SAT Mc Graw Hill 2011

Arithmetic Reasoning

342 MCGRAW-HILL’S SAT

Lesson 3: Numerical Reasoning Problems

The first statement, m< n< p< r,tells you that the al-

phabetical order is also the numerical order of the

numbers. The second statement, mnpr=0, tells you

that one of the numbers must be 0. (This is the zero

product property!) The third statement, m+n+p+r

=0, tells you that you must have at least one positive

and one negative, and all the numbers must “cancel

out.” This means that mcan’t be 0 because then none

of the numbers would be negative, and rcan’t be 0,

because then none of the numbers would be positive.

Thus, either nor pis 0. This means that both I and II

are necessarily true, so you can eliminate choices (A),

(B), and (D). The example m=−3, n=0, p=1, r= 2

shows that statement III is not necessarily true, so the

answer is (C).

Digit Problems

Some of the most common problems on the SAT

are numerical reasoningproblems, which ask

you to think about what happens to numbers

when you perform basic operations on them.

You just need to know the common numerical

and arithmetic rules and think logically.

Example:
If a+bis negative, which of the following CANNOT
be negative?
(A) ab (B)ab^2 (C)a^2 b
(D) a^2 b^2 (E) a−b

Start by thinking about what mightbe true about aand
band what mustbe true about aand b.First think of
possible values for aand b.−2 and 1 work, because
a+b=− 2 + 1 =−1. Notice that this proves that (A), (B),
and (E) are incorrect, because they can be negative:
(A) ab=(−2)(1) =−2, (B) ab^2 =(−2)(1)^2 =−2, and
(E) a−b=(−2) −(1) =−3. But (C) a^2 b=(−2)^2 (1) =4 is
positive,so does that mean the answer is (C)? Not so
fast! Your job is notto find which one can be positive,
but rather which cannot be negative.Notice that
(C) can be negative if aand bare, say, 1 and − 2
(notice that a+bis still negative, so those values
work): (C) a^2 b=(1)^2 (−2) =−2. Therefore, by process of
elimination, the answer is (D).

This question is much easier if you remember a sim-
ple fact: If xis a real number, then x^2 is never nega-
tive. If you don’t know this already, play around with
possible values of xuntil you see why this is true.
Then look at choice (D) a^2 b^2. a^2 can’t be negative, and
neither can b^2 , so a^2 b^2 can’t be negative.

Example:
If m< n< p< r, mnpr=0, and m+n+p+r=0,
then which of the following must be true?
I. If mand nare negative, then p=0.
II. np= 0
III.m+r= 0

(A) I only (B) II only

(C) I and II only (D) I and III only

(E) I, II, and III

You may see a question on the SAT like the one

below, where letters represent digits.Remember

that digits can only take the values 0, 1, 2, 3, 4,

5, 6, 7, 8, and 9.Also remember that you may

have to consider “carried” digits when looking

at a sum or product. Lastly, you may find it

best to work from left to right rather than

right to left.

Example:

1 BA

+ 8 B

21 1

If Aand Brepresent distinct digits in this addi-

tion problem, what is the value of A−B?

(A) 9(B) 7 (C) 2

(D) 7 (E) 9

Look at the left (hundreds) column first. Since the

sum has a 2 in the hundreds place, there must be a

carry of 1 from the tens place. Therefore, B+ 8 +

(carry from ones column, if any) =11. This means

B = 2 or 3. Trying each one shows that only

B=2 and A=9 works, giving 129 + 82 =211. There-

fore A−B= 9 − 2 =7, so the answer is (D).