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GMAT Prep Data Suffeciency problem

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Source: — Data Sufficiency |
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by niketdoshi123 » Sun Apr 22, 2012 9:51 pm
Statement 1
total no of customers in division B not given, only the combined service error rate of both divisions is given. So the service error rate , in number of service error per 100 customers, for division B can not be calculated.
Hence not sufficient
Statement 2
This statement only gives the no of customers in division B.
Hence not sufficient

Combining both the statements
This will give all the data to find the service error rate for division B.
Hence sufficient

so the answer is C.
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by Anurag@Gurome » Sun Apr 22, 2012 9:53 pm
krishna239455 wrote:This problem has confused me a lot.
As per me the first data is suffecient to answer the question.
But the correct answer is "C".
Pls help to solve this problem.
Statement (1) gives overall service error rate as 1.5 per 100 customers. But it doesn't give any information about the no. of customers in Div B; NOT sufficient.

Statement (2) gives the no. of customer in division B as 9,350. So, combining this with the main question we know the total no. of customers in division A and B. Also we know the no. of service errors in division A but we don't know the no. of service errors in division B; NOT sufficient.

Combining (1) and (2), we know the total no. of customers in division A and B. Also we know the no. of service errors in division A and the total service error rate for division A and B.
Hence the no. of service errors and service error rate per 100 customers in Division B can be calculated; SUFFICIENT.

The correct answer is C.
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by krishna239455 » Sun Apr 22, 2012 10:44 pm
Dear Anurag
Tell me where I am going wrong:
Statement 1: Total no of customers for Divn A and totoal no. of service erroros is given. Hence we can find error rate for A (ERa).
Overall error rate is 1.5.
Hence, ERa+ERb=1.5
ERb=1.5-ERa.
We can find error rate of "B" division.
Statement 1 suffecient.
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by transfer9858 » Sun Apr 29, 2012 5:02 pm
i thought this was A also? Since the error rate is 1.5 aren't we able to figure out by weighted average?
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