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04 Aug 2025, 10:43
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Minimum: 26
Maximum: 30
A family of five has an average (arithmetic mean) age of 20 years and a median age also 20 years. Each member of the family is a whole number of years old, and all have different ages. Also, the oldest member is 16 years older than the youngest member.
Select for Minimum the minimum possible age of the oldest member of the family, and select for Maximum the maximum possible age of the oldest member. Make only two selections, one in each column.
Select for Minimum the minimum possible age of the oldest member of the family, and select for Maximum the maximum possible age of the oldest member. Make only two selections, one in each column.
| Minimum | Maximum | |
| 21 | ||
| 24 | ||
| 25 | ||
| 26 | ||
| 28 | ||
| 30 |
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Official Answer
Minimum: 26
Maximum: 30
04 Aug 2025, 10:43
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Official Solution:
Let the ages of the family members be \(a\), \(b\), \(c\), \(d\), and \(e\). Given:
• \(a < b < c < d < e\)
• The average age is 20. Thus, the sum of the ages is 5 * 20 = 100.
• The median age \(c\) is 20.
• Also, since the oldest member is 16 years older than the youngest member, \(e = a + 16\).
Thus, we have:
• \(a < b < 20 < d < a + 16\)
• \(a + b + 20 + d + (a + 16) = 100\)
Minimizing \(e\):
Since the sum of the terms is fixed, to minimize \(e = a + 16\), we need to maximize \(b\) and \(d\).
• Given that \(b < 20\), the highest possible value of \(b\) is 19.
• Given that \(d < e\), and \(e = a + 16\), the highest possible value of \(d\) is \(a + 15\).
So we get:
\(a + 19 + 20 + (a + 15) + (a + 16) = 100\)
This gives \(a = 10\), and \(e = a + 16 = 26\).
The distribution of ages in this case would be: {10, 19, 20, 25, 26}
Maximizing \(e\):
Since the sum of the terms is fixed, to maximize \(e = a + 16\), we need to minimize \(b\) and \(d\).
• Given that \(a < b\), the least possible value of \(b\) is \(a + 1\).
• Given that \(20 < d\), the least possible value of \(d\) is 21.
So we get:
\(a + (a + 1) + 20 + 21 + (a + 16) = 100\)
This gives \(a = 14\), and \(e = a + 16 = 30\).
The distribution of ages in this case would be: {14, 15, 20, 21, 30}
Correct answer:
Minimum "26"
Maximum "30"
Bunuel
Let the ages of the family members be \(a\), \(b\), \(c\), \(d\), and \(e\). Given:
• \(a < b < c < d < e\)
• The average age is 20. Thus, the sum of the ages is 5 * 20 = 100.
• The median age \(c\) is 20.
• Also, since the oldest member is 16 years older than the youngest member, \(e = a + 16\).
Thus, we have:
• \(a < b < 20 < d < a + 16\)
• \(a + b + 20 + d + (a + 16) = 100\)
Minimizing \(e\):
Since the sum of the terms is fixed, to minimize \(e = a + 16\), we need to maximize \(b\) and \(d\).
• Given that \(b < 20\), the highest possible value of \(b\) is 19.
• Given that \(d < e\), and \(e = a + 16\), the highest possible value of \(d\) is \(a + 15\).
So we get:
\(a + 19 + 20 + (a + 15) + (a + 16) = 100\)
This gives \(a = 10\), and \(e = a + 16 = 26\).
The distribution of ages in this case would be: {10, 19, 20, 25, 26}
Maximizing \(e\):
Since the sum of the terms is fixed, to maximize \(e = a + 16\), we need to minimize \(b\) and \(d\).
• Given that \(a < b\), the least possible value of \(b\) is \(a + 1\).
• Given that \(20 < d\), the least possible value of \(d\) is 21.
So we get:
\(a + (a + 1) + 20 + 21 + (a + 16) = 100\)
This gives \(a = 14\), and \(e = a + 16 = 30\).
The distribution of ages in this case would be: {14, 15, 20, 21, 30}
Correct answer:
Minimum "26"
Maximum "30"
Paattaa
Joined: 24 Jul 2025
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07 Oct 2025, 11:40
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I like the solution - it’s helpful.
