The Official SAT Study Guide by CollegeBoard

Test Preparation
21

@VANURSEPRAC
No one can be “well-prepped” for the non-calculator part because this section is unique to the new SAT. Many students are having trouble with this sections for a variety of reasons, not just the bizarre language of the word problems. If you are having trouble with word problems, you should collect all the word problems from all the new official practice tests and create 2-3 variants (with different numbers) for each one. Practice those until you become good at cutting through excess verbiage.

One way in which new SAT word problems are more similar to ACT word problems is that many new questions include “dummy” information: information that is not useful to solving the problem. These questions test your ability to identify the needed information and to ignore the rest. The old SAT math had very few problems with excess information. Students who expect to use all the given information are going to be scratching their heads.

Here are some general hints that will work for many of these word problems:

(1) Remember that many or even most of the words are irrelevant to the math. Don’t waste time worrying about verbiage. Cut through verbiage to the math. Math consists of equations, variables, numbers, and units.

(2) Read the question part first (the part in front of the question mark). This tells you what you have to find out.

(3) If the question gives you one or more formulae, identify which formula you need to answer the question. Some problems give you more formulae than you need. Don’t worry about that. Your job is just to find which one you need and ignore the others.

(4) If the formula has constants, find the numerical values of the constants in the formula and substitute these values into the formula. Some problems give you more values than you need. Don’t worry about that. Your job is just to find which values you need and ignore the others.

(5) Check for changes in units. Convert all units to the same units.

(6) Solve the formula (with the substituted constants) for the missing variable. This is usually the answer to the question or what you need to determine the answer.

Questions that require you to write the equations are more similar to the old SAT math word questions. You define the variables, write the equation or equations, and solve for required unknown. Here too it is important to check for changes in units. All units have to be the same, and the answer has to be in the required units.

22

Here is an illustration of this procedure with an official-like problem:

v=v0-gt
h=(-1)/2 gt^2+v0 t
v^2=〖v0〗^2+2gh

A ball is shot straight upwards with an initial speed of 30 meters/second. The equations above describe the motion of projectiles near the surface of the earth under the influence of gravity, where v0 is the initial speed of the ball, v is the speed of the ball as it is moving up in the air, h is the height of the ball above the ground, t is the time elapsed since the ball was shot upwards, and g is the acceleration due to gravity (9.8 m/s2).

  1. How long will it take for the ball to reach its maximum height to the nearest tenth of a second?
  2. What is the maximum height the ball will reach to the nearest meter?

Solution:
Step 1:
Cut through the verbiage to the math.
Read the part in front of the question mark first:
How long will it take the ball to reach its maximum height?
This tells you that you need the height equation:

h=(-1)/2 gt^2+v0 t

Step 2: Look for the numerical values of the constants: v0 and g. Be careful for changes in units. There aren’t any here.
v0= 30 meters per second; g = 9.8 meters per second^2.

Substitute the numbers for v0 and g into the equation.

h=-4.9t^2+30t

Step 3: You have to find the maximum height and the time of the maximum height. Since this is quadratic function (a parabola), you need the coordinates of the vertex. On the calculator part, a fast way to find the coordinates of the vertex is to use Compete the Square with a TI-Nspire CX CAS to put the function in vertex form.

completeSquare(-4.9t^2-30t,t)
returns: 45.9184-4.9∙(t-3.06122)^2

The maximum height is 45.9184, or 46 to the nearest meter. The time is 3.06122, or 3.1 to the nearest tenth of a second.

If this were a non-calculator question, to find the vertex use x=-b/2a. (The numbers would be better than these if this were a non-calculator question.)

-b/2a =–30/2(–4.9)=30/9.8=3.1 (to the nearest tenth of second).

To find the y-coordinate, substitute x = 3.1 in the function:
–4.9(3.1)^2 + 30(3.1) = 46 (to the nearest meter).

23

Here is a different kind of hard word problem I made up following the model of problem 29 of the October 28 PSAT. (This was a calculator grid-in, but in my opinion the numbers are ok also for the non-calculator part.)

Philip sells Black Forest cakes in his bakery shop. At a price of $25, Philip sells 10 Black Forest cakes per week. For every $2.50 he decreases the Black Forest cake price, he sells 5 more Black Forest cakes per week. To maximize the revenue, how much, in dollars, should he charge for 1 cake? (Revenue = price per cake x number of cakes sold. Disregard the $ sign when gridding your answer.)

24

Step 1: Cut through verbiage:
The question: How much should Philip charge for 1 cake to maximize revenue?

This means you need an equation for the revenue in terms of how much Philip charges for 1 cake.

revenue = (charge for 1 cake) x (number of cakes sold)

But if you define the variable as

c = charge for 1 cake

it is going to be hard to express how many cakes he sells in terms of c.

It is better to define the variable as

Step 2:

n = the number of $2.50 price reductions

Then
charge for 1 cake = 25-2.50n

number of cakes sold = 10 + 5n

revenue = (charge for 1 cake)x(number of cakes sold) = (25-2.5n)(10+5n)

This is a quadratic function. To maximize it, we need to find the vertex.

Step 3 (calculator version): With a TI-Nspire CX CAS, use Complete the Square to put the function in vertex form.

completeSquare((25-2.5n)∙(10+5n),n)
returns 450-12.5∙(n-4)^2

At the vertex, n=4, so the sales price for 1 cake that maximizes revenue =25 – 2.50n = 25 – (2.50)(4) =15

Step 3 (non-calculator version): expand the function and use x = –b/2a.

(25 – 2.5n)(10 + 5n) = 250 + 100n – 12.5n^2
a = –12.5, b = 100

n =–b/2a = –100/(–25) = 4

sales price for 1 cake = (25 – 2.50n) = 25 –(4)(2.50) =15

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@Plotinus I would like to offer a quick way to get the time it takes for ball to reach maximum height when the ball is thrown straight up. At maximum height, the velocity becomes 0 just before the ball drops. So set v=0 in equation v=v0-gt. Then you have 0=v0-gt. Then gt=v0. gt=30. t=30/g. t=30/9.8 = 3.06122

26

@Plotinus, Thank you so much!

27

@BunnyBlue
Thanks for pointing out an alternate solution.

The question asks for the maximum height, not just the time of the maximum height, so you would have to go back and substitute the value for t into the quadratic height function.

If you have a CAS calculator, you can solve the problem in one step for both time and height with Complete the Square. This is going to be faster.

If you are doing the problem by hand, using x=-b/2a ism’t slower than using 0=v0-gt: it’s the same math. In both cases, you then have to substitute the obtained value for t into the h function, so it will be slower than using CompleteSquare. But both methods are good.

28

@plotinus Thank you!

29

@Plotinus Yes, thank you! Very helpful posts!