I need help with a riddle hoping someone can help

Mandy is 25 years older than Sandy

Sandy is 25 yrs younger than Mandy

Mandy's age is even

Sandy's age is odd

Together the sum of their ages is 63 how old is each sister?

anyone?

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- Oct 24th 2007, 04:58 PMjakRiddle help
I need help with a riddle hoping someone can help

Mandy is 25 years older than Sandy

Sandy is 25 yrs younger than Mandy

Mandy's age is even

Sandy's age is odd

Together the sum of their ages is 63 how old is each sister?

anyone? - Oct 24th 2007, 05:13 PMJhevon
aren't these two statements redundant?

Quote:

Mandy's age is even

Sandy's age is odd

Together the sum of their ages is 63 how old is each sister?

anyone?

since Sandy's age is odd, we can write it as: $\displaystyle 2n + 1$ for $\displaystyle n \in \mathbb{N}$ ...(i used the natural numbers instead of the integers since negative ages make no sense, but whatever).

since Mandy is 25 yrs older than Sandy, Mandy's age is given by:

$\displaystyle 2n + 1 + 25 = 2n + 26$

The sum of their ages is 63, thus we have:

$\displaystyle 2n + 1 + 2n + 26 = 63$

now solve for $\displaystyle n$ and you will be able to find their ages - Oct 24th 2007, 07:37 PMSoroban
Hello, jak!

What a silly problem!

Where did it come from?

Quote:

Mandy is 25 years older than Sandy.

Sandy is 25 yrs younger than Mandy. . . . . well, duh!

Mandy's age is even. . . . . Not relevant

Sandy's age is odd. . . . . . Who cares?

Together the sum of their ages is 63. . . . . Finally, something useful!

How old is each sister?

They're sisters? .Boy, that's really important!

(That's quite an age spread for siblings, isn't it?)

Let $\displaystyle S$ = Sandy's age.

Then $\displaystyle S + 25$ = Mandy's age.

"The sum of their ages is 63": . $\displaystyle S + (S + 25) \:=\:63$

- Oct 24th 2007, 08:02 PMJhevon
- Oct 25th 2007, 02:14 AMdravid2
sandy is 44 while mandy is 19

- Oct 25th 2007, 01:38 PMjak
sorry for wasting everyones time thanks for the response. After I posted it, I realized duh as many had suggested....

anyway thanks for the help - Nov 29th 2007, 01:11 PMivananeedshelponmath2hmm
- Nov 29th 2007, 09:14 PMJhevon