76 McGRAW-HILL’S SAT
3.C This question is asking what the remainder is
when 29, 21, and 64 are each divided by 3. When 29
is divided by 3 the remainder is 2; when 21 is divided
by 3 the remainder is 0; and when 64 is divided by 3
the remainder is 1.
(Chapter 7, Lesson 7: Divisibility)
4.B 1 & 2
Substitute using definition: 1(1 −1) +2(2 −1)
Simplify: 0 + 2 = 2
(Chapter 9, Lesson 1: New Symbol or Term Problems)
5.A In a triangle, any side must have a length that is
less than the sum of the two other lengths but greater
than the difference of the two other lengths. Therefore,
the third side must have a length between 15 − 9 =6 and
15 + 9 =24, so a length of 5 is impossible.
(Chapter 10, Lesson 2: Triangles)
6.B The volume of a cube is equal to s^3 , where sis
the length of one edge. If s^3 =64, then s=4, and so each
square face has an area of s^2 = 42 =16. Since a cube has
six faces, the total surface area is 6(16) =96.
(Chapter 10, Lesson 7: Volumes and 3-D Geometry)
7.D
Simplify: x+ 18 = 32
Subtract 18: x= 14
So the numbers are 2, 6, 10, and 14. The median is the
average of the two middle numbers: =8.
(Chapter 9, Lesson 2: Mean/Median/Mode Problems)
8.A Indicate the congruent sides with tick marks:
in a triangle, the angles across from equal sides are
equal; indicate this in the diagram. Your angles should
be marked as shown. Since the angles in a triangle have
a sum of 180°, y+y+ 180 −x= 180
Subtract 180: 2y−x= 0
Add x:2y=x
Divide by 2:
(Chapter 10, Lesson 2: Triangles)
9.D Notice that the graph is of all the points that
are more than one unit away from −1. The distance
from a point to −1 is ⎟x−(−1)⎟, or ⎟x+ 1 ⎟; if this dis-
tance is greater than one, then ⎟x+ 1 ⎟> 1.
(Chapter 8, Lesson 6: Inequalities, Absolute Value,
and Plugging In)
10.B “Must be true” kinds of questions are often best
answered by process of elimination with examples.
y
x
=
2()610
2
+
()x
x+++
= +++ =
2610
4
8261032,soBegin with a simple set of values, for instance a=0,
b=−1, and c=0. Notice that these values satisfy all
of the given information. This example clearly shows
that statement I need not be true, because 0 is not
greater than 0, and that statement III need not be
true, because (0)(0) is not greater than (−1)^2. This
leaves only statement II as a possibility, so the answer
must be (B).
(Chapter 6, Lesson 7: Thinking Logically and Checking)
(Chapter 9, Lesson 3: Numerical Reasoning Problems)11.E You have five choices for the first digit: 1, 3, 5,
7, and 9; ten choices for the middle digit (any digit will
do), and five choices for the last digit: 0, 2, 4, 6, and 8.
So the total number of possibilities is 5 × 10 × 5 =250.
(Chapter 9, Lesson 5: Counting Problems)12.B To find how many more seconds it will take
the machine to cut circle Athan circle B,you can
find the length of time it takes to cut each circle and
subtract them. The laser cuts the circumference of
each circle, so you must find that first. Circle Ahas
an area of 64π. Since the area of a circle is πr^2 , the
radius of the circle is 8. Since the area of circle Bis
16 π, its radius is 4. The circumference of a circle is
2 πr,so the circumference of Ais 2π(8) = 16 πand the
circumference of Bis 2π(4) = 8 π. The difference of
their radii is 16π− 8 π= 8 π. The time it takes to cut that
length is given by the formula time=distance/rate.(Chapter 10, Lesson 8: Circles)
(Chapter 9, Lesson 4: Rate Problems)13.E The slope of ACis rise/run==6/7.Therefore the slope of CBis −6/7. Using the slopeformula: =−6/7Simplify: =−6/7Cross-multiply: −6(8 −k) = 42
Divide by 6: − 8 +k= 7
Add 8: k= 15
(Chapter 10, Lesson 4: Coordinate Geometry)14.C m= 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 ×10. You
can factor even further in terms of primes: m= 1 × 2
× 3 × (2 × 2) × 5 ×(2 ×3) × 7 ×(2 × 2 ×2) ×(3 ×3) ×
(2 ×5). This shows that there are a maximum of eight
factors of 2, so the greatest power of 2 that is a factor
of mis 2^8.6
8( −k)(())
(
51
8
−−
−k)(())
(
51
81
−−
−)
()
(
8
3
8
3
ππcm
cm/second)= sec