Posts tagged with: OT Test 8 Section 4

Could you do Test 8 section 4 question 24 for me?

I can’t understand why the answer is D and not C.

It’s not A because we have no way of knowing what the people who didn’t vote thought.

It’s not B because we know nothing about the age of the people voting.

It’s not C because the votes are worth the same whether they come in via social media or text message. If you change the method by which some people voted, you don’t change their votes.

(By the way, we know contestant 2 won 70% of 30% of the votes, and 40% of 70% of the votes. That means she lost: 0.7\times 0.3 = 0.21 and 0.4\times 0.7 = 0.28, so contestant 2 won 21% + 28% = 49% of the vote.

It is D because we know contestant 2 won 70% of the social media vote, and choice D is basically asking whether that’s true. If 70% of social media voters voted for contestant 2, then social media voters were more likely to prefer candidate 2, period.

On a question like this, look for reasons to eliminate choices rather than looking for reasons to accept choices. Otherwise, sometimes it’s easy to get tricked by a choice that sounds pretty good at first (like C) but doesn’t hold up under scrutiny.

Practice test 8 Calculator #13

First, you can plug in on this one, so if you feel rusty on your exponent rules at all, that’s a good move. Especially on the calculator section. Say, for example, that you plug in 4 for a. Just enter it all into your calculator (you may need to be careful with parentheses in the exponent depending on the kind of calculator you have):

    \begin{align*}4^{-\frac{1}{2}}&=x\\0.5&=x\end{align*}

Now that you know x, plug 0.5 into each answer choice to see which one gives you 4.

A) \sqrt{0.5}\approx 0.707

B) -\sqrt{0.5}\approx -0.707

C) \frac{1}{0.5^2}=4

D) -\frac{1}{0.5^2}=-4

Obviously, C must be the answer.

To solve this algebraically, first start by squaring both sides. Raising a power to a power is the same as multiplying the powers, so that’ll get rid of the 1/2 on the left:

    \begin{align*}\left(a^{-\frac{1}{2}}\right)^2&=x^2\\a^{-1}&=x^2\end{align*}

Now raise both sides to the –1 power to get a truly alone. Remember that a negative exponent is the same as 1 over the positive exponent, so you can transform the right hand side from x^{-2} to \frac{1}{x^2} to finish the problem.

    \begin{align*}\left(a^{-1}\right)^{-1}&=\left(x^2\right)^{-1}\\a&=x^{-2}\\a&=\frac{1}{x^2}\end{align*}

Test 8 Section 4 #23

I think the easiest way to get this one is to plug in some numbers. We know that every 4 quarters is a year, so the correct answer will be the one that gives us the same result for 1 year as 4 quarters, 2 years as 8 quarters, 3 years as 12 quarters, etc.

At 1 year, the original equation works out thusly:

    \begin{align*}M&=\num{1800}(1.02)^t\\M&=\num{1800}(1.02)^1\end{align*}

Which answer choice(s) result in the same expression when q = 4? Only choice A does:

    \begin{align*}M&=1800(1.02)^\frac{q}{4}\\M&=1800(1.02)^\frac{4}{4}\\M&=1800(1.02)^1\end{align*}

If you’re concerned that another one might also work, just check with your calculator!

    \begin{align*}1800(1.02)^1&=1836\\1800(1.02)^\frac{4}{4}&=1836\\1800(1.02)^{4(4)}&\approx 2471\\1800(1.005)^{4(4)}&\approx 1950\\1800(1.082)^4&\approx 2467\end{align*}

Sure enough, only choice A gives you the same result after 4 quarters that the original equation gives you for 1 year.

Test 8 Section 4 #8

This is a good one to backsolve! Note that you’re asked for a value of x + 1 (as opposed to a value for x) so backsolving is really easy. Start with the easiest ones to try.

    \begin{align*}x+1&=\dfrac{2}{x+1}\\\\\text{C)     }\ \ \qquad 2&=\dfrac{2}{2}\\\\2&\ne1\\\\\text{D)     }\ \ \qquad 4&=\dfrac{2}{4}\\\\4&\ne 0.5\\\\\text{B)     }\qquad\sqrt{2}&=\dfrac{2}{\sqrt{2}}\\\\\sqrt{2}&=\dfrac{2}{\sqrt{2}}\times\dfrac{\sqrt{2}}{\sqrt{2}}\\\\\sqrt{2}&=\dfrac{2\sqrt{2}}{2}\\\\\sqrt{2}&=\sqrt{2}\end{align*}

Choice B looks good to me!

Test 8 Section 4 #27 please

Yeah, this is a pretty tricky one! The key to getting through it is recognizing that whenever you have a point on a line, you can plug the coordinates into the line’s equation. Because the question wants to know about \dfrac{r}{p}, we should start by getting our equations in terms of r and p.

We know (pr) is on yxb, so we substitute:

    \begin{align*}y&=x+b\\r&=p+b\end{align}

Likewise, we know that (2p, 5r) is on y = 2xb, so we substitute:

    \begin{align*}y&=2x+b\\5r&=2(2p)+b\\5r&=4p+b\end{align}

From there, we can use the fact that both equations have a b in them to our advantage. Solve each for b, then set them equal to each other:

    \begin{align*}5r&=4p+b\\5r-4p&=b\\\\r&=p+b\\r-p&=b\\\\5r-4p&=r-p\end{align}

Now just combine like terms and solve for \dfrac{r}{p}:

    \begin{align*}5r-4p&=r-p\\4r&=3p\\r&=\dfrac{3}{4}p\\\dfrac{r}{p}&=\dfrac{3}{4}\end{align}

 

Test 8 Section 4 #30

First, don’t be intimidated by all the visuals here. This one is not nearly as bad as it looks.

Start by identifying the maximum of the graph of f. Hopefully you agree with me that the greatest y-value the graph reaches is 3, which it reaches when x = 4. So we know that the maximum value of f is 3. Therefore, k = 3.

From there, all you need to do is read the table and find g(3). What does the table say in the g(x) column when it has a 3 in the x column? That’s right, g(3) = 6, so the answer is 6.

Hey Mike, could you please help me with practise test 8, section 4 question 38? If you simply plug in P as 152, it’s easy to get the answer. However, im not sure what “if P approximates the values in the table to within 10 micrograms per milliliter” exactly means? How does that statement help me to solve the question or change the numbers I’m inputting as P?

Thank you so much!

I think you’ve got the key here already: recognizing that P is supposed to approximate all the values in the graph/table, so start with the easy ones! The table tells you that from minute 0 to minute 5, the penicillin concentration goes from 200 to 152 micrograms per milliliter. As you said, plug those values into the formula:

    \begin{align*}P(t)&=200b^\dfrac{t}{5}\\\\152&=200b^\dfrac{5}{5}\\\\152&=200b^1\\\\\dfrac{152}{200}&=b\\0.76&=b\end{align*}

Now use your calculator to quickly test whether setting b=0.76 gives you values within 10 micrograms per milliliter to each of the values in the table.

200(0.76)^{\dfrac{5}{5}}=152\\\\200(0.76)^{\dfrac{10}{5}}=115.5\\\\200(0.76)^{\dfrac{15}{5}}=87.8\\\\200(0.76)^{\dfrac{20}{5}}=66.7

Indeed, each of those numbers is well within 10micrograms per milliliter of the values in the table. Now all we need to do is remember to round to nearest tenth! 0.76 rounded to the nearest tenth is 0.8, which is the answer.

Note: If you rounded to the nearest tenth first, you’d actually still be OK:

200(0.8)^{\dfrac{5}{5}}=160\\\\200(0.8)^{\dfrac{10}{5}}=128\\\\200(0.8)^{\dfrac{15}{5}}=102.4\\\\200(0.8)^{\dfrac{20}{5}}=81.9

That’s a little worse, but each of those values is still within 10 of the values in the table (OK, so 128 is exactly 10 from 118). Trying 0.7 or 0.9 would give you additional comfort that 0.8 is the right answer.

 

Can you explain question #24 in Practice Test #8 Section 4?

The key to this one is to eliminate every choice that makes an unjustified (i.e., not directly supported by the question) assumption.

You can eliminate choice A because we don’t know anything about the preferences of the viewers who didn’t vote. Generally speaking, don’t draw conclusions about what could have happened, draw conclusions about what DID happen.

You can eliminate choice B because the question mentions nothing about the ages of voters. Don’t let your own assumptions about who texts and who uses social media cloud your judgment!

You can eliminate choice C because it’s mathematically false (more on this below) but you should also lean towards eliminating it just because it’s using something that didn’t happen as a basis for its conclusion. As before, resist the urge to make conclusions about what could have happened when you only know what DID happen.

Choice D is totally supported by the data: 70% of social media voters preferred Contestant 2, and only 40% of text message voters preferred Contestant 2. Therefore, social media voters were more likely to prefer Contestant 2 than text message voters.

Now, as for why C isn’t mathematically true…let’s plug in! Say there were 100 voters. We know 30% of the votes came in via social media, so that’s 30 votes. The other 70 votes must have come in via text message.

Of the 30 social media votes, Contestant 2 got 70% of them, or 21 votes. Of the 70 text message votes, Contestant 2 got 40% of them, or 28 votes.

So of the 100 votes, Contestant 2 only got 21 + 28 = 49 votes! That’s not enough to win the contest.

Could you please explain Test #8, calculator-allowed section, number 28?

Thanks!

You have to evaluate both range and standard deviation here. The former you can calculate. The latter you just evaluate visually; You will never have to actually calculate a standard deviation on the SAT.

Let’s talk about range first. To calculate range you just subtract the smallest number in the set from the biggest number. In this case, that means that r_1=88-56=32 and r_2=112-80=32. The numbers in the second set are obviously higher, but the range in both sets is the same: the highest number is 32 greater than the lowest number. Therefore, you really only need to look at choices A and D—the only choices that say r_1=r_2.

Like range, standard deviation is a measure of variability—the bigger the standard deviation, the more spread out the values in a set are from the mean. Visually, you can look at both dot plots in this question and see that the distributions are quite different. In the first plot, there are a couple outliers, but generally the pulse rates are pretty tightly concentrated around the mean of 72.  In the second plot, the dots are much more spread out—if you picked a dot randomly in the second plot, you’d be just as likely to land on an extreme data point like 80 or 112 as you would on the mean of 96. Therefore, you can conclude that the standard deviations of the two sets are not the same (it will be bigger in the second plot). That makes choice D the only option.