JUANR
Posts: 837
Joined: Wed Dec 13, 2000 12:37 am

### A Math Problem

This is a simple math problem that my cousin asked me to solve and I must admit I have not been able to. (law is killing my math abilities)

This is it:

My age now is two times (the double) of the age that you had when I had your age. When you'll have my age, our ages combinated will be 63. How old are we?

This is not for any class or contest so there is no hurry, I'll keep trying myself.

Juan
SKBO
Bogotá: 2600 Metros Más Cerca De Las Estrellas; Vamos por los XVII Juegos Nacionales!!!!!!!!!

EGGD
Posts: 11880
Joined: Sat Feb 24, 2001 12:01 am

### RE: A Math Problem

The reason why you're having so much trouble with it is because its badly worded, its quite a simple problem and if it was worded correctly i'm sure you'd get it quickly!

JUANR
Posts: 837
Joined: Wed Dec 13, 2000 12:37 am

### RE: A Math Problem

EGGD

It is perfectly worded, I asked before posting.

Juan
SKBO
Bogotá: 2600 Metros Más Cerca De Las Estrellas; Vamos por los XVII Juegos Nacionales!!!!!!!!!

EGGD
Posts: 11880
Joined: Sat Feb 24, 2001 12:01 am

### RE: A Math Problem

Well, don't try and do it in your head, write everything down and work through it logically.

JUANR
Posts: 837
Joined: Wed Dec 13, 2000 12:37 am

### RE: A Math Problem

I am not trying I already have three A4 size sheets filled with ecuations on both sides, I had not post this if I haven't been trying this for hours.

Juan
SKBO
Bogotá: 2600 Metros Más Cerca De Las Estrellas; Vamos por los XVII Juegos Nacionales!!!!!!!!!

EGGD
Posts: 11880
Joined: Sat Feb 24, 2001 12:01 am

### RE: A Math Problem

wow, hours? Maybe you should just give up!

JUANR
Posts: 837
Joined: Wed Dec 13, 2000 12:37 am

### RE: A Math Problem

EGGD:

I never give up

Juan
SKBO
Bogotá: 2600 Metros Más Cerca De Las Estrellas; Vamos por los XVII Juegos Nacionales!!!!!!!!!

canuckpaxguy
Posts: 1482
Joined: Sat Sep 06, 2003 2:31 pm

### RE: A Math Problem

Ages are 21 and 42.
Am I missing a trick here?

Oh wait - I misred the question....gimmie a sec.
I must have "combinated" the numbers wrong.

G

[Edited 2004-12-01 00:05:17]

theCoz
Posts: 3933
Joined: Mon Aug 16, 2004 11:06 am

### RE: A Math Problem

2x + x = 63

3x = 63

x = 21

You are 42 yrs (2x)

The other is 21 yrs (x)

Arniepie
Posts: 1429
Joined: Thu Aug 04, 2005 11:00 pm

### RE: A Math Problem

25.2 years old and 12.6 years old
[edit post]

EGGD
Posts: 11880
Joined: Sat Feb 24, 2001 12:01 am

### RE: A Math Problem

No! As I said, it was worded really badly..

You should have let him work it out though, especially after all this time!

theCoz
Posts: 3933
Joined: Mon Aug 16, 2004 11:06 am

### RE: A Math Problem

it looks like more of a word problem to me.

JUANR
Posts: 837
Joined: Wed Dec 13, 2000 12:37 am

### RE: A Math Problem

No, I already came to those answers and they did not fit in, if the age of "I" is 2Y, then Y is not the current age of "YOU" 2Y is the age "YOU" had when "I" had the age "YOU" has now, so those answers in which "I" age is the double of "YOU" are wrong. Also, notice that 63 is not the current combinated age of both "I" and "YOU", it is the combinated age when "YOU" gets the age that "I" now has.

EGGD: it is perfectly written but If you have doubts then I'll post the original question in Spanish and see if you can write it down in english better, that would be more useful than keep on bashing.:

"La edad que yo tengo es el doble de la edad que tu tenías cuando yo tenía tu edad. Cuando tu tengas la edad que yo tengo, la suma de nuestras edades será 63. Qué edades tenemos tu y yo?

Juan
SKBO
Bogotá: 2600 Metros Más Cerca De Las Estrellas; Vamos por los XVII Juegos Nacionales!!!!!!!!!

Arniepie
Posts: 1429
Joined: Thu Aug 04, 2005 11:00 pm

### RE: A Math Problem

EGGD,

Sorry, couldn't help myself.

To make things up with you I'll give U another one, just logics and no real maths but a nice one anyway.

5 pirates on a ship have a 100 bars of gold.
Being pirates ,they want to share as little as possible.
All the pirates have a different age and the oldest pirate can make the first suggestion about splitting up the bars.
If a majority doesn't agree with the proposed deal they may shoot the oldest pirate , then the next in line (2nd oldest) can make a suggestion, and so on.

What does the oldest pirate propose so he can keep his life and also keep as many bars as possible for himself.

keep in mind that all pirates act logically and no emotions are involved.
[edit post]

EGGD
Posts: 11880
Joined: Sat Feb 24, 2001 12:01 am

### RE: A Math Problem

Juan - Firstly, there is no such word as 'combinated'. Secondly, there is no insinuation in the question as to whether it is all 3 ages combined or merely the concurrent ages. Although, the way the second sentence is constructed suggests that it is the latter and hence people come to the conclusion that the two ages are 42 and 21. For all intents and purposes, both answers are correct (unless i've skipped something huge!  ).

Arniepie
Posts: 1429
Joined: Thu Aug 04, 2005 11:00 pm

### RE: A Math Problem

Juan - Firstly, there is no such word as 'combinated'. Secondly, there is no insinuation in the question as to whether it is all 3 ages combined or merely the concurrent ages. Although, the way the second sentence is constructed suggests that it is the latter and hence people come to the conclusion that the two ages are 42 and 21. For all intents and purposes, both answers are correct (unless i've skipped something huge! ).

Actually the answer is never 21 and 42.
Juanr is specifically asking for the ages at the present time, the combined (iso combinated) age of 63 refers to an age in the future (When you'll have my age)

So to conclude

juanr's Cousin (=person A) age NOW is 2 times (=2x) the age of Juanr (=person B)
When B has the present age of A his age will be 2x, at the same moment A's age will be 2x + x ,he will have aged exactly x-years (the same as person B)
So at 63 their combined age will be 2x(=A)+2x+x(=B)=63
5x=63
x=12.6 (present age B)
2x=25.2 (present age A)

[edit post]

VirginFlyer
Posts: 4229
Joined: Sun Sep 10, 2000 12:27 pm

### RE: A Math Problem

I don't believe we have had the correct answer yet.

'My age' = 42; 'Your age' = 21 doesn't satisfy the final requirement: When you'll have my age, our ages combinated will be 63. In this case, when you are 42, I am 63, so the combined age is 105.

'My age' = 25.2; 'Your age' = 12.6 doesn't satisfy the first requirement: My age now is two times (the double) of the age that you had when I had your age. In this case, when I was your age (12.6), you were 0. Double of 0 is still 0.

The best way to do this is to write equations, and make sure they match up with exactly what is being said. Before writing equations, we should define some variables, noting that there are three time instants being specified - the past (the age that you had when I had your age), the present (the ages we are looking for) and the future (When you'll have my age). Here are the variables:

An = my age now
Ap = my age past
Af = my age future

Bn = your age now
Bp = your age past
Bf = your age future

Da = Age diffence between you and I.

All of these numbers should have a single solution if this is a sensible problem.

Now we take the first statement:

My age now is two times (the double) of the age that you had when I had your age

We express this as:

An = 2Bp

Now, because this occurs at a time when i was your present age, it was Da years ago. I.e. Bp = Bn - Da

An = 2(Bn - Da)

But we know that Bn = An - Da (because Da is the age difference)

An = 2((An - Da) - Da)

An = 2(An - 2Da)

An = 2An - 4Da

An = 4Da

This is a key equation.

Now, we look at the second statement:

When you'll have my age, our ages combinated will be 63

We express this as:

Af + Bf = 63

Because this occurs at a time when you are my present age, it is Da years in the future. I.e. Af = An + Da, Bf = An

An + Da + An = 63

2An + Da = 63

Now recalling the key equation An = 4Da

8Da + Da = 63

9Da = 63

Da = 63/9

Da = 7

We are now able to solve:

An = 4Da

An = 4 x 7

An = 28

Bn = An - Da

Bn = 28 - 7

Bn = 21

We should now check that this agrees with the statements:

My age now is two times (the double) of the age that you had when I had your age

When I was 21, you were 14. 28 is two times 14. Condition Satisfied

When you'll have my age, our ages combinated will be 63

When you are 28, I will be 35. 28 + 35 is 63. Condition Satisfied

Thus the answer to the problem is:

'My age' = 28, 'Your age' = 21

V/F

[Edited 2004-12-01 01:33:58]
"So powerful is the light of unity that it can illuminate the whole earth." - Bahá'u'lláh

yooyoo
Posts: 5686
Joined: Wed Nov 26, 2003 5:01 am

### RE: A Math Problem

The answer is C...always go for C.

Andreas
I am so smart, i am so smart... S-M-R-T... i mean S-M-A-R-T

damirc
Posts: 726
Joined: Fri Feb 13, 2004 8:43 am

### RE: A Math Problem

Arniepie ...

... well ... one solution would be (to keep the majority of votes) to keep 33 bars, and give 33 bars to the 2nd and the 3rd oldest pirate and the remaining gold bar to the 4th oldest pirate. The youngest pirate stays empty handed, but the majority would presumably accept such a solution.

What's your solution?

D.

ZKSUJ
Posts: 6823
Joined: Tue May 18, 2004 5:15 pm

### RE: A Math Problem

It sounds like a case of Simutaneous Equations.

JUANR
Posts: 837
Joined: Wed Dec 13, 2000 12:37 am

### RE: A Math Problem

VirginFlyers: Thank You Very Much!!! That is the right answer; I'll check the procedure later carefully in order to see where I was wrong.

My cousin is very happy about it.

Welcome to my RU list.

Juan
SKBO
Bogotá: 2600 Metros Más Cerca De Las Estrellas; Vamos por los XVII Juegos Nacionales!!!!!!!!!

Gdabski
Posts: 408
Joined: Wed Oct 17, 2001 8:17 pm

### RE: A Math Problem

Alternatively, to use less symbols, you can solve these simultaneous equations:

x - my current age
y - your current age
(x-y) - age difference now, time from when I had your age to present, time from present to when you'll have my age

My age now is two times of the age that you had when I had your age
x=2(y-(x-y))

When you'll have my age (be x years old) our ages combinated will be 63
x+(x+(x-y))=63

/Gdabski

[Edited 2004-12-01 18:35:44]

jfkaua
Posts: 972
Joined: Tue Aug 31, 2004 7:42 am

### RE: A Math Problem

ah we learned this freshman year at my school.. i owned at these things.. i did it similar to the way Gdabski does it
-----
on the pirate question.. when they vote, does the oldest count?

[Edited 2004-12-01 21:52:59]

Arniepie
Posts: 1429
Joined: Thu Aug 04, 2005 11:00 pm

### RE: A Math Problem

on the pirate question.. when they vote, does the oldest count?

All votes count offcourse.

Damirc,

Sorry it is not that easy.

[Edited 2004-12-02 00:42:52]
[edit post]

damirc
Posts: 726
Joined: Fri Feb 13, 2004 8:43 am

### RE: A Math Problem

Okay. I gave that pirate puzzle some more thought.

So let's go from the opposite end.

If only Pirate 5 is alive: he gets all the 100 bars - therefore he will only vote yes if he is given all of the 100 bars.

If Pirates 4 and 5 are alive: he will never win the majority with Pirate 5 since P5 can get a better deal in any case. Therefore Pirate 4 can maximally get 0 bars (in any case).

If Pirates 3, 4 and 5 are alive: P3 will take 50, P4 will take 50 (both will vote yes), and P5 will get 0.

P2, P3, P4 and P5 are alive: they need a majority vote therefore P2, P3 and P4 must vote yes (we already know that P5 will always vote with a no) - P4 will vote no if he gets less than 50 bars, since he can get 50 bars if he lets P2 die and splits the bounty with P3. P3 will do the same. Therefore P2 would get 0, P3 - 50 and P4 - 50, P5 - 0 (P2 will get to keep his life).

P1, P2, P3, P4 and P5 are alive: P1's priority is his life (requirement) and if he gets any goldbars, he'll be happy. P2 will be happy if he gets more than 0 bars. P3 or P4 will only vote yes if they get at least 50 bars. Therefore (if my assumptions are correct thus far):

P1 takes 48 bars and keeps his life. - He obviously votes yes.
P2 gets 1 bar (he's better off than if P1 dies). He votes yes.
P3 gets 51 bars (he's better off than if P1 and P2 die). He votes yes.
P4 gets 0 bars. He votes no.
P5 gets 0 bars. He votes no.

We have 3 yes votes and 2 no votes - the majority acceps the solution, and P1 keeps his life.

Arniepie: any better now?

D.

futureuapilot
Posts: 1329
Joined: Mon May 24, 2004 7:50 am

### RE: A Math Problem

Wow, with all the BN's and AF's and such, I'm surprised I didn't see all that many Airline abbreviations...
-Sam
The Pilot is the highest form of life on Earth!

HAWK21M
Posts: 29978
Joined: Fri Jan 05, 2001 10:05 pm

### RE: A Math Problem

42 & 21.
regds
MEL
I may not win often, but I damn well never lose!!!

USAFHummer
Posts: 10261
Joined: Thu May 18, 2000 12:22 pm

### RE: A Math Problem

Wow...good work VF!! Now why do I have a sudden urge to take a remedial math course?

Greg
Chief A.net college football stadium self-pic guru

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