## Guess my age

"My wife’s age", said John, "can be found by reversing the figures of my own age. She is my senior, and the difference between our ages is 1/11 of their sum".

How old is John?

*[submit your answers in the comment section below, and we'll unmask them in 1 day]*

marioberges| Profile February 25th, 2008 - 9:43 amShe is 54.

He is 45.

zxo| Profile February 25th, 2008 - 9:56 am(W-J)*11=W+J

10W=12J

W/J=6/5

So we know that John’s age must be a multiple of 5, and that the first digit must be less than the second (because his wife is older). Thus, we are left with 15, 25, 35, and 45. Of these, only 45 fits the requirement that W/J = 6/5.

Thus, John is 45 and his wife is 54.

mroth01| Profile February 25th, 2008 - 10:24 am45?

falwan| Profile February 25th, 2008 - 10:28 amJohn = 45;John’ Wife = 54…

54 – 45 = 9 ;;; 1/11(54 + 45 ) = 1/11(99) = 99/11=9.

Thanks.

falwan| Profile February 25th, 2008 - 10:29 amJohn = 45 ; John’ Wife = 54.

Shawn| PUZZLE GRANDMASTER | Profile February 25th, 2008 - 10:37 amJohn is 45.

When you reverse ages, the difference between them will always be 9.

F = female’s age

M = male’s age

(F – M) = 9

So : (F + M) / 11 = (F – M) = 9

and : (F + M) = 99

John is 45, wife is 54

sharhan| Profile February 25th, 2008 - 10:43 amThe wife is 54, John is 45…quite easy.

suineg| PUZZLE MASTER | Profile February 25th, 2008 - 11:13 amA=First digit of Jhon Age

B=Second digit of Jhon age

My wife age is the reverse of my digits and she is my senior:

BA>AB

the difference between our ages is 1/11 of their sum:

BA-AB= (AB+BA)/11—> 10BA/12=AB now this give you some possibilities, but the reverse of the digits gives you a hint:

there is a pattern: if the distance between A and B is 1 the difference between BA and AB is 9*1, if its 2 is 9*2 and so on, for example:

50 and 05, the distance is 5 and the difference is 9*5= 45= 50-05.

the difference has to be a multiple of 9 and AB= 10/12 BA:

the distance has to be one because the difference between BA and AB has to be 9 cant be greater because of the relation AB= 10/12BA

now you can see what number with the distance between his digits in one fufill the condition: 54—45

sarag| Profile February 25th, 2008 - 11:50 amJohn is 45 yo,

his wife is 54 yo.

jason| Profile February 25th, 2008 - 12:04 pmjohn is 45

theWizof6| Profile February 25th, 2008 - 12:06 pmJohn is 45

cdv02d| Profile February 25th, 2008 - 12:08 pmJohn is 45, making his wife 54.

foger1979| Profile February 25th, 2008 - 1:57 pmhe is 45

she is 54

Together they are 99

Difference is 9 or 1/11 of 99

Ari| Profile February 25th, 2008 - 2:05 pmJohn is 45.

pedstar| Profile February 25th, 2008 - 4:15 pmJohn is 45.

drew101010| Profile February 25th, 2008 - 5:17 pmJohns age is 45. Wifes age is 54. The sum of them is 99 and the difference 99. 9/99 is 1/11.

da Moose| Profile February 25th, 2008 - 6:34 pmJohn is 45.

jasc| Profile February 25th, 2008 - 6:56 pm20

jasc| Profile February 25th, 2008 - 7:02 pm^ got too excited, forgot about digits

John’s 45

djp| Profile February 25th, 2008 - 7:27 pmHe is 45.

sunshine1953| Profile February 25th, 2008 - 7:54 pmJohn is 45

mgillig| Profile February 25th, 2008 - 8:44 pmJohn is 45

doylerules| Profile February 25th, 2008 - 8:57 pmThe husband is 45. The wife is 54.

The equation that you use is:

wife’s age=husband’s age*1.2

This leaves only 18 possibilities because the answer must be an integer. I brute forced it from there!

T| Profile February 26th, 2008 - 2:33 amHer Age – His Age = (Her Age + His Age)/11

So testing ages for which the digits could be reversed, starting at a reasonable difference (54 and 45) you get the answers.

She is 54 he is 45.

GIGS1890| Profile February 26th, 2008 - 4:08 amJohns 54. Wife is 45.

difference of 9, sum of 99.

Got through basic Algebra (let one of the digits be a, the other b. John’s age is 10a+b, wife’s is 10b+a)

carlhazlewood| Profile February 26th, 2008 - 8:55 amHe’s 45.

RK| Profile February 26th, 2008 - 9:02 amVery good, John is 45. Several nice explanations above.

I know many of you have been asking for some difficult math-type puzzles; in about 2-3 weeks will be able to post the 1st really hard one.

pieman| Profile February 26th, 2008 - 6:09 pm45.

alachigh| Profile March 10th, 2008 - 11:55 am52