aggregate totals (it doesn't make any sense to average aggregate totals - the things for which weighted averages make sense are individual data points, such as test scores or temperatures). moreover, the things you're trying to average on the left side aren't parallel to the right-hand quantity, which is a rate per 100 customers rather than an aggregate total. see the posts above for what you actually can do on this problem. |
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| Re: wtd average |
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Ron Purewal
MGMAT STAFF
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btw, if you tried solving this equation yourself, you'd get a pretty sizable negative number for x. that fact in itself would confirm that this method isn't viable. |
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| In 1995 Division A of Company X had 4,850 customers |
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