Thanks to Dieter from Germany for the suggestion! This puzzle appeared in a German newspaper.
Math StackExchange
https://math.stackexchange.com/questions/2455403/number-of-people-to-be-found-knowing-constraints-on-mean-ages
Subscribe: https://www.youtube.com/user/MindYourDecisions?sub_confirmation=1
Send me suggestions by email (address at end of many videos). I may not reply but I do consider all ideas!
If you purchase through these links, I may be compensated for purchases made on Amazon. As an Amazon Associate I earn from qualifying purchases. This does not affect the price you pay.
If you purchase through these links, I may be compensated for purchases made on Amazon. As an Amazon Associate I earn from qualifying purchases. This does not affect the price you pay.
Book ratings are from January 2023.
My Books (worldwide links)
https://mindyourdecisions.com/blog/my-books/#worldwide
My Books (US links)
Mind Your Decisions: Five Book Compilation
https://amzn.to/2pbJ4wR
A collection of 5 books:
“The Joy of Game Theory” rated 4.3/5 stars on 290 reviews
http://amzn.to/1uQvA20
“The Irrationality Illusion: How To Make Smart Decisions And Overcome Bias” rated 4.1/5 stars on 33 reviews
http://amzn.to/1o3FaAg
“40 Paradoxes in Logic, Probability, and Game Theory” rated 4.2/5 stars on 54 reviews
http://amzn.to/1LOCI4U
“The Best Mental Math Tricks” rated 4.3/5 stars on 116 reviews
http://amzn.to/18maAdo
“Multiply Numbers By Drawing Lines” rated 4.4/5 stars on 37 reviews
http://amzn.to/XRm7M4
Mind Your Puzzles: Collection Of Volumes 1 To 3
https://amzn.to/2mMdrJr
A collection of 3 books:
“Math Puzzles Volume 1” rated 4.4/5 stars on 112 reviews
http://amzn.to/1GhUUSH
“Math Puzzles Volume 2” rated 4.2/5 stars on 33 reviews
http://amzn.to/1NKbyCs
“Math Puzzles Volume 3” rated 4.2/5 stars on 29 reviews
http://amzn.to/1NKbGlp
2017 Shorty Awards Nominee. Mind Your Decisions was nominated in the STEM category (Science, Technology, Engineering, and Math) along with eventual winner Bill Nye; finalists Adam Savage, Dr. Sandra Lee, Simone Giertz, Tim Peake, Unbox Therapy; and other nominees Elon Musk, Gizmoslip, Hope Jahren, Life Noggin, and Nerdwriter.
My Blog
https://mindyourdecisions.com/blog/
Twitter
https://twitter.com/preshtalwalkar
Instagram
https://www.instagram.com/preshtalwalkar/
Merch
https://teespring.com/stores/mind-your-decisions
Herr Dieter, gut problem. Ya!
As someone from India this was very easy then a typical maths question that comes in our exams…..
Antes de entrar Ann en la sala, la edad media es (m=s/p); siendo “p” el número de personas presentes y “s” la suma de sus edades →→ Si “a” es la edad de Anne → (pm+a)/(p+1)=(m+4) → a=m+4(p+1) →→ (pm+2a)/(p+2)=(m+7) → a=m+(7/2)(p+2) →→→ m+4(p+1) = m+(7/2)(p+2) → p=6
Gracias y un saludo cordial.
here's another method I used that seems a little easier:
A = m + 4(n+1)
the age of ann will be the initial average, plus 4 years for everyone now in the room, so that the average age gets raised by 4 years
A = (m+4) + 3(n+2)
the age of beth will be the new average (the initial average plus four years), plus 3 years for everyone now in the room
A = A
m + 4(n+1) = (m+4) + 3(n+2)
4n + 4 = 4 + 3n + 6
n = 10 – 4 = 6!!
Imagine the 1st twin have been born right before the new year…
If Ann and Beth not twins it Will be easier
i got a easier solution learnt from Singapore education.
Let x be the number of people originally.
Equating Ann excess to Beth excess from original average ,
4(x + 1) = 3(x + 2) + 4
Solving, we get x = 6
Wouldn't call this impossible, totally straightforward solution.
t … total age initially
n … number of people initially
a … age of Anne and her twin respectively
(t + a)/(n + 1) = t/n + 4
(t + 2a)/(n + 2) = (t + a)/(n + 1) + 3 = t/n + 7
remove fractions:
0 = (t + 4n)(n + 1) – (t + a)n = 4n² + (4 – a)n + t
0 = (t + 7n)(n + 2) – (t + 2a)n = 7n² + (14 – 2a)n + 2t
combine:
0 = 2•0 – 0 = n² – 6n = n(n – 6).
If n = 0, average doesn't change, so must n = 6.
There are n people in the room.
To raise the average age by 4, Ann has to be 4(n+1) years older than the original average.
To raise the average age by 3, Beth has to be 3(n+2) years older than the average with Ann there – in other words, Beth is 3(n+2)+4 years older than the original average.
Since Ann and Beth are twins, 4(n+1)=3(n+2)+4…. giving n=6
In addition, we can generalise:
Suppose Ann raises the average by x, and Beth raises the average by y:
The equation becomes x(n+1)=y(n+2)+x…. giving n=2y/(x-y)
(Obviously this will only make sense if (x-y) is a factor of 2y, so that there are a whole number of people!)
If Ann enters the room shortly before midnight/her birthday and her twin enters shortly after midnight, then there have been 5 people in the room originally. Is my math correct? ^^
From looking at the banner, 6 people. Their age is 0. The twins' age is 28. Other solutions exist, but always 6, just different ages.
No equations needed.
I got 6 😊 using the same method
Why “seemingly impossible”? This is easier than most MYD puzzles.
Since the age of the initial occupants (m) doesn’t matter, I solved by assuming a roomful of newborns. One 28-yr-old nurse walks in, then another.
Well I did some calculation starting with
S/n + 4 = (S + A)/(n+1) and
S/n + 7 = (S + 2A)/(n+2)
I Eliminated S and got a single equation, where A eliminated itself and I got, that there have to be n = 22/7 people. oO
I'd like to think that Ann and Beth are both 28.
The funniest part is that Ann is 28 years older than average age of people in the room. She's either elderly or "the meeting" is a bunch of toddlers
Ann’s age = Beth’s age = a
Average age of people initially in room = n
Number of people initially in room = x
nx + a = (n + 4)(x + 1) → a = n + 4x + 4
nx + 2a = (n + 7)(x + 2) → 2a = 2n + 7x + 14
2n + 7x + 14 = 2(n + 4x + 4) = 2n + 8x + 8 → 6 = x
There were 6 people in the room before Ann entered.
Good one. Solved it without watching video
Ya: Ann's age, Y0: average age of the people before Ann entered room, n:number of people before Ann entered room. Then we have;
Ya-Y0=4(n+1)
2(Ya-Y0)=7(n+2)
Or 8(n+1)=7(n+2)
i.e. n=6, and Ya-Y0=28. So there were 6 people before Ann entered the room, and she is 28 years older than them on average. 😊
I did it reasoning that Ann must be (n+1)x4 years older than the average person in order to increase the average age by 4 and Beth must be (n+2)x3 years older than the average which now included Ann's age.
Since all that matters is how much older the sisters are, you can think of everyone else in the room as being 0yrs old.
So a 28yr old enters a room of six 0yr olds the avg age is 28/7=4yrs
Another 28yr old enters and the avg age is 56/8=7yrs
Ann and Beth appear to be nurses looking after 6 toddlers.
A 1 year old baby
Thats not a riddle bro. That's a math word problem.
And a rocket surgery level type at that
Fun that what appears to be a critical variable is a red herring. I was feeling lazy and put it into excel and started playing, took about 2 minutes
I have written the age of Ann (A),: respectively the age of her twin sister (T) based on how they modify the value x for the average age of the n people present in the meeting room.
(n1+n2+…+n)/n=x
In other words, if we redistribute equally the total number of years to the n people, each of them gets a "share" of x years.
(n1+n2+…+n+A)/(n+1)=x+4
A must add to the initial sum of ages , a number of years encompassing x and (n+1) times the increase of 4 years to x:
A=x+4(n+1)
The average age after adding T looks like this:
(n1+n2+…+n+A+T)/(n+1+1)=x+4+3
Therefore T must enclose the "share"of (x+4) plus (n+2) times the increase of 3 years:
T=x+4+3(n+2)=x+4+3n+6=x+3n+10
Ann and her sister are twins, therefore:
A=T
x+4n+4=x+3n+10
n=6
"And that is the answer!" 😀
Thank you for this crunchy riddle! ☘️
Also, prove that at least three people in the room are under 18
It's 6
I think defining this "impossible' is an overkill.
An algebraic solution is simple and staiightforward.
Let X be a the number of people , Y is the sum of their ages and S the average age (before Ann's joining). Let's Z be Ann's age.
So, we have following 3 equations.
(1) Y/X = S so Y = SX
(2) (Y+Z)/(X+1) = S+4 so Y+Z = SX + 4X + 1S + 4
(3) (Y+2Z)/(X+2) = S+7 so Y + 2Z = SX + 7X + 2S + 14
Subtract 1 from 2 and 2 from 3
(2)-(1) Z = 4X + 1S + 4
(3) -(2) Z = 3X + 1S + 10
So that 4X + 4 = 3X + 10 hence X = 6
Notice that the general case where first age increase is M and second age increase is N resulted in
Z = MX +1S + M
Z = NX + 1S + M + 2N
Hence
X = 2N/(M-N)
so in our case it's 2*3/(4-3),
I got 6 with a slightly different method.
Let be:
E the age of Ann and Beth
M the average age of the original group
N their number.
Adding Ann to the group rises the average age by 4, so:
E = M + 4(N+1)
Because the age of Ann must be enough to rise by 4 the average age M of the N people plus herself
Adding Beth rises the age by 3, so:
E= (M+4) + 3(N+2)
Because, for the same reason, the age of Beth (that's still E) adds 3 to the new average age (that is N+4) of all the N original people, plus Ann, plus herself.
So:
E = M + 4N + 4
E = M + 4 + 3N + 6
E = (M +4) + 4N
E = (M + 4) + 3N + 6
(M +4) + 4N = (M + 4) + 3N + 6
4N = 3N + 6
N = 6
P.S. Ann and Beth are 28 y/o
Some observation: there are 2 equations, but 3 variables. Because nm cancels out constantly, instead of ending up with 3 sets of related 2-variable relations, the exercise ends on 1 solved variable and 2 related variables. n is 6 and for the others (not mentioned in the video) – twin's age is equal to average age + 28 (e.g. twins are 38 and there are 6x 10 yo).
The issue with trying to solve this from the thumbnail… the detail that it's a "meeting" and "a few people give Ann the evil eye" suggests that the others in the room are not newborn babies, something I hadn't factored in when trying to solve it.😂
Hey, that was fab!
I figured it was based on their ages being identical (being twins).
But I could only follow the path once both were = ?A.
It’s worth mentioning that only the number of ppl can be solved for not the average age or the twins ages as it’s indeterminate with 3 variables and 2 equations.
Let T0 be total age, number be N.
A = T0 / N. => T0 = A.N
Enter Ann (x).
T1 = T0 + x = (A+4) * (N+1)
=> AN+x=AN + A + 4N + 4
=> x = A + 4N + 4 ===A
Enter twin
T2 = T1 + x = (A+7).(N+2)
=> AN + 2x = AN + 2A + 7N + 14
=> 2x = 2A + 7N + 14 ===B
Solve eqs A and B for N.
and you'll get N = 6.
n=6, simple question writing down 2 equations and solving for n.
The age of Ann ist 28 years higher then the average age of the 6 people before entering the room.
"impossible riddle" and it's just an avarage tyt math question for Turkish students 😭😭😭