# A mother's age is $19$ years more than the sum of the ages of her sons

A mother's age is $19$ years more than the sum of the ages of her two sons. $5$ years ago, the mother's age was $4$ times than the sum of the ages of her two sons. What is the age of the older child?

• Let's say the sum of the ages of her sons is $x$, the mother's age will be $x+19$. $5$ years ago, $x-5$ = $x+19-5$. However, I believe that I've gone too wrong.

• What kind of methods can I use to solve this question?

I'll be waiting for your professional helps.

My Kindest Regards!

• @amWhy Can you explain why? – user533031 Mar 10 '18 at 17:28
• If the sons have ages $m,n$ and the mother has age $A$ (all today) then today we have $m+n+19=A$. Five years ago we had $A-5=4(m-5+n-5)=4(m+n)-40$ – lulu Mar 10 '18 at 17:30
• @amWhy I can take it from there! That was awesome and very very clear. However, is there any other method to solve this question? – user533031 Mar 10 '18 at 17:31
• Note: I don't see how to compute the age of the older child. It's easy to compute $x$ (or, in my notation, $m+n$) but I don't see anything that let's us get at $m,n$ separately. – lulu Mar 10 '18 at 17:34
• Nothing said about how many sons? – JustANoob Mar 10 '18 at 17:36

There isn't enough information to solve for the age of the older child, even if there are two children. I can only solve for the following:

$\text{Ages of sons combined} = x$

$\text{Mother's age}=y$

$$x+19=y$$ $$4(x-10)=y-5$$ $$4x-40=x+19-5$$

$$3x=54$$

$$x=\dfrac {54}{3}=18$$

You know the sums of the ages of the two sons is $18$. But, you need to find the age of one of the children to find the age of the other child.

Note that the younger child must be older than $5$ years but younger than $9$ years. For example, if the younger child is $7$ years old, then the older child is $11$ years old.

• Hold on, why this question doesn't have an integer solution? We could find $x = 18$ easily. – user533031 Mar 10 '18 at 17:53
• And you can know how years old the older child is by giving values. If we find $x = 13$, then the older child is $9$ and younger child is $8$. Am I wrong or not? – user533031 Mar 10 '18 at 17:55
• I think that's why, I hadn't written that she has two sons. What do we think now? – user533031 Mar 10 '18 at 17:57
• Yes, you are incorrect. One son, lets say is $z$ years, the other son $\frac {14}3-z$ years. How many values can you fit into $z$. Infintely many @Busi – user535339 Mar 10 '18 at 17:59
• However, she already has two son. By the way, we could find $x$ out and finally we give values for each son. See the new edit. – user533031 Mar 10 '18 at 18:00

\begin{align} M &= \text{mother’s current age in years} \\ S &= \text{older son’s current age in years} \\ s &= \text{younger son’s current age in years} \end{align}

The first statement translates to

$$M = 19+S+s$$

The second statement translates to

$$M-5 = 4\bigl( (S-5)+(s-5) \bigr)$$

We also know logically that any solutions, if they exist, should be restricted to

\begin{align} M &> S >0 \\ S &> s >0\\ M &> s >0 \end{align}

Can you take things from there? How does the number of variables compare to the number of equations?

hint

Today $$y=x+19$$ five years ago

$$y-5=4 (x-5n)$$ $$14=3x-20n$$ $n$ is the number of sons.

The sum $x$ must satisfy $x>5n$

thus $n=2$ and $x=18$ years . the mother has $37$ years .

• I didn't get your hint. – user533031 Mar 10 '18 at 17:52
• Your hint does not say anything about how to answer the question. There simply isn't enough information. – user535339 Mar 10 '18 at 17:53
• @Busi And now, you did. – hamam_Abdallah Mar 10 '18 at 18:21
• @Busi: Need the info about at least how different sons' ages are. For e.gxample., we can find that mother's age today is $37$ and sum of two sons' ages are $18$. Thus, five years ago that sum was $8$. Therefore, they should be $4$ possibilities: Either $(7,1)$, $(6,2)$, $(5,3)$, or $(4,4)$, if they are twins. Thus, today's ages would be: Either $(12,6)$, $(11,7)$, $(10,8)$, or $(9,9)$, the last one by assuming they are twins. – Mathew Mahindaratne Mar 10 '18 at 20:01