Posts tagged with: OT Test 1 Section 4

Test 1 Calculator OK #16

The graph here is definitely a linear relationship between h and C, so we’re going to want to figure out the slope and the C-intercept. Then we’ll use the standard y=mx+b slope-intercept formula, only for this problem it’ll be C=mh+b.

Slope is rise over run, or the change in the vertical variable divided by the change in the horizontal variable. Let’s pick two points from the line and plug them into the slope formula. How about (1, 8) and (3, 14)?

    \begin{align*}\text{slope}=\dfrac{C_2-C_1}{h_2-h_1}=\dfrac{14-8}{3-1}=3\end{align*}

Once we have the slope, we’re actually done because whoever wrote this question wasn’t very tricky with the answer choices. The C-intercept of this graph is obviously not zero, and only choice C has the slope we just calculated plus a non-zero C-intercept. Sure enough, a quick glance at the graph confirms that the C-intercept is 5, meaning the equation should be C=3h+5.

Whats the best and fastest way of doing Practise test 1, section 4, question 23?

Would it be calculating the ratio of all the options and then comparing it to the Human Resources ratio?

Thanks!

Yeah, that’s the best and fastest way, although you can save yourself some calculator keystrokes by rounding. This isn’t just a time saver: the more keystrokes you make, the more likely you are to make an error! Rounding to the nearest million (it’s already in thousands) will work just fine. So, my calculations would look like this:

Human resources 2007 to 2010: \dfrac{4051}{5921}=0.68

Agriculture/natural resources 2007 to 2010: \dfrac{374}{488}=0.77

Education 2007 to 2010: \dfrac{2165}{3008}=0.72

Highways and transportation 2007 to 2010: \dfrac{1468}{1774}=0.83

Public safety 2007 to 2010: \dfrac{263}{464}=0.57

There you have it: the Human resources and Education budgets have the closest ratios.

Hi Mike. Can you please explain Practise test 1, section 4, question 14? I figured out that the range was the largest change (but only by physically calculating the mean and median etc.)

Is there a better way to do this?

Also, I’d just like to say THANK YOU THANK YOU THANK YOU for this book it really is amazing!!

Well, you have to consider every answer choice, but there are a few mental shortcuts that I think are helpful.

The main thing to recognize right away is that the data is in order, so calculating the median is very easy. The middle value is 12, and if you remove that outlier 24, the median will STILL be 12. (It’s useful to remember that, in general, outliers have little to no effect on the median—that’s why you usually hear about median household income instead of mean household income, for example.) So I can cross of median without calculating it, and then I can cross off choice D as well.

The data being neatly sorted also helps one to think about the mean. Notice how nicely clustered the data is: how the 13s, 14s and 15s pretty nicely balance out the 11s, 10s, and 9s. That tells you that the mean might not be exactly the same as the median, but it should still be very close to 12. So without adding up all the values, I know the average of 20 of the values should be about 12, which means their sum should be about 240 (average times # of numbers equals sum). Adding the erroneous 24 value in there won’t change the average much: \dfrac{\approx 240+24}{21}\approx 12.57. (Note that while this estimate is not completely accurate, it doesn’t matter because you’re just looking for the biggest change, not the exact change.)

The range, on the other hand, will change a lot! The range with the 24 measure in there is 24-8 = 16. Take that 24 out and the range becomes 16 - 8 = 8. That’s by far the biggest change.

PS: Glad you like the Math Guide! 🙂

Test 1 #35 (calculator section)

You can find the formula for the volume of a right circular cylinder at the beginning of the section; it’s V=\pi r^2h. Since the question tells you the volume (72π cubic yards) and the height (8 yards), you can solve for the radius:

    \begin{align*}72\pi&=\pi r^2(8)\\9&=r^2\\3&=r\end{align*}

Now, be careful! The question asks for the diameter, not the radius. Even though they underline the word diameter, I bet a lot of people still miss this question by putting the radius.

If the radius is 3, then the diameter is 6.

Hello, can you please explain number 27 section 4 test 1?

Yeah, this is a really tricky one. The key to getting it right is recognizing just how much of the field the students are NOT covering! In a 10 ft by 10 ft field, there will be 100 1 ft by 1 ft squares. Look, below is a 10 by 10 grid with X’s marking places where the 10 students in this question might be.

The data in the table tells us that, on average, there are about 150 worms in each of the small boxes. Since there are 100 small boxes in the big field, we need to multiply 150 by 100 to get our answer for the approximate number of earthworms in the whole field.

150(100) = 15,000

Hi! Can you please explain number 25 section 4 test 1?

Yep! You’re given this equation: h=-4.9t^2+25t. That’s the height of a ball t seconds after it is launched. You need to know how much time goes by before the ball hits the ground. The key thing to remember is that h will equal zero when the ball hits the ground, so you’re really solving 0=-4.9t^2+25t for t.

Since this is the calculator section, you have lots of options at your disposal.

First, you can backsolve. Just plug each answer choice in for t until you get something close to zero as an answer. This is a little tricky because you won’t get exactly zero, but the question does say “approximately.” Below is my calculator’s screen after I tried choice C, recognized that that was probably too high, remembered that a falling ball will get closer and closer to 0 the longer it’s in the air, and then tried D, which turned out to be pretty close to zero.

t1s4-25a

Another way to go is to graph. This is my favorite way for this question, because you can just eyeball the graph and see the right answer—you don’t even need to use the zero function to get it exact. Look:

Pretty obvious that that parabola hits y = 0 around x = 5, right?

The last way you can go is to factor and solve algebraically.

0=-4.9t^2+25t

0=t(-4.9t+25)

Since you aren’t interested in the fact that the height was zero at time zero, solve 0=-4.9t+25.

0=-4.9t+25

-25=-4.9t

5.102...=t

Will you please work #20 in section 4, Test 1.

Sure! When you see percents, you should almost always think about plugging in 100. In this case, that works like a charm. Say the laptop originally cost $100. When it was put on sale for 20% off, the price went to $80. Buying an $80 laptop with 8% sales tax will cost 80+\frac{8}{100}(80)=86.4. Which answer choice gives you 100 when you plug in 86.4 for p? Only choice D.

If you want to know WHY that works, then look at the algebra. Forget about the tax for a moment. Alma bought the computer at a 20% discount. So if the computer originally cost x dollars, then the price she paid can be calculated thusly:

    \begin{align*}&x-\frac{20}{100}(x)\\=&x-0.2x\\=&0.8x\end{align*}

Now we need to add 8% sales tax to that. We do that by taking the price she paid and increasing it by 8%:

    \begin{align*}&0.8x + \frac{8}{100}(0.8x)\\=&0.8x+0.08(0.8x)\\=&0.8x(1+0.08)\\=&0.8x(1.08)\\=&(0.8)(1.08)x\end{align*}

That expression gives you the price Alma paid for the laptop—which the question calls p. So we can say:

    \begin{align*}p=(0.8)(1.08)x\end{align*}

Since the question asks you for the original price, x, in terms of p, solve for by dividing.

    \begin{align*}\dfrac{p}{(0.8)(1.08)}=x\end{align*}

Test 1 Section 4 #34

Lots of people miss this one because they don’t read carefully enough. The question asks for the total number of slots the station can sell on Tuesday AND Wednesday—two days! So we need to figure out how many 30-minute slots there are in a day, and then multiply by two.

There are 2 slots in an hour, and 24 hours in a day, so there are 48 slots in a day. Multiply that by 2 to cover Tuesday and Wednesday and you get the answer: 96. Below is how you’d set it up if you were being very careful with your units, which, incidentally, I strongly recommend.

\dfrac{2\text{ half-hour slots}}{1\text{ hour}}\times\dfrac{24\text{ hours}}{1\text{ day}}\times 2\text{ days}=96\text{ half-hour slots}

Test 1 Section 4 #36

You’re in the calculator section, so the fastest way to get this is by graphing. Just type the function in carefully, and graph it:

t1s4-36-1

Those vertical asymptotes around x = 3 tell you that 3 is probably the answer. Want to make sure? Look at your Table:

t1s4-36-3

Yep, the function is undefined at = 3.

Whenever I post a calculator-based solution, though, people always ask me how to solve without a calculator. Of course, there’s a way, but here is a good place for me to recommend to you that if you don’t have one of your own, you find a calculator you can use on test day (maybe you can borrow one from school) and take the time now to get used to using it. Anyway, the math:

    \begin{align*}h(x)&=\dfrac{1}{(x-5)^2+4(x-5)+4}\end{align*}

What they’re trying to show you there with the (x – 5)s is that you might want to consider them a whole value and simplify this by substituting for (x – 5). Let’s say (x – 5) = p for a moment. Then we can say:

    \begin{align*}h(x)&=\dfrac{1}{p^2+4p+4}\end{align*}

Of course, that’s a binomial square, so we can say:

    \begin{align*}h(x)&=\dfrac{1}{(p+2)^2}\end{align*}

Now reintroduce (x – 5) for p:

    \begin{align*}h(x)&=\dfrac{1}{(x-5+2)^2}\end{align*}

    \begin{align*}h(x)&=\dfrac{1}{(x-3)^2}\end{align*}

And that is why the function is undefined at x = 3: the denominator equals zero when x = 3.

Note that you can also simplify to h(x)=\dfrac{1}{(x-3)^2} if you expand the denominator rather than doing the substitution I just did. That’s not too much more labor intensive and is arguably more straightforward. That solution looks like this:

    \begin{align*}h(x)&=\dfrac{1}{(x-5)^2+4(x-5)+4}\end{align*}

    \begin{align*}h(x)&=\dfrac{1}{x^2-10x+25+4x-20+4}\end{align*}

    \begin{align*}h(x)&=\dfrac{1}{x^2-6x+9}\end{align*}

    \begin{align*}h(x)&=\dfrac{1}{(x-3)^2}\end{align*}

Test 1 Section 4 #30

To get this one, it’s helpful (but not necessary, if you’re nimble enough) to know the vertex form of a parabola, which is y=a(x-h)^2+k where (h,k) is the vertex of the parabola. If you know that, then you can recognize immediately that only choice D is even in that form! And, sure enough, choice D says y=(x-1)^2-16, which indicates a vertex at (1,-16). That’s consistent with the given figure.

Even if you don’t know the vertex form, though, you can get this if you read the question and figure carefully, as long as you know that the vertex of a parabola is its minimum or maximum. You need the coordinates of the vertex to show up in the equation as constants. Just by looking at the figure you can see that the x-coordinate is definitely 1, even if you’re not sure what the y-coordinate is because of the scale of the y-axis. Choice D is the only choice with a 1 in it as a constant! (You’re also in the calculator allowed section, so you are totally allowed to graph the original equation and find the exact vertex if you don’t trust your eyeballs).

But OK, we can do the algebra, too. To solve this algebraically, we need to convert from the given equation to the vertex form (above) by completing the square.

    \begin{align*}y&=x^2-2x-15\\y+15&=x^2-2x\\y+15+1&=x^2-2x+1\\y+16&=(x-1)^2\\y&=(x-1)^2-16\end{align*}

Test 1 Section 4 #28!

Now here’s a question I love because it’s got a really easy solution if you take a step back from it. You’re asked about the system of inequalities y\geq 2x+1 and y>\frac{1}{2}x-1. Solutions to a system of inequalities are the overlap of both individual solution regions, so really all you need to do to solve this is to recognize that the first inequality will have no solutions in Quadrant IV: it has a positive slope and a positive y-intercept.

If the first inequality has no solutions in Quadrant IV, then the system can have no solutions in Quadrant IV.

Below is an illustration of the full solution, which you can draw a less pretty version of based on the slopes and intercepts given. The blue region is the solution to the first inequality, and the red region is the solution to the second inequality. The overlapping (purple) region is the solution to the system of the two inequalities. See how the blue region never touches Quadrant IV? That’s why I’m saying that’s the only inequality you need to pay attention to if it’s the first one you think about.

Test 1 Sec4 #24 help!

For this one, you’re looking for the equation of a circle with a center of (0,4) that passes through \left(\dfrac{4}{3},5\right). The best way to go here is just to build the equation.

For this test, you should know the standard circle equation: A circle with center (h, k) and radius r has the equation (x-h)^2+(y-k)^2=r^2. You can fill in the left side immediately: The center of this circle is (0, 4), so the left side should be (x-0)^2+(y-4)^2=x^2+(y-4)^2.

To fill in the right side, you need to find r (or r^2). Easy enough:

\left(\dfrac{4}{3}\right)^2+1^2=r^2\\ \dfrac{16}{9}+1=r^2\\ \dfrac{16}{9}+\dfrac{9}{9}=r^2\\ \dfrac{25}{9}=r^2

So the answer you want is A.

A note for those who like quick and dirty solutions: Once you’ve got the left side of the equation, you know the answer must be either A or C. If you’re paying close attention, you’ll notice that the only difference between A and C is on the right side of the equation: A says r^2=\dfrac{25}{9} and C says r^2=\dfrac{5}{3}. If you’re in a hurry, that’s probably good enough—you know you’re looking for a squared value, and A provides it.

Another note: if you can’t remember the circle equation, you can still get this question right! Use your calculator to plug x=\dfrac{4}{3} and y=5 into each answer choice. Only one will work. 🙂

From the new blue book, test #1, section 4, #29:
For a polynomial p(x), the value of p(3) is −2. Which of the following must be true about p(x) ?
A) x − 5 is a factor of p(x).
B) x − 2 is a factor of p(x).
C) x + 2 is a factor of p(x).
D) The remainder when p(x) is divided by x − 3 is −2.

Yeah, this is a tricky one. I actually think the way many students will get it is by elimination. Given p(3) = –2, we know nothing about x – 5, x – 2, or x + 2. So, choice D might not be intuitive, but at least it’s related to p(3) somehow.

Remember that if p(3) = 0, then we know that x – 3 is a factor of p(x). Here’s the tricky bit: if p(3) = –2, then p(3) + 2 = 0. In other words, if we shift p(x) up 2, then we’ll have a polynomial that’s divisible by x – 3. If that’s the case, then the remainder when p(x) is divided by x – 3 must be –2.

Really, the best way to understand this is to see an example. Let’s make up a function: f(x)=(x-2)(x-3). We can FOIL that out: f(x)=x^2-5x+6. When we graph that, we see that there are zeros at x = 2 and x = 3. And because we made the function the way we did, we already know that it’s divisible by x – 2 and x – 3.

function1

Now, what happens when we shift that function down by 2? f(x)-2=x^2-5x+6-2=x^2-5x+4. Let’s call that g(x). Obviously now we don’t have zeros at x = 2 and x = 3. We have –2s instead: g(2) = g(3) = –2.

function2

What happens when we divide g(x) by x – 2?

     \begin{align*} &x-3\\ x-2|&\overline{x^2-5x+4}\\ &\underline{x^2-2x}\\ &\:\:\:\:\:\:{-3x+4}\\ &\:\:\:\:\:\:\underline{-3x+6}\\ &\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:-2 \end{align*}

That’s right—we get x – 3 with a remainder of –2.