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805+ (Hard)|   Math Related|         
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GMAT Club Olympics 2025 (Day 8): A family of five has an average 10 Jul 2025, 04:55
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a+b+c+d+20=100
a+b+c+d=80
d-a=16 So d= a+16
a+b+c+a+16= 80
2a+b+c= 80-16
2a+b+c=64
Now looking at the options To get the maximum old age we have to minimize the rest of the terms meaning the 3rd age is 20 and the 4th age 21. From here we can try a few options greater than 21 starting with 30 which is extreme. If the 5th person is 30, the first person is 14 so 14,15,20,21,30 which yields a mean of 20 so Max is 30
To get the minimum age of the oldest person we max the rest meaning the second person has to be 19 so a,19,20,d,e. Looking at the choices obviously it cannot be 21 given there has to be another person between 3rd and 5th person. So try 24 meaning 8,19,20_24...the missing person is 29 which is incorrect based on the constraints. The next choice we try is 26 so 10,19,20_26 meaning the 4th person is 25 which perfectly fits the constraints
So Max 30 years
Min 26 years
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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GMAT Club Olympics 2025 (Day 8): A family of five has an average 10 Jul 2025, 05:09
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Different ages so:
x < y < 20 < z < x+16

y is between x+1 and 19
z is between 21 and x+15

average=20 -> the 5 ages adds up 20*5=100

If y=x+1 and z=21
x+x+1+20+21+x+16=100
3x=42
x=14 -> x+16=30 maximum

If y=19 and z=x+15
x+19+20+x+15+x+16=100
3x=30
x=10 -> x+16=26 minimum

The right answers:
Minimum=26
Maximum=30
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GMAT Club Olympics 2025 (Day 8): A family of five has an average 10 Jul 2025, 05:26
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s+x2+20+x4+s+16 = 100
=>2s+x2+x4=64

For min, take x2, x4 as high as possible. That mean, x2 = 19, x4 = s+15. Plugging this, we get s+16 = 26.

For max, take x2, x4 as low as possible. That means, x2 = s+1, x4 = 21. Plugging these, we get s+16 = 30.
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GMAT Club Olympics 2025 (Day 8): A family of five has an average 10 Jul 2025, 05:30
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Option D and Option F are the correct answers.

Lets try to understand the question first and then we will try to solve for it.

So the question starts by telling us that "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". Then the question asks us to find the minimum and maximum possible age of the oldest member of the family.

First lets assume some values which will help us in calculation.
Lets assume the age of youngest member = a
the age of second youngest member = b
the age of oldest member = y
the age of second oldest member = x
Now the question tells us that the youngest member is 16 year younger than the oldest member which would mean that: y = a+16.

As the average age of the family is 20 then the combine total age of the family will be 100.

From here we can derive an equation i.e.: a+b+20+x+y = 100
a+b+x+y = 80
a+b+x+a+16 = 80, as the question told us that youngest member is 16 year younger than the oldest member which would mean that: y = a+16.
2a+b+x = 64

Now we have used up all the information which we got from the question. Now to solve further lets take the help of options available in the question.

We can not consider Option A wile solving because as mention in the question all the family member have different age which is a whole number and the median age i.e. the third oldest member is 20 years old and Option A gives us the value which is 21 and if oldest member is 21 then their is no value left for the second oldest member.

Option B: Lets assume y = 24, then a = 8 now we put this value in the question which we got.
=>2a+b+x = 64
=>16+b+x = 64
=>b+x = 48
We know that x could be x≤23, Even if we take x = 23 then from here we will get b = 25 which is not possible as then it will become greatest value which can not be true. Eliminated


Option C: Lets assume y = 25, then a = 9 now we put this value in the question which we got.
=>2a+b+x = 64
=>18+b+x = 64
=>b+x = 46
We know that x could be x≤24, if we take x = 24 then the value of b will be 22 which would be more than the median i.e. the third oldest member which is not true. Eliminated


Option D: Lets assume y = 26, then a = 10 now we put this value in the question which we got.
=>2a+b+x = 64
=>20+b+x = 64
=>b+x = 44
We know that the maximum value which x can take is 25, which will give us that b≤19. These values satisfies the conditions. Selected

So the minimum value of the oldest member is 26 (Option D)

Option E: Lets assume y = 28, then a = 12 now we put this value in the question which we got.
=>2a+b+x = 64
=>24+b+x = 64
=>b+x = 40
The maximum value which x can take is 27, which will give us b≤13. This satisfies the condition but lets check whether the next option can also give us the answer or not.


Option F: Lets assume y = 30, then a = 14 now we put this value in the question which we got.
=>2a+b+x = 64
=>28+b+x = 64
=>b+x = 36
Now in this case the only value of b and x which satisfies all the conditions are 15 and 21 respectively. So here we can say that this option satisfies all the conditions and gives the Maximum value of the oldest member i.e. 30. Selected

So after solving and checking each of the above options we can finally conclude that the minimum and maximum possible values of the oldest member of the family is 26 (Option D) and 30 (Option F) respectively.

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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GMAT Club Olympics 2025 (Day 8): A family of five has an average 10 Jul 2025, 05:34
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The ages of the members are:
youngest
second
20
fourth
oldest=youngest+16

mean=20
total=mean*5=100

youngest+second+20+fourth+youngest+16 = 100
2*youngest+second+fourth = 64

maximum value for second: 19
maximum value for fourth: youngest+15
2*youngest+19+youngest+15 = 64
youngest=10
minimum oldest=youngest+16=26

minimum value for second: youngest+1
minimum value for fourth: 21
2*youngest+youngest+1+21 = 64
youngest=14
maximum oldest=youngest+16=30

Correct answers: Minimum=26 and Maximum=30
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GMAT Club Olympics 2025 (Day 8): A family of five has an average 10 Jul 2025, 05:52
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There are five family members with different age, and everyone is whole number years old

AM = 20
Median = 20

arranging family number from youngest to oldest the person in the middle will have age 20

suppose youngest is y years old then oldest will be y+16 years old

for minimum possible age of the oldest

y , 19, 20, y+15, y+16 take mean of all

(3y + 70)/5 = 20

y = 10

oldest = 26 years old

for max possible age of the oldest

y, y+1, 20, 21, y+16

(3y+58)/5 = 20

oldest = 30

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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GMAT Club Olympics 2025 (Day 8): A family of five has an average 10 Jul 2025, 06:51
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The total of their ages is = 20*5 = 100
Median age is 20
If youngest age is x, oldest age is x+16
All ages are distinct.

To find the minimum possible age of the oldest person, maximize all other ages. Ages in ascending order for this case ->
x, 19, 20, x+15, x+16
x+19+20+(x+15)+(x+16) = 100
3x = 100-70
x = 10
Therefore, x+16 = 10+16 = 26

To find the maximum possible age of the oldest person, minimize all other ages. Ages in ascending order for this case ->
x, x+1, 20, 21, x+16
x+(x+1)+20+21+(x+16) = 100
3x = 100-58
x = 14
Therefore, x+16 = 14+16 = 30
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GMAT Club Olympics 2025 (Day 8): A family of five has an average 10 Jul 2025, 07:14
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Y, _ , Median, _, Y+16
above is the order of the ages with Median = 20 and sum of ages = mean*5 = 20*5 = 100


1. to find minimum possible age of oldest member.
To mimimize oldest member's age, we have to maximise the other members' ages which can be done by :

Y , 19, 20, Y+15, Y+16
then adding the ages:
Y + 19+20+Y+15+Y+16 = 100
3Y+ 70 = 100
3Y = 30
Y = 10
Oldest member's age = 10+16= 26


So minimum age of oldest member = 26


1. to find maximum possible age of oldest member.
To maximize oldest member's age, we have to minimize the other members' ages which can be done by :

Y , Y+1 , 20, 21, Y+16
then adding the ages:
Y + Y+1+20+21+Y+16 = 100
3Y+ 58 = 100
3Y = 42
Y = 14
Oldest member's age = 14+16= 30

So maximum age of oldest member = 30
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GMAT Club Olympics 2025 (Day 8): A family of five has an average 10 Jul 2025, 07:22
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Total of family ages = 20*5 = 100
_ _ 20 _ _
Let youngest member’s age = a1
Oldest member’s age = a1+16 = a5

To minimize age of oldest member, minimize a and maximize all others
Since all members have diff ages, a1<a2<a3<a4<a5
Max a2 = 20-1 = 19
Max a4 = a1+15
a1+19+20+a1+15+a1+16 = 100
3a1 + 70 = 100
3a1 = 30
a1= 10
Min a5 = 26

To maximize age of oldest member, maximize a1 and minimize all others
Min a2= a+1
Min a4 = 21
a1+a1+1+20+21+a1+16 = 100
3a1+58 = 100
3a1 = 42
Max a1 = 14
Max a5 = 30

Min, Max = 26,30
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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GMAT Club Olympics 2025 (Day 8): A family of five has an average 10 Jul 2025, 07:26
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5 family members
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.
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

So,
Average = 20
=> Total = 100

Median = 20
=> 3rd person is of age 20

Let the ages of all 5 people be a,b,c,d,e
Also,
a < b < c < d <e
e = a + 16
c = 20

=> a < b < 20 < d < a + 16
Now,
Total = a + b + c + d + e = 100
=> a + b + 20 + d + a + 16 = 100
=> 2a + b + d = 64

Here we need to remember that,
a < b < 20 < d < e which is the main conditions to be satisfied
Lets try the given options 1 by 1,

e = 21, => a = 5
2*(5) + b + d = 64
b + d = 54 (not possible since e = 21 has to be the oldest and above condition cant be satisfied)

e = 24, => a = 8
2*(8) + b + d = 64
b + d = 48 (not possible since e = 24 has to be the oldest and above condition cant be satisfied)

e = 25, => a = 9
2*(9) + b + d = 64
b + d = 46 (not possible since e = 25 has to be the oldest and above condition cant be satisfied)

e = 26, => a = 10
2*(10) + b + d = 64
b + d = 44 (possible since if b = 19 and d = 25, e = 26 will be the oldest and above conditions are also satisfied)

e = 28, => a = 12
2*(12) + b + d = 64
b + d = 40 (possible since if b = 17 and d = 23, e = 28 will be the oldest and above conditions are also satisfied)

e = 30, => a = 14
2*(14) + b + d = 64
b + d = 36 (possible since if b = 15 and d = 21, e = 30 will be the oldest and above conditions are also satisfied)

So, from this we find out possible values which satisfy above conditions are 26,28 and 30

=> Minimum = 26 and Maximum = 30
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GMAT Club Olympics 2025 (Day 8): A family of five has an average 10 Jul 2025, 07:41
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n = 5
mean = 20
median = 20
All have different ages
Oldest = youngest + 16

Let the age of youngest be x

x __ 20 __ x+ 16
Sum of ages = 100
Now check options one by one
If x+16 = 30, x = 14
The other two members can have ages 15 and 21.

As we keep trying options we see that x+16 can be 26 but not lesser than that.

Minimum: 26
Maximum: 30
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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GMAT Club Olympics 2025 (Day 8): A family of five has an average 10 Jul 2025, 07:51
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We know avg. age and median both are 20.

In order in minimise the age of oldest person we need to maximise youngest person.

We also know oldest is 16 years older than youngest.

So we can also say let youngest=x then oldest will be x+16

Since we need to minimise the oldest

x,x+1,20,x+15,x+16 will be age of all people.
We know average is 20 so we know sum is 100

x+x+1+20+x+15+x+16=100
4x+52=100
4x=48
x=12
youngest when maximised is 12 years

Therefore min age of the oldest person is 12+16=28 years

We can only see 30 as maximum age that is greater then minimum hence 30 is maximum age.

Min: 28, Max 30
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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