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Minimum: 26
Maximum: 30
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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.
| Minimum | Maximum | |
| 21 | ||
| 24 | ||
| 25 | ||
| 26 | ||
| 28 | ||
| 30 |
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Official Answer
Minimum: 26
Maximum: 30
09 Jul 2025, 08:30
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Bunuel
GMAT Club Official Explanation:
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}
General Discussion
09 Jul 2025, 08:33
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Here's a brief breakdown:
Let the ages be A1 < A2 < A3 < A4 < A5
Given:
Average = 20 => A1 + A2 + A3 + A4 + A5 =100
Median = 20 => A3 =20
All ages are distinct whole numbers.
A5 = A1 +16
Substitute A3 =20 and A5 = A1 +16 into the sum:
A1 +A2 +20+A4 +(A1 +16)=100
2A1 + A2 + A4 = 64
Also, we know A1 < A2 < 20 < A4 < A5
Minimum Possible Age of the Oldest Member (A5 ):
To minimize A5 , we need to minimize A1 (since A5 = A1 +16).
To minimize A1
, we need to maximize A2 and A4 in the equation 2A1 + A2 + A4 = 64.
Maximum possible A2 is 19 (since A2 < 20 and is a whole number).
Maximum possible A4 (to allow A5 to be small) would be A5 −1=(A1 +16)−1= A1 +15.
Substitute A2 =19 and A4 = A1 +15 into 2A1 + A2 + A4 =64:
2A1 +19+(A1 +15)=64
3A1 +34 = 64
3A1 =30 => A1 =10.
This gives A5 =10+16=26.
Ages: 10, 19, 20, 25, 26 (valid set).
So, the minimum possible A5 is 26.
Maximum Possible Age of the Oldest Member (A5 ):
To maximize A5 , we need to maximize A1
To maximize A1
, we need to minimize A2 and A4 in the equation 2A1 + A2 + A4 = 64.
Minimum possible A2 is A1 +1 (since A1 <A2).
Minimum possible A4 is 21 (since A4 >20).
Substitute A2 = A1 +1 and A4 =21 into 2A1 + A2 + A4 =64:
2A1 + (A1+1)+21=64
3A1 +22=64
3A1 =42 --> A1 =14.
This gives A5 =14+16=30.
Ages: 14, 15, 20, 21, 30 (valid set).
So, the maximum possible A5 is 30.
Let the ages be A1 < A2 < A3 < A4 < A5
Given:
Average = 20 => A1 + A2 + A3 + A4 + A5 =100
Median = 20 => A3 =20
All ages are distinct whole numbers.
A5 = A1 +16
Substitute A3 =20 and A5 = A1 +16 into the sum:
A1 +A2 +20+A4 +(A1 +16)=100
2A1 + A2 + A4 = 64
Also, we know A1 < A2 < 20 < A4 < A5
Minimum Possible Age of the Oldest Member (A5 ):
To minimize A5 , we need to minimize A1 (since A5 = A1 +16).
To minimize A1
, we need to maximize A2 and A4 in the equation 2A1 + A2 + A4 = 64.
Maximum possible A2 is 19 (since A2 < 20 and is a whole number).
Maximum possible A4 (to allow A5 to be small) would be A5 −1=(A1 +16)−1= A1 +15.
Substitute A2 =19 and A4 = A1 +15 into 2A1 + A2 + A4 =64:
2A1 +19+(A1 +15)=64
3A1 +34 = 64
3A1 =30 => A1 =10.
This gives A5 =10+16=26.
Ages: 10, 19, 20, 25, 26 (valid set).
So, the minimum possible A5 is 26.
Maximum Possible Age of the Oldest Member (A5 ):
To maximize A5 , we need to maximize A1
To maximize A1
, we need to minimize A2 and A4 in the equation 2A1 + A2 + A4 = 64.
Minimum possible A2 is A1 +1 (since A1 <A2).
Minimum possible A4 is 21 (since A4 >20).
Substitute A2 = A1 +1 and A4 =21 into 2A1 + A2 + A4 =64:
2A1 + (A1+1)+21=64
3A1 +22=64
3A1 =42 --> A1 =14.
This gives A5 =14+16=30.
Ages: 14, 15, 20, 21, 30 (valid set).
So, the maximum possible A5 is 30.
