Topic awaiting preservation: Mission Impossible II?

I've spoken to 3 people this morning, if you multiply their ages by mine the product is 2450. If you add their ages the answer is double your age.

One of these people i spoke with celebrated an important birthday this year, I celebrated the same bday five years ago.

How old are you and how old am I?

Can anyone here give this 'literary' impossible question a shot?

It was in a psychometric testing interview exam that I did and I had NO clue.

Hmm, at first, I was thinking it was a poorly worded question, but after re-reading it a few times, I think I have an idea...

There's 5 people you're concerned with here:
The author/person giving you the question ('I' and 'mine' references), who we'll call: q
The three people they talked to this morning: x, y, z
You, the person reading/answering the question: a

I could only get about halfway through without checting, or assuming something about the question. (And it still might be completely wrong.)

The first equations you get are:
1) (x + y + z)q = 2450
2) (x + y + z) = 2a

One of those three (let's pick x) is 5 years older than the author, so we also have:
3) x = q - 5

Plugging that into 1 and 2 gives us:
4) ( (q - 5) + y + z )q = 2450
5) ( (q - 5) + y + z ) = 2a

Now here was my cheat/assumption, which might very well be right, but I kinda have my doubts.
You are one of the three people the author spoke to, so you are either y or z. Plugging you in for y, gives:

6) ( (q - 5) + a + z )q = 2450
7) ( (q - 5) + a + z ) = 2a

And here's where the numbers get ugly because you put your age in for a, so everyone will get a different answer from here on (except of course, people that are the same age should get the same numbers). For me, I'm 21 for another couple weeks, so here's what I get:

8) ( (q - 5) + 21 + z )q = 2450
7) ( (q - 5) + 21 + z ) = 2(21) = 42 = (q + 16 + z)

9) (q^2 - 5q) + 21q + zq = (q^2 + 16q + zq) = 2450
10) (q + z) = 42 - 16 = 26

Solving 10) for z, gives:
11) z = 26 - q

Plugging that into 9:
12) ( q^2 + 16q + (26 - q)q ) = 2450 = ( q^2 + 16q + 26q - q^2 ) = 42q = 2450
and, q = 58 1/3

Plugging that into 11) gives: z = 26 - 58 1/3 = -32 1/3
Plugging q into 3) gives: x = 58 1/3 - 5 = 53 1/3

So, after all is said and done, this is what I get:
a (me) = 21
q (author) = 58 1/3
x = 53 1/3
y = 21
z = -32 1/3

That's my shot, but I wouldn't bet the farm on it for a few reasons...
53 1/3 doesn't seem like a particularly "important" birthday
-32 1/3 is usually a bad answer when dealing with birthdays
However, that's the first thing I come up with on a Monday morning, so maybe it'll be enough for someone else to figure it out with...

I'm gonna say 18 and 25 respectfully, I put no logic or thought behind it, except I'm 25, and besides the Dwarf used an umbrella to press teh elevator buttons, and the walls are clear because green houses are made from glass!

Cheers,

no need for assumptions Lord_Fuku .. your problem was going perfect until your assumption .. and according to my understanding we aren't one of the 3 individuals he spoke with .. the only problem with the question IMHO .. is that the age of the author will vary due to the fact that if he is asking for your age then from these calculations the authors age will vary per an individuals age

but after doing my calculations I got the author to be 45.37 .. but I'll round it down to 45


.::. cEll .::. 513

Lord_Fukutoku,

Shouldn't equation 1) be xyzq = 2450?

.



-- not necessarily stoned... just beautiful.

Nah technically as written it would be

xq + yq + zq = 2450

but that can be reduced to

( x + y + x )q



.:[ Never resist a perfect moment ]:.

"multiply their ages by mine" implies x*y*z*q to me, no addition, especially since it says "the product is 2450" and not "the sum".


 

[edit: you're right, it's x * y * z * q.]
It's unsolveable. Even if I assume my own age as 18, you need one more equation, as you gain the result:

y * z * (41 - y - z) * ((41 - y - z) - 5) = 2450

If I had one more, you could remove either y or z and solve. So no, the problem cannot be solved without assumptions.

The reason for this is you need as many equations as unknowns to solve. If we make no assumptions, with have 3 equations and 5 unknowns. We need to assume two things in order to solve it. If we assume our own age, and assume we're one of the three original people, it's doable.

(Edited by Iron Wallaby on 06-13-2005 20:52)

I wasn't sure about whether it was supposed to be x*y*z*q, or (x + y + z)q... The way it's worded is kinda ambiguous to me. At first I thought xyzq, but the math was a bit easier the other way, which is why I picked that one
Either way though, you need one more piece of information, otherwise you're stuck with 2 unknowns at best, unless you assume you're one of the three (which the more I think about it, the more I think might be the case).

The only other thing that might help is the bit about "One of these people i spoke with celebrated an important birthday this year"... which could be a variety of ages:
- 0, just born
- 1, maybe not that important
- 15, quinceanera
- 18, no longer a minor
- 21, legal drinking age
- 24 or 25, car insurance rates become more reasonable
- 40, over the hill
- 65, retired
- 100, really old
Plus a bunch of others...

The problem there is that most of those are culture or location specific...

i am 22 *poing*
x=22
they are all the same age (and could be according to the problem or even approximately according to reality)
3y=2x
each one of the is 14,7 years old *poing*
y=14,7

xy=2450
you are 166,7 years old *poing*

you could live in a very weird country where 161 and 14 are ages considered important for the same weird reason. And chances are, we have different notions of "important birthdays" anyway, so it's safe to assume so.

*poing*


you are very old, you shouldn't type so much kaboi

~this is not a signature~

The way I read the problem, you get an equation (cubuc I think) that looks like it's unsolvable, but if you solve for the 0s you will get two answers that are not likely and one that is.

I worked on it for a while yesterday, but didn't get an answer. Don't really have time to do anything with it today.

.



-- not necessarily stoned... just beautiful.

Hi Guys, been away for the last couple of days.

Clarification: There are 3 people (yzx) then theres q and finally theres you (a)

In total there 5 people of which you need to find out your age (a) and (q)s age.

I've tried all sort of equations but it get complex everytime. Looks to me like Lord_Fukutoku was on the right track.

Well, let's see... I suppose we can assume American culture, then? The ones I think likely, then, are:

- 18, Legal Adult
- 21, Legal Drinking Age (least likely, though, as it's not that important to most, since they either break the law anyway or don't care)
- 40, Over the Hill

I suppose it can be tried with any and see what's the most reasonable. If I had to bet, I'd guess 18. We can then assume that x = 18 and q = 23. This simplifies it to a solveable position, I think. (I'll try it later)

---
Website

(Edited by Iron Wallaby on 06-15-2005 16:08)

We're given:
xyzq = 2450
x + y + z = 2a
q = x + 5

Now, previously I said that I would guess 18 as the important birthday... well, I changed my mind. I think, to be reasonable, we need to deal with integer math (how do you handle someone's age as a fraction? Floor, ceil, round?). So that means that 2450/q should be integer. Which means that q ought to be in the set:
{ 7, 10, 14, 25, 35, 49, 50, 70, 98, ... } (beyond this, he's older than 120, which is obscenely unlikely; and even though 1, 2, 4, and 5 are technically in the set, if they were in then x would be nonpositive.)

Furthermore, we get the following equation:

yz = 2450 / (q^2 - 5q))

What numbers work then? q = { 7, 10 }. Getting closer.

If q = 7, then x = 2 and yz = 175.
If q = 10, then x = 5 and yz = 49, meaning that y = 7 and z = 7. However, 5 + 7 + 7 is not divisible by 2 (second eq.) so we throw it away.

So yz = 175. What factors can we have? y = 5, z = 35; and y = 7, z = 25. Thus, a can be either 21 or 17.

So, assuming my equations are right, you're definitely 7. I'm either 21 or 17. Therefore, because it's ambiguous and since 2 isn't an important birthday, I think that the equation is wrong.

So what about xq + yq + zq = 2450?

---
Website

(Edited by Iron Wallaby on 06-15-2005 17:26)

So, here's what we're given:

(x + y + z) * q = 2450
(x + y + z) / 2 = a
x = q - 5
q = { 7, 10, 14, 25, 35, 49, 50, 70, 98 }

Thus, a = 2450 / (2 * q).

The only possible solutions are q = { 25, 35, 49 }, since everything else results in a noninteger a or an a out of the range of a reasonable life expectancy.

If you're 25, the special birthday is 20 and I'm 49.
If you're 35, the special birthday is 30 and I'm 35.
If you're 49, the special birthday is 44 and I'm 25.

None of those birthdays sound especially special to me. So this one is even more ambiguous.

---
Website

I assumed "important birthday" meant 10, 20, 30, 40, etc. Kinda ambiguous (which of course is the point of the puzzle).

[edit]Just tried working it out - it seems a little strange since the average of x, y, z and q's ages would be 612.5...[/edit]

this post will self-destruct in 10...

(Edited by Danaan on 06-15-2005 17:47)

quote:

---

[edit]Just tried working it out - it seems a little strange since the average of x, y, z and q's ages would be 612.5...[/edit]

---

That would be if it were (x + y + z + q) = 2450
Since you multiply q by each of x,y,z , the average drops significantly.
Ballpark numbers: say q is 20, that makes x+y+z ~ 120, divide by 3 is 40, so 20, 40, 40, and 40 are reasonable ages. It just becaomes a problem when you try to make the numbers work exactly according to the little bit of information given.

quote:

---

- 40, over the hill

---

The nice thing about being over the hill is that I can take my feet off the pedals and coast. Wheeeeee!

Hey kaboi...really tuff question there...
(xyz)m=2450
(x+y+z)=2q
there has to be some missin info...this thing keeps on looping....my trash can is full!!!....i bet u'll get the answer on radio soon.

Your 21 and im 7 years

The three other ages are 35, 5, 2. I've brainstormed this with my former math professor and these figures add up neatly.

The important birthday was meant to throw us off, I think.

1: (xyz)m=2450 : m=2450/(xyz)

2: (x+y+z)=2n : n=x+y+z/2

3: x=m-5

m=2450/(m-5)yz >> 2450/myz-5yz=m >> m^2=2450/-4yz

n=(m-5+y+z)/2 >> ((m-5+y+z)/2)^2=n^2

(25+m^2+y^2+z^2-10m+2my+2mz+2yz)4=4n^2 ....

ya...it goes on...there r still 2 many unknowns....it wont work out...try another way

I did a lot of guessing with the numbers, and seeing your answer helped a lot.
I will represent th question teller as "y", the listener as "m", and the other individuals as "a", "b", and "c".
This is my answer:
y=5
m=25
a=1
b=14
c=35

In other words, (abc)y=2450 == 1*14*35*5=2450
In part: a+b+c=2m == (1+14+35)/2=25

This works because the individual turning one has had their first birthday, a "significant" b-day, and you simply haven't had your birthday (turned 6) yet this year.

I realize this is an older post, but I'm new to Asylum and I enjoyed hurting my head with this thread.

___________________________________
Quidquid Latine Dictum Sit, Altum Viditur ~
Whatever is said in Latin sounds profound