Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Apr 1612:00 PM EDT
-11:59 PM EDT
You’re first to know: Save $200 on GMAT prep starting tomorrow, 4/13. Get 6 official practice tests and study with real questions. Be ready to save!
Apr 2610:00 AM EDT
-11:00 AM EDT
The Target Test Prep course represents a quantum leap forward in GMAT preparation, a radical reinterpretation of the way that students should study. Try before you buy with a 5-day, full-access trial of the course for FREE!
Apr 2711:00 AM EDT
-12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
09 Jul 2025, 08:00
Kudos
Bookmarks
Minimum: 26
Maximum: 30
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
52% (02:48) correct
48%
(03:12)
wrong
based on 289
sessions
History
Date
Time
Result
Not Attempted Yet
|
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 |
ShowHide Answer
Official Answer
Minimum: 26
Maximum: 30
09 Jul 2025, 08:30
Kudos
Bookmarks
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
Kudos
Bookmarks
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.
