Chapter 14 Practice Test 4 With Answers - Mcgraw-Hill'S Psat/nmsqt Page 36
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45508
M
GRAW-HILL’S PSAT/NMSQT
C
originally 458 þ 158 ¼ 608. There are 1808 in a line, so
10. D
The figure in the diagram has 6 sides. So the
x þ y ¼ 180
sum of the angles must be (6 2 2)1808 ¼ 4(1808) ¼ 720
Substitute for y: x þ 60 ¼ 180
x ¼ 120
Subtract 60:
w
o
o
96
(Chapter 11 Lesson 1: Lines and Angles)
x
o
5. E
The container contains 6 gallons of water.
a
o
z
Since there are 4 pints in 1 gallon, there are
o
o
92
6 Â 4 ¼ 24 pints in the container. Since there are 2
cups in 1 pint, there are 24 Â 2 ¼ 48 cups in the con-
y
tainer. If 6 cups are poured out, this means that there
o
are 48 2 6 ¼ 42 remaining of the original 48 cups.
42
7
¼
There are 3608 in a circle: a þ 92 ¼ 360
48
8
Subtract 94:
a ¼ 268
(Chapter 8 Lesson 4: Ratios and Proportions)
There are 7208 in a hexagon (6 sides), so:
y þ a þ z þ 96 þ w þ x ¼ 720
6. D
To find the perimeter, first find the length of
Substitute for a:
y þ 268 þ z þ 96 þ w þ x ¼ 720
each side of the triangle. The two legs are easy: the
Combine like terms:
y þ 364 þ z þ w þ x ¼ 720
one along the x-axis is 3, and the one along the y-
Subtract 364:
y þ z þ w þ x ¼ 356
axis is 4. The hypotenuse is a bit more complex,
y þ z þ w þ x
Divide by 4:
¼ 89
and you need to use the Pythagorean theorem to
4
find its length:
2
2
2
a
þ b
¼ c
(Chapter 11 Lesson 2: Triangles)
2
2
2
Substitute:
3
þ 4
¼ c
2
Combine:
25 ¼ c
3
3
11. B
a
þ b
¼ 26
Take square root:
c ¼ 5
3
3
Stack and add:
À b
¼ 28
þa
3
2a
¼ 54
To find the perimeter, add the three sides:
3
Divide by 2:
a
¼ 27
3 þ 4 þ 5 ¼ 12.
a ¼ 3
Take cube root:
(Chapter 11 Lesson 5: Areas and Perimeters)
(Chapter 9 Lesson 2: Systems)
7. D
When b is quadrupled, it becomes 4b. Three
times the value of p is 3p. When 4b is subtracted
12. C
Consider the side of length 6 to be the base,
from 3p, the difference is (3p 2 4b). When that differ-
and “attach” the side of length 12. Notice that the tri-
ence is subtracted from y, the result is:
angle has the greatest possible height when the two
y 2 (3p 2 4b)
sides form a right angle. Therefore the greatest pos-
y À 3p þ 4b
Distribute:
1
sible area of such a triangle is
(6)(10) ¼ 30, and the
(Chapter 9 Lesson 1: Solving Equations)
2
minimum possible area is 0. Roman numerals I and
II are possible.
8. C
The quickest way to solve this problem is to
(Chapter 11 Lesson 2: Triangles)
come up with an example for p that fits the rules in
the problem. When p is divided by 5, the remainder
13. C
You are told that (6 ¤ 3) A (4 A 2) ¼ 12 ¤ 2
is 4. Let’s say that p ¼ 9.
and asked to find the value of (4 ¤ 3) A 6. First you
9 4 5 ¼ 1 remainder 4
have to figure out what ¤ and A represent. This will
To find the remainder when p þ 3 is divided by 5,
rely on some trial and error: Let’s say the ¤ symbol
substitute in your value for p and solve.
represents a minus sign:
p þ 3 ¼ 9 þ 3 ¼ 12
(6 ¤ 3) A (4 A 2) ¼ 12 ¤ 2
What is the remainder when 12 is divided by 5?
Substitute À for ¤: (6 À 3) A (4 A 2) ¼ (12 À 2)
Simplify:
3 A (4 A 2) ¼ 10
R2
12 4 5 ¼ 2
If A ¼ þ:
3 þ (4 þ 2) ¼ 9 = 10
If A ¼ 4 :
3 4 (4 4 2) ¼ 1:5 = 10
(Chapter
10
Lesson
3:
Numerical
Reasoning
If A ¼ Â:
3 Â (4 Â 2) ¼ 24 = 10
Problems)
So, ¤ = 2
9. E
Use the fundamental counting principle to
solve this problem. 4 Â 3 Â 6 Â 2 ¼ 144
(Chapter 10 Lesson 5: Counting Problems)
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