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Double Matrix method

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Double Matrix method

by Mo2men » Thu Feb 02, 2017 3:45 am
Jack took an exam of 45 questions, how many questions did he answer correctly?

(1) The number he answered incorrectly is twice the number of questions he did not attempt at all.
(2) The total number of questions he answered incorrectly or did not attempt is 3/9 of all questions.

How I can use double matrix method? How statement 2 is interpreted when using 'or'?

I feel it is illogical to say un-attempted question is either correct or incorrect.

Thanks
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Source: — Data Sufficiency |
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by GMATGuruNY » Thu Feb 02, 2017 5:11 am
Mo2men wrote:Jack took an exam of 45 questions, how many questions did he answer correctly?

(1) The number he answered incorrectly is twice the number of questions he did not attempt at all.
(2) The total number of questions he answered incorrectly or did not attempt is 3/9 of all questions.
For each question, there are 3 options:
Correct, Incorrect, Not Attempted.

Let C = correct, I = incorrect, N = not attempted.

Since the total number of questions = 45, we get:
C + I + N = 45.

Statement 1:
Case 1: N=1, I=2, C = 45-1-2 = 42.
Case 2: N=2, I=4, C = 45-2-4 = 39.
Since C can be different values, INSUFFICIENT.

Statement 2:
Since I + N constitute 3/9 of the 45 questions, C must constitute 6/9 of the 45 questions:
C = (6/9)45 = 30.
SUFFICIENT.

The correct answer is B.
Last edited by GMATGuruNY on Thu Feb 02, 2017 5:30 am, edited 1 time in total.
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by GMATGuruNY » Thu Feb 02, 2017 5:16 am
Mo2men wrote:How I can use double matrix method?
Because there is no overlap between correct and incorrect -- a question cannot be answered BOTH correctly and incorrectly -- I would not use a double-matrix.
How statement 2 is interpreted when using 'or'?
incorrect or not attempted = (total incorrect) + (total not attempted).
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by Mo2men » Thu Feb 02, 2017 5:23 am
GMATGuruNY wrote:
For each question, there are 3 options:
Correct, Incorrect, Not Attempted.

Let C = correct, I = incorrect, N = not attempted.

Since the total number of questions = 45, we get:
C + I + N = 45.

Statement 1:
Case 1: N=1, I=2, C = 45-1-2 = 42.
Case 2: N=2, I=4, C = 45-2-4 = 39.
Since C can be different values, INSUFFICIENT.

Statement 2:
Since I + N constitute 3/9 of the 45 questions, C must constitute 2/9 of the 45 questions:
C = (2/9)45 = 10.
SUFFICIENT.

The correct answer is B.
Thanks for your response.

But i do not understand how you arrive to 'C must constitute 2/9 of the 45 questions'.

I think I+N= 3/9 *45 =15

So C= 30

Can you please clarify??
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by GMATGuruNY » Thu Feb 02, 2017 5:33 am
Mo2men wrote:I think I+N= 3/9 *45 =15

So C= 30

Can you please clarify??
Good catch.
My post above has been corrected to read as follows:
Since I + N constitute 3/9 of the 45 questions, C must constitute 6/9 of the 45 questions:
C = (6/9)45 = 30.
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by Mo2men » Thu Feb 02, 2017 5:41 am
GMATGuruNY wrote:
Mo2men wrote: How statement 2 is interpreted when using 'or'?
incorrect or not attempted = (total incorrect) + (total not attempted).
If we use the word 'and' instead of 'or', does it make any difference?

Thanks for support
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by GMATGuruNY » Thu Feb 02, 2017 5:46 am
Mo2men wrote:
GMATGuruNY wrote:
Mo2men wrote: How statement 2 is interpreted when using 'or'?
incorrect or not attempted = (total incorrect) + (total not attempted).
If we use the word 'and' instead of 'or', does it make any difference?

Thanks for support
For a question to be answered incorrectly, it must be ATTEMPTED.
Thus, it is not possible for a question to be incorrect AND not attempted.
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by Mo2men » Thu Feb 02, 2017 5:49 am
GMATGuruNY wrote:
For a question to be answered incorrectly, it must be ATTEMPTED.
Thus, it is not possible for a question to be incorrect AND not attempted.
Thanks a lot.

good catch. It is certainly illogical.
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