# Challenging Distance, Speed, and Time Word Problem

Two people, $$A$$ and $$B$$, walk from P to Q and back. $$A$$ starts 1 hour after $$B$$, overtakes him 2 miles from Q, meets him 32 minutes later, and arrives at P when $$B$$ is 4 miles off. Find the distance from P to Q.

I really cannot think of anything to solve this problem. I tried drawing a diagram to model the situation, but that didn't seem to help very much. The problem I am having is translating the English to mathematical language; which is really what makes me struggle with word problems in general.

Even if I could just have the equations to describe the problem, that would help tremendously; my problem is forming the equations, but solving them is easy for me. If I could also get some tips for how to form equations based on the word problems, that would be much help. Thanks.

• Please say more about your attempts at solving the problem. What you said is too vague to really see what you are thinking. Also: Is one to assume that each walker goes at a constant rate? (If not, the question seems unanswerable.) Mar 23 at 1:28
• @paw88789 I think the question is assuming they are going at a constant rate. Mar 23 at 1:49
• @boojum would B's rate be $\frac{x}{92}$ ? Mar 23 at 1:56
• I guess I had thought A and B met at Q due to the "meets him 32 minutes later" line; since the only way they can meet after A has overtaken B is if A stops for B. Mar 23 at 2:08
• So the total time B went from P to Q is constituted by 60 minutes, the time it took for A to overtake B, and the other 32 minutes? Mar 23 at 2:11

Let $$d$$ be the distance from P to Q. Let $$r_a$$ and $$r_b$$ be the walking rates of A and B respectively (in miles per hour).

The key facts are:

(1) B takes one hour longer than A to walk $$d-2$$ miles.

(2) In 32 minutes, the sum total distance walked by A and B is $$4$$ miles.

(3) B takes one hour longer to walk $$2d-4$$ miles than it takes A to walk $$2d$$ miles.

From (1) you get (4): $$\frac{d-2}{r_b}=\frac{d-2}{r_a}+1$$.

From (2) you get (5): $$\frac{32}{60}\cdot(r_a+r_b)=4$$

From (3) you get (6): $$\frac{2d-4}{r_b}=\frac{2d}{r_a}+1$$

This gives you three equations in three unknowns for you to solve, which you said would be the easy part for you. [It may be helpful to note that the left side of equation (6) is exactly twice the left side of equation (4).]