Up Your Score (37 page)

However, we also need to add the breaks. Since there are 16 laps, the fish must take 4 breaks:

4 × 5 seconds = 20 seconds

Answer: 8 minutes, 20 seconds = 8 ⅓ minutes = (C)

Enough of this math stuff—the SAT requires fast thinking in difficult situations. So here’s a scenario—you have three seconds to come up with the appropriate response.

A psychotic iguana with a bottle of dishwashing detergent is chasing you and gaining every second. The soft grass you’re running on barefoot suddenly ends and you’re faced with the option of treading on a minefield strewn with broken glass or walking across a river of flowing lava. Which do you choose?

Answer: Neither. What are you, stupid? I’d much rather face an iguana (even a psychotic one with detergent) than deal with a minefield or hot lava. So remember, if you really don’t like
any
of your choices, then “it cannot be determined from the information given” may be an option.

JaJa says: A lot of the SAT’s math is really not that advanced. It just requires fast thinking. So practice!

The rule for equations is
do it to both sides
. We don’t care what
it
is, but do it to both sides. That keeps everything nice and equal. So if your lover is your enemy:

(lover = enemy)

and you want to kill your enemy, you must kill your lover, too:

(kill your lover = kill your enemy)

There. Nice and equal. Well, equal, anyway. So if you have

x
+ 36 = 40

and you want to solve for
x
, then do it like this. Subtract 36 from the left
and
subtract 36 from the right:

x
+ 36 − 36 = 40 − 36, which means that
x
= 4.

Now check it by substituting 4 for
x
in the original expression:

Examples:

1. 12
x
= 24

Divide both sides by 12 to get
x
= 2.

2. 3
x
+ 4 = 28

Subtract 4 from both sides to get 3
x
= 24.

Divide both sides by 3 to get
x
= 8.

3. Solve for fish:

3(fish + grapefruit) = college

Divide both sides by 3:

Subtract grapefruit from both sides:

Okay. “Number 3 was a moronic question,” you might say to yourself. Yes. It was. But they ask similar questions on the SAT just to see if you know these rules. They use
x
,
y
, and
z
more than they use fish and grapefruit, but it’s the same basic idea.

4.
x
−
y
= 17

What is
y
+ 12?

(A)
x

(B)
x
+ 5

(C)
x
− 5

(D)
x
− 7

(E) It cannot be determined from the information given.

Answer:
x
−
y
= 17

x
−
y
+
y
= 17 +
y

x
=
y
+ 17

x
− 5 =
y
+ 17 − 5

x
− 5 =
y
+ 12

(C) is correct.

5.
x
+ 3 −
y
= 10

Solve for
x
in terms of
y
.

(A)
y
+ 7

(B)
y
+ 3

(C)
y
− 7

(D) 10 −
y

(E) 13 −
y

Answer:
x
+ 3 −
y
= 10

x
+ 3 −
y
+
y
= 10 +
y

x
+ 3 = 10 +
y

x
+ 3 − 3 = 10 +
y
− 3

x
= 7 +
y

(A) is correct.

When faced with complicated algebraic expressions, test takers can get confused or flustered. While simple problems should be solved algebraically, more complicated problems can be solved in another way. Instead of working through the problem, consider substituting numbers for variables. Choose numbers that are easy to work with, such as 1, 2, or 10. Also try a negative number such as −1 or −2. Zero is useful, too.

Example:

Last year, a town had a population of 2,000 +
x
. If the population increased by 25 people this year, which of the following expressions represents this year’s population?

(A) 2,000 + 25
x

(B) 2,025 +
x

(C) 5,000 + 25
x

(D) 5,025 + 25
x

(E) 5,250 +
x

While this can and should be solved using pure algebra, substituting the number 1 for
x
is an easy alternative. By doing this, the population was 2,001 and then increased by 25, becoming 2,026. By substituting 1 for
x
in the answer choices, you find that choice (B) is correct.

More complicated example:

4
^x
+ 4
^x
+ 4
^x
+ 4
^x
=

(A) 4
<sup>x
+ 1</sup>

(B) 4
<sup>x
+ 2</sup>

(C) 4
<sup>x
+ 4</sup>

(D) 4
^4
x

(E) 4
^x4

This problem can be solved algebraically, by saying that 4
^x
+ 4
^x
+ 4
^x
+ 4
^x
= 4 · 4
^x
= 4
^x
+1
. If, however, you get stuck or have a difficult time following this method, you can substitute a number for
x
instead. By saying that
x
= 2, 4
²
+ 4
²
+ 4
²
+ 4
²
= 64 = 4
³
. Substituting 2 for
x,
we see that 4
<sup>x
+ 1</sup>
= 4
^{2 + 1}
= 4
³
. Therefore, the answer is 4
<sup>x
+ 1</sup>
or (A).

Whenever you have to factor, you should always ask yourself these five questions to get the thing factored.

1.
Is there anything in common?

Example: In 3
x
²
+ 6
x
, there’s a 3
x
that both terms have in common.

3
x
²
+ 6
x
= 3
x
(
x
+ 2)

2.
Is this a difference of two squares?

If both terms are perfect squares, they are factorable.

Example:
x
²
– 36 = (
x
− 6) (
x
+ 6)

3.
Is this a trinominal?

Sometimes, an expression with three terms can be factored into two expressions with two terms each. (For an explanation of this, see below.)

Example:
x
²
–
x
– 12 = (
x
+ 3) (
x
− 4)

4.
Is this the sum or difference of two cubes?

If both terms are cubes, you can factor them based on these formulas:

x
³
+
y
³
= (
x
+
y
)(
x
²
−
xy
+
y
²
)

x
³
–
y
³
= (
x
–
y
) (
x
²
+
xy
+
y
²
)

Example: 8
– y
³
= (2–
y
)(4 + 2
y
+
y
²
)

5.
Can I group this?

Grouping means to rearrange the terms so that a common factor can be pulled out.

Example: 3
x
²
+ 6
x
+ 5
xy
+ 10
y
= 3
x
(
x
+ 2) + 5
y
(
x
+ 2) = (3
x
+ 5
y
) (
x
+ 2)