1. No category

Enter keyword(s)
Home | Registration | Test Prep | Scores | College Planning | Financial Aid | Career Planning | Student Blog | FAQs
Sample Tests
English Test
Set 1
Questions / Answers
Set 2
Questions / Answers
Set 3
Questions / Answers
Set 4
Questions / Answers
Set 5
Questions / Answers
Mathematics Test
Set 1
Questions / Answers
Set 2
Questions / Answers
Set 3
Questions / Answers
Set 4
Questions / Answers
Set 5
Questions / Answers
Reading Test
Set 1
Questions / Answers
Set 2
Questions / Answers
Set 3
Questions / Answers
Set 4
Questions / Answers
ACT Assessment Sample Question Answer Key and
Question Explanations
Set 4: Mathematics
1. The correct answer is C.
5a = 25 so a =
= 5; a2 = 5 · 5 = 25
A. If a2 = 5, then a =
B. If a2 = 20, then a =
and 5a = 5
25 so a2 5.
=
and 5a = 5
25 so a2
=
D. If a2 = 125, then a =
and 5a = 5
E. If a2 = 625, then a =
= 25 and 5a = 5(25) = 125
=
20.
25 so a2
25 so a2
125.
625.
2. The correct answer is H.
30% of $1,000,000 is 0.3($1,000,000) = $300,000.
When split equally among 5 group members, each gets ($300,000) = $60,000.
F. If each got $30,000, the group's share would be 5($30,000) = $150,000, but that's only
15% of $1,000,000.
G. If each got $50,000, the group's share would be 5($50,000) = $250,000, but that's only
25% of $1,000,000.
Science Test
Set 1
Questions / Answers
Set 2
Questions / Answers
Set 3
Questions / Answers
Set 4
Questions / Answers
Set 5
Questions / Answers
Writing Test
Sample Essay 1
Sample Essay 2
Sample Essay 3
Sample Essay 4
Return to Test Prep
J. If each got $200,000, the group's share would be 5($200,000) = $1,000,000, but that's
100% of $1,000,000.
K. If each got $300,000, the group's share would be 5($300,000) = $1,500,000, and that's
more than the amount of sales generated.
3. The correct answer is B.
42 – k(4) – 12 = 0, so 4 – 4k = 0, k = 1 when x = 4, x2 – x – 12 = 42 – 4 – 12 = 0.
A. If k = 7, then x2 – 7x – 12 should be 0 when x = 4. But 42 – 7(4) – 12 = –24
C. If k = –1, then x2 + x – 12 should be 0 when x = 4. But 42 + 4 – 12 = 8
0.
0.
Test Descriptions
D. If k = –3, then x2 + 3x – 12 should be 0 when x = 4. But 42 + 3(4) – 12 = 16
0.
E. If k = –7, then x2 + 7x – 12 should be 0 when x = 4. But 42 + 7(4) – 12 = 32
0.
4. The correct answer is H.
=
=
F. Incorrect because
= only when x =
, since
=
G. Incorrect because
= only when x = , since
=
J. Incorrect because
= 6 only when x = –3, since
=
K. Incorrect because
= 8 only when x = – , since
=
=
= =
=6
=
=
=8
5. The correct answer is E.
=
, so 5(u + 4) = (u – 2)15, 5u + 20 = 15u – 30, 10u = 50, u = 5.
Check:
A. If u = –3,
= and
=
=
=
= –1 and
=
= 15 and –1
=
= – and
=
=
15, so –3 doesn't satisfy the
equation.
B. If u = –1,
= 5 and –
5, so –1 doesn't satisfy the
equation.
C. If u = 3,
=
= 5 and
=
=
and 5
, so 3 doesn't satisfy the equation.
D. If u = 4,
=
= and
=
=
and
, so 4 doesn't satisfy the equation.
6. The correct answer is H.
Parallel lines have equal slopes; kx + 2y = 8,
so 2y = –kx + 8, y = x + 4.
18x + ky = 12, so ky = –18x + 12, y =
=
x+
. The slopes must be equal, so
, –k2 = –36, k = ±6. A positive value is required, so k = 6.
F. If k = 0, the lines are 2y = 8 or y = 4, which is horizontal, and 18x = 12 or x = , which is
vertical. A horizontal line cannot be parallel to a vertical line.
G. If k = 2, the lines are 2x + 2y = 8 or y = 4 – x, which has slope –1, and 18x + 2y = 12 or
y = 6 – 9x, which has slope –9. –1 –9, so the lines cannot be parallel.
J. If k = 8, the lines are 8x + 2y = 8 or y = 4 – 4x, which has slope –4, and 18x + 8y = 12 or
y = – x, which has slope – . –4 – , so the lines cannot be parallel.
K. If k = 12, the lines are 12x + 2y = 8 or y = 4 – 6x, which has slope –6, and 18x + 12y =
12 or y = 1 – x, which has slope – . –6 – , so the lines cannot be parallel.
7. The correct answer is E.
A = r2 so A =
2
=
A. In order for to be the area, the radius would have to be
radius is inch, not
inch.
, since
2=
, but the
B. In order for to be the area, the radius would have to be
radius is inch, not
inch.
, since
2=
, but the
C. In order for to be the area, the radius would have to be
radius is inch, not
inch.
, since
2=
, but the
D. In order for to be the area, the radius would have to be
radius is inch, not
inch.
, since
2=
, but the
8. The correct answer is J.
Since sin x = cos(90° – x) for all x, if sin x = cos y, then Y = 90° – X and X + Y = X + (90
X) = 90°. Conversely, if X + Y = 90°, Y = 90° – X, cos Y = cos(90° – X) = cos 90° cos X +
sin 90° sin X = 0 · cos X + 1 · sin X = sin X.
F. If X + Y = 30°, then Y = 30° – X and
sin X = cos(30° – X)
sin X = cos 30° cos X + sin 30° sin X
sin X = cos X + sin X
sin X =
cos X
=
tan X =
, so X = 60°, Y = –30°.
But both X and Y must be between 0° and 90°, so
X + Y 30°.
G. If X + Y = 45°, then Y = 45° – X and
sin X = cos(45° – X)
sin X = cos 45° cos X + sin 45° sin X
sin X =
cos X + sin X
1–
sin X =
cos X
=
tan X =
+ 1, so X = 67.5°, Y = –22.5°.
But both X and Y must be between 0° and 90°,
so X + Y 45°.
H. If X + Y = 60°, then Y = 60° – X and
sin X = cos(60° – X)
sin X = cos 60° cos X + sin 60° sin X
sin X = cos X + sin X
1–
sin X = cos X
=
tan X = 2 + , so X = 75°, Y = –15°.
But both X and Y must be between 0° and 90°, so
X + Y 60°.
K. If X + Y = 135°, then Y = 135° – X and
sin X = cos(135° – X)
sin X = cos 135° cos X + sin 135° sin X
sin X =
cos X + sin X
1–
sin X =
cos X
=
tan X = –( + 1), so X = 112.5°, Y = 22.5°.
But both X and Y must be between 0° and 90°, so
X + Y 135°.
9. The correct answer is A.
The triangle is a 30°-60°-90° triangle so the sides are in the ratio 1:
So =
, 2 = x , x = 2.
:2.
B. If x were , then, according to the ratios for 30°-60°-90° triangles, the hypotenuse
would be 2 feet long. But this is the length of a leg, and the length of the hypotenuse
must be greater than the lengths of the legs.
C–E. Since these values all exceed 2 , none can be correct because each would imply
that 60° is less than 30°, since the smaller angle must be opposite the smaller side.
10. The correct answer is F.
–2x > 4 so – (–2x) < – (4), x < –2 and the graph is all points to the left of –2.
G, H, and K are not correct because each includes x = 0; –2(0) = 0 and 0
J is not correct because J includes x = 3; –2(3) = –6 and –6
4.
4.
11. The correct answer is D.
Lines parallel to 2x – y = 5 have the same slope as 2x – y = 5. The slope-intercept form for
2x – y = 5 is y = 2x – 5, so the slope is 2. The desired line has slope 2 and passes through
(0,0). Its equation is then y = 2x or 2x – y = 0.
A. The line with equation –2x + y = 5 does not go through (0,0), since –2(0) + 0 5.
B. The line with equation 2x + y = 5 does not go through (0,0), since 2(0) + 0 5.
C. The slope of the line with equation x – y = 0 is 1, since the slope-intercept form is y =
So the line with equation x – y = 0 is not parallel to the line 2x – y = 5.
E. The slope of the line with equation 2x + y = 0 is –2, since the slope-intercept form is
y = –2x. So the line with equation 2x + y = 0 is not parallel to the line 2x – y = 5.
© 2006 by ACT, Inc. All rights reserved.
ACT Corporate Home | Contact Us | Site Index