Posts tagged with: OT Test 3 Section 4

Can you help me with Question #29 in Practice Test 3, Section 4? I found the answer by subtracting multiples of 9 from 122 to find one that was divisible by 5. Is there a better way?

 
The way I like to go here is to set up equations. If we say left-handed females are x and left-handed males are y, then we can fill in the table like this:

Now we have a system we can solve!

x+y=18
5x+9y=122

We eventually want a number of right-handed females, so let’s solve for x

5x+9(18-x)=122
5x+162-9x=122
40=4x
10=x

Remembering that x is the number of left-handed females, now we just need to multiply by 5 to get the right-handed females: 50.

The last trick here is that the question asks for a probability that a right-handed student being picked at random is female. There are 50 female right-handed students and 122 right-handed students overall, so the probability is \dfrac{50}{122}\approx 0.410.

Could you please explain SAT Past paper 3, section 4, question 12? I don’t really understand why the double root is considered a ‘distinct’ zero.

Thank you for your time!

Sure. Would it make you feel better if “distinct” were replaced with “unique”? Those terms are interchangeable here.

When the question says the function has “five distinct zeros,” that really just means there are five values that 1) don’t equal each other, and 2) make the function equal zero.

In other words, f(a) = 0, f(b) = 0, f(c) = 0, f(d) = 0, and f(e) = 0 for unique values abcd, and e.

In OTHER other words, the function f touches the x-axis five times. Sidestepping all of the above, all you really need to do is count the number of times each graph touches the x-axis.

Choices A and B touch the x-axis four times, and choice C touches the x-axis six times. Only choice D touches it five times.

Test 3 Section 4 #30

This one is yucky. Let’s attack it by substituting. First, we’re told that b=c-\frac{1}{2}, so let’s reflect that (I’m going to use the decimal version of \frac{1}{2} for simplicity):

    \begin{align*}3x+(c-0.5)&=5x-7\\3y+c&=5y-7\end{align*}

Now, let’s solve both equations for c:

    \begin{align*}3x+(c-0.5)&=5x-7\\c-0.5&=2x-7\\c&=2x-6.5\\\\3y+c&=5y-7\\c&=2y-7\end{align*}

Because we now have two different equations with c isolated, we can substitute again here to get an equation that only contains xs and ys.

    \begin{align*}2x-6.5&=2y-7\\2x&=2y-0.5\\x&=y-0.25\end{align*}

That translates to “x is y minus \frac{1}{4},” so the answer is A.

 

Test 3 Section 4 #26

Fun fact and major shortcut: when a line passes through the origin, the coordinates of each point are in the same ratio. For example, if you knew a line passed through (3, 6) and (30, 60), you’d know that line passed through the origin. That’s because when a line passes through the origin, its equation will simply be ymx + 0, or simply ymx. Put another way, y will be directly proportional to x.

We can use that fact to solve quickly:

    \begin{align*}\frac{2}{k}&=\frac{k}{32}\\\\k^2&=64\\k&=\pm 8\end{align*}

Of course, because there are no negative answer choices, we choose C for 8.

The other way to go here would be to backsolve by just trying to draw the points. You can cross off A right away because any line that goes through (2, 0) and (0, 32) can’t also go through the origin! Even a rough drawing should convince you that the line containing(2, 16) and (16, 32) doesn’t go through the origin. Things get a little trickier with (2, 4) and (4, 32) because it might look CLOSE, but hopefully the right answer looks even closer.

Test 3 Section 4 #24

Of course, we could use algebra to solve this (and I will down below), but because the answer choices are numbers that can easily be dropped back into the word problem, my recommended strategy is backsolving.

Start by trying C; assume 23 students in the class. If Mr. Kohl gives each student 3 ml, he’ll have 5 ml left over. 3(23)+5=74. For Mr. Kohl to give each student 4 ml, he’ll need an additional 21 ml. 4(23)=92. Is that 21 higher than 74? It is not—it’s only 18 more.

That suggests we need even more kids in the class, so the answer is almost certainly D, which says there are 26 students. If Mr. Kohl gives each student 3 ml, he’ll have 5 ml left over. 3(26)+5=83. For Mr. Kohl to give each student 4 ml, he’ll need an additional 21 ml. 4(26)=104. Is that 21 higher than 83? Yes! D is the answer.

Now, the algebra. What we want to do here is come up with two expressions that are both equal to n, the amount of solution Mr. Kohl has. I’ll use x for the number of students he has.

If Mr. Kohl gives each student 3 ml, he’ll have 5 ml left over. n=3x+5

In order to give each student 4 ml, he will need an additional 21 ml. n=4x-21

Now that we have those equations, we can solve by substitution.

    \begin{align*}3x+5&=4x-21\\5&=x-21\\26&=x\end{align*}

official practice test 3 section 4 q.no.35 how to solve?

The question tells you that the store’s average after 10 ratings is 75. That means that the first 10 ratings have a sum of 750, because \text{average}=\dfrac{\text{sum}}{\text{number of values}}, \text{average}\times\text{number of values}=\text{sum}. In this case, 75\times 10=750.

You need the first 20 ratings to have an average of 85, which means you need the sum of the first 20 ratings to be 20\times 85=1700. So far so good?

Now, we know that the first 10 ratings added to 750, so we need the next 10 ratings to add to 1700-750=950.

Since we want to know the smallest possible value for the 11th rating, let’s assume that the 12th-20th ratings are the highest they can be, 100. In other words, let’s assume there are 9 perfect 100 ratings, and set the 11th rating equal to x.

    \begin{align*}x+9(100)&=950\\x+900&=950\\x&=50\end{align*}

 

Hello Mike! Could you please tell me, what is the answer of Question 16, Section 4, Test 3? Thank you veery much for your help and time.
Katerina

The answer is -2, and here’s why. They’re asking you to find a value of x for which f(x)+g(x)=0. That can only happen when f(x)=-g(x). So you’re looking for a place on the graph where f curve and the g curve are the exact same distance from the x-axis.

The vertices of the graphs of f(x) and g(x) are (-2,-2) and (-2,2), respectively. In other words, when x=-2, the graphs of f(x) and g(x) are the same distances from the x-axis. In other words, f(-2)=-2, g(-2)=2, and therefore f(-2)+g(-2)=0

 

Can you please explain Test 3 Section 4 #23?

Yep! The key to getting this one is recognizing that when they tell you that \sin \left(a^{\circ}\right)=\cos \left(b^{\circ}\right), they’re telling you that a+b=90. That’s just a little trig fact you should keep in mind when you’re taking the SAT. It can be derived pretty easily—just picture any right triangle with acute angles measuring a^{\circ} and b^{\circ}:

Of course, since it’s a right triangle, a+b=90. And if you do your SOH-CAH-TOA, you’ll see that \sin \left(a^{\circ}\right)=\cos \left(b^{\circ}\right) and \cos \left(a^{\circ}\right)=\sin \left(b^{\circ}\right).

Anyway, back to the question. Once we recognize that a+b=90, the rest is just substitution.

    \begin{align*}a+b&=90\\(4k-22)+(6k-13)&=90\\10k-35&=90\\10k&=125\\k&=12.5\end{align*}

Test 3 Section 4 #36

This is the calculator section; graphing will probably help you visualize what’s going on here. Don’t worry about the inequalities for now—just graph the lines. You’ll have to zoom out a bunch to see the intersection, but here’s what it looks like on my calculator:

The inequalities tell you that y is less than or equal to both of those lines, and you’re looking for the greatest possible y-value in the solution set. In other words, you’re looking for the highest point that’s BELOW OR ON both lines. With the graph in front of you, it should be easier to see that the point you need has to be the intersection of the two lines. So use your calculator to find that:

There you have it—the answer is 750.

Pro-tip: if you get another question like this, with two linear inequalities where you’re asked for the greatest (or least) possible value of one of the coordinates in the solution set, you’re looking for the intersection of the lines.

Once you know that, you may choose to just solve a problem like this algebraically next time:

–15x + 3000 = 5x
3000 = 20x
150 = x

If x = 150, at the intersection, then y ≤ 5(150); y ≤ 750.

Test 3 Section 4 #34

Remember that circle problems are often ratio problems, and this is definitely one of those cases. Generally, we can say that \dfrac{A_{\text{sector}}}{A_{\text{circle}}}=\dfrac{\text{central angle measure}}{360^\circ\text{ or }2\pi}.

Since we’re given the central angle in radians and no other information, we just need to figure out the ratio of the central angle to 2\pi.

    \begin{align*}\dfrac{\left(\dfrac{5\pi}{4}\right)}{2\pi}=\dfrac{5\pi}{8\pi}=\dfrac{5}{8}\end{align*}

You can either grid in that fraction or its decimal equivalent: .625.

how can you use plug-in for #13 on page 607?

To get this one by plugging in, pick values for all the variables on the right side of the equals sign: tv, and k. For example, say t = 2, v = 3, and k = 5. Use your calculator to see what that gives you for h:

    \begin{align*}h&=-16(2)^2+(3)(2)+5\\h&=-53\end{align*}

Now plug –53 in for h, 2 in for t, and 5 in for k in each answer choice. The one that gives you 3 (our original value for v) is the right answer.

Choice D does the trick:

    \begin{align*}v&=\dfrac{h-k}{t}+16t\\3&=\dfrac{-53-5}{2}+16(2)\\3&=-29+32\\3&=3\end{align*}

Note that I’m not saying you need to do the question this way, just that it CAN be done this way. On a more complicated algebraic manipulation question, you might be happy to have a method like this in your back pocket.

Because I know someone will ask, here’s the algebra:

    \begin{align*}h&=-16t^2+vt+k\\h+16t^2-k&=vt\\\dfrac{h-k+16t^2}{t}&=v\\\dfrac{h-k}{t}+\dfrac{16t^2}{t}&=v\\\dfrac{h-k}{t}+16t&=v\end{align*}

Will you please answer question # 27, Test 3, section 4. Is it possible to use real numbers in your example? Thanks!

Sure! The thing you want to remember when you’re plugging in numbers is that you’re doing it to make your life easier, so pick numbers that will really make your life easier. In this case, that means I’m going to say that the original rectangle is a 10 by 10 square, so its area is 100. Obviously, it’s very easy to work with 100 when dealing with percents, and it’s also easy to take different percents of 10 to increase/decrease side lengths.

So, yeah. Here’s our original rectangle:

Now we need to increase its length by 10 percent (10 percent of 10 is 1, which means the length increases from 10 to 11), and decrease its width by p percent. What should we do for p? Well, why not backsolve?

Say p = 20, like answer choice C says. (I’m picking that because it’s in the middle, and also because 20 is the easiest choice to work with so why not try it first.) If p = 20, then we decrease the width by 2, taking it from 10 to 8. So here’s the new rectangle:

What’s the difference between the original area of 100 and the new area of 88? 12. And here’s where we pat ourselves on the back for picking numbers that gave us 100 as our starting area: 12 is what percent of 100? 12 percent! So the first choice we tried, C, is right. Awesome!

(Whenever I illustrate backsolving and the first choice I try is the right choice, I feel compelled to show what a wrong choice would look like. So let’s quickly look at choice B. We’re still increasing the length by 10 percent, so the length is still 11. The width is now decreased by 15 percent, making it 8.5. The new area is 11\times 8.5=93.5. Is that a 12 percent decrease from the original area of 100? No, 93.5 is too big! So if we didn’t already know the answer, we’d know that the next choice we should try should be smaller.)

Q#22, S#4, SAT Test#3

This one is tricky! Here’s why: they talk about three numbers, but they really only give you enough information to write two equations, and they only ask for the value of one of the three numbers. So it looks way more complicated than it really is.

Let’s come up with some equations. First, the sum of three numbers is 855. Easy enough:

x+y+z=855

And then the second thing they tell you: x is 50% more than the sum of the other two numbers.

(y+z)\times\dfrac{50}{100}+(y+z)=x

That simplifies to (or maybe you prefer shorthand and just started by writing this) the following:

1.5(y+z)=x

Now we can solve by substituting or eliminating. I think substitution is a little easier here since we already have an x = equation:

1.5(y+z)+y+z=855

1.5(y+z)+1(y+z)=855

2.5(y+z) = 855

y+z=\dfrac{855}{2.5}

y+z=342

Now that you’ve got y + z, all you need to do is solve for x.

x+342=855

x=513

(Note that, while I didn’t do this in my explanation because I think it might have been confusing, you can also just use a single letter (like a) instead of yz to represent the sum of the other two numbers, since we never need to split them apart.)

Question #21, section #4, SAT test #3?

This question asks about exponential growth. In exponential growth, an initial amount is multiplied, over and over again, by a growth factor. Most savings accounts you could go to a bank today and get will provide exponential growth because they pay interest on the amount in your account when the calculate interest, rather than just paying you a flat amount each year. So each successive period, you get paid the same interest rate on slightly more money than you did last time.

To see why only choice C provides exponential growth, let’s say I have $100 and I invest in each of the accounts described in the answer choices.

In choice A, 2% of the initial value is added each year. 2% of $100 is $2, so I basically get $2 per year, forever. That’s linear growth.

In choice B, 1.5% of the initial value and an extra $100 is added each year. That’s a great deal for me only investing $100 up front, but it’s still not exponential growth: I get $1.50 + $100 = $101.50 added per year—the same value added to my account each year, forever. That’s linear growth.

In choice C, 1% of the CURRENT VALUE is added each year. So the first year, I get $1, which gives me $101. The second year, I’ll get $1.01, because that’s 1% of $101. Then I’ll have $102.01. The year after that, I’ll get slightly more again. Each year, I get ever so slightly more. I’m not getting rich quick with this deal, but it is exponential growth.

In choice D, again, I’m just getting the same amount added every year. That’s linear growth.

(The general equation you should have in your mind when you’re dealing with exponential growth is y=a(1+r)^x, where y is the current value of whatever is growing, a is its initial value, r is its growth rate in decimal form, and x is the number of periods of compounding. The formula can get more complicated than that, but probably won’t on the SAT.)

Please explain question #8, section #4, SAT test #3! Thanks Mike!

The equation you’re given is y = 0.56x + 27.2, where y is the average number of students per classroom and x is the number of years since 2000. You’re asked what the number 0.56 represents in the equation.

Before you get too lost in the details of the particular model, recognize that y = 0.56x + 27.2 is really just a line in ymxb form! So without even thinking about average students per classroom or years, we know that 0.56 represents the line’s slope, or rate of change. Which answer choices refer to rates of change?

Really, only choice C does. A talks about a total number, B talks about an average number, and D talks about a difference (which is just one number minus another). Only C talks about an INCREASE, which is the kind of word we’re looking for to describe slope. So the answer must be C.