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A six-page spreadsheet is constructed so that each row displ

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by vinni.k » Wed Aug 22, 2018 7:39 am

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A six-page spreadsheet is constructed so that each row displays one number, and all numbers are consecutive beginning at 1 (so line 1 contains the number 1; line 2 contains the number 2; etc.). If each page has the same number of rows, and a number is chosen at random from one particular page of the document, what is the probability that the number will be divisible by 3?

(1) Each page has 72 rows of numbers.

(2) The number will be chosen from page 2.

OA is A

Got the question but not sure if my working was correct: 1/6 * 24/72 (selecting 1 page out of 6, and there are 24 rows of multiples of 3: 3,6,9.....72 out of 72 rows)
Please correct me if i am wrong.

Thanks
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Source: — Data Sufficiency |

by Jay@ManhattanReview » Wed Aug 22, 2018 10:28 pm
vinni.k wrote:A six-page spreadsheet is constructed so that each row displays one number, and all numbers are consecutive beginning at 1 (so line 1 contains the number 1; line 2 contains the number 2; etc.). If each page has the same number of rows, and a number is chosen at random from one particular page of the document, what is the probability that the number will be divisible by 3?

(1) Each page has 72 rows of numbers.

(2) The number will be chosen from page 2.

OA is A

Got the question but not sure if my working was correct: 1/6 * 24/72 (selecting 1 page out of 6, and there are 24 rows of multiples of 3: 3,6,9.....72 out of 72 rows)
Please correct me if i am wrong.

Thanks
1/6 should be 6/6. Since 1/6 implies that only a specific page is qualified, it is not correct. We know that all the six pages are identical, so each of the six pages qualifies. Thus, the probability that the number will be divisible by 3 = 6/6*24/72 = 1/3.

The correct answer: A

Hope this helps!

-Jay
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by vinni.k » Thu Aug 23, 2018 7:50 am
Thanks Jay <i class="em em-grinning"></i>
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