SAT Mc Graw Hill 2011

256 MCGRAW-HILL’S SAT

Keep Your Options Open

There are often many good ways to solve an

SAT math problem. Consider different strate-

gies. This gives you a way to check your work.

If two different methods give you the same an-

swer, you’re probably right!

Numerical Analysis—Plugging In

Let’s come back to the problem we saw in Lesson 4:

If 3x^2 + 5x+ y = 8 and x≠0, then what is the value

of?

Back in Lesson 4 we solved this using substitution,
an algebraic method. Now we’ll use a numerical method.
Notice that the equation contains two unknowns. This
means that we can probably find solutions by guessing
and checking. Notice that the equation works if x= 0 and
y= 8. But—darn it—the problem says x≠0! No worries—
notice that x= 1 and y= 0 also work. (Check and see.)
Now all we have to do is plug those numbers in for x

and y:. Same answer,

whole different approach!

“Plugging in” works in two common situations:

when you have more unknowns than equations

and when the answer choices contain un-

knowns.Always check that your numbers sat-

isfy the conditions of the problem. Then solve

the problem numerically,and write down the

answer. If the answer choices contain un-

knowns, plug the values into every choice and

eliminate those that don’t give the right answer.

If more than one choice gives the right answer,

plug in again with different numbers.

If 3m= mn+ 1, then what is the value of min terms
of n?

(A) n+ 1
(B) n− 2

(C)

(D)

(E)

Because the choices contain unknowns, you can
plug in. Pick a simple number for mto start, such as 1.
Plugging into the equation gives 3 = n+ 1, which has the

2

3 +n

1

3 +n

1

3 −n

16 2

35

16 2 0

31 51

16

8

22 2

−

+

=

−

+

==

y

xx

()

() ()

16 2

352

−

+

y

xx

solution n= 2. Now notice that the question asks for m,

which is 1. Write that down and circle it. Now substitute

2 fornin the choices and see what you get:

(A) 3

(B) 0

(C) 1

(D) 1/5

(E) 2/5

Only (C) gives the right answer.

Algebraic Analysis

You can also solve the problem above algebraically:

3 m= mn+ 1

Subtract mn:3m−mn= 1

Factor: m(3 −n) = 1

Divide by (3 −n): m =

Testing the Choices

Some SAT math questions can be solved just by

“testing” the choices. Since numerical choices

are usually given in order, start by testing

choice (C). If (C) is too big, then (D) and (E) are

too big, also, leaving you with just (A) and (B).

This means that you have only one more test to

do, at most, until you find the answer.

If 3(2)n+1 – 3(2)n= 24, what is the value of n?

(A) 2

(B) 3

(C) 4

(D) 5

(E) 6

Here you can take an algebraicor a numericalap-

proach. That is, you can solve the equation for nor you

can “test” the choices to see if they work. For this les-

son, we’ll try the “testing” strategy. Since the choices

are given in ascending order, we’ll start with the mid-

dle number, (4). Substituting 4 for ngives us 3(2)^5 –

3(2)^4 on the left side, which equals 48, not 24. (It’s okay

to use your calculator!) Since that doesn’t work, we

can eliminate choice (C). But since it’s clearly too big,

we can also rule out choices (D) and (E). That’s why

we start with (C)—even if it doesn’t work, we still narrow

down our choices as much as possible. Now just test ei-

ther (A) or (B). Notice that (B) gives us 3(2)^4 – 3(2)^3 ,

which equals 24, the right answer.

Now try solving the problem algebraically, and see

if it’s any easier!

1

3 −n

Lesson 6: Finding Alternatives