Dr. Steve's SAT Math Problem Thread for the Revised SAT

Test Preparation SAT Preparation
1

Since my SAT Math Problems Thread became quite popular, I thought I would start a thread of math problems for the March 2016 revised SAT. I will be providing my own questions. I have created these to be similar to the material that his already been released by the College Board. Let’s start by warming up with a Level 1 question.

Level 1 “Heart of Algebra”

  1. Joseph joins a gym that charges $79.99 per month plus tax for a premium membership. A tax of 6% is applied to the monthly fee. Joseph is also charged a one-time initiation fee of $95 as soon as he joins. There is no contract so that Joseph can cancel at any time without having to pay a penalty. Which of the following represents Joseph’s total charge, in dollars, if he keeps his membership for *t* months?

    (A) 1.06(79.99 + 95)t/i 1.06(79.99t* + 95)
    © 1.06(79.99t) + 95
    (D) (79.99+.06t) + 95

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C. Total dollars = 95 + 79.99(1+0.06)t = 95 + 79.99*1.06t.

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That’s good. Looks like you solved it using a straightforward algebraic method. Here are more details:

Since the monthly membership fee is 79.99 dollars, and the tax is 6%, the total monthly fee, with tax, is 1.06(79.99) dollars per month. It follows that the total monthly fee for t months is 1.06(79.99t). Finally, we add in the one-time initiation fee to get 1.06(79.99t) + 95, choice ©.

Notes: (1) 6% can be written either as the decimal .06 or the fraction 6/100.

To change a percent to a decimal, simply divide by 100, or equivalently, move the decimal point two places to the left, adding in zeros if necessary. Note that an integer has a “hidden” decimal point right after the number. In other words, 6 can be written as 6., so when we move the decimal point two places to the left we get .06 (we had to add in a zero as a placeholder).

To change a percent to a fraction, simply place the number in front of the percent symbol (%) over 100.

(2) Since the tax is 6%, it follows that the tax for $79.99 is .06(79.99) or (6/100)(79.99) dollars.
It follows that the total monthly fee, including tax, is 79.99 + .06(79.99) dollars.

We can use the distributive property to simplify this expression as follows:
79.99 + .06(79.99) = 1(79.99) + .06(79.99) = 1.06(79.99)

(4) In note (2) we saw that one way to get the total monthly fee, including tax, is to add the amount of tax to the untaxed amount. A quicker way is to simply multiply the monthly fee by 1.06. A justification for why this works is given in the last line of note (2).

(5) If you need to pay a certain dollar amount more than once, simply multiply by the number of times you need to pay.

For example, if you need to pay 100 dollars five times, then the final result is that you pay 100⋅5 = 500 dollars. More generally, if you need to pay 100 dollars t times, then the final result is that you pay 100t dollars.

In this problem we want to pay the monthly fee t times. Since the monthly fee is 1.06(79.99), the final result is 1.06(79.99)t, or equivalently 1.06(79.99t)

(6) Don’t forget to add on the one-time initiation fee to 1.06(79.99t) to get 1.06(79.99t) + 95 dollars.

(7) If the algebra is confusing to you, you can also solve this problem by picking a number. Try setting t = 2, for example.

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@DrSteve seems like the only tricky part to the above question is not picking (B) 1.06(79.99t + 95) (e.g. if the $95 fee was also taxed). I nearly did that when looking at the question.

I’m not sure what I think about the new SAT, but it’s not really of any concern to me…(hopefully)

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I agree. This more about arithmetic.

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Based upon what the College Board has released, math on the revised SAT will no longer be very tricky. The same strategies that we know for the SAT will still work (plugging in, picking numbers, trying simple operations, etc.), but students that are strong in algebra will be rewarded (on the current SAT algebra can almost always be avoided).

This new SAT is more ACT-like than it is SAT-like, but in my opinion it will be a harder test than the ACT.

The simple Level 1 problem I just gave is a perfect example. Students that are strong in being able to set up an algebraic equation from a “real world” situation will certainly be rewarded here, but this problem could still be solved by picking numbers.

I will give a full “picking numbers” solution in my next post for those students that are still not comfortable with this strategy.

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Totally off-topic, I hope no one minds, but I was working with the Dr’s 28 SAT Math Lessons last night and would like to say that its a really excellent book. Perfect for a strong student to use to review right before a test.

Question for the Dr: Do you think you’ve seen enough released questions for the new exam to be confident that these new examples will be sufficiently relevant for that exam?

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@CHD Let me answer the question this way. I am confident that every question I have created so far is consistent with what the College Board believes will be on the exam. There is a good chance that the types of questions will evolve over the first year once the College Board has actual data and feedback, but I don’t think it will be a rapid evolution (does the expression “rapid evolution” makes sense - a little help from the linguists out there). For example I am 90% certain that there will be a question similar to the one I just posted in this thread on the March SAT based upon the research I have been doing for the last month.

If I have time to keep up with this thread, I think that students reading it will have an advantage going into the March SAT (and even the October PSAT for National Merit candidates).

I think that you’ll see that some of the question types are significantly different from what’s on the current SAT (but some will be similar).

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That is an easy problem. One of the sample problems required finding the area of a regular hexagon. My impression is that it will be more “practical” and applicable to higher education, emphasizing algebra and data analysis. It seems less IQ test, but it will by tricky in some ways, requiring you to apply the math to practical word problems.

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Yes. This is a Level 1 problem - so it would be considered easy, but I personally think it’s harder than most Level 1 math problems on the current SAT.

I’ll give a problem that requires finding the area of a regular hexagon next. But first let’s solve the problem I gave by picking numbers. Note that I will be using a calculator for this solution which may not be allowed if this problem is in the “no calculator” section:

Let’s choose a value for t, say t = 2, so that Joseph keeps his gym membership for 2 months.

Now 6% of 79.99 is 4.80 (to the nearest cent). So each month of membership, including tax, is 79.99 + 4.80 = 84.79 dollars. It follows that 2 months of membership, with tax, is 2⋅84.79 = 169.58 dollars. When we add the initiation fee we get 169.58 + 95 = 264.58 dollars.

Put a nice big, dark circle around the number 264.58 so you can find it easily later. We now substitute t = 2 into each answer choice and use our calculator:

(A) 1.06(79.99 + 95)*2 ≈ 370.98
(B) 1.06(79.99*2 + 95) ≈ 270.28
(C) 1.06(79.99*2) + 95 ≈ 264.58
(D) (79.99 + .06*2) + 95 = 175.11

Since choices (A), (B), and (D) came out incorrect, we can eliminate them. Therefore the answer is choice ©.

**Important note:/b is not the correct answer simply because it came out to 264.58. It is correct because all three of the other choices did not come out correct.

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I am a good math person but not particularly mathy. For example I do not like to solve math puzzles and such but I was able to take math classes quiet above calculus and do very well. In the old SAT once a while I would see a question and think “how can I solve this?”. I worked all the questions that are released for the new SAT and I never had to ask that. I knew exactly and immediately what needs to be done. My only problem is racing the clock because I can get a bit slow with calculations. It seems to me that students that have been through a rigorous Algebra 2 class should do well. Can a student that did not have a rigorous math education overcome and self study for the test? I do not know. My advice right now would be for students to take Algebra 2 very seriously and try to do the best they can in the class.

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The new SAT seems more heavily focused on algebra and analyzing data. Not a whole lot of geometry or discrete math (counting/combinatorics, number theory). Not sure what I feel about that, although I do feel that most HS geometry and discrete math curricula are currently very poor.

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As promised, here is a problem involving a regular hexagon:

Level 4 Geometry (calculator required)

The head of a copper “hexagon head screw bolt” has the shape of a cylinder with a hole shaped like a regular hexagon. The cylindrical head is 2 cm thick with a base diameter of 3 cm. The hexagonal hole is only half the thickness of the entire head, and each side of a hexagonal cross section has a length of 1 inch. Given that the density of copper is 8.96 grams per cubic cm, and density is mass divided by volume, find the mass of the head to the nearest gram.

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They are trying to make it relevant to what you would need for college science and social science, hence all the algebra and data analysis. It is pretty word problem oriented and still pretty different from ACT math. More how to apply math than mathy questions.

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@DrSteve I have a question: What’s a screw bolt? I got 39.96 g, but I think I visualized the shape wrong.

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@DrSteve you mean 1 cm, right? Otherwise the dimensions make no sense.

I got a mass of 103.39 g, which rounds to 103 g.

@Synonyms if you know what an Allen wrench (sometimes called hex set) is, or have screwed a bolt in using one, you’ll know what we’re talking about. Google “hexagonal screw bolt.”

The above problem seems more computational than anything though; you do have to work with a couple volumes and such, but in real life, I would probably take out the kitchen scale…

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@MITer94 Thanks for the tip. I got 103.39, too.

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@MITer Yes I mean 1 cm of course - thanks for catching that typo, and yes your answer is correct.

@Synonyms My original problem has a picture above it, but I don’t believe I can post pictures on CC. On the SAT, this problem would certainly have a picture above it, so you probably wouldn’t be confused by the picture. MITer described it pretty well.

I’ll post a complete explanation soon.

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Here is a detailed solution:

We first compute the volume of the head. There are two parts to the volume.

The bottom half of the head is a cylinder with height 2/2 = 1 cm and base radius 3/2. It follows that the volume is V = πr^2 h = π(3/2)^2 (1) = 9π/4 cm^3.

The top half of the head consists of the same cylinder as the bottom half, but this time we have to subtract off the volume of a hexagonal prism. The regular hexagonal face can be divided into 6 equilateral triangles, each with area A = (s^2 √3)/4 = (1^2 √3)/4 = √3/4. So the volume of the hexagonal prism is V = Bh = ((6√3)/4)(1) = (3√3)/2 cm^3 and the volume of the top half of the head is 9π/4 – (3√3)/2 cm^3

It follows that the total volume of the head is 9π/4 + (9π/4 – (3√3)/2) = 18π/4 – (3√3)/2 = (9π – 3√3)/2 cm^3.

Finally, D = M/V ⇒ 8.96 = M/((9π – 3√3)/2) ⇒ M = 8.96 ⋅ (9π – 3√3)/2 ≈ 103.39 grams.

To the nearest gram, the answer is 103.

**Notes:/b The radius of a circle is 1/2 the diameter, or r = (1/2)d.

In this problem the base diameter of the cylinder is 3 cm. It follows that the base radius of the cylinder is 3/2 cm or 1.5 cm.

(2) The volume of a cylinder is V = πr^2 h where r is the base radius of the cylinder and h is the height of the cylinder.

For example, the bottom half of the screw is a cylinder with base radius 3/2 cm and height 1 cm. So the volume is V = π(3/2)^2 (1) = 9π/4 cm^3.

(3) A regular polygon is a polygon with all sides equal in length, and all angles equal in measure.The total number of degrees in the interior of an n-sided polygon is (n – 2) ∙ 180

For example, a six-sided polygon (or hexagon) has (6 – 2) · 180 = 4 · 180 = 720 degrees in its interior. Therefore each angle of a regular hexagon has 720/6 = 120 degrees.

(4) For those of us that do not like to memorize formulas, there is a quick visual way to determine the total number of degrees in the interior of an n-sided polygon. Simply split the polygon up into triangles and quadrilaterals by drawing nonintersecting line segments between vertices. Then add 180 degrees for each triangle and 360 degrees for each quadrilateral.

Since a hexagon can be split up into 2 triangles and 1 quadrilateral, a hexagon has 2(180) + 360 = 720 degrees. This is the same number we got from the formula.

(5) If we draw segments from the center of the hexagon to each vertex of the hexagon, we see that the central angles formed must add up to 360 degrees. Therefore each central angle is 60 degrees.

In general, the number of degrees in a central angle of an n-sided polygon is 360/n.

(6) Each of the segments mentioned in note (5) is a radius of the circumscribed circle of this hexagon, and therefore they are all congruent. This means that each triangle is isosceles, and so the measure of each of the other two angles of any of these triangles is (180 – 60)/2 = 60. Therefore each of these triangles is equilateral. This fact is worth committing to memory.

(7) The area of an equilateral triangle with side length s is A = √3/4 s^2 (see note (8) below). It follows that the area of an equilateral triangle with side length 1 is √3/4 (1)^2 = √3/4.

(8) Most students do not know the formula for the area of an equilateral triangle, so here is a quick derivation.

Start by drawing a picture of an equilateral triangle with side length s, and draw an altitude from a vertex to the opposite base. Note that an altitude of an equilateral triangle is the same as the median and angle bisector (this is in fact true for any isosceles triangle).

So we get two 30, 60, 90 right triangles with a leg of length s/2 and hypotenuse of length s.

We can find h by recalling that the side opposite the 60 degree angle has length √3 times the length of the side opposite the 30 degree angle. So h = (√3 s)/2.

Alternatively, we can use the Pythagorean Theorem to find h: h^2 = s^2 – (s/2)^2 = s^2-s^2/4 = (3s^2)/4 . So h = (√3 s)/2.

It follows that the area of the triangle is A = 1/2 (s/2 + s/2)((√3 s)/2) = 1/2 s((√3 s)/2) = √3/4 s^2.

(9) The volume of a prism is V = Bh where B is the area of the base of the prism and h is the height of the prism.

In this problem we have a hexagonal prism with B = (6√3)/4 = (3√3)/2 and h=1. It follows that the volume of this prism is V = ((3√3)/2)(1) = (3√3)/2.

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Level 2 “Problem Solving and Data Analysis”

A chemist has a supply of 5.2 liter bottles of a certain solvent that must be shipped to a central warehouse. The warehouse can accept the solvent at the rate of 3 hectoliters per minute for a maximum of 8 hours per day. If 1 hectoliter equals 100 liters, what is the maximum number of bottles that the warehouse could receive from the chemist each day?

(A) 461
(B) 462
© 27,692
(D) 83,200