CHAPTER 2 / DIAGNOSTIC SAT 73
4.C Since the prime factorization of 98 is 2 × 7 ×7,
and since the greatest common factor of mand 98 is
a prime number, that greatest common factor must
be 2 or 7. Since it is not even, it must be 7.
(Chapter 7, Lesson 7: Divisibility)
(Chapter 8, Lesson 5: Factoring)
(Chapter 9, Lesson 1: New Symbol or Term Problems)
5.D ⎟k−0.5⎟< 10
Translate: −10 < k−0.5 < 10
Add 0.5: −9.5 < k< 10.5
The smallest possible integer value for kis −9 and the
greatest is 10. The total number of integers between
−9 and 10, inclusive, is 10 −(−9) + 1 =20.
(Chapter 6, Lesson 2: Analyzing Problems)
(Chapter 8, Lesson 6: Inequalities, Absolute Value,
and Plugging In)
(1 hour) to get to work, she must be going 50 miles/
hour. If she increases her speed by 20% for the
trip home, then her speed coming home is (1.20)
(50 miles/hour) =60 miles/hour. To travel 50 miles
at 60 miles/hour will take her (50 miles)/(60 mph) =
5/6 hour, which is 5/6(60 minutes) =50 minutes.
(Chapter 9, Lesson 4: Rate Problems)9.0.4 Remember that “percent” means “divided by
100,” so 0.5 percent of 80 means 0.5 ÷ 100 × 80 =0.4
(Chapter 7, Lesson 5: Percents)10..333 or 1/3 Just pick three consecutive odd inte-
gers, like 1, 3, and 5. Since dis the middle of these,
d=3. Since sis the sum of these, s= 1 + 3 + 5 =9. So
ddivided by sis 3/9 or 1/3.
(Chapter 9, Lesson 3: Numerical Reasoning Problems)- 30 4/9 of c^2 is 24
Translate: (4/9)(c^2 ) = 24
Multiply by 5/4: (5/4)(4/9)(c^2 ) =(5/4)(24)
Simplify: (5/9)(c^2 ) = 30
(Chapter 6, Lesson 4: Simplifying Problems)
(Chapter 8, Lesson 1: Solving Equations) - 60 The sum of the four angles in a quadrilateral
is 360°. The sum of the parts in the ratio is 3 + 4 + 5 +
6 =18. Therefore the angles are 3/18, 4/18, 5/18, and
6/18 of the whole, which is 360°. So the smallest angle
measures (3/18)(360°) = 60 °.
(Chapter 10, Lesson 2: Triangles)
(Chapter 7, Lesson 4: Ratios and Proportions) - 28 Subtract the equations: 5a+ 6 b= 13
−(4a+ 5 b=9)
a+b= 4
Multiply by 7: 7 a+ 7 b= 28
(Chapter 6, Lesson 4: Simplifying Problems)
(Chapter 8, Lesson 2: Systems) - 6 Simply substituting m=3 in the equation gives
. The quickest way to simplify this expression
is to multiply both the numerator and the denominatorby the common denominator, 8. This givesIf you happen to be an algebra jock, you might notice
that you can simplify the original expression by multi-
plying numerator and denominator by the common de-
nominator m^2 – 1, which is equivalent to (m– 1)(m+1).
This simplifies the complex expression to just 2m,
which equals 6 when m=3.24
1
6
+
=.
1
4
+
1
2
1
8
yx
1y=f(x)–1 56.A Since fis a quadratic function, its graph is a
parabola with a vertical axis of symmetry through its
vertex, which in this case is the line x=2. This means
that, for any given point on the graph, its reflection
over x=2 is also on the graph. Notice from the given
graph that the value of f(5) is about 2.5, as shown
above. If we reflect this point, (5, 2.5) over the axis of
symmetry, we get the point (–1, 2.5). In other words,
f(5) =f(–1), so k=–1.
7.C It helps to know the perfect squares and the
perfect cubes. The first seven perfect squares greater
than 1 are 4, 9, 16, 25, 36, 49, and 64. The first three
perfect cubes are 8, 27, and 64. Clearly, the only inte-
ger between 1 and 100 that is both a perfect cube and
a perfect square is 64 = 43 = 82. Therefore m=4 and
n=8, so m+n= 4 + 8 =12.
(Chapter 8, Lesson 4: Working with Roots)
8.B This is a rate problem, so remember the basic
rate formula: distance=rate×time.Start by picking a
value for the distance from Amanda’s home to work.
No matter what distance you choose, the final answer
will be the same, so choose a distance that’s easy to
calculate with, like 50 miles. If it takes her 60 minutes