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In 1995 Division A of company had 4850 customers. If there

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jananijayakumar
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In 1995 Division A of company had 4850 customers. If there [#permalink] New post   20 Aug 2010, 01:58
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In 1995 Division A of company had 4850 customers. If there were 86 service errors in Division A that year, what was service-error rate, in number of service errors per 100 customers, for Division B of Company X in 1995?

(1) In 1995 the overall service-error rate for Division A and B combined was 1,5 errors per 100 customers.
(2) In 1995 Division B had 9,350 customers, none of whom were customers of Division A

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Bunuel
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Re: In 1995 Division A of company had 4850 customers. If there [#permalink] New post   20 Aug 2010, 04:39
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In 1995 Division A of the company had 4850 customers, If there were 86 servise errors in Division A that year,what was servise error rate, in number of servise errors per 100 customers in Division B of the company in 1995?

Question \(\frac{x}{B}100=?\) Where x is the servise errors in Division B and B is # of customers of division B.

(1) In 1995 the overall service error rate for Division A and B combined was 1,5 errors per 100 customers.

\(\frac{86+x}{4850+B}=\frac{1.5}{100}\), two variables one equation - can not solve for variables. Also can not get the ratio needed. Not sufficient.

(2) IN 1995 division B had 9350 customers, non of whom were customers for division A

B=9350, clearly insufficient.

(1)+(2) We know the value of B, hence we can calculate x, from (1) and the fraction \(\frac{x}{B}\). Sufficient.

Answer: C.
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Re: In 1995 Division A of company had 4850 customers. If there [#permalink] New post   17 Nov 2010, 08:24
Rephrase the question:

Is there sufficient info to know the ratio of errors to customers in division B, if the error ratio for division A is already known?

Statement 1: The Ratio for the two divisions combined is insufficient, because I don't know how much customer overlap there is.

Statement 2: The number of customers in Division B is insufficient because I don't know how many errors there are.

However statement 2 adds that there is no customer overlap, so if I combine the two statements there is sufficient info.

The answer is C
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Re: In 1995 Division A of company had 4850 customers. If there [#permalink] New post   17 Nov 2010, 08:34
Bunuel wrote:
[b]
(1) In 1995 the overall service error rate for Division A and B combined was 1,5 errors per 100 customers.

\(\frac{86+x}{4850+B}=\frac{1.5}{100}\), two variables one equation - can not solve for variables. Also can not get the ratio needed. Not sufficient.


In a ratio question, two variables and one equation, can be solved. The issue here is that you cannot simply add the customers of division a and b if you don't know how many overlap
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Re: In 1995 Division A of company had 4850 customers. If there [#permalink] New post   17 Nov 2010, 12:14
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dwiesenfeld wrote:
Bunuel wrote:
[b]
(1) In 1995 the overall service error rate for Division A and B combined was 1,5 errors per 100 customers.

\(\frac{86+x}{4850+B}=\frac{1.5}{100}\), two variables one equation - can not solve for variables. Also can not get the ratio needed. Not sufficient.


In a ratio question, two variables and one equation, can be solved. The issue here is that you cannot simply add the customers of division a and b if you don't know how many overlap


Two variables and one equation can be solved for ratio of the variables only when there is no constant term getting added or subtracted.
An equation like \(x + 4850B = 1.5x\) can be solved for ratio of x and B. But an equation like \(\frac{86+x}{4850+B}=\frac{1.5}{100}\) cannot be.
The issue here is that you do not have the number of customers of division B. Since the average is weighted, you need to have the number of customers of B to get the error rate of division B. (and in case there is an overlap, then you also need the number of services which were provided together by division A and B and also the overlap in the errors)
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Re: In 1995 Division A of company had 4850 customers. If there [#permalink] New post   29 Nov 2017, 07:39
Bunuel wrote:
In 1995 Division A of the company had 4850 customers, If there were 86 servise errors in Division A that year,what was servise error rate, in number of servise errors per 100 customers in Division B of the company in 1995?

Question \(\frac{x}{B}100=?\) Where x is the servise errors in Division B and B is # of customers of division B.

(1) In 1995 the overall service error rate for Division A and B combined was 1,5 errors per 100 customers.

\(\frac{86+x}{4850+B}=\frac{1.5}{100}\), two variables one equation - can not solve for variables. Also can not get the ratio needed. Not sufficient.

(2) IN 1995 division B had 9350 customers, non of whom were customers for division A

B=9350, clearly insufficient.

(1)+(2) We know the value of B, hence we can calculate x, from (1) and the fraction \(\frac{x}{B}\). Sufficient.

Answer: C.


Hi Bunuel
The overall service error rate for Division A and B combined was 1,5 errors per 100 customers. Can we just plus the (error rate A/customer A) + (error rate B/customer B)?
Could you please explain?
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Re: In 1995 Division A of company had 4850 customers. If there [#permalink] New post   07 Jan 2018, 19:58
akara2500 wrote:
Bunuel wrote:
In 1995 Division A of the company had 4850 customers, If there were 86 servise errors in Division A that year,what was servise error rate, in number of servise errors per 100 customers in Division B of the company in 1995?

Question \(\frac{x}{B}100=?\) Where x is the servise errors in Division B and B is # of customers of division B.

(1) In 1995 the overall service error rate for Division A and B combined was 1,5 errors per 100 customers.

\(\frac{86+x}{4850+B}=\frac{1.5}{100}\), two variables one equation - can not solve for variables. Also can not get the ratio needed. Not sufficient.

(2) IN 1995 division B had 9350 customers, non of whom were customers for division A

B=9350, clearly insufficient.

(1)+(2) We know the value of B, hence we can calculate x, from (1) and the fraction \(\frac{x}{B}\). Sufficient.

Answer: C.


Hi Bunuel
The overall service error rate for Division A and B combined was 1,5 errors per 100 customers. Can we just plus the (error rate A/customer A) + (error rate B/customer B)?
Could you please explain?


Hello Bunuel,
In addition to the above question by akara2500 , please explain how the rephrased question is\(\frac{x}{B}100\) and not \(\frac{x}{100B}\) ? Shouldn't the "service errors per 100 customers " convert to \(errors/100\) ? Doesn't \(\frac{x}{B}100\) mean error rate per 100 customers ?
Please correct me if I'm wrong.
Thank you.
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Re: In 1995 Division A of company had 4850 customers. If there [#permalink] New post   07 Jan 2018, 20:29
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TaN1213 wrote:
akara2500 wrote:
Bunuel wrote:
In 1995 Division A of the company had 4850 customers, If there were 86 servise errors in Division A that year,what was servise error rate, in number of servise errors per 100 customers in Division B of the company in 1995?

Question \(\frac{x}{B}100=?\) Where x is the servise errors in Division B and B is # of customers of division B.

(1) In 1995 the overall service error rate for Division A and B combined was 1,5 errors per 100 customers.

\(\frac{86+x}{4850+B}=\frac{1.5}{100}\), two variables one equation - can not solve for variables. Also can not get the ratio needed. Not sufficient.

(2) IN 1995 division B had 9350 customers, non of whom were customers for division A

B=9350, clearly insufficient.

(1)+(2) We know the value of B, hence we can calculate x, from (1) and the fraction \(\frac{x}{B}\). Sufficient.

Answer: C.


Hi Bunuel
The overall service error rate for Division A and B combined was 1,5 errors per 100 customers. Can we just plus the (error rate A/customer A) + (error rate B/customer B)?
Could you please explain?


Hello Bunuel,
In addition to the above question by akara2500 , please explain how the rephrased question is\(\frac{x}{B}100\) and not \(\frac{x}{100B}\) ? Shouldn't the "service errors per 100 customers " convert to \(errors/100\) ? Doesn't \(\frac{x}{B}100\) mean error rate per 100 customers ?
Please correct me if I'm wrong.
Thank you.


You could answer your question if you plugged some numbers. Say x = 10 and B = 50. What would be the number of service errors per 100 customers?
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In 1995 Division A of company had 4850 customers. If there [#permalink] New post   07 Jan 2018, 22:35
Bunuel wrote:
TaN1213 wrote:

Hello Bunuel,
In addition to the above question by akara2500 , please explain how the rephrased question is\(\frac{x}{B}100\) and not \(\frac{x}{100B}\) ? Shouldn't the "service errors per 100 customers " convert to \(errors/100\) ? Doesn't \(\frac{x}{B}100\) mean error rate per 100 customers ?
Please correct me if I'm wrong.
Thank you.


You could answer your question if you plugged some numbers. Say x = 10 and B = 50. What would be the number of service errors per 100 customers?


In this case I am getting 20 and .002 (fraction) in respective cases . If # of service errors cannot be in fraction, then we should not get fraction when we put x= 3 and B=7 in
\(\frac{x}{B}100\). But we are getting a fraction of 42.8 in this case.
I get confused with such wording as X per Y abc. Eg. : cost of painting is $2 per 1/5 of pavement
Please provide with your insights
Thank You.
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Bunuel
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Re: In 1995 Division A of company had 4850 customers. If there [#permalink] New post   07 Jan 2018, 22:52
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TaN1213 wrote:
Bunuel wrote:
TaN1213 wrote:

Hello Bunuel,
In addition to the above question by akara2500 , please explain how the rephrased question is \(\frac{x}{B}100\) and not \(\frac{x}{100B}\) ? Shouldn't the "service errors per 100 customers " convert to \(errors/100\) ? Doesn't \(\frac{x}{B}100\) mean error rate per 100 customers ?
Please correct me if I'm wrong.
Thank you.


You could answer your question if you plugged some numbers. Say x = 10 and B = 50. What would be the number of service errors per 100 customers?


In this case I am getting 20 and .002 (fraction) in respective cases . If # of service errors cannot be in fraction, then we should not get fraction when we put x= 3 and B=7 in
\(\frac{x}{B}100\). But we are getting a fraction of 42.8 in this case.
I get confused with such wording as X per Y abc. Eg. : cost of painting is $2 per 1/5 of pavement
Please provide with your insights
Thank You.


The number of service errors per 100 customers is basically the percentage and should be calculated as given in the solution: \(\frac{x}{B}100\). For example, x = 10 and B = 50, then \(\frac{x}{B}100=20\), which means that the number of service errors per 100 is 20.
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Re: In 1995 Division A of company had 4850 customers. If there [#permalink] New post   22 Feb 2021, 02:05
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jananijayakumar wrote:
In 1995 Division A of company had 4850 customers. If there were 86 service errors in Division A that year, what was service-error rate, in number of service errors per 100 customers, for Division B of Company X in 1995?

(1) In 1995 the overall service-error rate for Division A and B combined was 1,5 errors per 100 customers.
(2) In 1995 Division B had 9,350 customers, none of whom were customers of Division A

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Responding to a pm:

You can use weighted average here but do not do the actual calculations.

Note that this is the weighted average formula:

Wa/Wb = (Ca - Cavg)/(Cavg - Cb)

Here the weight is the number of customers because rate given is per 100 customers. We need the number of customers for each division or the ratio of the number of customers.
We know Ca = 8600/4850 (rate per 100 customers). We know Wa = 4850
We don't know Wb, Cavg and we need to find Cb.

Stmnt 1 gives us Cavg = 1.5

Stmnt 2 gives us Wb = 9350

Using both, we can find Cb.
Answer (C)
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Re: In 1995 Division A of company had 4850 customers. If there [#permalink] New post   11 Jun 2023, 03:42
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Re: In 1995 Division A of company had 4850 customers. If there [#permalink]
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