대교 ACT

DIRECTIONS: Solve each problem, choose the correct answer, and then fill in the corresponding oval on your answer document.

Do not linger over problems that take too much time. Solve as many as you can; then return to the others in the time you have left for this test.

You are permitted to use a calculator on this test. You may use your calculator for any problems you choose,

but some of the problems may best be done without using a calculator.

Note: Unless otherwise stated, all of the following should be assumed.

1 Illustrative figures are NOT necessarily drawn to scale.
2. Geometric figures lie in a plane.
3. The word line indicates a straight line.
4. The word average indicates arithmetic mean.

1.	Which of the following is divisible by 3 (with no remainder)? A. 2,725 B. 4,210 C. 4,482 D. 6,203 E. 8,105

2.	A particle travels 1 10 ⁸ centimeters per second in a straight line for 4 10 ^-6 seconds. How many centimeters has it traveled? F. 2.5 10 ² G. 2.5 10 ¹³ H. 4 10 ² J. 4 10 ^-14 K. 4 10 ^-48

3.	The triangle below is isosceles and is drawn to scale. What is the measure of N ? A. 22^o B. 68^o C. 78^o D. 79^o E. 89^o

4.	In the figure below, and is 10 units long. What is the area, in square inches, of ABC ? F. 12.5 G. 25 H. J. 50 K. Cannot be determined from the given information

5.	If 7y = 2 x - 5, then x = ? A. 5y + 5 B. y - 5 C. y + 5 D. E.

6.	Which of the following statements completely describes the solution set for 3(x - 4) = 3 x - 12 ? F. x = 3 only G. x = 0 only H. x = -12 only J. There are no solutions for x. K. All real numbers are solutions for x.

7.	When graphed in the (x,y) coordinate plane, at what point do the lines x + y = 5 and y = 7 intersect? A. (-2,0) B. (-2,7) C. (0,7) D. (2,5) E. (5,7)

8.	The area of a trapezoid is h(b₁ + b₂), where h is the altitude, and b₁ and b₂ are the lengths of the parallel bases. If a trapezoid has an altitude of 5 inches, an area of 55 square inches, and one base 13 inches long, what is the length, in inches, of its other base? F. 9.0 G. 16.8 H. 19.4 J. 45.0 K. 97.0

9.	If you have gone 4.8 miles in 24 minutes, what was your average speed, in miles per hour? A. 5.0 B. 10.0 C. 12.0 D. 19.2 E. 50.0

10.	If a and b are any real numbers such that 0 < a < 1 < b, which of the following must be true of the value of ab ? F. 0 < ab < a G. 0 < ab < 1 H. a < ab < 1 J. a < ab < b K. b < ab

11.	In right triangle ABC below, what is the sine of A ? A. B. C. D. E.

12.	The graph of y = is shown below. Among the following, which is the best representation of y = ? F. J. G. K. H.

Set 2: Mathematics

1. The correct answer is C.

A rule for divisibility by 3 is: if the sum of the digits of a number is divisible by 3, then so is the number; 4 + 4 + 8 + 2 = 18, which is divisible by 3.

A. For 2,725, 2 + 7 + 2 + 5 = 16, which is not divisible by 3.
B. For 4,210, 4 + 2 + 1 + 0 = 7, which is not divisible by 3.
D. For 6,203, 6 + 2 + 0 + 3 = 11, which is not divisible by 3.
E. For 8,105, 8 + 1 + 0 + 5 = 14, which is not divisible by 3.

2. The correct answer is H.

Using D = rt, D = = 4 10 ^{8 - 6} = 4 10 ² cm

F. If D = 2.5 10 ², then t = = 2.5 10 ^-6, but the particle traveled for 4 10 ^-6 sec.

G. If D = 2.5 10 ¹³, then t = = 2.5 10 ⁵, but the particle traveled for 4 10 ^-6 sec.

J. If D = 4 10 ^-14, then t = = 4 10 ^-22, but the particle traveled for 4 10 ^-6 sec.

K. If D = 4 10 ^-48, then t = = 4 10 ^-56, but the particle traveled for 4 10 ^-6 sec.

3. The correct answer is D.

From the figure, , so M N as base angles of an isosceles triangle.
Then, 22^o + 2m(N) = 180^o; m(N) = = 79^o.

A. If the measure of N were 22^o, then N and L are the congruent base angles of the isosceles triangle. This means , which is clearly not the case from the figure.
B. If the measure of N were 68^o, the measure of M would also be 68^o and the angle sum for MLN would be 68^o + 68^o + 22^o = 158^o. But the angle sum must be 180^o.
C. If the measure of N were 78^o, the measure of M would also be 78^o and the angle sum for MLN would be 78^o + 78^o + 22^o = 178^o. But the angle sum must be 180^o.
E. If the measure of N were 89^o, the measure of M would also be 89^o and the angle sum for MLN would be 89^o + 89^o + 22^o = 200^o. But the angle sum must be 180^o.

4. The correct answer is K.

It is impossible to find the area of the triangle without more information, because there are infinitely many isosceles triangles having a base that is 10 units long.

F. If AD = 2.5, then the area is (10)(2.5) or 12.5, but the length of is not given as 2.5.

G. If AD = 5, then the area is (10)(5) or 25, but the length of is not given as 5.
H. If AD = 5, then the area is (10)(5) or 25, but the length of is not given as 5.

J. If AD = 10, then the area is (10)(10) or 50, but the length of is not given as 10.

5. The correct answer is E.

7y = 2x - 5, so 2x = 7y + 5, x = .

A. If x = 5y + 5, then 2x - 5 = 2(5y + 5) - 5 = 10y + 10 - 5 = 10y + 5,
but 10y + 5 7y for all y, so x 5y + 5.

B. If x = y - 5, then 2x - 5 = 2 - 5 = 7y - 10 - 5 = 7y - 15,
but 7y - 15 7y for all y, so x y - 5.

C. If x = y + 5, then 2x - 5 = 2 - 5 = 7y + 10 - 5 = 7y + 5,
but 7y + 5 7y for all y, so x y + 5.

D. If x = , then 2x - 5 = 2 - 5 = 7y - 5 - 5 = 7y - 10,
but 7y - 10 7y for all y, so x .

6. The correct answer is K.

3(x - 4) = 3x - 12 so the equation becomes 3x - 12 = 3x - 12, which is true for all real numbers.

F. 3 is not the only solution because 0 is also a solution; if x = 0, 3(0 - 4) = 3(-4) = -12
and 3(0) - 12 = -12.
G. 0 is not the only solution because 3 is also a solution; if x = 3, 3(3 - 4) = 3(-1) = -3
and 3(3) - 12 = 9 - 12 = -3.
H. -12 is not the only solution because 3 is also a solution; if x = 3, 3(3 - 4) = 3(-1) = -3
and 3(3) - 12 = 9 - 12 = -3.
J. Since 3(0 - 4) = 3(-4) = -12 and 3(0) - 12 = -12, x = 0 is a solution so 3(x - 4) = 3x - 12 has at least one solution.

7. The correct answer is B.

If x + y = 5 and y = 7, then x + 7 = 5 and x = -2. The point of intersection is (-2,7).
Check: -2 + 7 = 5 and 7 = 7, so (-2,7) is on both lines.

A, C, D, E cannot be correct since two distinct lines intersect in at most one point,
and (-2,7) is the point of intersection, so none of the points in A, C, D, and E can be on both lines.

8. The correct answer is F.

Substituting into the formula, 55 = (5)(13 + b), so 110 = 5(13 + b), 110 = 65 + 5 b, 5 b = 45, b = 9.

G. If the other base were 16.8, the area would be (5)(13 + 16.8) = 74.5, not 55.
H. If the other base were 19.4, the area would be (5)(13 + 19.4) = 81, not 55.
J. If the other base were 45.0, the area would be (5)(13 + 45) = 145, not 55.
K. If the other base were 97.0, the area would be (5)(13 + 97) = 275, not 55.

9. The correct answer is C.

t = 24 min = = hr

A. If r = 5, then d = rt = 5 = 2, not 4.8
B. If r = 10, then d = rt = 10 = 4, not 4.8
D. If r = 19.2, then d = rt = 19.2 = 7.68, not 4.8
E. If r = 50, then d = rt = 50 = 20, not 4.8

10. The correct answer is J.

If all members of 0 < a < 1 < b are multiplied by a, which is positive since 0 < a,
the inequality is 0 < a²< a < ab, so a< ab.
If all members of 0 < a < 1 < b are multiplied by b, which is also positive since 0 < b by transitivity,
the inequality is 0 < ab < b < b², so ab < b.
Combining gives a < ab < b.

Let a = , b = 2. These are valid values since 0 < < 1 < 2. Then ab = 1.

F. 1 , so F isn't true.
G. 1 1, so G isn't true.
H. 1 1, so H isn't true.
K. 2 1, so K isn't true.

11. The correct answer is D.

, then sin A = .
For a given right triangle, the sine of each angle is unique, so sin A has no other value.

A. sin A because , since 7(25) = 175, (24)(24) = 576 and 175 576.
B. sin A because .
C. sin A because > 1 and the sine ratio never exceeds 1.
E. sin A because > 1 and the sine ratio never exceeds 1.

12. The correct answer is K.

When x > 7, = x - 7, so the graph of y = will be identical to the graph of y = for all x > 7. This eliminates G

and J.

When x < 7, = -(x - 7), so the graph of y = will be the reflection about the x-axis of the graph of y = for all x < 7. The graph in F

is identical to the graph of y = for all x, so F can be eliminated. The graph in H

is the original graph reflected about the line x = 7 for x < 7, so H can be eliminated. The graph in K

is identical to the original graph for x > 7 and is the reflection about the x-axis of the original graph for x < 7.