When Bob was same age as Brooke, Phil was $20$. And when Brooke was same age as Phill, Bob was $14$. How old is Brooke now?

Question

When Bob was same age as Brooke, Phil was $20$. And when Brooke was same age as Phill, Bob was $14$. How old is Brooke now?

Even if that's a question, your tips are more important for me. Honestly, I'm afraid of these questions that seem complex. As I mean, what do we have to do when we're solving the age word problems? So, am trying to find a method that is useful, easy, not complex. As seen, I've no idea about the question.

With my kindest regards!

Answer 1

When dealing with such questions, the method is basically always the same : state that $x = $ something, $y=$ something, ect...Then, translate the sentences from English to maths.

Here : $x=$ age of Bob, $y=$ age of Brooke, $z=$ age of Phil.

When Bob was the same age of Brooke, Phil was $20$.

This translates into : $z = 20 + (x-y)$. Indeed, Bob was the same age than Brooke $x-y$ years ago.

Now it's your turn ; how would you translate :

When Brooke was same age as Phill, Bob was 14.

Once you figured it out, hopefully you'll be able to work out the solution.

Answer 2

A big difficulty here is that the problem tricks you into using the same name for different quantities. "Phil's age at the time when Bob had the age that Brooke does now" is not the same as "Phil's age at the time when Phil had the age that Brooke does now" (which is just Brooke's age).

I'm going to use krirkrirk's notation, slightly clarified: $$x = \text{Bob's age now,}\quad y = \text{Brooke's age now,} \quad z = \text{Phil's age now.}$$

So let's translate your first assumption into an equation:

"When Bob has the age that Brooke has now…" …when is that exactly? It's not now — we can't assume $x=y$. Rather, we're looking at a time $a$ years in the future (or $-a$ years in the past), when Bob's age will be $x+a$; and our condition becomes (1) $x+a=y$.

"…Phil is 20". Note that this is the time $a$ years in the future (or $-a$ in the past), when Phil's age will be (or was) $z+a$. And we're now being told that that age is 20, so: (2) $z+a=20$.

At this point, we can eliminate $a$ as a variable (we don't really care when this is happening, so we don't need $a$ around). We can do this by solving for $a$ in (1) and substituting into (2), or vice versa, or by taking the difference of the two equations. Doing the last, we get $\boxed{x - z = y - 20}$ (the other procedures will yield an equivalent condition).

(Note that we could wait to do this, but then we'd be dealing with more equations in more unknowns.)

Exercise: Translate the second condition to a similar equation.

Now we have two equations to deal with in three unknowns, which you may have been told is bad. And it is, if we need to know exactly what $x$, $y$, and $z$ are. But we don't; we only care about the value of $y$. And if you did the exercise correctly, you should be able to eliminate $x-z$ the same way we eliminated $a$, and end up with a single solvable equation in $y$.