Posts filed under: Q and A

Hi Mike! Your SAT book is amazing. Thank you. My daughter is prepping and is on Test #7 of the official tests. Do you have the techniques and concepts for each question as your book only goes to Test #6. I remember reading you had supplement, but now can’t find the link. Please let me know. Thank you!

Thanks, I’m glad you and your daughter like the book! I do have breakdowns for tests 7 and 8 available in the Math Guide Owners section of this site, link here. If you aren’t yet registered, you can get access here.

A question about composite functions:
If f(x) = √x and g(x) = x^2, and you are solving for f(g(x)), ordinarily you solve for g(x) and then plug that value into f(x) to solve for f(g(x)). But what if x is a negative number? When you square it, you’ll get a positive value and then when you take the square root of that to solve for g(f(x)), the final answer will be positive only. Is that correct?


Yes, that’s correct. And to that I’ll add that if you look at g(f(x)), a negative number will not work.

In your graphing calculator, have a look at y=\sqrt{x^2} and compare it to y=\left(\sqrt{x}\right)^2. They’re the same when x is positive, but only one exists when it’s negative.

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!


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

How do you solve number 2 on page 28 with the plugging in method?

Here’s the question (from the Plugging In chapter in the Math Guide):

The way to go here is to plug in values to evaluate choices, which makes this a bit of a hybrid between a plugging in question and a backsolving question. Use the answer choices to guide your plugging in.

For example, choice A tells us to try x < 0. That turns out to be a good idea. Say x = –2 and y equals, oh, I dunno, 3. That fits the conditions: xy and –3xy. Can you come up with any values of x greater than 0 where that would work? No? Well, maybe A is correct, then! But just for the moment, let’s try to eliminate the other choices to reassure ourselves.

What about choice B, though? Wouldn’t x = –2 still work if y = –1? Sure would, so we can eliminate that one.

Can we plug in values that would eliminate choice C? Sure can! In fact, x = –2 and y = 3, which we just used above when we were considering choice A, show that choice C doesn’t need to be true.

Does choice D have to be true? Well, it’s true when x = –2 and y = 3, but what if x = –2 and y = 5 instead? Then xy would be true, and –3xy would be true, but (–2)^2 is not greater than 5, so choice D doesn’t need to be true.

By trying a few quick numbers informed by the answer choices, we can eliminate B, C, and D, leaving only choice A standing. We’re done!

Test 4, section 4, question #27. Why is the answer D instead of C? Please explain this.

The scatterplot shows (my emphasis) “the relative housing cost and the population density for several large US cities in the year 2005.” Note that several is not all, so we can’t use this data to say never. If you plotted all US cities, you very well might see some dots below 61%.

A line of best fit is a predictive tool, but it does not tell you anything definitively.

The following expression represents the concentration, in microbubbles per milliliter, of the microbubble suspension remaining in a client t seconds after the technician has begun the ultrasound process.

100,000,000 ⋅ 0.5^(t/120)

Which expression shows the multiple of the concentration remaining each minute as a constant or coefficient?

You didn’t give me answer choices, but it seems like the key here is converting seconds to minutes. If you have seconds and you want minutes, you divide by 60, so you’re looking for something that has \dfrac{t}{60} in it.

Off the top of my head, maybe this?

100,000,000 \times 0.5^{\left(\frac{t}{60}\times\frac{1}{2}\right)}

Thomas is making a sign in the shape of a regular hexagon with 4-inch sides, which he will cut out from a rectangular sheet of metal. What is the sum of the areas of the four triangles that will be removed from the rectangle?

So it’ll look like this:

It’s helpful just to know that a regular hexagon’s interior angles all measure 120°, but you can also calculate that using (n-2)\times 180^\circ:

\dfrac{(6-2)\times 180^\circ}{6}=120^\circ

That means that the four triangles you’re cutting off the rectangle are each 30°-60°-90° triangles with 4-inch hypotenuses.

Those will have legs of 2 and 2\sqrt{3}, and therefore areas of \dfrac{1}{2}(2)(2\sqrt{3})=2\sqrt{3}. Since there are four such rectangles, the total area you’re cutting off is 8\sqrt{3}

After doing some practice problems on Rational Equations on Khan Academy, I was wondering would the equation be undefined if after factoring the equation we obtain cancellable factors on the denominator and the numerator? For ex: ( (2x-1) (x+4) ) / (x+4) would -4 be a plausible argument? I see the x+4 factor can be cancelled together with the one in the numerator.

I don’t believe you’ll see this tested on the SAT, but the way you’d typically handle this is to say that the expression is equivalent to 2x – 1 for all x ≠ –4. Does that help?

There are five houses on each side of a street, as shown in the figure above. No two houses next to each other on the same side of the street and no two houses directly across from each other on opposite sides of the street can be painted the same color. If the houses labeled G are painted gray, how many of the seven remaining houses cannot be painted gray?

Without knowing what the figure looks like, I can’t say, but the process to follow to answer this is first to cross out all houses that can’t be gray given the rules. That means any house next to a G or across from a G gets crossed out.

To make sure you aren’t missing any, do each G one at a time: pick a house with a G in it, then cross off the house to the left, the house to the right, and the house across. Then move to the next G, etc. Since the question asks how many houses cannot be gray, count all the houses you crossed out, and that’s your answer.

One note: the current SAT doesn’t really ask questions like this. Are you sure you’re studying from up-to-date materials?

Here’s a problem from Khan Academy’s SAT practice section. Please explain this one for me.

A marine aquarium has a small tank and a large tank, each containing only red and blue fish. In each tank, the ratio of red fish to blue fish is 3 to 4. The ratio of fish in the large tank to fish in the small tank is 46 to 5. What is the ratio of blue fish in the small tank to red fish in the large tank?

A ratio of 3 red to 4 blue means the total must be a multiple of 7, so it might help you to first multiply the totals in each tank by 7. The 46:5 ratio for large tank to small tank becomes 322:35. Let’s just pretend those are the actual numbers of fish in the tank.

In the small tank, 4/7 of the 35 fish in the tank are blue, so 20 fish are blue.

In the large tank, 3/7 of the 322 fish are red, so 138 fish are red.

Therefore, the ratio of blue fish in the small tank to red fish in the big tank is 20:138, which simplifies to 10:69.

Note that while multiplying everything by 7 was helpful (in my opinion) in making the math more intuitive, you don’t need to. You can simply enter the following in your calculator and ask it to convert to a fraction:

\left(\dfrac{4}{7}\times 5\right)\div\left(\dfrac{3}{7}\times 46\right)

I’m having a lot of trouble with this problem, could you please break it down for me? It is number 2 on the functions practice questions. “If f(x-1)=x+1 for all values of x, which of the following is equal to f(x+1)?”
A. x+3
B. x+2
C. x-1
D. x-3

In the solutions in the back of the book, I have two approaches. I’m assuming you’ve already looked at those and neither went all the way for you, so here’s another look.

We know that function notation means, generally, that the bit in parentheses goes through some series of mathematical operations, and a result is spit out. So, for example, stuff can be added or subtracted, or the argument could be multiplied, divided, raised to a power, etc.

In this particular problem, we know that f(x-1) goes through some process that spits out an end result of x+1. Here’s the crux: what could happen to (x-1) to make it equal x+1? No multiplying or raising to powers necessary! The only thing you need to do to turn (x-1) into x+1 is add 2!

That’s how we know that the function f just adds 2.



Therefore, we can figure out what f(x+1) must be:


Does that help?

The midpoint formula tells you that the a segment with endpoints (a, b) and (c, d) will have a midpoint at ((c)/2,(d)/2)

So we know that (+ 9)/2 = 6 and (5 + y)/2 = 3. We can solve those!

(+ 9)/2 = 6
+ 9 = 12
x = 3

(5 + y)/2 = 3
5 + y = 6
y = 1

Therefore, x + y = 4.

from Tumblr

Practice Test 6 Calculator Section # 29

All you need to know about average speed is that it’s equal to \dfrac{\text{distance traveled}}{\text{time it took to travel that distance}}.

You’re given a formula for the distance traveled: when the car’s been traveling t seconds, it’s gone a distance of 16t\sqrt{t}. That formula contains all you need to calculate the average speed.

\text{distance traveled}=16t\sqrt{t}

\text{time it took to travel that distance}=t

\text{average speed}&=\dfrac{\text{distance traveled}}{\text{time it took to travel that distance}}=\dfrac{16t\sqrt{t}}{t}=16\sqrt{t}


I wanted to ask for your assistance on an SAT Practice Test question? It’s a question on systems of equations. The question I’m referring to is question 11 on the no-calculator math section of Practice Test #4.

Thank you in advance :’)

Sure thing. I’ve actually already posted a solution to that one here.

(You can check which official questions I’ve posted solutions for here.)

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.