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Sample Tests
English Test
Set 1
Questions / Answers
Set 2
Questions / Answers
Set 3
Questions / Answers
Set 4
Questions / Answers
Set 5
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Mathematics Test
Set 1
Questions / Answers
Set 2
Questions / Answers
Set 3
Questions / Answers
Set 4
Questions / Answers
Set 5
Questions / Answers
ACT Assessment Sample Question Answer Key and
Question Explanations
Set 3: Mathematics
1. The correct answer is A.
B.
C.
D.
E.
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=
=
=
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2. The correct answer is H.
Reading Test
Set 1
Questions / Answers
Set 2
Questions / Answers
Set 3
Questions / Answers
Set 4
Questions / Answers
Science Test
Set 1
Questions / Answers
Set 2
Questions / Answers
Set 3
Questions / Answers
Set 4
Questions / Answers
Set 5
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Writing Test
Sample Essay 1
Sample Essay 2
Sample Essay 3
Sample Essay 4
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AXZ is an exterior angle of XZY.
By the exterior angle theorem, 45o + m( XZY) = 130o, so
m( XZY) = 130o - 45o = 85o.
Or, by definition of supplementary angles, m( YXZ) + 130o = 180o, so m( YXZ) = 180o
130o = 50o.
Then for XZY, the angle sum is 45o + 50o + m( XZY),
and 180o = 95o + m( XZY) and m( XZY) = 180o - 95o = 85o.
F. If m( XZY) were 45o, then XZY would have angle sum
50o + 45o + 45o = 140o 180o so m( XZY) 45o.
G. If m( XZY) were 60o, then XZY would have angle sum
50o + 45o + 60o = 155o 180o so m( XZY) 60o.
J. If m( XZY) were 95o, then XZY would have angle sum
50o + 45o + 95o = 190o 180o so m( XZY) 95o.
Test Descriptions
K. If m( XZY) were 105o, then XZY would have angle sum
50o + 45o + 105o = 200o 180o so m( XZY) 105o.
3. The correct answer is B.
5y = -3x + 10
(5y) = (-3x + 10)
y=- x+2
This is the slope-intercept form, y = mx + b, where m is the slope, so the slope is - .
A. (0,2) is on the line. If the slope were -3, then (1,-1) would be on the line
since
= -3, but 5(-1)
-3(1) + 10 since -5
7. So (1,-1) isn't on the line, and the
slope isn't -3.
C. (0,2) is on the line. If the slope were , then (3,7) would be on the line since
, but 5(7)
-3(3) + 10 since 35
1. So (3,7) isn't on the line, and the slope isn't .
D. (0,2) is on the line. If the slope were 2, then (1,4) would be on the line since
2, but 5(4)
-3(1) + 10 since 20
since 60
=
7. So (1,4) isn't on the line, and the slope isn't 2.
E. (0,2) is on the line. If the slope were 10, then (1,12) would be on the line since
= 10, but 5(12)
=
=
-3(1) + 10
7. So (1,12) isn't on the line, and the slope isn't 10.
4. The correct answer is J.
x2 - 7x + 6 = 6, so x2 - 7x = 0, x(x - 7) = 0, so x = 0 or x = 7.
To check, if x = 0, 0 - 0 + 6 = 6. If x = 7, 72 - 7(7) + 6 = 6.
So 0 and 7 both satisfy x2 - 7x + 6 = 6 and the lesser of the two values is 0.
F, G, H, and K cannot be correct because x2 - 7x + 6 = 6 is a quadratic equation and
quadratic equations have at most two real solutions. Both 0 and 7 have been shown to be
solutions, so no other numbers can be solutions.
5. The correct answer is E.
If x is the number of people, then
is the size of the serving, so
if each serving is
at least cup.
So
4 3x, 75 3x, 25 x, so x 25, and any number of people 25 or fewer can be
served with a serving size of at least cup.
If x > 25, x > (25), x > 18 .
So for more than 25 people to be served the minimum amount of cup, the box would
have to hold more than 18 cups. Therefore, the maximum number of people who could
be served at least cup is 25.
6. The correct answer is K.
If n is one integer, then 34 - n is the other. To maximize n(34 - n), consider y = x(34 - x) =
34x - x2 = -(x2 - 34x) =
-(x2 - 34x + 289) + 289 = -(x - 17)2 + 289.
The graph is a parabola that turns downward and has maximum point (17,289). Applying
this to the task of maximizing n(34 - n) for even integers n, the closest even value to 17 is
16 (or 18), so n = 16 and 34 - n = 18 (or n = 18 and 34 - n = 16). The product is 288. So the
even integers are 16 and 18. Their sum is 34, and their product is maximum.
F-J are not correct because each is less than 288.
7. The correct answer is C.
The graph must have slope 3 and y-intercept 1. This graph has a positive slope and a
positive y-intercept.
A.
The line goes through (0,0), so it can't have y-intercept 1.
B.
The line crosses the y-axis at a point below the x-axis, so the y-intercept
can't be 1.
D.
The line has a negative slope, so the slope can't be 3.
E.
The line has a negative slope, so the slope can't be 3.
8. The correct answer is J.
2x + 3x + 5x = 180, 10x = 180, x = 18, so the smallest angle has measure 2(18) = 36o.
F. If the smallest angle is 18o, the others are 27o and 45o, but then the angle sum of the
triangle is 18o + 27o + 45o = 90o, not 180o.
G. If the smallest angle is 20o, the others are 30o and 50o, but then the angle sum of the
triangle is 20o + 30o + 50o = 100o, not 180o.
H. If the smallest angle is 30o, the others are 45o and 75o, but then the angle sum of the
triangle is 30o + 45o + 75o = 150o, not 180o.
K. If the smallest angle is 40o, the others are 60o and 100o, but then the angle sum of the
triangle is 40o + 60o + 100o = 200o, not 180o.
9. The correct answer is C.
If p% play a musical instrument, then (100 - p)% don't; (100 - p)% is
,
since "%" means "out of 100." The number of students who do not play an instrument is
n or
.
If n = 100 and p = 30, then 30 play an instrument and 70 don't.
A. np = 100(30) = 3,000, not 70
B. .01 np = .01(100)(30) = 30, not 70
D.
=
= -290,000, not 70
E. 100(1 - p)n = 100(1 - 30)(100) = -290,000, not 70
10. The correct answer is G.
If x + 2y = 1, then x = 1 - 2y.
Substituting into 2x + y = 5, 2(1 - 2y) + y = 5,
2 - 4y + y = 5, 2 - 3y = 5, -3y = 3, y = -1.
Then x = 1 - 2(-1) = 1 + 2 = 3.
Check: If x = 3 and y = -1, 3 + 2(-1) = 3 - 2 = 1 and 2(3) + (-1) = 6 - 1 = 5, so the solution
for the system is x = 3 and y = -1 and then x + y = 2.
F, H, J, K. x + 2y = 1 and 2x + y = 5 are different nonparallel lines because the slope of the
first line is - and the slope of the second line is -2. Different nonparallel lines intersect in
at most one point and that point is (3,-1). The sum of any two numbers is unique, so none
of the other answers can be correct.
11. The correct answer is C.
or 60; 63 60.
Or,
is about 7,
is about 9, so
is about 63. So 63 is the best approximation.
12. The correct answer is J.
The maximum value for sin X, where X is any function of x, is 1 so sin 3x 1 for all x.
Then multiplying by 4, 4 sin 3x 4, so the maximum value of 4 sin 3x is 4.
F, G, H. When x = , 4 sin 3x = 4 sin
= 4 sin = 4 1 = 4. Since the values in F-H
are less than 4, none can be the maximum value of 4 sin 3x.
K. If 12 were a value of 4 sin 3x, then 12 = 4 sin 3x for at least one value of x. But then 3 =
sin 3x for at least one value of x. But sin 3x 1 for all x, so 12 can't be a value of 4 sin 3
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