A Neat problem: If his age is added to hers, result is 91...

[Deleted member 4993]

If you add the age of a man to the age of his wife, the result is 91.
He is now twice as old as she was when he was as old as she is now. How old is the man, and how old is his wife?

 

[galactus]

Re: A Neat problem

These age problems can tie ones brain in a knot.

Anyway, is this correct SK?.

Let h=husband's age and w=wife's age.

h+w=91

The tricky part:

h=2(w-(h-w))=2(2w-h)=4w-2h

Then we solve the little system and get he is 52 and she is 39.

Seems correct. Is it SK, Is it, huh, is it, huh, is it?. :lol:

 

[Deleted member 4993]

Re: A Neat problem

These age problems can tie ones brain in a knot. <<< That is correct

Anyway, is this correct SK?.

Let h=husband's age and w=wife's age.

h+w=91

The tricky part:

h=2(w-(h-w))=2(2w-h)=4w-2h

Then we solve the little system and get he is 52 and she is 39.

Seems correct. Is it SK, Is it, huh, is it, huh, is it?. :lol:

I did it in a slightly different way - to get the same result.

of course the first equation is:

M + W = 91.......................................(1)

Now I assume "Y" years ago - He is now twice as old as she was

then

M = 2(W - Y)

M - 2W + 2Y = 0..............................................(2)

Then also "Y" years ago - he was as old as she is now

then

W = M - Y

M - W - Y = 0.................................................(3)

then (2) + 2 * (3) gets

3M - 4W = 0

Then of course rest follows......

 

[Denis]

Re: A Neat problem

My turn, Sir Khan!

- "1/5 of C" years ago,
- B was half of what A will be,
- when A is 6 years more than 5 times what D was,
- when D was 1/4 of what A was.
OK: find the ages of A B C D......
gee, almost forgot:
3 years ago, B was what D will be in 13 years;
oh ya: everybody is over 1 and under 50.
TIDBIT: B is a Star Trek fan; and he was watching it
when A was born and when C and D were hospitalized
for what was suspected as the Mad Cow disease.
Whoooops...I see I goofed: darn it, there's 2
age possibilities; well then, find both!