Gifted Issues Discussion homepage Forums General Discussion Alternate approach to algebra problem

Sometimes I only see the obvious so am looking to see if any of you math experts have a quicker approach (other than the two obvious to me) to getting the final answer to the following problem once you set up the equation. You can use factoring covered in the final third/quarter of Algebra I or substitute the five answer choices in the equation. Since DS likes to calculate in his head, it is almost safer to just try the five answer choices, starting with the likeliest ones. The issue is that DS has the conceptual understanding to set up the equation from the problem, but tends to do all calculations in his head so may arrive at the wrong answer when he can't use a calculator.

Here's the problem:

In calm weather, an aircraft can fly from one city to another 200 miles north of the first and back in exactly 2 hours. In a steady north wind, the round trip takes 5 minutes longer. The speed of the wind, in miles per hour, is
A. 8
B. 20
C. 32
D. 35
E. 40

Here's a Solution:
If you let the wind speed be w miles per hour, then the time in hours required for the aircraft to make the roundtrip in the wind is:

25/12 = 200/(200 + w) + 200/(200 - w)

You can obviously use factoring concepts covered in Algebra I(multiply both sides by the product of (200 + w) and (200 - w)) to solve for w. It is not that difficult but susceptible to miscalculations when done in your head. Since you can probably guess at the correct range, it almost may be quicker/safer to substitue the likely answer once you set up the equation.

My question is whether there is a less calculation heavy way to solve this problem. I have been making my living with words so my math is somewhat rusty and I only see the most obvious approaches.

My algebra is obviously even rustier because I don't see where the 25/12 comes from.

Perhaps you forgot your basic science?

time = distance/speed

25/12 is the 2 hours and 5 minutes it takes to travel with the north wind.

One trick for this problem is to keep the 200 in the equation as if it was a variable whenever possible (like x or y) until the last few lines... This saves evaluating out the 200 quite a number of times.

First I start by diving by 25.. and turn it into:

1/12 = 8/(200 + w) + 8/(200-w)

At this point don't evaluate the 200 with anything till you get..

200(200-12*16) = w^2
200(200-192) = w^2
200(8) = w^2
1600 = w^2

This puzzled me at first as well. This is a time it takes in hours just written as a fraction. 1/12 is 5 minutes. 24/12 is 2 hours.

Obviously part of the problem is setting up the equation to start with (and understanding that).

Once you get to an equation like yours, I personally think it's easier to simplify it with the variable (and you might as well keep going and solve it). The effort of manipulating an expression with a variable is often barely different to manipulating it after substituting a particular number for the variable, so doing the latter 5 times is almost certainly a waste of time.

A slight variation of "multiply both sides by the product of (200 + w) and (200 - w)" would be to put the RHS over a common denominator. And I wouldn't call that "factoring".

As far as a general strategy for multi choice problems that amount to solving an equation: On the one hand there might be situations where it's better to actually solve it then select the correct answer, and on the other hand there might be situations where it's better to substitute in each of the given choices. I couldn't give any rule of thumb for this but would say that someone who could consistently choose the better strategy would be at a higher level than the level being tested.

Also don't just think about how to get through a test. You want your DS to be able to do the algebra (whether or not it is necessary for this particular question) and if he's not fluent in doing it mentally he should build up his fluency on paper (i.e. "walk before you run (or fly)"). You have to think what's better for long term learning.

Thanks - that results in smaller numbers through the steps, which would make it less cumbersome to do in your head. Of course, it would matter less if he could use a calculator.

Interestingly, your approach made me think of another way to use smaller numbers - by keeping 200 squared until the last step:

200^2 - (24/25)(200^2) = w^2

(200^2)/25 = w^2

200/5 = w

Seting up the equation is not a problem for DS - he learned d=rt in the 4th grade science unit on motions.

I was trying to use a shorthand - you are correct that it is not actually factoring but once you covered the factoring concepts in Algebra I, you would learn to remove the variable from the denominator by multiplying by (200 + w)(200 - w) to solve for w.

In this particular case, because the numbers in the equation are small and I can sense which numbers are more likely, it actually takes less time to arrive at the correct answer by substitution than by solving for w.

I see your point about long term learning, but this situation actually has nothing to do with that. In a real school situation, he would either need to lay out many of the steps for essay problems or simply use his calculator to check his answer to avoid careless calculation errors for multiple choice questions. This problem is in the context of no calculator plus time contraint.

Yup you made it better.

Doing this is very helpful when taking math tests & classes. Not as necessarily as helpful in the real world, but does avoid making calculation mistakes. And I think a LOT of students want to evaluate their numbers out too early, and it often takes seeing how much easier such questions are to do this way to help break the habit. I think it comes from all the drill kids get in arithmetic.

This is particularly helpful in standardized math tests & questions. I tell my kids, that for this kind of test if the numbers start getting particularly painful to handle you have probably made a minor mistake OR there is a simpler way to solve it.

Other option for this problem if you are trying to figure out how to solve it the quickest & most accurate. A hybrid of the solve vs. plug in. Personally I find plugging into the original equation a bit tricky what with figuring out all those LCD. But if you get the problem partly done it's a bit more obvious.

LOL. That reminds me of DD, who is not very mathy but really loves food. Everytime I needed to explain a math concept to her in more depth, she gets it right away if there's cookies or pizza or other favorite foods involved rather than the original objects. I think that she perks up and pays attention.