Joan, Kylie, Lillian, Miriam all celebrate their birthdays

[pck2103]

Can anyone help me solve this problem, I think that I am off to the right start, where you convert all ages to a common person, but not sure where to go from there, thanks for the help!!

Joan, Kylie, Lillian, and Miriam all celebrate their birthdays today. Joan is 2 years younger than Kylie, Kylie is 3 years older than Lillian, and Miriam is one year older than Joan. Which of the following could be the combined age of all four women today?
51
52
53
54
55

 

[Deleted member 4993]

Re: Age Problem, HELP!!!

Can anyone help me solve this problem, I think that I am off to the right start, where you convert all ages to a common person, but not sure where to go from there, thanks for the help!!

Joan, Kylie, Lillian, and Miriam all celebrate their birthdays today. Joan is 2 years younger than Kylie, Kylie is 3 years older than Lillian, and Miriam is one year older than Joan. Which of the following could be the combined age of all four women today?
51
52
53
54
55

Please share with us your work/thoughts (more than you did)

To start
Joan is 2 years younger than Kylie, -------> J = K - 2

Kylie is 3 years older than Lillian, and ----> K = L + 3

Miriam is one year older than Joan---------> M = J + 1

You need to find -

J + K + L + M = ??? What would be the next step?

 

[Loren]

Re: Age Problem, HELP!!!

Another hint---Each of their ages must be a whole number.

 

[pck2103]

Re: Age Problem, HELP!!!

I converted them all into K. So J = K-2, L = K-3, M = K-1, and K = K, all together I got...
K-2 + K-3 +K -1 + K = 4K -6
4k -6 = an integer..... this is where i get stuck.

 

[Deleted member 4993]

Re: Age Problem, HELP!!!

I converted them all into K. So J = K-2, L = K-3, M = K-1, and K = K, all together I got...
K-2 + K-3 +K -1 + K = 4K -6
4k -6 = an integer..... this is where i get stuck.

So now you see that sum must be an even number.

And

If you add 6 to the sum - the result must be divisible by 4.

 

[pck2103]

Re: Age Problem, HELP!!!

and 54 +6 is divisible by 4. is there a more mathematical way to do that or should I have just tried that even numbers listed, like i did.

 

[Loren]

Re: Age Problem, HELP!!!

>>is there a more mathematical way to do that?

I don't know. But, using substitution, I solved for Lillian's age and came up with L= (x-6)/4 where x is the sum of their ages. Then I substituted in the various possible answers for x to see which one gave me a whole number.

 

[masters]

Re: Age Problem, HELP!!!

and 54 +6 is divisible by 4. is there a more mathematical way to do that or should I have just tried that even numbers listed, like i did.

Let L = Lillian's age

Let L+3 = Kylie's age (Kylie is 3 years older than Lillian)

Let L+1 = Joan's age (Joan is 2 years younger than Kylie: L+3-1)

Let L+2 = Miriam's age (Miriam is 1 year older than Joan: L+1+1)

Add the ages:

\(\displaystyle 4L + 6 = \not{51} \ \ \not{52} \ \ \not{53} \ \ 54 \ \ \not{55}\)

All sums are eliminated except 54. 54 is the only sum that wil make L a positive integer.

\(\displaystyle 4L + 6 = 54\)

\(\displaystyle 4L=48\)

\(\displaystyle L=12\)

Ages:

\(\displaystyle \begin{array}{ccccc}Lillian & L & 12 \\ Kylie & L+3 & 15 \\ Joan & L+1 & 13 \\ Miriam & L+2 & 14 \\ Total & = & 54 \\ \end{array}\)