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GMAT Club Olympics 2025 (Day 8): A family of five has an average 09 Jul 2025, 10:29
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Min Value:
x+19+20+(x+15)+(x+16)=100
x=26

Max value:
x+(x+1)+20+21+(x+16)=100
x=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.
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GMAT Club Olympics 2025 (Day 8): A family of five has an average 09 Jul 2025, 10:30
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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.
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given: range = 16, median = 20, average = 20

Maximum age of oldest member => assume the biggest option that is 30 which is 10 away from the mean and the youngest member will be 30-16 = 14 which is 6 away from the mean Heece it is possible answer as second youngest and second oldest can be adjusted to give a mean of 20.

For the minimum we take lowest option which is 21 which cannot be as we cannot have the the mean of 20
Lets consider next option which is 24 and the youngest will be 24-16 = 8 which is 12 away from the mean of 0 hence again not possible to achieve a mean of 120 with this.
Then we go for 25 which will have youngest as = 25-16 = 9 which is again not possible to achieve with a mean of 20.
Now go for 26 which will have youngest as = 25-16 = 10 which is possible as the second youngest can be 19 and second highest can be 25 to balance out the mean to 20. Hence the answer.
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GMAT Club Olympics 2025 (Day 8): A family of five has an average 09 Jul 2025, 10:35
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Column 1: For the oldest person's age to be minimum, we have to maximize others' ages as much as possible. We know that the median is 20, and the sum of all ages is \(20*5=100\). If we assign a for the age of the oldest person, we can write all ages as below:

\((a-16),19,20,(a-1),a\)
This way, we maximize each one except for the oldest; all ages are different numbers, and 20 was the median.
Now we can use the total to see if a could be an integer, greater than the rest.
\((a-16)+19+20+(a-1)+a=100\), then \(3a=78\), which will result in \(a=26\), which is in line with all the conditions.

So the answer for the first column is 26.


Column 2: For the oldest person's age to be maximum, we have to minimize the other's age as much as possible. We can write all ages as below:

\((a-16),(a-15),20,21,a\)
This way, we minimize each one except for the oldest; all ages are different numbers, and 20 was the median. So, we can use total again to calculate a.
\((a-16)+(a-15)+20+21+a=100\), then \(3a=90\), which will result in \(a=30\), which is in line with all the conditions.

So the answer for the second column is 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.
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GMAT Club Olympics 2025 (Day 8): A family of five has an average 09 Jul 2025, 10:39
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sum of ages of 5 members=5*20=100
let 's'-smallest so, oldest=s+16, given all ages are different
for minimum age of oldest: we have to find s smallest value to satisfy all conditions
so ages are; s 19 20 s+15 s+16
=>3s+70=100 => s=10 oldest minimum age=s+16=26
for maximum age of oldest: we need to find s largest value to satisfy all conditions
so ages are; s s+1 20 21 s+16
=>3s+58=100 => s=14 oldest maximum age=14+16=30
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GMAT Club Olympics 2025 (Day 8): A family of five has an average 09 Jul 2025, 10:46
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So we got 5 people, avg is 20, median is 20.
Let's assume:
A B 20 C A+16
And A < B < 20 < C < A+16.
The problem says the age diff between max and min is 16 years.
So we can play around with B and D.

A + B + 20 + C + A + 16 = 100
2A + B + C = 64
B + C = 64 - 2A

Range for B: (A+1, 19)
Range for C: (21, A+15)

Min of B+C = A+1+21 = A+22
Max of B+C = 19+A+15 = 34+A

Min of B+C = Max of (64-2A)
A+22 = 64-2A
A(max) = 42/3 = 14

Max of B+C = Min of (34+A=64-2A)
A(min) = 30/3 = 10

When:
A(max) = 14, then A+16=30
A(min) = 10, then A+16=26
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GMAT Club Olympics 2025 (Day 8): A family of five has an average 09 Jul 2025, 11:10
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Family size = 5
Average age = 20 --> total age = 5 * 20 = 100
Median age = 20 --> middle person’s age = 20
All ages different whole numbers
Oldest = youngest + 16
Let youngest = y
Oldest = y + 16
Ages in order: y, a, 20, b, y + 16 (all distinct integers)
Sum: y + a + 20 + b + (y + 16) = 100
Simplify: 2y + a + b + 36 = 100
So, a + b + 2y = 64
To minimize oldest (y + 16), minimize y:
Try y = 8 --> oldest = 24 (minimum possible)
To maximize oldest, maximize y:
Try y = 12 --> oldest = 28 (maximum possible)
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GMAT Club Olympics 2025 (Day 8): A family of five has an average 09 Jul 2025, 11:34
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Sum of ages=100
x _ 20 _ x+16

(1) Maximum

14, 15, 20, 21, 30

(2) Mimimum

10, 19, 20, 25, 26
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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.
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GMAT Club Olympics 2025 (Day 8): A family of five has an average 09 Jul 2025, 11:51
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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.

5 Member, Avg = 20, sum = 100, Median mean A3 = 20

A1 + A2 +20 +A4 +A5 = 100
if A1 = x, then A5 = X+16
x + A2 +20 +A4 +x+16 = 100

to minimize, Maximize A2 and A4, and age are different, reduce 1 to next highest value
x + 19 +20 +x+15+x+16 = 100
3x+70=100
X = 10, A5 = 26


to maximize, Minimize A2 and A4, and age are different, add 1 to the lowest value
x + (X+1 ) +20 +21 +x+16 = 100
3x+58 = 100
3x = 42
X = 14
X+16 = 30
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GMAT Club Olympics 2025 (Day 8): A family of five has an average 09 Jul 2025, 11:52
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The five family members ages, listed from youngest to oldest, have an average of 20, so their total age is 100. The third age is 20 because it's the median. Also, the fifth age is 16 years older than the first age. To find the smallest possible age for the oldest member, we made the second and fourth ages as large as possible without violating the different whole numbers rule or the median. With the third age at 20, the second age could be at most 19, and the fourth age could be at most one less than the fifth age. Applying these limits showed the oldest member's minimum age is 26 years old. To find the largest possible age for the oldest member (the fifth value), we made the second and fourth ages as small as possible. The fourth age had to be at least 21 (one more than the third age), and the second age had to be one more than the first age. Using these minimum limits, the oldest member's maximum age is 30 years old.

Regards,
Lucas

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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.
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GMAT Club Olympics 2025 (Day 8): A family of five has an average 09 Jul 2025, 11:57
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Average = 20
Total sum of age = 100

Median = 20

a b 20 d e where a<b<20<d<e
also e = a+16
All have different ages

For e to be minimum, we need to maximize rest all
b = 19
d = e-1 = a+15

=> a+19+20+a+15+a+16 = 100
=> 3a + 70 = 100
=> a = 10

So ages are 10 19 20 25 26

For e to be Maximum, we need to minimize rest all
b=a+1
d=21

=> a+a+1+20+21+a+16 = 100
=> 3a+58 = 100
=> a = 14

So ages are 14 15 20 21 30

Hence 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.
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GMAT Club Olympics 2025 (Day 8): A family of five has an average 09 Jul 2025, 12:03
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Bunuel
 


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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.
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Let the five family members be A, B, C, D, E respectively. Let the order be from minimum to maximum in the same order mentioned above.

All the five different members have distinct values. The median C = 20 . Since average = 20. Sum = 100.

A + B + C + D + E = 100

A + B + D + E = 80

E = 16 + A

If E = 21, or 24 or 25. Then A = 5 or 8 or 9 respectively. So D becomes greater than E which is not possible.

If E = 26, then C = 20, A = 26-16 = 10. Sum = 56 . Remaining B + D = 100 -56 = 44.

To make E minimum, D and B has to be maximum. So, D = 25 and B = 19.

So, the values are A = 10, B = 19, C =20, D = 25 and E = 26.

E ( minimum ) = 26

We need to find E maximum. If E = 30, then A = 30-16 = 14 , C =20.

For E to be maximum, the value of D and B should be less. So D can take a value just greater than C, which is 21. And, B is a value just greater than A , that is 15.

So, the values are A = 14, B =15, C =20, D = 21, and E =30

E ( maximum) =30 .

Therefore, the MAXIMUM and MINIMUM values of E are 26 and 30 respectively.
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Avg Age = 20 ; Total of Ages = 100 ; Median = 20
Based upon the info, let the ages be : a < b < 20 < c < (a+16)

-> a + b + 20 + c + a + 16 = 100
-> 2a + b + c = 64
-> b + c = 64 - 2a

To minimize e,
-> first maximize b to 19 (since b < 20)
-> the maximize c to e-1 -> c = a + 15
Now, 19 + a + 15 = 64 - 2a
-> 3a = 30
-> a = 10
-> e = a+16 -> 26

To maximize e,
-> first minimize c to 21 (since c > 20)
-> then minimize b to a+1
Now, a + 1 + 21 = 64 - 2a
-> 3a = 42
-> a = 14
-> e = a+16 -> 30

Final Ans : Min Age = 26 and Max Age = 30
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Given conditions,

1) They all have different ages (a1, a2, a3, a4, a5)
2) each member age is a whole number
3) Average age is 20 => sum of their ages = 20×5=100.
a1 +a2 + a3+ a4+ a5 = 100
4) Median age is 20
Since there are 5 people, median would be arranged as
a1 < a2 < a3 < a4 < a5
the middle age, a3 =20
5) Oldest is 16 years older than youngest=> a5 = a1 + 16

we have find,
a) Minimum the minimum possible age of the oldest member
b) Maximum the maximum possible age of the oldest member

let's solve now,
as per condition (4), a1 < a2 < a3(20) < a4 < a5
as per condition (3), a1 +a2 + a3+ a4+ a5 = 100
=> a1 +a2 + 20 + a4+ a5 = 100
=> a1 +a2 + a4 + a5 = 80
as per condition (5), a5 = a1 + 16, subsitute in above equation
a1 +a2 + a4 + a1 + 16 = 80
2* a1 + a2 + a4 = 64 --->equation 1

now lets find (a) Minimum the minimum possible age of the oldest member

based on conditions, as a3 = 20 => a2<20
So, let a2 =19.

as per condition (5), a5 = a1 + 16 and as per condition (4)
a4<a5

a4 ≤a1 +15 and a4 >20, so let a4 ≥21.

Now, substitute a2=19 in equation 1
2* a1 + a2 + a4 = 64
2*a1+a4=45

a4 ≤a1 +15
a4=45−2a1

45−2a1 ≤a1 +15
45−15≤3a1
30≤3a1
10≤a1

So, the smallest possible integer value for a1 = 10
a2 =19
a3=20 (Given)
a4=45−2(10)=45−20=25
a5=a1 +16=10+16=26

So, the minimum possible age of the oldest member (a5) = 26

now lets find (b) Maximum the maximum possible age of the oldest member

As per equation, 2* a1 + a2 + a4 = 64

consider a2 and a4 to be small values

a2>a1 => a2 = a1 + 1
a4>20 => a4 = 21

subsitute above in equation 1

2a1 + (a1 + 1) + 21 = 64
3a1 + 22 = 64
3a1 = 42
a1 = 14

then a2 = 15, a3 = 20( Given), a4 = 21, a5 = a1 + 16 = 30

So, the maximum possible age of the oldest member (a5) is 30
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[i]
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.

Minimum the minimum possible age of the oldest member of the family = ?

Maximum the maximum possible age of the oldest member = ?

If the average age of 5 people is 20 => total sum is 20 x 5 = 100
if the medium age is 20 + everyone has a different age then : youngest < _ < 20 < _ < oldest

If minimum age would be 25 for the oldest => youngest would be 9 => 25 + 9 + 20 = 54
Then the two remaining numbers of age need to add up to 46 => not possible

if min age would be 26 then youngest would be 10 => 26 + 10 +20 = 56
then the two remaining numbers need to add up to 44 => e.g. 25 and 19 => ok

If maximum age would be 30 then youngest would be 14 => 14 + 20 + 30 = 64
then two remaining numbers of age need to add up to 36 => 15 and 21 => ok

Minimum age = 26
Maximum age = 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.
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Let's assume we have five members

m1 m2 m3 m4 m5

Age of m3 = 20

Total age = 20*5 = 100

Max age of m5 =

m1 + m1 + 1 + 20 + 21 + m1 + 16 = 100

3m1 + 58 = 100

3m1 = 42

m1 = 14

Age of m3 = 14 + 16 = 30

Min age of m5

m1 + 19 + 20 + m1 + 15 + m1 + 16

3m1 + 70 = 100

m1 = 10

Age of m3 = 10 + 16 = 26

Answer

Minimum = 26
Maximum = 30
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Question Analysis: n = 5 members, ages are five different integers | avg = 20 -> sum of ages = 100 | med = 20, so there are 2 members younger and 2 older than 20 years old. let's call youngest person's age X, so oldest will be X+16.

Finding max and min X+16:
Minium X+16 will occur when all other ages (a1 to a4) are maximum. a3 is 20, so a2 will be 19 and a4 will be X+15 (1 year less than the oldest) -> X + 19 + 20 + (X+15) + (X+16) = 100 -> X+16 = 26

Maximum X+16 will occur when all other ages (a1 to a4) are minimum. a2 will be X+1 and a4 will be 21 -> X + (X+1) + 20 + 21 + (X+16) = 100 -> X+16 = 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.
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GMAT Club Olympics 2025 (Day 8): A family of five has an average 09 Jul 2025, 14:06
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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.
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The age of the oldest member will be maximum when the age of the second oldest member is minimum and the age of the two youngest member is as small as possible

By the calculations shown above, the maximum age of the oldest member = 30 years

The age of the oldest member will be minimum when the age of the second oldest member is close to maximum and the age of the two youngest member is as large as possible.

By the calculations shown above, the minimum age of the oldest member = 26 years

Minimum = 26
Maximum = 30
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GMAT Club Olympics 2025 (Day 8): A family of five has an average 09 Jul 2025, 14:15
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Let the ages of the five family members be a1,a2,a3,a4,a5.

Avg age: a1+a2+a3+a4+a5=100
Median age: a3=20
oldest member is 16 years older than youngest member: a5=a1+16
Now putting values of a3 and a5 in eqn:
a1+a2+a4+(a1+16)=100-20
2a1+a2+a4=64

Minimum Possible Age of the Oldest Member (a5):
To minimize a5, we need to minimize a1
From 2a1+a3+a4=64. minimizing a2 and a4 will minimize a1
[*]a2 must be less than a3=20. So, the maximum possible value for a2 is 19.
[*]a4 must be greater than a3=20. So, the minimum possible value for a4 is 21.

So, 2a1+a2+a4=64
2a1+19+21=64
a1=12

Hence, a5=12+16 =28


Maximum Possible Age of the Oldest Member (a5):
To maximize a5, we need to maximize a1

[*]a2 must be greater than a1 and less than a3=20. The minimum possible value for a2 is a1+1.
[*]a4 must be greater than a3=20. The minimum possible value for a4 is 21.

In, 2a1+(a1+1)+21=64
3a1+22=64
3a1=42
a1=14
So a5=14+16 => 30


The minimum possible age of the oldest member is 28
the maximum possible age of the oldest member is 30.
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GMAT Club Olympics 2025 (Day 8): A family of five has an average 09 Jul 2025, 17:03
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N=5,mean=20 ,total age=100, median=20
third person is 20year old
oldest=16+eldest
x+a+20+b+x+16=100
2x+a+b=64
if we put 10 in x
20+19+25=64
so , minimum age can be 26
to max it lets keep x as 14
28+14+21=64
both these satisfy
So,min:26
max:30
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GMAT Club Olympics 2025 (Day 8): A family of five has an average 09 Jul 2025, 19:34
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Solution:

The average age of five family members is 20 years, and the median age is also 20 years. This tells us that the set of values in a set is in an arithmetic sequence. Also, it is mentioned that the oldest member = 16+youngest member.

Sum = 5*20=100. We can now find out the last and first numbers in the set by applying this formula: Sum = Number/2(last no.+first no.)

100= 5/2((16+Y)+Y)= Youngest member age = 12 years and the oldest would be 28 years.

Now, applying this formula, we will be able to determine the common difference. Last No. = First No. + (N-1)d

28 = 12 +(5-1)d = 4(common difference). So, the sequence follows 12,15,20,24.28.

The minimum possible age of the oldest member of the family = 24

The maximum possible age of the oldest member = 28

Bunuel
 


This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

 



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.
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