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11 Feb 2025, 01:30
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A
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Difficulty:
15%
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Question Stats:
82% (01:34) correct
18%
(01:33)
wrong
based on 62
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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.
(1) Each page has 72 rows of numbers.
(2) The number will be chosen from page 2.
11 Feb 2025, 01:57
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Let number of rows = n
Number of pages = 6
Total rows = 6n
P = Number divisible by 3 from 6n/6n
We need to know n
Statement 1: n = 72
We can find how many numbers in 6*72 are divisible by 3, to calculate the probability
Sufficient
Eliminate B,C,E
Statement 2:
Number chosen from Page 2, doesn't help us to find how many numbers in n are divisible by 3
Insufficient
Answer is A
Number of pages = 6
Total rows = 6n
P = Number divisible by 3 from 6n/6n
We need to know n
Statement 1: n = 72
We can find how many numbers in 6*72 are divisible by 3, to calculate the probability
Sufficient
Eliminate B,C,E
Statement 2:
Number chosen from Page 2, doesn't help us to find how many numbers in n are divisible by 3
Insufficient
Answer is A
Bunuel
11 Feb 2025, 08:47
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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?
Each page of spreadsheet looks like:
(1) Each page has 72 rows of numbers.
This gives us that each page has (72-3)/3 + 1 = 24 multiples of 3
Hence probability of picking multiple of 3 will be 24/72
Hence sufficient
(2) The number will be chosen from page 2.
Does not give us any information about the sample space.
Hence insufficient
IMO Ans A
Each page of spreadsheet looks like:
| 1 | - | - |
| 2 | - | - |
| 3 | - | - |
| . | - | - |
| n | - | - |
(1) Each page has 72 rows of numbers.
This gives us that each page has (72-3)/3 + 1 = 24 multiples of 3
Hence probability of picking multiple of 3 will be 24/72
Hence sufficient
(2) The number will be chosen from page 2.
Does not give us any information about the sample space.
Hence insufficient
IMO Ans A
