Algebra Released Questions 42-64

Standard 19.0
42. Toni is solving this equation by completing the
square. ax2 +bx +c = 0 (where a≥ 0)
• Step 1: ax2 +bx = - c
•
•
•
Step 2: x2 + b/ax = - c/a
Step 3: ?
Which should be Step 3 in the solution?
a) x 2   c  b x
b a
b) x  b   c
a
ax
b
b
c
b
c) x  x 
 
a
2a
a 2a
2
d)
2
b
c  b 
 b 
x  x
   

a
a  2a 
 2a 
2
2
• The key to answering this question is
knowing the formula for “completing the
square”
• The next step in completing the square is
2
b
to add   to both sides of the equation.
 2a 
• By doing this, you will form a perfect
square on the left side of the equation.
• At this point, solve the problem for the
question is asking for step 3. ∴ “d” is the
answer since it is the only one adding
(b/2)2 to each side.
• The next two steps are to find the square
root of each side and solve the equation.
Standard 19.0
43.Four steps derive the quadratic formula are
shown below.
What is the correct order for these steps?
a) I, IV, II, III
b) I, III, IV, II
c) II, IV, I, III
d) II, III, I, IV
• You must first look over what they give you
and what they ask you. The question is
telling you that the quadratic formula is
where you end up at.
• The question is also helping you on the test.
Notice steps I and IV were in the previous
question for “completing the square”. Use
this to your advantage. Knowing those two
steps and knowing the actual formula of
•  b  b 2  4ac is very close to step III (or
2a
• where you want to end up at) you can
assume step II is 3rd. ∴ “a” is the answer
Standard 20.0
44. Which is one of the solutions to the
equation 2x2 – x – 4 = 0?
a)
c)
1
 33
4
1 33
4
b)
1
  33
4
d)
 1 33
4
• Using the quadratic formula, solve.
2
2x  x  4  0
2
 ( 1)  1  (4)(2)( 4)
(2)(2)
1  1  ( 8)( 4)
4
1  1  32
4
• ∴ “c” is the answer
1  33
4
Standard 20.0
45. Which state best explains why there is
no real solution to the quadratic
equation
2x2 + x + 7 = 0
a)
b)
c)
d)
The value of 12- 4 • 2 • 7 is positive.
The value of 12- 4 • 2 • 7 is equal to 0.
The value of 12- 4 • 2 • 7 is negative.
The value of 12- 4 • 2 • 7 is not a perfect
square.
•
•
•
•
•
1st, this is a tricky question so you want to take it
slow. You are dealing with a quadratic equation
which means they are asking for what is inside
the radical sign “√ “ in the quadratic formula.
Taking what is inside “b²- 4ac”, you plug in the
numbers given to solve. This step has already
been done for you in the choices given.
Begin eliminating what you know. If you find
the value of the numbers given, “a” and “b” are
false statements ruling them out.
“c” and “d” are true statements (answer is -55).
Now, if the discriminant is a negative number
your answer is always no real solution.
∴ “c” is the answer
Standard 20.0
46. What is the solution set of the quadratic
equation 8x2 + 2x + 1 = 0?
a)
c)
1 1 
 , 
2 4
b)
1
 1  7 1  7 
,


8
8


d) No real solution
2 ,1  2

• Using the quadratic formula, you would
plug in the values of “a”, “b”, and “c”.
•When you begin to solve you will notice the
discriminant (b² - 4ac) is negative, this
equation has no solution in the real number
system.
 2  2 2  ( 4)(8)(1)
( 2)(8)
2
16
∴ “d” is the answer
4  32
 2   28
16
Standard 21.0
47.The graph of the equation y = x2 – 3x – 4 is
shown below.
For what value of
x is y = 0?
a)
b)
c)
d)
x = -1 only
x = -4 only
x = -1 and x = 4
x = 1 and x = - 4
• The question is simply asking for the “x”
value if y = 0. In this case, the parabola
intersects “x” in two places. Both places
will be the value of “x”.
• Since the line intersects “x” in two places
look at “c” and “d” first. At points (-1, 0)
and (4, 0), notice “y” equals 0 and the line
intersects “x”.
• Therefore (∴) x = -1 and x = 4 and your
Standard 21.0
48. Which best represents the graph of
y = - x2 + 3?
•
•
If the coefficient of x² is positive, the parabola
(the “u” shaped line) opens upward, if it is
negative it opens downward. In this case,
its negative so “c” and “d” are eliminated.
Finding the vertex (curve of the parabola).
b
2

y  ax  bx  c
2a
2
0
y  1(0)  3

2(1)
y3
x0
• In this type of problem, if you are missing
the “bx”, then “c” (3) will be your yintercept.
• ∴ “b” is the answer
Standard 22.0
49. How many times does the graph of y =
2x2 – 2x + 3 intersect the x-axis?
a)
b)
c)
d)
none
one
two
three
2
2

2( 2)
2
4
1
x
2
y
y
y
y
1
1
2   2   3
2
2
1
2   1  3
4
1
1 3
2
1
2
2
• With no negative sign in front of the
coefficient 2 (2x2 – 2x + 3) the parabola opens
upward. With “y” being a positive number,
the parabola will not intersect the x-axis.
• ∴ “a” is the answer
Standard 23.0
50. An object that is projected straight
downward with initial velocity v feet per
second travels a distances s = vt + 16t2,
where t = time in seconds. If Ramon is
standing on a balcony 84 feet above the
ground and throws a penny straight down
with an initial velocity of 10 feet per second,
in how many seconds will it reach the
ground?
a) 2 seconds
b) 3 seconds
c) 6 seconds
d) 8 seconds
• They give you the formula s= vt + 16t².
• You need to read and dissect the problem
and place it into the formula.
• “s” = 84 feet. Remember it tells you “t”
represents seconds.
• “vt” = 10 ft per second 10/1 (“t” is unknown)
• “16t²” = unknown seconds that the “vt”
increased by time it hits the ground.
8t  21  0
8t  21
0  16t 2  10t  84
5
0  2(8t 2  5t  42)
t  2
8
0  2(8t  21)(t  2)
84  10t  16t 2
t 20
t 22  02
t2
• ∴ , since you don’t go back in time (negative
number), your answer is 2 seconds or “a”.
Standard 23.0
51. The height of a triangle is 4 inches
greater than twice its base. The area of
the triangle is 168 square inches. What is
the base of the triangle?
a) 7 in.
b) 8 in.
c) 12 in.
d) 14 in.
• Using the formula A=½bh, plug in the given
to solve
• Set it up as a quadratic equation equal to 0.
1
168  b(4  2b)
2
2  168  b(2b  4)
336  2b 2  4b
0  2b 2  4b  336
0  2(b 2  2b  168)
0  2(b  14)(b  12)
b  14  0
b  14  14  14  0
b  14
b  12  0
b  12  12  12  0
b  12
• -14 is not an answer nor does it make sense
to be an answer. 12, however is given.
• ∴ “c” is the answer
Standard 12.0
52. What is x 2  4 xy  4 y 2 reduced to
2
3
xy

6
y
lowest terms.
a)
x  2y
3
b)
x  2y
3y
c)
x  2y
3
d)
x  2y
3y
x  4 xy  4 y
3 xy  6 y 2
x  2 y x  2 y 
3 yx  2 y 
x  2 y 
3y
2
2
•1st, factor the numerator.
•Next, reduce the denominator by factoring out what
is common, 3y.
•Finally, reduce what you have in common in the
numerator and denominator.
Standard 12.0
53. Simplify
terms.
a)
3 x  1
2x 1
b)
3 x  3
2x 1
c)
32 x  3
4 x  1
d)
3 x  3
2x 1
6 x 2  21x  9 to lowest
4x2 1
6 x  21x  9
2
4x 1
2
3( 2 x  7 x  3)
2 x  12 x  1
3( 2 x  1)( x  3)
2 x  12 x  1
3( x  3)
( 2 x  1)
2
•After factoring you are left with
Standard 12.0
54. What is
terms.
a)
c)
x2
x 1
x2
x 1
2
x  4 x  4 reduced to lowest
2
x  3x  2
b)
d)
x2
x 1
x2
x 1
• UNFOIL or factor the numerator and
denominator
• Reduce
• ∴ “a” is the answer
x  4x  4
2
x  3x  2
 x  2  x  2 
 x  2  x  1
x  2
( x  1)
2
Standard 13.0
55.
7z  7z
z 4
 3
2
4z  8
z  2z  z
2
a) 7 z  2 
4( z  1)
c)
7 z ( z  1)
4( z  2)
2
b)
7 ( z  2)
4( z  1)
d) 7 z ( z  1)
4( z  2)
• Factor, reduce, then multiply the
(numerator •numerator) and the
(denominator • denominator)
7 z

 7z
( z  4)
 3
2
(4 z  8) ( z  2 z  z )
7 z ( z  1) ( z  2)(z  2)

4( z  2) z ( z  1)(z  1)
2
2
7 z  2 
4( z  1)
• ∴ “a” is the answer
Standard 13.0
56. Which fraction equals the product?
 x  5  2 x  3 



 3x  2  x  5 
a)
c)
2x  3
3x  2
x  25
2
6 x  5x  6
2
b)
d)
3x  2
4x  3
2 x  7 x  15
2
3x  13x  10
2
• To solve this problem you use FOIL for the
numerators and denominators
• Numerators = (x + 5)(2x – 3)
• Denominator = (3x + 2)(x – 5)
•
Numerator
•
(x + 5)(2x – 3)
• 2x² - 3x + 10x – 15
•
2x² + 7x – 15
•
∴ “d” is the answer
Denominator
(3x + 2)(x – 5)
3x² - 15x + 2x - 10
3x² - 13x - 10
Standard 13.0
57.
x  8 x  16 2 x  8
 2
x3
x 9
2
2( x  4)
a)
2
( x  3)(x  3)
2
c)
( x  4)( x  3)
2
b)
d)
2( x  3)( x  3)
x4
( x  4)(x  3)
2( x  3)
2
x
2

 8 x  16
2x  8
 2
( x  3)
x 9
x  4( x  4) 
( x  3)
2( x  4)
( x  3)(x  3)
x  4( x  4)  ( x  3)(x  3)
( x  3)
2( x  4)
( x  4)( x  3)
2
∴ “c” is the answer.
Standard 15.0
58. A pharmacist mixed some 10% saline
solution with some 15% saline solution
to obtain 100 mL of a 12% saline
solution. How much of the 10% saline
solution did the pharmacist use in the
mixture?
a)
b)
c)
d)
60 mL
45 mL
40 mL
25 mL
• Create a chart to keep track of all the
information then set up equation to solve.
Amount of
Solution (mL)
Percent
Saline
Amount of
Saline
10% Solution
x
10%
.1x
15% Solution
100 – x
15%
15 - .15x
12% Solution
100
12%
12
15  .15x  .1x  12
15  .05x  12
15  15  .05x  12  15
• The answer is 60mL
• ∴ “a” is the answer
 .05x  3
 .05x
3

 .05x  .05
x  60
Standard 15.0
59. Andy’s average driving speed for a 4hour trip was 45 miles per hour. During
the first 3 hours he drove 40 miles per
hour. What was his average speed for
the last hour of his trip?
a)
b)
c)
d)
50 miles per hour
60 miles per hour
65 miles per hour
70 miles per hour
• Setup a chart to keep track of the given.
Part of trip
Rate
Time
Distance
First 3 hours
40
3
120 miles
Total Trip
45
4
180 miles
• Next, use what you know. The total trip is
180 miles by using Rate •Time = Distance
(rt=d). Also using the same formula, Andy
drove for 120 miles. To find the average
speed for miles per hour you subtract the
two distances.
(45  4)  (40  3)  x
180 120  x
60  x
• ∴ “b” is the answer
Standard 15.0
60. One pipe can fill a tank in 20 minutes,
while another takes 30 minutes to fill the
same tank. How long would it take the
two pipes together to fill the tank?
a)
b)
c)
d)
50 min
25 min
15 min
12 min
• 1st, reduce your choices with what you
know. You can eliminate “a” and “b” since
one pipe alone can fill up the tank in 20
minutes, thus two tanks will not take as
long. Then set up as fractions.
1 t ank
1 t ank

x
20 minut es 30 minut es
 1 3  1 2
 
 x

 20 3   30 2 
3
2

x
60 60
5
1
 Reduce 
x
60
12
• 1 tank per 12 minutes. ∴ “d” is the answer
Standard 15.0
61.Two airplanes left the same airport
traveling in opposite directions. If one
airplane averages 400 miles per hour and
the other airplane averages 250 miles per
hour, in how many hours will the distance
between the two planes be 1625 miles?
a)
b)
c)
d)
2.5
4
5
10.8
•
In finding distance the formula is rt=d (rate •
time = distance)
Rate
Time
Distance
Airplane 1
400
t
400t
Airplane 2
250
t
250t
400t  250t  1625
650t  1625
650t 1625

650 650
t  2 .5
•
2.5 hours would be your answer. ∴ “a” is the
Standard 16.0
62. Which relation is a function?
a)
b)
c)
d)
{(-1, 3), (-2, 6), (0, 0), (-2,-2)}
{(-2, -2), (0, 0), (1, 1), ( 2, 2)}
{(4, 0), (4, 1), (4, 2), (4, 3)}
{(7, 4), (8, 8), (10, 8), (10, 10)}
• Knowing what a “relation” and a “function”
are is key.
• A relation is grouping of inputs (x) with
outputs (y) normally shown as ordered
pairs (x, y).
• A function is a relation however, each
input has only one output. No inputs will
be used twice.
• Notice in “a”, -2 is used twice, “c” 4 is used
more than once, and “d” 10 is used twice.
Only “b” does not have a repeating input.
Perform the vertical line test to verify.
• ∴ “b” is the answer
Standard 17.0
63. For which equation graphed below are
all the y-values negative?
• This is the type of problem you want to do
quickly and not over think it.
• Notice the question asked for the y-values
that are all negative. To solve, you look at
the lines in the graphs. In short, if the line
goes above or on the x-axis it is not the
answer. Thus, answers “b”, “c”, and “d” all
go past the x-axis. When they do this y≥ 0
(y is greater than or equal to zero) making
those three choices incorrect.
• ∴ “a” is the answer
Standard 17.0
64. What is the domain of the function shown
on the graph below?
a)
b)
c)
d)
{-1, -2, -3, -4}
{-1, -2, -4, -5}
{1, 2, 3, 4}
{1, 2, 4, 5}
• This problem is clearly a trick trying to
determine if you know the difference
between domain (it’s “x”) and range (it’s
“y”). (x, y) or (domain, range)
• Now, knowing that domain is “x”, and all
the points are to the right of zero (positive
numbers), answers “a” and “b” are
eliminated (negative numbers)
• Then use what you see. There is a point at
(5, -2). Answer “c” has no 5 while “d” does.
• ∴ “d” is the answer