In 1995 Division A of Company X had 4,850 customers

[Guest]

The problem is attached.

I think i get the answer to this conceptually. Can someone please walk me through the math involved in coming up with the answer to this problem?


The answer is C.

Thanks.

[GMAT 2007]

We need service error rate/100 customers in Division B. To calculate the rate we need two things: -

1) No. of customer in Div B in 1995
2) Service errors in B in 1995

From question stem we know the no. of customers in Div A(4850) and no. of service errors(86) in Div A

Statement (1) - Gives overall service error rate i.e. 1.5/100 customers. But it doesn't give any information about no. of customers in Div B. Hence insufficient.

Statement (2) - Gives no. of customers in Div B (9850). From question stem and (2) we know the total no. of customers in A & B also the no. of service errors in A. Still service errros in B is missing. Hence insufficient

Combine (1) & (2)

We get total service error rate for A & B, also we know the total no. of customers and no. of service errors in Div A

Hence no. of service errors and service error rate/100 customers in Div B can be calculated.

Hence sufficient.

GMAT 2007

[StaceyKoprince]

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.

[Guest]

Yes, the step by step method would be helpful to understanding the concept with more clarity.

Thanks.

[StaceyKoprince]

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.)

[Guest]

The problem is attached.

I think i get the answer to this conceptually. Can someone please walk me through the math involved in coming up with the answer to this problem?


The answer is C.

Thanks.

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.

[RonPurewal]

two things.

thing number one: BAD algebra error

Statement 1 says.

(86/4850) + (Errors in B/Total # of customers in B)= 1.5/100

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

(Errors in B/Total # of customers in B) = .015 - .0177

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!

[guest612]

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.

[RonPurewal]

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.

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.

[RonPurewal]

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.

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.

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.

[bgussin]

two things.



thing number one: BAD algebra error

Statement 1 says.

(86/4850) + (Errors in B/Total # of customers in B)= 1.5/100

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

(Errors in B/Total # of customers in B) = .015 - .0177

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!

I still feel uneasy about this problem. I can clearly see why the math above wouldn't work out, because you get a neg number. However, to save time on the test and avoid actually doing the math, I want to understand the concept better. I can't put my finger on how this is conceptually different than a "combined work" problem. In CW problem, we say Rate1 + Rate2 = RateCombined. There the rates are described as #jobs/time. Here, in this problem, why wouldn't the combined error rate simply be the combination of the individual error rates, described as #errors/customer? Again, I understand the math clearly doesn't work out under this assumption, but I'm looking for a conceptual reason why this problem differs from a common combined work problem. Is a combined error rate somehow different than an overall error rate? Thanks.

[RonPurewal]

Here, in this problem, why wouldn't the combined error rate simply be the combination of the individual error rates, described as #errors/customer?

the problem here is that your question is too vague. notice that you're using the word "combination". you may or may not be doing this on purpose, but, essentially, you're saying "combination" (a vague word that can mean a number of different things) so that you don't have to commit yourself to a more narrowly defined word, such as "sum".

unfortunately, not all rates "combine" in the same way.

for instance, SOME rates are additive. for instance, if machine A burns 5 discs per hour and machine B burns 10 discs per hour, then the two machines working together burn 5 + 10 = 15 discs per hour.
on the other hand, OTHER rates may not necessarily be additive. for instance, if machine A burns 5 discs per hour on the day shift and 10 discs per hour on the night shift, then machine A's rate for the whole period is somewhere between 5 and 10 discs per hour (depending on the lengths of the two shifts). nothing in this problem is 5 + 10 = 15 discs/hour. (this is closer to what's happening in the posted problem.)

you have to use the context of the problem to find out exactly what "combine the rates" means in each case. they don't all fit the same mold.

[kgopalncc]

Can you solve the problem like this?

From question we know:
Division A has 4850 customers and 86 service errors.

We want to know the service error rate for every 100 customers in Division B.

4850/100 = 48.50

Service Error for Div. A is 86/48.50

St 1) Can be set up as (86 + Y)/(48.5 + X) = 1.5
Where X= Customers in Div. B / 100
Y= Service Errors in Div. B
We cannot solve this equation: INSUFF

St 2) Tells us the number of Customer in Div. B is 9350 but gives no info on Service Errors for Div. B : INSUFF

Together) We know (86 + Y)/(48.5 + X)= 1.5
and we know X= 9350/100 = 93.50

We can stop here since we know we can solve for Y and ultimately calculate Service Error rate for Div. B.

[jlucero]

Can you solve the problem like this?

From question we know:
Division A has 4850 customers and 86 service errors.

We want to know the service error rate for every 100 customers in Division B.

4850/100 = 48.50

Service Error for Div. A is 86/48.50

St 1) Can be set up as (86 + Y)/(48.5 + X) = 1.5
Where X= Customers in Div. B / 100
Y= Service Errors in Div. B
We cannot solve this equation: INSUFF

St 2) Tells us the number of Customer in Div. B is 9350 but gives no info on Service Errors for Div. B : INSUFF

Together) We know (86 + Y)/(48.5 + X)= 1.5
and we know X= 9350/100 = 93.50

We can stop here since we know we can solve for Y and ultimately calculate Service Error rate for Div. B.

You can solve it this way as long as you are careful that x needs to be divided by 100 at the end. At least to me, it seems easier to convert each of these expressions to a fraction over 100:

errors/customers = error rate
(86 + Y)/(4850 + X)= 1.5/100

Notice that the math is equivalent, as you are moving the 100 from the right side to the left side of the expression and then dividing 4850 and X both by 100. Again, the math works out the same way, and if you 100% understand your logic then stick with it, but students will often find a way that works and not understand why it works and that leads to errors on future questions.