All data in a certain school are recorded either on computer or paper

Question

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%

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

Shobhit7 · Accepted Answer

Let total data be 100

Computer:
Let inaccurate data be x
Therefore, inaccurate data on both computer + paper = 0.8x
And, inaccurate data only on computer = 0.2x

Paper:
Let inaccurate data be y
Therefore, inaccurate data on both computer + paper = 0.48y
And, inaccurate data only on paper = 0.52y

Now, according to the question stem: 
0.8x = 0.48y = 12
So, x= 15 and y= 25

So, sum total of inaccurate data = 0.2x + 0.52y + 12 = 3 + 13 + 12 = 28
Hence, accurate data = 100 - 28 = 72.
Correct option : C

Tan2017 · Answer

To solve this question we need to break down the information in the question

All data in a certain school are recorded either on computer or paper. - only 2 possibilities ie the summation of the records on paper and computer. Also only 2 categories of data is available- Accurate or Inaccurate.
Thus we assume the total no of records as 100

Out of all data, inaccurate data, both on computer and paper, make up 12%. - If the total records is 100, then inaccurate records is only 12

Out of inaccurate records on the computer, data on paper make up 80%. - This implies of the total inaccurate records on computer the record on both paper and computer makes up 80%====> 12/80*100=15
Total inaccurate records on computer=15

And, out of the inaccurate data on paper, records on computer account for 48%. - Similarly Total inaccurate records on paper=12/48*100=25

Total inaccurate records=15+25-12=28

Therefore total accurate record either on computer or paper=100-28=72

IMO=C

KUDOS please, if you like the post or if it helps :-)

Schnauss · Answer

Could you please explain why you are dividing 12/80 and 12/48 to back out the real number of both mistakes? I couldn't think to do that based on the language. Is there a link to using this technique or similar problems?

Thanks in advance!

Tan2017 · Answer

I have used a combination of unitary method and venn diagrams to arrive at the solution

Based on my initial assumption, the total no of inaccurate data on both computer and paper equals 12

However carefully read this statement "Out of inaccurate records on the computer, data on paper make up 80%" - This implies:-
1) 80% is both computer and paper
2) 20% is only computer

Now we already know inaccurate records both computer and paper equals 12

When there are 80 records on both computer and paper, the total no of records on computer are 100
Therefore If the actual records on both computer and paper are 12, then the actual total no of records on computer = 12*100/80

Hope this helps! I haven't come across any similar questions, will let you know if i do so !

stne · Answer

Dear Moderator, 
Found this PS question , in the DS section of overlapping problems , Hope you will look into this. Thank you.

Bunuel · Answer

__________________________
Moved to PS forum. Thank you.

Archit3110 · Answer

GMATinsight  : sir could you please share solution on how to solve this question using matrix system...

bpdulog · Answer

Is record a subset of data? The wording is confusing for this one

Archit3110 · Answer

Yes it is really confusing.. usually such questions where 2 details are given can be solved using 2x2 matrix.. but in this case I'm not at all sure how to go ahead...,,,,??

Posted from my mobile device

AkshdeepS · Answer

Bunuel  : Please help . Not able to understand the question itself.

ScottTargetTestPrep · Answer

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

RJ7X0DefiningMyX · Answer

Here's my explanation for the solution to this not-so-difficult question, but rather a bit complex-to-understand Q (at least was for me):

PLEASE HIT KUDOS IF YOU LIKE MY SOLUTION

MathRevolution · Answer

Let the total records be 100.

Total inaccurate data: 12% = 12

Total inaccurate (12)= 80% (Computer + Paper)

=>  \frac{(12*100) }{ 80}  = 15

Total inaccurate (12)= 48% (Paper + Computer)

=>  \frac{(12*100) }{ 48}  = 25

Total inaccurate records: 25 + 15 - 12 = 28.

Total accurate: 100 - 28 = 72

Answer C

AndiTT · Answer

I agree, it is confusing as hell... if not wrong. It states that "all data in a certain school are recorded EITHER on computer or paper". So I thought a record could be EITHER on computer OR paper, never both. And the second "dimension" is correct / incorrect. So naturally, I would draw a 2x2 matrix and solve the problem. But obviously that does not work at all. Because there are records that are both on computer and paper. Once you understand that, you can draw a Venn-diagram with 3 circles, its easy from there. But you would never come to that point when you think about the very first sentence. Or am I wrong about that wording there? I´m not a native, so please explain! Would be happy to actually understand it!

Usernamevisible · Answer

The trick is to recognize this as an overlapping sets problem. Let C represent inaccurate records on the computer, P represent inaccurate records on paper, and C and P represent records that are inaccurate on both computer and paper. The statement "Out of inaccurate records on the computer, data on paper make up 80%" means that (C and P) / C = 80%. Similarly, the statement "Out of the inaccurate data on paper, records on computer account for 48%" means that (C and P) / P = 48%. Finally, we are told that records that are inaccurate on both computer and paper make up 12% of all data.

---

Step 1. Find inaccurate computer records. Since 12% is 80% of all inaccurate computer records, C = 12 / 0.8 = 15%.

---

Step 2. Find inaccurate paper records. Since 12% is 48% of all inaccurate paper records, P = 12 / 0.48 = 25%.

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Step 3. Find all inaccurate records. Use inclusion-exclusion: C union P = C + P - (C and P) = 15 + 25 - 12 = 28%. So, 28% of all records are inaccurate.

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Step 4. Find accurate records. Accurate records = 100% - 28% = 72%. Therefore, the correct answer is (C).

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Exam takeaway: Whenever you see wording like "Out of A, B make up x%," think of it as "(A and B) is x% of A." Similarly, "Out of B, A account for y%" means "(A and B) is y% of B." Once you identify the intersection, calculate the size of each set using the given percentages, then apply inclusion-exclusion to find the union. Finally, subtract the union from 100% if the question asks for the complement.

All data in a certain school are recorded either on computer or paper

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All data in a certain school are recorded either on computer or paper

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