634 MCGRAW-HILL’S SAT
- C The rest of the sentence describes how the
processes of death change a formerly living body. In
saying that he beheld the corruption of death succeed
to the blooming cheek of life, he is saying that death
and decay have replaced or defeated life. - E The narrator reveals his sense of privilege in
this discovery by stating that he is alone(line 83)
among the many men of genius(line 81) who had
studied this topic before.
Section 7
- C 16 is equal to 2(7) +2, so it is two more than a
multiple of 7.
(Chapter 7, Lesson 7: Divisibility) - E Five oranges can be bought for 5¢ more than
the price of four, which is 4(20¢) +5¢ = 85¢. $3.40 is
equivalent to 4(.85), so it will buy 4(5) = 20 oranges.
(Chapter 8, Lesson 7: Word Problems) - A If ris positive, then –ris negative. If you add
another negative, then the result will be even more
negative.
(Chapter 7, Lesson 6: Negatives) - D Twelve less than the product of 3 and x+1 can
be represented as 3(x+1) − 12
Distribute: 3 x+ 3 − 12
Simplify: 3 x− 9
(Chapter 8, Lesson 7: Word Problems) - C The square has an area of 200 × 200 =40,000
square feet. 40,000 ÷5,000 =8, so this will require
eight bags of seed at $25 apiece. 8 ×$25 =$200.
(Chapter 10, Lesson 5: Areas and Perimeters) - C Analyzing the right angle shows that x+y=90.
Since the sum of the angles in a triangle is always 180°,
x+y+w= 180
Substitute x+y=90: 90 +w= 180
Subtract 90: w= 90
(Chapter 10, Lesson 2: Triangles) - D If the numbers have a product of 0, then at
least one must equal 0. Call the numbers x, y,and 0.
The problem also says that x+y=7 and x−y=11.
Add the equations: x+y= 7
+(x−y=11)
2 x= 18
Divide by 2: x= 9
Plug back in, solve for y: 9 +y= 7
Subtract 9: y=− 2
So the least of the numbers is −2.
(Chapter 8, Lesson 2: Systems)
If you prefer to look at it as a “combination” problem,
the number of triangles is the number of ways of
choosing three things from a set of four, or 4 C 3 =4.- B The only way that abcwould not be a multiple
of 4 is if none of the three numbers is a multiple of 4
andno two of them are even (because the product of
two evens is always a multiple of 4). One simple ex-
ample is a=1, b=2, and c=3. This example rules out
choices (A), (C), (D), and (E).
(Chapter 9, Lesson 3: Numerical Reasoning
Problems) - A A large percent change from 2002 to 2003 is
represented by a point in which the y-coordinate is
much greater than the x-coordinate. Point Arepre-
sents a change from 30 in 2002 to 70 in 2003, which
is a percent change of (70 −30)/30 ×100% =133%.
(Chapter 7, Lesson 5: Percents)
(Chapter 11, Lesson 5: Data Analysis) - E If both classes have 100 students, then class B
had 30 students participate in 2002 and 50 in 2003, for
a total of 80. Class Ehad 80 in 2002 and 60 in 2003, for
a total of 140. The difference, then, is 140 − 80 =60.
(Chapter 11, Lesson 5: Data Analysis) - E If class Dhas 120 students, then 80% of 120, or
96 students participated in 2002. If the same number
participated from class C, then 96 is 60% of the num-
ber of students in class C. If the number of students in
class Cis x, then .60x=96. Divide by .6: x=160.
(Chapter 11, Lesson 5: Data Analysis) - E Substitute x=−1 into the equation to find c.
Simplify: 1 =− 4 +c
Add 4: 5 =c
So the equation is x^2 = 4 x+ 5
Subtract (4x+5): x^2 − 4 x− 5 = 0
Factor the quadratic (remember that since x=−1 is
a solution, (x+1) must be a factor): x^2 − 4 x− 5 =
(x+ 1)(x−5)
Therefore (x+1)(x−5) = 0
So the solutions are x=1 and x= 5
(Chapter 8, Lesson 5: Factoring) - A You can draw a diagram and see that there
are only four possible triangles: