SAT Practice 3
- D Start with the simplest odd values for mand n: m
=n=1. (There’s no reason why mand ncan’t equal the
same number!) Notice that m^2 +n^2 = 12 + 12 =2, which
isn’t divisible by 4, so statement II is not necessarily
true, and you can eliminate choices (C) and (E). Next,
notice that m^2 and n^2 must both be odd, so m^2 +n^2 must
be even, so statement I is necessarily true, and you can
eliminate choice (A). (m+n)^2 must be a multiple of 4
because m+nmust be even (odd +odd =even), so it is
a multiple of 2. When it is squared, it becomes a multi-
ple of 4. So III is true, and the answer is (D). - D Trial and error should show that A =3. If A is less
than 3, the product is too small. If A is greater than 3,
the product is too large. Since 633 × 8 =5,064, B =6. - A 500 ÷ 9 =55.55, so there are 55 multiples of 9 be-
tween 1 and 500. 500 ÷ 7 =71.43, so there are 71 mul-
tiples of 7 between 1 and 500. So j+k= 55 + 71 =126. - 12 Trial and error shows that the only way to
write 60 as the product of four integers, each
greater than 1, is 2 × 2 × 3 ×5. Their sum is
2 + 2 + 3 + 5 =12. - C Do the prime factorization:
1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 = 1 × 2 × 3 ×(2 ×2)
× 5 ×(2 ×3) × 7 ×(2 × 2 ×2) ×(3 ×3)Since there are seven factors of 2, the greatest
power of 2 that is a factor is 2^7.- B p+pq=4(p−pq)
Distribute: p+pq= 4 p− 4 pq
Divide by p: 1 +q= 4 − 4 q
(This is okay as long as pis anything but 0.)
Add 4q: 1 + 5 q= 4
Subtract 1: 5 q= 3
Divide by 5: q=3/5
Because pcan have many possible values but q
can only equal 3/5, (B)qis the only expression that
has only one possible value. - A acannot be even, because an even number
times any other integer yields an even number,
but a(b(c+d) +e) is odd. - A The only three different positive integers that
have a sum of 7 are 1, 2, and 4. The only three dif-
ferent positive integers that have a sum of 9 are 1,
3, and 5 or 1, 2, and 6. But 1 + 2 +6 doesn’t work,
since that would have twonumbers in common
with the first set, but it may only have one (C).
Since (C) is the only number they may have in
common, it must be 1. - C The only solution is 188 + 987 =1,175, so
A +B +C = 1 + 8 + 5 =14.
CHAPTER 9 / SPECIAL MATH PROBLEMS 345
Concept Review 3
- They are reciprocals, so their product is 1.
- They are opposites, so their sum is 0.
- Odd ×even =even
- Even ×even =even
- Odd ×odd =odd
- Even +even =even
- Odd +even =odd
- Odd +odd =even
- If nis odd, (−1)n=−1.
- If nis even, (−1)n=1.
- If x+y=0 and x≠0, then x/y=−1.
- Dividing by xis the same as multiplying by 1/x.
- Subtracting (x+1) is the same as adding −x−1.
- When a number is multiplied by its reciprocal,
the result is 1. - When a number and its opposite are added, the
result is 0. - When a number (other than 0) is divided by its
opposite, the result is −1. - It gets bigger.
- It gets smaller.
- It gets smaller (more negative).
- No. If xis 0, then −xis equal to x,and if xis neg-
ative, then −xis greater than x. - No. If xis between 0 and 1, then x^2 is smaller than
x.And if xis 0 or 1, then they are the same. If xis
negative, then x^2 is positive, and therefore greater
than x. - No. If xis between 0 and 1, then x^3 is smaller than
x^2. And if xis 0 or 1, then they are the same. If x
is negative, then x^2 is positive, and therefore
greater than x^3. - greater than 1.
- greater than 1.
- between 0 and 1.
Answer Key 3: Numerical Reasoning Problems