336 MCGRAW-HILL’S SAT
Answer Key 1: New Symbol or Term Problems
- y^4 =− 32
Translate: 4 y4–1=− 32
Simplify and divide by 4: y^3 =− 8
Take the cube root: y=− 2 - If Ωw=12, then wmust be a number whose dis-
tinct prime factors add up to 12. The prime num-
bers less than 12 are 2, 3, 5, 7, and 11. Which of
these have a sum of 12? (Remember you can’t re-
peat any, because it says the numbers have to be
distinct.) A little trial and error shows that the
only possibilities are 5 and 7, or 2, 3, and 7. The
smallest numbers with these factors are 5 × 7 = 35
and 2 × 3 × 7 =42. Since the question asks for the
leastsuch number, the answer is 35.
SAT Practice 1
- B 2 3 x= 12
Translate: (2)(3) +(3)(x) +(2)(x) = 12
Simplify: 6 + 5 x= 12
Subtract 6: 5 x= 6
Divide by 5: x=6/5 - D If x#y=1, then (x^2 /y^2 ) =1, which means x^2 =y^2.
Notice that x=−1 and y=1 is one possible solu-
tion, which means that
(A) x=y
(B)x=⏐y⏐
(E)xand yare both positive
is notnecessarily true. Another simple
solution is x=1 and y=1, which means that
(C)x=−y
is not necessarily true, leaving only
(D) as an answer. - B All of the consecutive “double times” are
1 hour and 1 minute apart except for 12:12 and
1:01, which are only 49 minutes apart. - A If^2 ⁄ 3 and yare complementary, then the sum
of their reciprocals is 1:^3 ⁄ 2 +1/y= 1
Subtract^3 ⁄ 2 :1/y=−1/2
Take the reciprocal of both sides: y=− 2
5. 5 The “double” symbol means you simply per-
form the operation twice. Start with 5, then $5 =
1/5. Therefore, $$5 =$(1/5) =1/(1/5) =5.
6. E
Multiply by 4:
Square both sides:
Plug in x=36 to the original and see that it works.- C If ais 12, which is even, then [12] = 1 + 2 =3 is
odd, which means that statement III is not
necessarily true. (Notice that this eliminates choices
(D) and (E).) Statement I is true because [10a] will
always equal [a] because 10ahas the same digits
as a,but with an extra 0 at the end, which con-
tributes nothing to the sum of digits. Therefore,
[10a] < [a] +1 is always true. Notice that this leaves
only choice (C) as a possibility. To check statement
II, though (just to be sure!), notice that the biggest
sum of digits that you can get if ais less than 1,000
is from 999. [999] = 9 + 9 + 9 =27; therefore,
[[999]] =[27] = 2 + 7 =9. It’s possible to get a slightly
bigger value for [[a]] if ais, say, 991: [[991]] =
[19] =10, but you can see that [[a]] will never ap-
proach 20. - C Since 13 is odd, 13& = 13 − 3 =10. Therefore,
13&& =10&. Since 10 is even, 10& =2(10) =20.
◊= =
=
=
x
xx
x4
15
6
36
.
Concept Review 1
§−4.5 =− 5
§−1.5 +§1.5 =− 2 + 1 =− 1
5.
6.
7.
Simplify:Take the cube root:
Square:yyyy
y@2 64
64
4
16
213=()=
()=
=
=
+xxx^2201
0@ =()=+919 3
11
@kk k
( −)=() =−+83 8 64
4
@ =()=