Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

A jet plane, flying 165 mph faster than a propeller plane, travels 2970 miles in 3 hours less time than the propeller plane takes to fly the same distance. How fast does each plane fly?

Is there a formula I can use to start this problem? Specifically, what can I do to begin finding the MPH of one of the planes?

share|cite|improve this question
For this type of problem, you are creating the formula. In this case, you have a couple unknowns: the total amount of time travelled and the speed. If $x$ is the speed of the faster plane and $y$ is the speed of the slower plane, then $xt_x - yt_y=0$. Can you come up with two more equations relating $x$ and $y$, and $t_x$ and $t_y$? – abiessu Sep 11 '13 at 23:32
Well, maybe. But before I try, can I ask what the smaller x to the right of the "t" indicates? (Sorry, I am so slow when it comes to this stuff). – Robin Sep 11 '13 at 23:34
Or, what it means in the context of the formula? – Robin Sep 11 '13 at 23:34
$t_x$ indicates the time it takes the plane associated with speed $x$ to fly the given distance. – abiessu Sep 11 '13 at 23:35
So what can I do to begin finding the MPH of one of the planes? – Robin Sep 11 '13 at 23:40

Distance = Rate $\times$ Time.

So, for the propeller plane, $2970 = x * y$, where $x$ is how fast the propeller plane flies, and $y$ is how long it flies for.

Then, for the jet plane, we know $2970 = (x + 165) * (y - 3)$.

If we use the first equation, we could represent $x$ in terms of $y$, or vice versa. Let's represent $x$ in terms of $y$ as $x = \frac{2970}{y}$. Then, for the jet plane, we know $2970 = (\frac{2970}{y} + 165)*(y-3)$.

Hence, $\frac{2970}{y-3} = \frac{2970}{y} + 165$. We could make this easier by saying, $\frac{2970}{y} + 165 = \frac{2970}{y} + \frac{165*y}{y} = \frac{2970 + 165*y}{y}$

This leads us to cross-multiplication of $\frac{2970}{y-3} = \frac{2970 + 165*y}{y}$. Hence, $2970*y = 2970 + 165*y*(y-3) \implies 2970*y = 2970*y - 2970*3+165*y^2 - 165*3*y$. Hence, 0 = -2970*3 + 165*y^2 - 165*3*y. From this point, you could use the quadratic equation to figure out y, and then use 2970 = x*y to solve for x.

share|cite|improve this answer
How would I set up the quadratic equation if for instance what would normally just be "165y" is now "165y^2"? – Robin Sep 11 '13 at 23:59
Well, the quadratic equation would be $165*y^2 - 165*3*y - 2970*3$, hence you just plug in the respective coefficients into $\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. – David Sep 12 '13 at 0:07
Argh, I'm sorry. I don't get it. I'll keep working with it. – Robin Sep 12 '13 at 0:15
Do you understand that the coefficients are 165, -165*3, -2970*3 in order? – David Sep 12 '13 at 0:16
Yes, and I got the solutions 9 and -6 but I'm not sure where those would be relevant in continuing to find the solution... – Robin Sep 12 '13 at 0:17

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.