Daily Math Problems Quarter I

By Evan Hughes

90 math problems designed to challenge you and help you hone your math skills. Each problem is designed to test your limits, and explanations are given with every problem, as well as tips for tackling arithmetical obstacles to make them simpler than they may seem. Also provided are formula sheets including all properties and formula that may be needed for any of the problems. Topics range from simple fractions through basic calculus. Real-world situations are presented in many of the problems to help demonstrate how math can be used outside of the school room.

• Language: English
• Publisher: Evan Hughes
• Release date: May 3, 2016
• ISBN: 9781311868695

Evan Hughes

Born in Texas but raised in Germany, Evan Hughes grew up in the land of fairy tales. His previous works include articles in the "Stuttgarter Algemeine" and several poetry anthologies, including "Patchwork" and "Musings". Most recently, one of his fairy tales can be found in the anthology "Merry Sorrows (Un)Happy Endings." Poetry infuses his fiction, giving it a unique musical rhythm.Evan holds a Bachelor of Arts degree in Physics from Pitzer College, and is the author of a series of articles on physics for laypeople. He has observed an extra-solar planet using Pomona College's 1m telescope at Table Mountain Observatory, and made use of the Very Large Array in New Mexico to study the Fornax A radiogalaxy.

Book preview

Contained in this book you will find 90 mathematical problems to challenge your skills and reasoning abilities. Each problem is presented in a way to test your ability to reason out situations, manipulate mathematical constructs, and use deductive and inductive reasoning to aid in conquering arithmetical obstacles.

This is not a text book. Nor is it intended as a substitute for a text book. However, you can certainly use it to supplement your mathematical education In fact, I encourage you to do so.

For every problem, I present explanations wherein I attempt to break each one done and make it as simple as possible. Don’t worry if you don’t know the techniques or formulae needed for any given problem – I’ll explain and provide them so that you might be able to master those techniques yourself.

These problems also are not presented in any particular order. One day you may be working on fractions or exponents, and the next tackling a situation involving the physics of gravity. They were originally a series of daily articles, and are presented in the order in which they were originally posted. Feel free to approach them in the sequence given, or, if you prefer, peruse them and try to do them in a progressive sequence of your choosing.

Formula Sheets

Problem of the Day 1

One of the beauties of mathematics, is that there are frequently multiple ways to approach any given problem, all of which will work. But this beauty also serves as a drawback – some ways will prove far more difficult, or at least time-consuming, than other ways. Couple this with the tendency of many people to go on what’s called math auto-pilot, which is letting habit take over and just doing what you’ve always done without thinking, and it is quite easy to waste time on a problem that could be finished much, much more quickly. We see this frequently on some standardized tests, like the SAT, which has problems which are specifically designed to make you waste your time. In this light, let’s look at what I consider to be a great example of this.

Here’s the problem. Try doing it yourself before reading on, where I address the different methods by which you can approach it. If you use the fastest method, you should be able to complete it in under 5 seconds.

In it, we’re multiplying six fractions. The tendency of many people, especially if they go on math auto-pilot, is to go straight to the multiplication. Let’s go ahead and do this, and see how many steps it’s going to take to finish the problem this way. First, we multiply the top numbers, or the numerators, so we get 2x x 3 x 4 x 5 x 6. Only the coefficients multiply, leaving the variable alone, so we get 720x. In the denominators, or numbers on the bottom, we get 2 x 3 x 4 x 5 x 6 x 7y. Again, only the coefficients multiply, so we get 5040y. Giving us the fraction 720x/5040y.

But we’re clearly not done. It looks like we need to simplify the fraction. At this point, you can do one of three things. Some people would do the long division, but that’s going to give a decimal, whose fractional equivalent isn’t obvious. On standardized tests, the answers available for this problem are usually going to be in fractions, so this isn’t a desirable method. We could factor both the numerator and denominator, find the greatest common factor of the two, then divide both by it (resulting in the need for long division again), and get the final answer. This is the MOST time consuming method. The third available is the one I suggest when you’re simplifying fractions, which is to just apply the first common factor that comes to mind for each one, and repeat until you get to the simplest form of the fraction. Here, I simplified first by 2, then 3, then 2 again, then 3 again, then 10, and by 2 once more. There were 6 total steps of simplification this way. Still faster than finding the greatest common factor, but, including the multiplication step, the seven total steps of this method were just far too slow.

The fastest method, the easiest one, and the one I recommend. SIMPLIFY BEFORE YOU MULTIPLY. I don’t know how many times I have to remind my students to do this, but it will always save you work and time. We remember that, as long as we’re multiplying fractions, when we see a number in the top of one of the fractions multiplied AND in the bottom of one, we can just cancel both of them. Cross the pair out. The two numbers don’t exist anymore. Here, we see 2 paired, 3, 4, 5, and 6. Holy cow, look at that, EVERYTHING cancels except the 1, x, and 7y. Our final answer, with ZERO multiplication, is x/7y (we can ignore the 1 since anything times one is just itself). Doing it this way, you can complete this problem well under 5 seconds, depending on how fast you can cross out numbers.

So, that’s it for our problem of the day. Remember to not let yourself go on math auto-pilot when attacking your problems – a second taken to give it a look and think of the fastest way to do it can save you minutes (or more) in the long run.

Problem of the Day 2

Let’s do some Algebra today. We’re going to tackle a problem with two variables. Here’s the problem:

Typically, whenever you have more than one variable, you need at least one equation for every variable in order to obtain a solution. Too few equations, and the system is unsolvable. So for our problem above, we typically need two equations to solve, since it has two variables. Which might make this one seem unsolveable. Or is it?

Another consideration is that we see it involves quadratics. A typical rule for polynomials, is that there exists one possible solution for every degree of the polynomial. The degree is equal to the highest exponent. So, for quadratics, or polynomials of degree two, we see there are 2 possible solutions. Let’s keep that in mind as we attempt to solve this problem.

There are two different approaches to this problem, and I’m going to start with the less obvious one first (and with reason, as we’ll see later on). The very first thing I’m going to do, is I’m going to divide both sides by the entire right side (x²-y²), which effectively gives me a rational expression equaling one.

Next, I recognize two things. First, the numerator, or top of my fraction, is a perfect square, so I’ll factor it as such. Second, the denominator, or bottom of my fraction, is a difference of squares, so I’ll also factor it accordingly.

I then see that at least one (x+y) term exists in both the numerator and denominator, so I can cancel one each, leaving us with a much simpler fraction. The quadratic has become linear.

For my next step, I’ll multiply both sides by (x-y), effectively moving it back to the right-hand side of the equation. I’ll then subtract x from both sides, making it go away, leaving me with y=-y. The only way this can be true, is if y=0. So we have one solution for the equation the exists everywhere along the line y=0.

Are we done yet? Let’s take another approach and see. This time, we see that there is a positive x² on both sides of the equation, so it effectively cancels and goes away when we subtract x² from both sides. I’ll then add y² to both sides, moving it to the left side of the equation.

I can now see that I can factor 2y out of all three existing terms, so I’ll do that.

This leaves me with x+y=0, which means we have a solution along the line x=-y. So we have another solution set for the equation. BUT, this means we now also have a unique solution for the equation. Where the two solution sets cross, we have a single point that solves the equation - in the solution set x=-y, where y=0 (as from the first solution set), so does x. So we can say this equation has two solutions sets - y=0 and x=-y, which intersect at the point (0,0).

What is our lesson learned from this problem? While there may be multiple ways to approach a mathematical problem, they will not always produce the same solution. This doesn’t make one way wrong. What it means, is that, with more complex situations, we cannot always let ourselves rest when we find a solution after using one technique. We need to check our work, and try other possible methods to see if there indeed IS another solution, especially when we’ve got a system where we know there could be more possible answers (as in this quadratic where we knew from the start there could be two possible answers). In other words, don’t get cocky, or the math could bite you.

Problem of the Day 3

Today’s problem comes from a physical situation I’ve been examining. In it, I ended up with an odd looking fractional expression, which I needed to simplify to determine if I’d gotten my units correct. Here it is:

I have to admit, when I first looked got it down to this, I stared at it for quite some time trying to determine what my next step would be. How could I make sense of this? Then it hit me. First, I need to remember that negative exponents follow the same rule as positive exponents, so I should let the negative exponents throw me. Second, how could I break these apart? I see a difference of squares in the numerator. Can I take advantage of that? Absolutely. In order to do so, my first step is to factor an s-1 out of both terms on the top. And since that’s just 1/s, I’m going to go ahead and move it to the denominator while I’m at it.

Now that I’ve done that, I have a difference of squares in both the top and bottom. Just because the exponents are negative, and that the powers are two and 4 in both, doesn’t mean it doesn’t work the same. It does. So let’s factor both.

Now that we’ve done that, we can see that we have a like term in both the numerator and denominator. Yes, the powers are in different order, but because both those terms are addition of positive values, their order doesn’t matter. They’re the same term. So we’ll cancel one in top and bottom.

Now I’m going to try to get the remaining terms to look the same. To do that, I’m going to factor a negative sign (-1, effectively) out of the parenthetical in the denominator, giving me a negative s-2 and positive s-1, just like in the numerator. I can reorder those accordingly.

Now I can cancel those remaining two like terms in the numerator and denominator, leaving me with the simplest form of the fraction. Which turns out to be the kind of units I wanted, by the way! And who would have thought upon initial inspection of the fraction that we would have gotten something so simple! This problem was entirely about two things. Don’t let negative exponents throw you – they work the same as positive exponents, as far as the algebra’s concerned. And recognize those difference of squares when they’re present, and take advantage of them to help you find ways to simplify your situation.

Problem of the Day 4

Today’s problem is short and sweet. It tests your ability to group and deal with exponents. Here it is:

Now some people, as soon as they see a variable in the exponent, freeze up and stop thinking. Don’t fall into that trap. Just remember the math still works exactly the same, and look for ways to change the expression to your advantage. You *could* use logarithm rules to solve this equation, but you don’t need to do so. Frequently, when you’re handling exponential equations, their solutions are much easier and never need to involve logarithms. This is one of those cases.

First thing we should do, is regroup everything. I see that, on the left side, not only do I have a common base, but I have the exact same term 4 times. So instead of worrying about how do I add exponential terms, I’m going to change this into a multiplication problem. Since I have 4 of them, I simply rewrite it so. Further, I see that 4 is a power of two, and all my exponentials have a base 2, so I’m going to write that 4 as a power of 2, like so:

Now I’ll divide both sides by 2², eliminating it from the left side. On the right, we have the rule for dividing exponential terms with common bases, where we simply subtract the denominator’s exponent from the numerator to get our result: 8-2=6, so we end up with 2⁶. And now that we have one term on either side of the equation, both with the same base, and no coefficients, the base no longer matters. We can ignore it completely. Our solution is x=6.

And, just to make sure we did our math right, we can always test it. 2⁶=64. 64+64+64+64=256. And 2⁸ is indeed 256. Beautiful!

So remember, when solving exponential equations, you don’t always need logarithms. It’s much easier if you can find a way to rewrite everything so that you can just ignore the bases.

Problem of the Day 5

Today, let’s tackle a problem that really demonstrates how math is useful in the real world. Kids, especially teenagers, frequently ask why they need to learn all that math, and how is it ever going to be of use to them in real life. Well, this situation is like one someone just might run into in handicrafts, an area where many people don’t think they need any mathematical skills.

Here’s the situation.

A painter is painting a logo for a company. The painter needs to know how much paint he needs. The company wants the blue circle to have a diameter of 2 meters. The painter has already figured out that every can of paint he uses can cover exactly one square meter. So how many cans of blue paint does he need? And purple paint? The logo is comprised of three shapes – a solid blue circle inscribed inside a black square outline inscribed inside a solid purple circle. The image below is what it looks like. See if you can complete the problem on your own before looking for the answers down below.

The first thing we need to do is figure out the dimensions of all our shapes. If the blue circle has a diameter of 2 meters, its radius is half that, or 1 meter. The sides of the square must be exactly the same length as the diameter of the blue circle inscribed in it (if you don’t believe me, shift it until the side is going straight through the center of the circle and you’ll see it’s true), so it’s got sides of 2 meters each.

From that, we can figure out the diameter of the purple circle. It turns out its diameter is the same as the diagonal of the square. And that we can determine using the good ol’ Pythagorean Theorem.

What’s even better, is the sides of the square are all the same, so in the Pythagorean theorem for the square, a=b=s. So from just that one piece of information, the radius of the blue circle, we’re able to find the radius of the purple circle, which is half its diameter, or half the diagonal of the square, and that comes to √2.

So now that we know the radii of both circles, we can figure out their areas. Reminding ourselves of the formula for the area of a circle, we have the following. Note that I used the subscript b for the blue circle and the subscript p for the purple circle.

But looking at the logo, we see that the painter doesn’t need enough purple paint to paint the entire purple circle. He only needs enough to paint the area of it outside the blue circle. Which is the same thing is if we took the whole purple circle and took out the part of it the blue circle covers. In other words, we