Posts filed under: Q and A

4, 7, 3, 4,….
In the sequence above, the first term is 4, the second term is 7, and each term after the second term is the nonnegative difference between the previous two terms. If the nth term is the first term of the sequence that is equal to zero, what is the value of n?

Okay I know this number can be solved through first principles(finding each number in the sequence manually) but I can’t help but wonder if there’s a certain algebraic formula or method one can utilize to solve it.

 

I’m sure I’ll get notation wrong since I’m just shooting from the hip here, but the terms can be represented algebraically like this. For terms x_1through x_n, where n>2:
x_1=4, x_2=7, x_n=|x_{n-1}-x{n-2}|.

But even once I’ve got the above, I’d still rather just list out terms until I get a zero than try to figure out how to program my calculator to iterate that. (This is cake with a spreadsheet, not that you can use one on whatever kind of test this is from…)

On page 28, question 8, when I looked at the solutions, the number 1 was plugged in to solve the problem. However, on the beginning of the book the author said to never plug 0 and 1…Anyhow, my question is, if never plugging 1 is suggested, how did you know that plugging 1 in would work in your favor for the question(btw I got it right but I didn’t plug 1 in, I thought of other numbers but it did take me longer to figure out the numbers that’d add up to 11)

 
The danger of plugging in 1 or 0, in general, is that those are the multiplicative and additive identities, respectively. That’s a fancy way of saying that multiplying by 1 or adding 0 doesn’t change anything. There’s also the fact that multiplying by 0 turns everything to 0. When you plug in something that doesn’t change anything, or that changes everything to zero, you’re a little less likely to eliminate all the wrong choices, which means you’ll end up having to plug in something else to finish the job anyway. That’s why I advise to avoid plugging in 1 or 0 as a general guideline. The more practice you get plugging in, the more you’ll start to get comfortable seeing when you can ignore that guideline.

The distinction here is that the question tells you that 2x+3y=11, so you’re not randomly choosing 1 as a number to plug in. You’re choosing it within the constraints of the question, and you’re choosing it as part of an ordered pair that satisfies the equation! In the solution, I choose x=3 and y=1, but actually you could also choose x=1 and y=3, or x=2.5 and y=2, etc.

Does that help?

a = 4800 – 6t
b = 5400 – 8t

In the system of equations above, a and b represent the distance, in meters, two marathon runners are from the finish line after running for four hours and t seconds. How far will runner a be from the finish line when runner b passes her?

A. 300 meters
B. 500 meters
C. 100 meters
D. 3000 meters

 

Runner b will pass runner a when they are the same distance from the finish line, so set those distance expressions equal.

4800-6t=5400-8t
2t=600
t=300

But be careful, t is not a measure of distance, it’s time! The runners will meet at 300 seconds from the 4 hour mark. To find the distance they’ll be from the finish line at that time, plug 300 back into one of the original equations.

a=4800-6(300)
a=3000

So the answer is D.

A rideshare app charges $2 per trip plus $0.4 per mile. A competitor charges $1 for the first 6 miles plus $0.5 per mile for every additional mile. For what length trip would the two services charge the same amount?

A. 10 miles
B. 18 miles
C. 20 miles
D. 40 miles

Can you craft an algebraic equation to solve this directly – or is plugging in the answers the way to go?

 

Starting from the answer choices works nicely here, and you know I love doing that, but since you asked, of course an equation is also viable. Say x is the number of miles in the trip. The first app charges 2+0.4x, and the second app (this is the tricky one) charges 1+0.5(x-6). That’s only for a trip greater than 6 miles, really, but all the answer choices are greater than 6, so we’re good here. Now just set them equal and solve!

2+0.4x=1+0.5(x-6)
2+0.4x=1+0.5x-3
2+0.4x=-2+0.5x
4=0.1x
40=x

This may be a little advanced for the SAT, but complex numbers sometimes show up –as do cubic polynomials– so hopefully you can address this for me! TIA!

Which of the following could be the full set of complex roots of a cubic polynomial with real coefficients?

A. { 0, 1, i}
B. {1, i, 2i}
C. {2, i}
D. {3, 2 + i, 2 – i}

 

I don’t feel like I’m qualified to teach a lesson on this. I know that a cubic polynomial must have 1 or 3 real roots, but I’m too far removed from the class I learned that in to reconstruct the proof here. (A lesson that’s always worth relearning in SAT prep: not all high school math is SAT math.) However, I can still help with this question because it’s multiple choice and happens to lend itself to one of my favorite approaches: we can backsolve!

If you translate the answer choices to what the factors of the polynomial would be, you can multiply those and see which one ends up with all real coefficients. I gravitated right to choice D, because that 2 + i and 2 – i conjugate looks like it’ll cancel out nicely. Let’s see!

We know that a cubic polynomial with roots 3, 2 + i, 2 – i would have the factored form: y=(x-3)(x-(2+i))(x-(2-i)). Now FOIL the complex factors:

y=(x-3)(x-(2+i))(x-(2-i))
y=(x-3)(x^2-(2+i)x-(2-i)x+(2+i)(2-i))
y=(x-3)(x^2-2x-ix-2x+ix+(4-i^2)) <-- All the i terms will go away here!
y=(x-3)(x^2-4x+(4-(-1)))
y=(x-3)(x^2-4x+(4+1))
y=(x-3)(x^2-4x+5)

See how the i terms cancel out? That won’t happen in any other choice, which means the other choices will result in coefficients that aren’t real. We don’t even need to fully expand this one to know it’s the right move.

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!

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.

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)