POSTTEST: Mathematical Reasoning 157

POSTTEST

Answers and Solutions

1. –3 and 3 Since the term with x 1is missing,
11. B 8 x 2 + 24 x + 18 = 2(4 x 2 + 12 x + 9) =
solve for x 2:
8 x 2 2+ 24 x + 18 = 2(4 x 2 + 12 x + 9) =
2 2 4x 36
4 x – 36 + 36 = 20 + 36 → 4 x = 36 → = → x 2 2(2
= 9x + 3)(2 x + 3) = 2(2 x + 3)2
2 2 4x 36 2 4 4
4 x – 36 + 36 = 0 + 36 → 4 x = 36 → = →x =9 2(2 x + 3)(2 x + 3) = 2(2 x + 3)2
4 4
4 x 2 36
36 + 36 = 0 + 36 → 4 x 2 = 36 → = → x2 = 9 12. D Factor and cancel:
2 4 4
4 x 36
36 → 4 x 2 = 36 → = → x2 = 9 x 2 – 9 x + 20 ( x – 4 )( x – 5)
4 4 = = x–4
x –5 x –5
Now use the square root property:
2 x 2 – 9 x + 20 ( x – 4 )( x – 5)
x = ± 9 → x = ±3 = = x–4
x –5 x –5
136 8 1 x – 9 x + 20 ( x – 4 )( x – 5)
2
2. B =4 =4 = = x–4
32 32 4 x –5 x –5
13. A 3 x – 11 < 5 x – 7 → 3 x – 11 – 5 x < 5 x – 7 – 5 x
31 3 2 5 3 8 2 15 16 < 531 x – 7 → 3 x – 11 – 5 x < 5 x – 7 – 5 x →
3. 11 7 + 4 = 11 and + = • + • = 3 x+– 11 =
40 8 5 5 8 8 5 40 40 40 –2 x – 11 < –7 → –2 x – 11 + 11 < –7 + 11 →
3 2 5 3 8 2 15 16 31 –2 x – 11 < –7 → –2 x – 11 + 11 < –7 + 11 →
+ = • + • = + = –2 x 4
8 5 5 8 8 5 40 40 40 –2 x < 4 → > → x > –2
4. 37.503 –2 –2
–2 x 4
875 7 –2 x < 4 → > → x > –2
5. C 0.875 = = –2 –2
1000 8
14. B Rearranging before dividing:
6. 5 and 2 A function has only one f(x) value
for each x. The missing numbers x –2 y 5 x3y5
−3 2
= 2 2
= xy 3
must match those already in the x y x y
chart.
15. A 4 x 2 – 7 x – 15 = 15 – 15 → 4 x 2 – 7 x – 15 = 0 →
x –2 –1 0 1 2 1 5 4–2 x 2 – 7 x – 15 = 15 – 15 → 4 x 2 – 7 x – 15 = 0 →
f(x) 5 3 0 2 6 2 7 5 ( 4 x + 5)( x – 3) = 0 → 4 x + 5 = 0 or x – 3 = 0
( 4 x + 5)( x – 3) = 0 → 4 x + 5 = 0 or x – 3 = 0 →
7. (6, 8) and Any point on the top half of the 4 x + 5 – 5 = 0 – 5 or x – 3 + 3 = 0 + 3 → 4 x =
(6, –8) circle and the point directly below 4 x + 5 – 5 = 0 – 5 or x – 3 + 3 = 0 + 3 → 4 x = –5
4 x –5
it on the lower half4 x of
+ 5the = 0 – 5 or x – 3 + 3 = 0 + 3 → 4 x = –5 or x = 3 →
– 5circle = or x = 3 → x =
4 4
will su!ce; for instance (–8, 6) and 4 x – 5 5
or x = 3 → = or x = 3 → x = – or x = 3
(–8, –6), or (0, 10) and (0, –10) also 4 4 4
work. 4 x –5 5
or x = 3 → = or x = 3 → x = – or x = 3
4 4 4
8. A f(5) = 2(5)2 + 7(5) + 3 = 2 • 25 + 35 + 3 = 50 + 38 = 88
f(5) = 2(5)2 + 7(5) + 3 = 2 • 25 + 35 + 3 = 50 + 38 = 88 x2y 4 xy 3
16. C Rearranging "rst, = .
xyz 3 z 3 z6
9. –8 and 15 15 – (–8) = 15 + 8 = 23 = 23 17. A Since the path forms a right
triangle, use the Pythagorean
⎛ –1⎞ ⎛ 1⎞ 1 1
– + theorem: a2 + b2 = c2. In this case,
10. A ⎜x ⎟⎜x ⎟ = x
3 3 3 3
= x0 = 1 42 + 32 = c2 ! 16 + 9 = c2 !
⎝ ⎠⎝ ⎠ 25 = c2 ! c = 25 = 5.
 158 POSTTEST: Mathematical Reasoning

POSTTEST

18. D She has 84 # 5 = 420 total points after 26. Erlene Convert both scores to decimals
"ve tests. She needs 85 # 6 = 510 total to compare. Erlene’s 86% = 0.86.
points to raise her average to an 85. Bob’s 42/50 = 0.84. Erlene has the
510 – 420 = 90. higher score.

4x2 9 9 0.004634 0.01 + 0.009 + 0.0023 + 0.0012 +

27.
19. B 4x2 – 9 = 0 → 4x2 = 9 → = → x2 = → 0.00067 = 0.02317. Divide by 5 to
4 4 4
4x2 9 9 get 0.004634.
– 9 = 0 → 4x2 = 9 → = → x2 = →
4 4 4 28. B Hexagons have six sides, so
9 3 p = 6 × 17 = 102 .
x2 = ± →x=±
4 2 29. B The area of a triangle is given by
2 9 3 1 1
x =± →x=± A = bh = • 10 • 5 = 5 • 5 = 25.
4 2 2 2
20. B The least common denominator 30. B Use the Pythagorean theorem.
is$12xy.
a2 + 122 = 202 →
2 x – 7 7 y + 3 3y 2 x – 7 2 x 7 y + 3 a2 + 144 = 400 → a2 + 144 – 144 = 400 – 144 →
– = • – • =
4x 6y 3y 4x 2x 6y
a2 + 144 = 400 → a2 + 144 – 144 = 400 – 144 →
2 x – 7 7 y + 3 3y 2 x – 7 2 x 7 y + 3
– = • – • = a2 = 256 → a2 = 256 → a = 16
4x 6y 3y 4x 2x 6y
a2 = 256 → a2 = 256 → a = 16
6 xy – 21y 14 xy + 6 x 6 xy – 21y – 14 xy – 6 x
– = =
12 xy 12 xy 12 xy 31. the square
y – 21y 14 xy + 6 x 6 xy – 21y – 14 xy – 6 x Since the diameter of the circle is equal to the
– = =
12 xy 12 xy 12 xy length of a side of the square, the circle "ts
–8 xy – 21y – 6 x 6 x + 8 xy + 21y within the square, so it is smaller. Or you can
=– calculate the areas, using a circle radius of 15.
12 xy 12 xy
32. D The %our/sugar ratio of 4:1 is for one
7 49 7 49 batch, and 4:1::20:5.
21. 28 = → 4n • = 4n • → 7n = 4 • 49 →
4 n 4 n
7 49 7n 4 • 49
→ 4n • = 4n • → 7n = 4 • 49 → = = 28A
→ n = 4 • 7 → n 33. By the properties of exponents,
4 n 7 7 5
7n 4 • 49
( 4)
5
= → n = 4 • 7 → n = 28 42 = = 25 = 32 .
7 7
22. C Although the ratio of sides is 1:2, we 34. D Multiplying, 127, 000 × 0.09 = 11, 430 .
square sides to "nd area, so the ratio of Subtracting 127, 000 – 11, 430 = 115, 570
areas is 1:4. residents not on farms.

23. line 2 35. A Multiplying 127, 000 × 0.17 = 21, 590

who are above the average age.
24. –! The highest-degree term rules the end
behavior. 36. C The trials are independent, so we
multiply, but the number of balls
19 95 19 changes.
25. =
2000 10, 000 2000 5 4 20 2
(reds ) × ( whites) = =
15 14 210 21
 POSTTEST: Mathematical Reasoning 159

POSTTEST

37. D (2 x + 5) = (2 x + 5)(2 x + 5) = 4 x 2 + 20 x + 2543.

2
y

x + 5)
2
= (2 x + 5)(2 x + 5) = 4 x 2 + 20 x + 25 10

38. –10 Since there is a greater-than-or-equal- 8

to sign, the number –10 is part of the 6
solution, along with all the integers
4
from there to positive in"nity.
2
1 3 4
39. 0.06 < 21% < 33 % < < 0.47 < x
3 8 5 –10 –8 –6 –4 –2 2 4 6 8 10
–2
40. C Solve for y to put the equation in the –4
slope-intercept form, y = mx + b :
–6
3 x – 5 y = 15 → 3 x – 5 y – 3 x = 15 – 3 x →
–8
3 x – 5 y = 15 → 3 x – 5 y – 3 x = 15 – 3 x →
–5 y –3 x + 15 3
–5 y = –3 x + 15 → = → y = x –3 –10
–5 –5 5
–5 y –3 x + 15 3
–5 y = –3 x + 15 → = → y = x –3
–5 –5 5
44. A increase = 1500 – 1250 = 250; percent
–5 y –3 x + 15 3 3
+ 15 → = → y = x – 3 , so m = 250 1
–5 –5 5 5 increase = = = 20%
1250 5
41. A
45. y = –7 x + 21
(2 x + 13) + (6 x + 3) = 2 x + 6 x + 13 + 3 = 8 x + 16
–7 – 7 –14
2 x + 13) + (6 x + 3) = 2 x + 6 x + 13 + 3 = 8 x + 16 The slope is m = = = –7. Using
4–2 2
= 2 x + 6 x + 13 + 3 = 8 x + 16
y = –7 x + b and the coordinates of the point
(2, 7), 7 = –7 • 2 + b → 7 = –14 + b → 21 = b.
42. C There are 12 # 11 # 10 choices of
women, but because the order they 46. C When the curve intersects the y-axis,
are chosen does not matter, divide x = 0. Substituting x = 0 results in
that by the number of ways to arrange y = 6 • 02 + 5 • 0 – 6 = –6.
three people (3 # 2 # 1). There are 6 #
5 # 4 choices of men, but because the
order they are chosen does not matter,
divide that by the number of ways to
arrange three people (3 # 2 # 1). Put it
all together to get (12 # 11 # 10) ÷ (3 #
2 # 1) # (6 # 5 # 4) ÷ (3 # 2 # 1) = 4400.