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04 Aug 2025, 09:01
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\(p\): \(\frac{2}{5}\)
\(q\): \(\frac{2}{25}\)
A computer is randomly generating an integer \(n\) between 1 and 600, inclusive.
Select for p the probability that \(n(n + 1)\) is divisible by 5, and select for q the probability that \(n(n − 1)\) is divisible by 25.
Select for p the probability that \(n(n + 1)\) is divisible by 5, and select for q the probability that \(n(n − 1)\) is divisible by 25.
| \(p\) | \(q\) | |
| \(\frac{1}{25}\) | ||
| \(\frac{2}{25}\) | ||
| \(\frac{4}{25}\) | ||
| \(\frac{1}{5}\) | ||
| \(\frac{2}{5}\) | ||
| \(\frac{3}{5}\) |
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Official Answer
\(p\): \(\frac{2}{5}\)
\(q\): \(\frac{2}{25}\)
04 Aug 2025, 09:01
Kudos
Bookmarks
Official Solution:
• For \(n(n + 1)\) to be divisible by 5, either \(n\) or \(n + 1\) must be a multiple of 5.
In every group of five consecutive numbers, such as {1, 2, 3, 4, 5}, {6, 7, 8, 9, 10}, ..., {596, 597, 598, 599, 600}, exactly two values of \(n\) satisfy this condition: one where \(n\) is divisible by 5, and one where \(n + 1\) is divisible by 5. Therefore, the overall probability is \(\frac{2}{5}\).
• Similarly, for \(n(n - 1)\) to be divisible by 25, either \(n\) or \(n - 1\) must be a multiple of 25.
In every group of 25 consecutive numbers, exactly two values of \(n\) satisfy this condition: one where \(n - 1\) is divisible by 25, and one where \(n\) is divisible by 25. For example, in the group {1, 2, ..., 25}, \(n(n - 1)\) is divisible by 25 when \(n = 1\) or \(n = 25\). In the group {576, 577, ..., 600}, it happens when \(n = 576\) or \(n = 600\). Therefore, the overall probability is \(\frac{2}{25}\).
Correct answer:
\(p\) "\(\frac{2}{5}\)"
\(q\) "\(\frac{2}{25}\)"
Bunuel
• For \(n(n + 1)\) to be divisible by 5, either \(n\) or \(n + 1\) must be a multiple of 5.
In every group of five consecutive numbers, such as {1, 2, 3, 4, 5}, {6, 7, 8, 9, 10}, ..., {596, 597, 598, 599, 600}, exactly two values of \(n\) satisfy this condition: one where \(n\) is divisible by 5, and one where \(n + 1\) is divisible by 5. Therefore, the overall probability is \(\frac{2}{5}\).
• Similarly, for \(n(n - 1)\) to be divisible by 25, either \(n\) or \(n - 1\) must be a multiple of 25.
In every group of 25 consecutive numbers, exactly two values of \(n\) satisfy this condition: one where \(n - 1\) is divisible by 25, and one where \(n\) is divisible by 25. For example, in the group {1, 2, ..., 25}, \(n(n - 1)\) is divisible by 25 when \(n = 1\) or \(n = 25\). In the group {576, 577, ..., 600}, it happens when \(n = 576\) or \(n = 600\). Therefore, the overall probability is \(\frac{2}{25}\).
Correct answer:
\(p\) "\(\frac{2}{5}\)"
\(q\) "\(\frac{2}{25}\)"
