Mathematics Level IC/IIC Subject Tests
219
ARCO
■
SAT II Subject Tests
w w w . p e t e r s o n s . c o m / a r c o
39. The correct answer is (B). One solution to this problem is to recognize that since the question asks
about
different orders, this is an appropriate problem for the formula used to calculate permutations:
3! = 3
× 2 × 1 = 6. Even if you didn’t recall the formula, don’t panic. You
should be able to count the
number of possibilities on your fingers:
1.
AJK
2.
AKJ
3.
KAJ
4.
KJA
5.
JAK
6.
JKA
And that’s all there is to it.
40. The correct answer is (A). Remember that the rules of exponents can be applied only to terms of
like bases. Here, the numerator has a base of 8, and the denominator a base of 2. Before we can
manipulate
the expression, it will be necessary to change one or the other term. There are several
different routes available to us. For example:
8
2
x
= (8)
2
x
= (2
× 2 × 2)
2
x
= (2
2
x
)(2
2
x
)(2
2
x
) = 2
2
x
+ 2
x
+ 2
x
= 2
6
x
Now we can complete our division:
2
6
x
÷ 2
4
x
= 2
6
x
– 4
x
= 2
2
x
Or:
(8)
2
x
= (2
3
)
2
x
= (2)
(3)(2
x)
= 2
6
x
And complete the division as shown above. Or you could choose to work with the denominator. I
think, however, that the best approach to this problem
is just to assume a value for x. Say
x = 1:
If we substitute 1 for
x into the answer choices, the correct choice will generate the value 4:
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