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1
Practice Calculus Readiness Test
Instructions:
• Read each problem carefully. Then work the problem on a separate sheet
of pap er and click on the box next to the correct choice. If you change your
mind, just click on a different choice.
• Use the navigational buttons at the bottom of e ach page to go to the next
or previous page.
• A calculator is not required for any questions on this test.
• This practice test consists of 25 problems. Click on “Begin Quiz”, then
begin.
1. Money in a bank triples every 8 years. If $100 is deposited today, what
will its value be after 32 years?
$8, 500 $8, 100 $1, 600 $400
2. The y-coordinate of the point of intersection of the graph of
−x + 4y = −50 and x + y = 20 is
6 0 −14 −6
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Begin Quiz
2
3. The rectangular box shown below has a square base and a closed top.
The height is twice the length of one side of the base. Its surface area
in terms of x is
20x 8x + 2x
2
10x
2
6x
4. If 2
13
is approximately equal to 8000, then, of the following, which best
approximates 2
26
?
640, 000 6, 400, 000 64, 000, 000 8000
13
5. 2
−5
· 64
2/3
=
512
1
512
1
1
2
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3
6. If f is a function whose graph is the parabola sketched below then
f(x) < 0 whenever
x < 1 or x > 3 x < 1
x > 3 1 < x < 3
7. If log
2
(x − 6) = 6 then x =
70 64 58
6
log
2
6
+ 6
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4
8. A function f is even if f(−x) = f(x) for each x in the domain of f . Of
the following, which best represents the graph of an even function?
(a) (b) (c) (d)
9. If
(2x − 3)(x + 5)
x − 7
= 0 then x =
5, 7, −
3
2
5 or
3
2
−5, 7, or
3
2
−5 or
3
2
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5
10. Of the following, which best represents the graph of x
2
+ y
2
−2y = 0?
(a) (b) (c) (d)
11. If f(x) =
5x+3
2x+3
then f(n + 1) =
8
5
5n + 3
2n + 3
+ 1
5n + 8
2n + 5
5n + 4
2n + 4
12. The slope of the line that goes through the points (−5, 4) and (3, −12)
is
−
1
2
8 −2 4
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6
13. Find all solutions to the equation 3x
2
= 4x + 1.
4/3, 1/3
2+
√
7
3
,
2−
√
7
3
4+3
√
2
6
,
4−3
√
2
6
2+
√
2
3
,
2−
√
2
3
14. In a standard coordinate system, the graph of the equation y = −3x+7
is
a line falling to the right a line rising to the right
a horizontal line not a line
15. The inequality |x − 4| ≤ 8 is equivalent to
−4 ≤ x ≤ 12 −12 ≤ x ≤ 4
−12 ≤ x ≤ 12 x ≤ 12
16. The quantity a − b is a factor of how many of the following?
a
2
− b
2
a
2
+ b
2
a
3
− b
3
a
3
+ b
3
one only two only three only four
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7
17. In the figure shown below, what is the distance between the points P
and Q?
11 7 6 5
18. The length of a c ertain rectangle is 6 meters more than twice its width.
What is the perime ter of the rectangle if the area of the rectangle is
260 square meters?
54 meters 60 meters 66 meters 72 meters
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8
19. What is the area of the rectangle s hown in the figure below? (Note:
The figure is not drawn to scale.)
3 27 31 39
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9
20. A rectangle R has width x and length y. A rectangle S is formed from
R by multiplying each of the sides of the rectangle R by 4 as s hown in
the figure below. What is the area of the portion of S lying outside R?
(Note: The figure is not drawn to scale.)
16xy 15xy 4xy x
4
y
4
21. What is the radian measure of an angle whose degree measure is 240
◦
?
π
3
2π
3
3π
4
4π
3
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10
22. csc(30
◦
) =
2
2
√
3
√
2
√
3
23. For which values of x in the interval 0 ≤ x ≤ 2π does
(sin x − 1)(sin x − 5) = 0?
π
2
only 1 and 5 π 0 and 2π
24. In the figure below, if sin R =
5
8
and r = 2, then what is q?
16
5
5
4
5
5
16
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11
25. In the right triangle shown in the figure below, tan θ =
x x
√
x
2
− 1 x
2
+ 1
√
x
2
− 1
• Click on “End Quiz” to have the computer grade your test. Then click
on “Correct My Answers” to see which questions you got wrong.
• Click on the green dots to see detailed solutions for each problem.
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End Quiz
Score:
Solutions to Practice Tests 12
Solutions to Practice Calculus Readiness Test
Solution to Question 1: In 8 years the value will be 3×100 = 300. Eight
years later (after a total of 16 years), the value will be 3 ×300 = 900. Eight
years after that (after a total of 24 years) the value will be 3 ×900 = 2700.
And eight years after that (after a total of 32 years) the value will be
3 × 2700 = 8100. Return
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Solutions to Practice Tests 13
Solution to Question 2: Adding the two equations we obtain 5y = −30
or y = −6. Return
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Solutions to Practice Tests 14
Solution to Question 3: The areas of the top and bottom are each
x ×x = x
2
. The area of each of the four sides is x ×2x = 2x
2
. So the total
surface area is 2(x
2
) + 4(2x
2
) = 2x
2
+ 8x
2
= 10x
2
. Return
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Solutions to Practice Tests 15
Solution to Question 4: 2
26
=
2
13
2
= (8000)
2
= 64, 000, 000.
Return
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Solutions to Practice Tests 16
Solution to Question 5: 2
−5
=
1
2
5
=
1
32
and 64
2/3
=
3
√
64
2
= (4)
2
=
16, so 2
−5
· 64
2/3
=
1
32
· 16 =
16
32
=
1
2
. Return
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Solutions to Practice Tests 17
Solution to Question 6: The graph is below the x axis for 1 < x < 3.
Return
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Solutions to Practice Tests 18
Solution to Question 7: log
2
(x − 6) = 6 ⇒ 2
6
= x − 6 ⇒ x = 2
6
+ 6 =
64 + 6 = 70. Return
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Solutions to Practice Tests 19
Solution to Question 8: If f (x) = f(−x) for all x, the the graph of f
must be symmetric with respect to the y axis. The only graph that has
this symmetry is (c). Return
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Solutions to Practice Tests 20
Solution to Question 9:
(2x − 3)(x + 5)
x − 7
= 0 ⇒ 2x − 3 = 0 or x + 5 =
0 ⇒ x =
3
2
or x = −5. Return
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Solutions to Practice Tests 21
Solution to Question 10: Any equation of the form Ax
2
+ Ay
2
+ Bx +
Cy+D = 0 is a circle, which narrows the choices to (a) and (c). Completing
the square, we see that x
2
+ y
2
− 2y + 1 = 1 ⇒ x
2
+ (y − 1)
2
= 1. We
know that (x −h)
2
+ (y −k)
2
= r
2
describes a circle with center (h, k) and
radius r. Hence our circle has radius (0, 1) and radius 1, which is choice
(c). Return
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Solutions to Practice Tests 22
Solution to Question 11:
f(n + 1) =
5(n + 1) + 3
2(n + 1) + 3
=
5n + 5 + 3
2n + 2 + 3
=
5n + 8
2n + 5
Return
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Solutions to Practice Tests 23
Solution to Question 12: The slope is
y
2
− y
1
x
2
− x
1
=
−12 − 4
3 − (−5)
=
−16
8
= −2
Return
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Solutions to Practice Tests 24
Solution to Question 13: We use the quadratic formula to solve the
equation 3x
2
− 4x − 1 = 0 with a = 3, b = −4, and c = −1. So
x =
−b ±
√
b
2
− 4ac
2a
=
4 ±
√
16 + 12
6
=
4 ±
√
28
6
=
4 ± 2
√
7
6
=
2 ±
√
7
3
So the solutions are
2+
√
7
3
and
2−
√
7
3
. Return
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Solutions to Practice Tests 25
Solution to Question 14: The graph of y = mx + b is a straight line.
Since the slope m = −3 is negative, the graph of this line is falling to the
right. Return
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Solutions to Practice Tests 26
Solution to Question 15:
|x − 4| ≤ 8 ⇒ −8 ≤ x − 4 ≤ 8 ⇒ (−8) + 4 ≤ x ≤ 8 + 4 ⇒ −4 ≤ x ≤ 12
Return
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Solutions to Practice Tests 27
Solution to Question 16: Recall that a
2
−b
2
= (a −b )(a + b), a
3
−b
3
=
(a − b)(a
2
+ ab + b
2
), and a
3
+ b
3
= (a + b)(a
2
−ab + b
2
) (a
2
+ b
2
does not
factor). So a + b is a factor of a
2
− b
2
and a
3
+ b
3
only. Return
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Solutions to Practice Tests 28
Solution to Question 17: The distance from P to R is 1 − (−2) = 3,
and the distance from R to Q is 7 −3 = 4. By the Pythagorean Theorem,
the distance from P to Q is
√
3
2
+ 4
2
=
√
9 + 16 =
√
25 = 5. Return
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Solutions to Practice Tests 29
Solution to Question 18: If we let w be the width of the rectangle, then
its length is l = 2w + 6, so the area is l × w = (2w + 6)w = 2w
2
+ 6w.
Since the area is 260, we have 2w
2
+ 6w = 260 ⇒ 2w
2
+ 6w − 260 = 0 ⇒
w
2
+ 3w − 130 = 0 ⇒ (w + 13)(w − 10) = 0 ⇒ w = 10, −13. Since w must
be positive , the width of the rectangle must be 10 m ete rs, so the length
is 2w + 6 = 26 meters, and so the perimeter is 2l + 2w = 20 + 52 = 72
meters. Return
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Solutions to Practice Tests 30
Solution to Question 19: The width of the rectangle is 5 − 2 = 3. The
height of the rectangle is f(2) = 2
3
− 2 + 7 = 8 − 2 + 7 = 13. So the area
is 3 × 13 = 39. Return
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Solutions to Practice Tests 31
Solution to Question 20: The area of rectangle S is (4x)(4y) = 16xy.
The area of rectangle R is xy. So the area of the portion of S lying outside
R is 16xy − xy = 15xy. Return
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Solutions to Practice Tests 32
Solution to Question 21: Since π radians equals 180 degrees, we convert
from degrees to radians by multiplying by
π
180
:
240 ·
π
180
=
4π
3
Return
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Solutions to Practice Tests 33
Solution to Question 22: Recall that sin(30
◦
) = 1/2, so csc(30
◦
) =
1
1/2
= 2 Return
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Solutions to Practice Tests 34
Solution to Question 23:
(sin x − 1)(sin x − 5) = 0 ⇒ sin x = 1 or sin x = 5
Since sin x is always b e tween −1 and 1, there are no values of x for
which sin x = 5. The only value of x in the interval 0 ≤ x ≤ 2π for w hich
sin x = 1 is x =
π
2
. Return
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Solutions to Practice Tests 35
Solution to Question 24:
5
8
= sin R =
r
q
=
2
q
⇒ 16 = 5q ⇒ q =
16
5
Return
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Solutions to Practice Tests 36
Solution to Question 25: By the Pythagorean Theorem, the third side of
the triangle has length
√
x
2
− 1. So tan θ =
opposite
adjacent
=
√
x
2
−1
1
=
√
x
2
− 1.
Return
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