Chapter 14 Practice Test 4 With Answers - Mcgraw-Hill'S Psat/nmsqt Page 37
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45CHAPTER 14 / PRACTICE PSAT 4
509
10
24
Let’s say the ¤ symbol represents a multiplication
Plug in the known values and solve:
¼
24 þ y
15
sign.
Cross-multiply: 10(24 þ y) ¼ 15(24)
Distribute:
240 þ 10y ¼ 360
Substitute  for ¤: (6  3) A (4 A 2) ¼ (12  2)
Subtract 240:
10y ¼ 120
Simplify:
18 A (4 A 2) ¼ 24
Divide by 10:
y ¼ 12
If A ¼ þ:
18 þ (4 þ 2) ¼ 24
If A ¼ 4 :
18 4 (4 4 2) ¼ 9 = 24
Now, find the perimeter of the shaded region:
If A ¼ À:
18 À (4 À 2) ¼ 16 = 24
13 þ 15 þ 10 þ 12 ¼ 50
(Chapter 11 Lesson 6: Similar Figures)
Now that we know that ¤ ¼ Â and A ¼ þ, we must
find the value of:
(4 ¤ 3) A 6
2
15. B
y
þ by þ c ¼ 0
Substitute for ¤ and A:
(4 Â 3) þ 6
12 þ 6 ¼ 18
Simplify:
2
Substitute 4 for y:
4
þ b(4) þ c ¼ 0
(Chapter 10 Lesson 1: New Symbol and Term
Simplify:
16 þ 4b þ c ¼ 0
Problems)
2
Substitute À3 for y: (À3)
þ À3b þ c ¼ 0
Simplify:
9 À 3b þ c ¼ 0
14. C
This question tests your understanding of
Stack and subtract:
16 þ 4b þ c ¼ 0
similar triangles. (Because the triangles are both
À7 À 7b ¼ 0
right triangles and they share angle C, they are
Add 7b:
À7 ¼ 7b
similar because this means the three different
angles of the two triangles are the same.)
Divide by 7:
À1 ¼ b
First find the length of DC using the Pythagorean
theorem.
Alternatively: If y ¼ 4 and y ¼ 23 are solutions,
2
2
2
(DC)
þ (BD)
¼ (BC)
then it follows that
(y 2 4)(y þ 3) ¼ 0
2
2
2
Substitute:
(DC)
þ (10)
¼ (26)
2
2
FOIL:
y
þ 3y À 4y À 12 ¼ 0
Simplify:
(DC)
þ 100 ¼ 676
2
2
Combine like terms:
y
À y À 12 ¼ 0
Subtract 100:
(DC)
¼ 576
2
Compare to original:
y
þ by þ c ¼ 0
Take the square root:
DC ¼ 24
Therefore, b ¼ 21 and c ¼ 212
You must set up ratios to solve this question.
(Chapter 9 Lesson 1: Solving Equations)
BD
DC
BC
¼
¼
AE
EC
AC
16. A
Set up equations to represent the infor-
mation:
To solve this problem, you need to find the values of
There are twice as many red jellybeans as orange
AB and ED.
jellybeans: r ¼ 2o
There are three times as many orange jellybeans as
A
yellow jellybeans: o ¼ 3y
x
There are four times as many yellow jellybeans as
B
green jellybeans: y ¼ 4g
15
26
Of all the colors, green is the least common, so start
10
by assuming you have just one green jellybean. If
E
D
C
y
there is 1 green, there are 4(1) ¼ 4 yellows; there
24
are 3(4) ¼ 12 oranges; there are 2(12) ¼ 24 reds. If
there
is
1
green
jellybean,
there
are
1 þ 4 þ 12 þ 24 ¼ 41 jellybeans. The probability that
10
26
Plug in the known values and solve:
¼
you pick a yellow jellybean, choosing at random, is
26 þ x
15
4 out of 41 jellybeans.
Cross-multiply: 10(26 þ x) ¼ 15(26)
Alternatively, this can be solved by assuming
Distribute:
260 þ 10x ¼ 390
you have x green jellybeans. This means you would
Subtract 260:
10x ¼ 130
have 4(x) ¼ 4x yellow jellybeans. You would have
Divide by 10:
x ¼ 13
3(4x) ¼ 12x orange jellybeans, and 2(12x) ¼ 24x red
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