All data in a certain school are recorded either on

[VJesus12]

All data in a certain school are recorded either on computer or paper. Out of inaccurate records on the computer, data on paper make up 80%. And, out of the inaccurate data on paper, records on computer account for 48%. Out of all data, inaccurate data, both on computer and paper, make up 12%. What is the percent of accurate data among all data without regarding the record method-either on computer or paper?

A. 45%
B. 56%
C. 72%
D. 86%
E. 90%

[spoiler]OA=C[/spoiler]

Source: Manhattan GMAT

[Scott@TargetTestPrep]

All data in a certain school are recorded either on computer or paper. Out of inaccurate records on the computer, data on paper make up 80%. And, out of the inaccurate data on paper, records on computer account for 48%. Out of all data, inaccurate data, both on computer and paper, make up 12%. What is the percent of accurate data among all data without regarding the record method-either on computer or paper?

A. 45%
B. 56%
C. 72%
D. 86%
E. 90%

[spoiler]OA=C[/spoiler]

Source: Manhattan GMAT

Let c = the number of inaccurate records on computer and p = the number inaccurate records on paper. So we have:

0.8c = 0.48p

80c = 48p

5c = 3p

Now, we can let c = 30, so p = 50, and hence 0.8 x 30 = 24 (or 0.48 x 50 = 24) errors are made on both computer and paper. Letting n = the total number of records, we have:

24 = 0.12n

n = 24/0.12 = 2400/12 = 200

So there are a total of 200 records, and we need to determine the number of accurate records regardless of the recording method.

50 errors are made on the paper, which included 24 that are made on the computer. However, since 30 errors are made on the computer, there must be 6 errors made on the computer that are not made on paper. Therefore, there are a total of 50 + 6 = 56 erroneous records out of the total 200 records. In other words, there are 200 - 56 = 144 accurate records, or 144/200 = 72/100 = 72% of the records are accurate.

Answer: C