1998 Income Taxes
[1] UIL HIGH SCHOOLMATHEMATICS MAGIC : BOOK 4
BASIC MATHEMATICS
6. The speed zone in front of Turtle High School is 25 mph. How many feet per second is that?
(A) 17 [pic] (B) 22 (C) 27 [pic] (D) 36 [pic] (E) 41 [pic]
Solution : [pic] = 36 [pic]
Note : To convert miles per hour to feet per second, multiply the miles
per hour by [pic].
25[pic] = [pic] = [pic] = [pic] = 36 [pic]
8. The average of five numbers is 55. If two of the numbers are each decreased by 5, two of the
others are each increased by 11 and the remaining number stays the same, the new average is :
(A) 52.6 (B) 53.8 (C) 55 (D) 56.2 (E) 57.4
Total of five numbers = 5(55) = 275
New average = [pic] = 57.4
10. Coach Friesan’s calculator team made scores of 350, 322, 294, and 266 on a practice test. The
difference in the mean and the median of his team’s scores was :
(A) 0 (B) 1 (C) 4 (D) 5 (E) 9
Solution : mean = [pic] = 308
median = [pic] = 308
mean - median = 308 - 308 = 0
18. The sum of 36 consecutive odd integers is 6[pic]. What is the median?
(A) 1296 (B) 864 (C) 216 (D) 108 (E) 36
Since there are 36 integers the median is the average of the two middle terms (average of 18th and
19th terms). Since the integers are consecutive odd integers, the medians will be the average of
the integers.
Solution : [pic] = 216
19. If four days before tomorrow is Thursday, what is three days after yesterday?
(A) Saturday (B) Sunday (C) Monday (D) Tuesday (E) Wednesday
Solution : Start by labeling a location called “today”. To the right of this label “tomorrow” and
to the left of today, label “yesterday”. Now read the problem to label other locations.
[pic] [pic] [pic]
Locate a spot that is four days before tomorrow and call this location Thursday.
[pic] [pic] [pic] [pic] [pic]
Locate a spot that is three days after yesterday.
[pic] [pic] [pic] [pic] [pic] [pic]
Begin labeling the days beginning with Thursday.
[pic] [pic] [pic] [pic] [pic] [pic]
[pic]
Notice that the day you are looking for is five days after Thursday, which is
Tuesday.
30. If a - b = 16, then b - a - 3 equals :
(A) - 19 (B) 4 (C) - 18 (D) - 12 (E) None of these
If a - b = 16, then - (a - b) = b - a = - 16.
B - A - 3 = - 16 - 3 = - 19
31. The sum of three consecutive odd integers is always divisible by
(A) 2 (B) 3 (C) 5 (D) 6 (E) 2 and 3
Let x = the smallest odd integer.
x + (x + 2) + (x + 4) = 3x + 6 = 3(x + 6)
Since the sum of any three consecutive odd integers is 3(x + 6), then this sum is divisible by 3.
ALGEBRA 1
4. Candy Kaine wants to mix some sour jawbreakers worth 75¢ a pound with some sweet jawbreakers
worth $1.25 a pound. How many pounds of sweet jawbreakers will she need to make a 10 pound
mixture of sweet and sour jawbreakers that sell for $1,00 a pound?
(A) 5 (B) 10 (C) 15 (D) 20 (E) 25
Let x = number of pounds of sweet jawbreakers needed ; 20 - x = number of pounds of sour
Jawbreakers needed.
.75(20 - x) + 1.25(x) = 1.00(20)
15 - .75x + 1.25x = 20 ; .5x = 5 ; x = [pic] = 10
5. The roots of the equation 2x[pic] - x[pic] - 5x - 2 = 0 are – 1, - .5, and R. Find R.
(A) 2 (B) 1 (C) .5 (D) - 1 (E) - 2
Note : If ax[pic] + bx[pic] + cx + d = 0, then the product of the roots is - [pic].
Solution : - [pic] = (- 1)(- .5)R ; - 1 = .5R ; R = 2
22. Les Doe applies for a 24 month bank loan in the amount of $2000. He borrows three-fifths of it at
5% simple interest and the rest at 3% simple interest. How much is his total payback at the end of
the 24 month period?
(A) $2168.00 (B) $2873.60 (C) $2120.00 (D) $2400.40 (E) $2473.50
Solution : I = PRT
I = [pic] + [pic] = 168
Total payback = $2000 + $168 = $2168.00
24. If 2y - [pic] = [pic] + 7y, then y equals __________.
(A) - [pic] (B) - [pic] (C) - [pic] (D) [pic] (E) [pic]
Solution : 2y - [pic] = [pic] + 7y
- [pic] - [pic] = 7y - 2y
5y = - [pic] - [pic]
5y = - [pic] - [pic] ; 5y = - [pic] ; y = [pic] = - [pic]
27. U. R. Broak is going to borrow $6000 to pay off some debts. He will have to pay back the $6000
plus the amount of simple interest in equal monthly payments. The credit union offers an interest
rate of 5% for 4 years. The bank offers an interest rate of 4% for 5 years. How much less will
the monthly payments be if he borrows the money from the bank?
(A) $0.00 (B) $5.00 (C) $12.00 (D) $30.00 (E) $40.00
Note : Interest = Principal x Rate x Time
Credit Union : I = 6000(.05)(4) = 1200; Total amount owed = 6000 + 1200 = 7200 ;
Monthly payments = [pic] = 150.00
Bank : 6000(.04)(5) = 1200 ; Total amount owed = 6000 + 1200 = 7200 ;
Monthly payments = [pic] = 120.00
$150.00 - $120.00 = $30.00
30. A veterinarian surveys 26 of his customers. The compiled results show that 14 customers have dogs, 10
have cats, 5 have fish, 4 have dogs and cats, 3 have dogs and fish, and 1 has cats and fish. No
customers has all three kinds of pets. How many customers have none of these pets?
(A) 2 (B) 3 (C) 4 (D) 5 (E) 6
Dogs only = 14 - 4 - 3 = 7
Cats only = 10 - 4 - 1 = 5
Fish only = 5 - 3 - 1 = 1
Dogs/Cats = 4
Cats/Fish = 1
Dogs/Fish = 3
Total = 7 + 5 + 1 + 4 + 1 + 3 = 21
Customers that have none of the pets = 26 - 21 = 5
33. P, Q, and R are the solutions to 8x[pic] - 2x[pic] - 5x = 0. Find P + Q + R + PQR.
(A) - 1[pic] (B) [pic] (C) 1[pic] (D) 7 (E) 23
Note : P + Q + R = sum of the roots and PQR = product of the roots
If ax[pic] + bx[pic] + cx + d = 0, then the sum of the roots = - [pic] and the product of the roots =
- [pic].
(P + Q + R) + PQR = - [pic] + [pic] = [pic]
34. The zeros of the polynomial p(x) = x[pic] - 5x + 2 are precisely the dimensions of a rectangle in
inches. The area of the rectangle is :
(A) 2 sq. in. (B) 4 sq. in. (C) 5 sq. in. (D) 9 sq. in. (E) 10 sq. in.
Assume that the zeros of the polynomial are r and t. The area of the rectangle = rt. Note :
rt = the product of the roots. If ax[pic] + bx + c = 0, then the sum of the roots = - [pic].
Area = - [pic] = 5
37. The radius is a bone in the human that connects the elbow and the wrist. A woman with a radius 22 cm
long, should be about 160 cm tall, and a woman with a radius 26 cm long should be about 174 cm tall.
About how long should the radius be of a woman who is 157.5 cm tall?
(A) 21.242 cm (B) 20.986 cm (C) 21.048 cm (D) 21.286 cm (E) None of these
Organize the data given as being coordinates on a line. You have the coordinates (x, 157.5),
(22, 160), and (26,174). x = the length of the radius of the woman who is 157.5 cm tall.
Assume that the relationship of the points are linear. Find the slope between points.
[pic] = [pic]
[pic] = [pic] ; 14(22 - x) = 4(2.5) ; 308 - 14x = 10 ; - 14x = - 298 ; x = 21.286
38. The number of solutions in positive integers of x + 2y = 13 is :
(A) 8 (B) 7 (C) 6 (D) 4 (E) None of these
Solve x + 2y = 13 for y.
2y = - x + 13 ; y = - [pic]x + [pic].
Looking at the equation, you will notice that substituting an odd positive integer will result in y
be a positive integer. Begin be letting x = 1, 3, 5, …
If x = 1, y = 6 ; if x = 3, y = 5 ; if x = 5, y = 4 ; if x = 7, y = 3, if x = 9, y = 2 ;
if x = 11, y = 1, if x = 13, y = 0 (Note : y is not positive).
Solutions are : (1, 6), (3, 5), (5, 4), (7, 3), (9, 2), and (11, 1). There are 6 solutions.
39. If you have three coins in your pocket totaling 35 cents and one of those coins falls out of your pocket,
what is the probability that you will still have 30 cents in your pocket?
(A) [pic] (B) [pic] (C) [pic] (D) [pic] (E) None of these
35¢ = 1 quarter + 1 nickel + 1 nickel
There are 3 coins. Thus, there are 3 possible outcomes. The best way to approach this problem is
to think, “The only way that a coin will fall out and you will not have 30 cents is if the quarter falls
out”. The probability of that happening is [pic]. To find the probability of a coin falling out and still
having 30 cents = 1 - [pic] = [pic].
GEOMETRY
3. Rene drew [pic] using the coordinates (1, 2), (2, - 2) and (5, 1). Find the area of Rene’s
triangle.
(A) 7.5 units[pic] (B) 5.5[pic] units[pic] (C) 7.5 units[pic] (D) 6 units[pic] (E) .5[pic] units[pic]
Set up a 3 by 3 matrix with the first column consisting 1’s, the second column consisting of the x-coordinate of
each vertex, The third column consisting of the y-coordinate of each vertex.
Area = [pic] = [pic][10 - (- 5)] = 7.5
6. Suppose D, E, and F are points on circle O such that segment MD is tangent to circle O and segment
ME is a secant to circle O through point F. If MD = 4, EF = x and MF = 2, then ME equals :
(A) 6 (B) 8 (C) 10 (D) 12 (E) 14
Sketch a circle with center O. From a point M drawn a tangent segment to circle O touching at point D.
From point M drawn a secant segment going through point F and ending on the circle on point E.
Segment MD is a tangent segment and segment ME is a secant segment. MF is the external segment of
the secant.
The tangent is the geometric mean between the external segment of the secant and the secant.
[pic] = [pic] ; 2(x + 2) = 4(4) ; 2x + 4 = 16 ; 2x = 12 ; x = 6
The secant ME has length x + 2 which is 6 + 2 = 8.
9. A scalene triangle has side lengths of 6”, 9”, and 11”. Find the area (nearest square inch).
(A) 39 in[pic] (B) 27 in[pic] (C) 26 in[pic] (D) 24 in[pic] (E) 13 in[pic]
Solution : The area of a scalene triangle with sides a, b, and c is equal to [pic],
where s = [pic].
S = [pic] = 13
A = [pic] = 26.9815 Answer : 27
25. The area of a regular hexagon is 54[pic] sq. cm. Find the perimeter of the hexagon.
(A) 54 cm (B) 9[pic] cm (C) 108 cm (D) 18[pic] cm (E) 36 cm
If you draw segments from the center of the hexagon to each vertex you will subdivide the hexagon
Into three equilateral triangles. The area of an equilateral triangle is A = [pic].
6[pic] = 54[pic] ; s[pic][pic] = [pic] ; s[pic] = [pic] ; s[pic] = 36 ; s = 6
Perimeter = 6(6) = 36
26. For a wheel rolling on level ground the number of rotations it makes varies inversely to its diameter.
If a wheel with a radius of 13.5 inches makes approximately 747 revolutions in rolling 1 mile, what
would the radius of a wheel need to be in order to make approximately 100 revolutions in rolling
1 mile? (nearest inch)
(A) 55 in (B) 75 in (C) 87 in (D) 95 in (E) 101 in
R = [pic] ; 747 = [pic] ; m = 2(13.5)(747)
R = [pic] ; 100 = [pic] ; 100d = 2(13.5)(747) ; d = [pic] = 201.690
Radius = [pic] = 100.845 ; radius is approximately 101 inches
28, Robin Stern enters the archery contest. The target is a circle with a diameter of 3 feet. The bull’s-eye is
a circle with a radius of 6 inches in the center of the target. What are the odds of hitting the bull’s-eye
if the arrow is equally likely to hit any spot on the target?
(A) [pic] (B) [pic] (C) [pic] (D) [pic] [pic] (E) [pic]
If the diameter of the target is 3 feet, its diameter in inches is 36 inches. The radius of the target is
18 inches. The area of the target is 18[pic]π = 324π. The area of the bull’s-eye is 6[pic]π = 36π.
The odds of hitting the bull’s-eye =
[pic] = [pic] = [pic] = [pic] = [pic]
ALGEBRA 2
4. The first five terms of an infinite arithmetic sequence is 6 [pic], A, B, C, 12 , … .
Find A + B + C.
(A) 9 [pic] (B) 14 [pic] (C) 18 [pic] (D) 25 (E) 28 [pic]
Let d = the common difference
6 [pic] + 4d = 12 [pic] ; 4d = 6 [pic] ; d = [pic] = [pic]
A = 6 [pic] + [pic] = 6 [pic] + [pic] = 6 [pic] = 7 [pic]
B = 7 [pic] + [pic] = 7 [pic] = 9 [pic]
C = 9 [pic] + [pic] = 9 [pic] = 10 [pic]
A + B + C = 7 [pic] + 9 [pic] + 10 [pic] = 26 [pic] = 28 [pic] = 28 [pic]
This problem could have been solved as follows :
B = [pic] = [pic] = 9 [pic]
Since B is the middle term, it is the average of 6 [pic] + A + B + C + 12. The sum of these
Five terms is equal to 5[pic] = 45 [pic] = 46 [pic]. A + B + C = 46 [pic] - (6 [pic] + 12 [pic]) =
46 [pic] - 18 [pic] = 28 [pic]
12. The probability of scoring above 150 on this mathematics test is 32%. If 112 students scored
above 150, then how many students scored 150 or less on this test?
(A) 238 (B) 336 (C) 224 (D) 262 (E) 182
If probability of scoring above 150 on this mathematics test is 32%, then the probability can be
considered to be [pic] and the Odds are [pic].
[pic] = [pic], where x = how many students scored 150 or less on this test.
32x = 68(112) ; x = [pic] = 238
22. Phil Harmonic drove to a concert and back home via the same route. His speed going to the
concert was 70 mph. His speed returning from the concert was 55 mph. What was Phil’s
average speed for the trip? (nearest tenth)
(A) 60.4 mph (B) 61.0 mph (C) 61.6 mph (D) 62.0 mph (E) 62.5 mph
Distance = Rate x Time
Let D = distance traveled ; Rate = [pic] = [pic] = [pic] =
2D[pic] = 2[pic] = 61.6
25. Let a = .333… in base six and b = .444… in base six. Find [pic] in base ten.
(A) 1.333… (B) .125 (C) 4.5 (D) .222… (E) .75
Solution : .333… (base 6) = [pic] + [pic] + [pic] + … = 3[[pic] + [pic] + [pic] + …] =
3[[pic]] = 3[[pic]] = 3[[pic]] = [pic]
.444… (base 6) = [pic] + [pic] + [pic] + … = 4[pic] =
4[pic] = 4[pic] = 4[pic] = [pic]
[pic] = [pic] = [pic] = .75
39. The product of the harmonic mean of 4 and 16 and the arithmetic mean of 4 and 16 is :
(A) 80 (B) 64 (C) 51.2 (D) 20 (E) 4
Solution : arithmetic mean = [pic] = 8
harmonic mean = [pic] = [pic] = [pic] = 8
arithmetic mean x harmonic mean = 8(8) = 64
41. Find m + n if [pic] x [pic] = [pic]
(A) - 28 (B) - 16 (C) - 11 (D) 22 (E) 32
Solution : 3m - 5n = 4 ; 2m - 4n = 7
4[3m - 5n = 4] - 5[2m - 4n = 7]
2m = - 19 ; m = - 9.5
If 2m - 4n = 7 and m = - 9.5, then 2(- 9.5) - 4n = 7.
- 19 - 4n = 7 ; - 4n = 26 ; n = - 6.5
m + n = - 9.5 + (- 6.5) = - 16
46. Which of the following numbers is considered to be a “deficient” number?
(A) 476 (B) 364 (C) 268 (D) 196 (E) 28
Use Number Sense shortcut for finding the sum of the positive integral divisors of each number.
If the sum of the positive integral divisors is less than twice the number, the number is deficient.
If the sum of the positive integral divisors is equal to twice the number, the number is perfect.
If the sum of the positive integral divisors is more than twice the number, the number is abundant.
476 = 2[pic](7)(17)
Sum of positive integral divisors : [pic] = 7(8)(18) = 1008
Note : If the exponent of a base is 1, simply increase the number by 1. If you notice
the work shown the (7)(17) became (8)(18). Keep this in mind on future problems.
1008 > 2(476) so 476 is an abundant number.
364 = 2[pic](7)(13)
Sum of positive integral divisors : [pic] = (7)(8)(14) = 784
784 > 2(364) so 364 is an abundant number.
268 = 2[pic](67)
Sum of positive integral divisors : [pic] = 7(68) = 476
476 < 2(268) so 268 is a deficient number.
196 = 2[pic](7[pic])
Sum of the positive integral divisors : [pic] = 7(57) = 399
399 > 2(196) so 196 is an abundant number.
28 = 2[pic](7)
Sum of the positive integral divisors : [pic] = (7)(8) = 56
56 = 2(28) so 28 is a perfect number.
47. Simplify : [pic]
(A) 1 (B) n[pic] + n - 2 (C) n + 2 (D) n[pic] - 3n - 2 (E) n + 1
Solution : [pic]
[pic]
(n - 1)(n + 2) = n[pic] + n - 2
54. If a[pic] = 2, a[pic] = 5, a[pic] = a[pic] - a[pic], where n ≥ 3, then a[pic] equals :
(A) - 5 (B) - 3 (C) - 2 (D) 2 (E) 5
Solution : a[pic] = 2, a[pic] = 5,
a[pic] = a[pic] - a[pic] = a[pic] - a[pic] = 5 - 2 = 3
a[pic] = a[pic] - a[pic] = a[pic] - a[pic] = 3 - 5 = - 2
Note the first four terms of the sequence : 2, 5, 3, - 2,
If you notice the pattern you will find that the fifth term = fourth term – third term,
the sixth term = fifth term – fourth term, etc.
Write the first 13 terms : 2, 5, 3, - 2, -5, - 3, 2, 5, 3, - 2, -5, - 3, 2
a[pic] = 2
55. R and S are the roots of x[pic] + 4x - 6 = 0. Find R[pic] + 4R[pic]S + 6R[pic]S[pic] + 4RS[pic] + S[pic].
(A) - 576 (B) - 48 (C) 256 (D) 1296 (E) 4096
Note : R[pic] + 4R[pic]S + 6R[pic]S[pic] + 4RS[pic] + S[pic] = (R + S)[pic] ; (R + S)[pic] can be
interpreted as being the sum of the roots raised to the fourth power.
If ax[pic] + bx + c = 0, then the sum of the roots = - [pic].
(R + S)[pic] = [pic] = (- 4)[pic] = 256
71. A bowl contains four ping pong balls numbered 1, 2, 3, and 4, Two balls are selected at random
without replacement. What is the probability that the average of the two numbers selected is less
than or equal to 2?
(A) 16[pic]% (B) 20% (C) 33[pic]% (D) 40% (E) 50%
Find all of the possible sums.
1 + 2 = 3 ; 1 + 3 = 4 ; 1 + 4 = 5
2 + 3 = 5 ; 2 + 4 = 6 ; 3 + 4 = 7
Possible sums are 3, 4, 5, 5, 6, and 7. For the average of the two numbers selected to be less
than or equal to 2, the sum must be 4 or less. The only sums that are less than or equal to 4 are 3
and 4. The probability is 2 out of 6 or [pic] which is equal to 33[pic]%.
73. Which of the following numbers is a “deficient” number?
(A) 992 (B) 840 (C) 666 (D) 524 (E) 496
Memorize the first three perfect numbers : 6, 28, and 496. Be aware that a multiple of a perfect
number is abundant. For example, multiples of 6 are abundant : 12, 24, 36, 48, 60, 72, …
Multiples of 28 are also abundant : 28, 56, 84, 112, 168, …, Multiples of 496 are also abundant :
496, 992, …
Notice that : (E) 496 is a perfect number. (A) 992 is a multiple of 496 so it is also abundant.
(B) 840 = 28(30) so it is also abundant. (C) 666 is a multiple of 6 so it is abundant. The only
answer left is (D) 524,
Keep in mind that a number is divisible is 6 if it is even and a multiple of 3. Thus you can use this
Information to eliminate some of the answer choices.
74. If x - y = - 2 and 5xy = 6, then x[pic] + 3xy + y[pic] = ?
(A) - 12 (B) - 3 (C) 4 (D) 5.2 (E) 10
(x - y)[pic] = x[pic] - 2xy + y[pic]
To find x[pic] + 3xy + y[pic] you must add 5xy to x[pic] - 2xy + y[pic]. Since 5xy = 6, then
(x - y)[pic] + 5xy = x[pic] - 2xy + y[pic] + 5xy = (- 2)[pic] + 6 = 4 + 6 = 10
ADVANCED MATH
3. [pic] is an obtuse isosceles triangle such that m[pic] is 104° and EF is 14 cm. Find the
area of [pic] to the nearest integer.
(A) 118 cm[pic] (B) 95 cm[pic] (C) 77 cm[pic] (D) 60 cm[pic] (E) 48 cm[pic]
Sketch and label [pic]. Since the triangle is isosceles and EF = 14, then DF = 14. Use the
Following formula to find the area of the triangle : Area = [pic]absinC
Area = [pic] = 95
4. A box contains 5 green balls, 4 blue balls, and 3 red balls. Two balls are randomly selected, one at
a time, without replacement. What is the probability that both are blue?
(A) [pic] (B) [pic] (C) [pic] (D) [pic] (E) [pic]
[pic] = [pic] = [pic]
11. The area (in square units) of the region bounded by y = 0, x = 1, and y = x[pic] is :
(A) [pic] (B) [pic] (C) [pic] (D) 1 (E) 1 [pic]
Solution : [pic] = [pic] = [pic] - [pic] = [pic]
12. A regular deck of 52 cards is shuffled and the top two cards are dealt face up. What is the
probability that both cards are prime numbers?
(A) [pic] (B) [pic] (C) [pic] (D) [pic] (E) [pic]
The cards that are prime numbers are the 2, 3, 5 and 7. Since there are four suits (hearts,
diamonds, clubs, and spades), that means that there are 16 cards which have a prime number.
[pic] = [pic] = [pic]
17. The football team at Lyttal U. has 11 lineman, 9 running backs/receivers, and 2 quarterbacks.
How many 11 member squads can they form if each squad contains 7 lineman, 3 backs/receivers,
and 1 quarterback?
(A) 416 (B) 828 (C) 28,380 (D) 55,440 (E) 705,432
[pic] = [pic] = (330)(84)(2) = 55,440
18. Sir Ve Orr puts a blue dot on a map. He puts a red dot on the map 25 cm due east of the blue
dot and a green dot 5 cm from the blue dot on a heading of 315°. How far is the green dot from
the red dot? (nearest tenth of a cm)
(A) 20.0 cm (B) 21.8 cm (C) 25.5 cm (D) 27.5 cm (E) 28.8 cm
Make sketch depicting location of the dots. Connect the dots to form a triangle with vertices
G(green), R(red), and B(blue). The measure of [pic] = 135°.
Use the Laws of Cosines, Let x = distance between green dot and red dot.
x[pic] = 5[pic] + 25[pic] - 2(5)(25)cos135°
x = [pic] = 28.7537 ; Answer ; 28.8
20. [pic] = __________ + C, where C is some arbitrary constant.
(A) - [pic] (B) [pic] (C) - [pic] (D) [pic] (E) - [pic]
Solution : [pic] = [pic] = [pic] = - [pic]
25. How many asymptotes does this function have? f(x) = [pic].
(A) 4 (B) 3 (C) 2 (D) 1 (E) 0
Solution : f(x) = [pic] = [pic]
Vertical asymptotes : (x - 3)(x + 2) = 0
x - 3 = 0 and x + 2 = 0
x = 3 and x = - 2
Horizontal asymptote : y = lim[pic][pic] = 1
Y = 1
Slant asymptotes : None (If the degree in the numerator is one
greater than the degree in the denominator there is
a slant asymptote.
There are 3 asymptotes.
33. The highest average monthly temperature for Miller’s View is 78.5° F and occurs in July. The
lowest average monthly temperature occurs in January and is 43.5°. The average monthly
temperature of Miller’s View varies sinusoidally with the month. What would be the predicted
average temperature for March? (nearest tenth)
(A) 49.3°F (B) 52.3°F (C) 55.2°F (D) 58.1°F (E) 61.0°F
Sketch graph of sinusoid using the information given. Label the x-axis from 0 to 13. Let
1 represent January, 7 represent July, and 13 represent January (Note : March will be 3 on
the x-axis). The standard sinusoidal equation is y = C + AcosB(x - D). In the graph,
C = 61°, A = 17.5, B = [pic] and D = 7.
Let T = temperature.
T = 61 + 17.5cos[pic](x - 7)
To find the average temperature in March, let x = 3.
T = 61 + 175cos[pic](3 - 7) = 52.25° ; Answer : 52.3°F
34. Alley and Lane are in the finals of a bowling tournament. The winner receives $100 and the
runner-up receives $60. Alley, last year’s champion, has a 70% probability of winning. What
is Alley’s mathematical expectation?
(A) $88.00 (B) $76.00 (C) $70.00 (D) $60.00 (E) $42.00
Solution : .70(100) + .30(60) = 88.00
36. What is the probability that a factor of 100 is a Fibonacci number?
(A) [pic] (B) [pic] (C) [pic] (D) [pic] (E) [pic]
Fibonacci sequence : 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, …
Factors of 100 : 1, 2, 4, 5, 10, 20, 25, 50, 100
Probability = [pic] = [pic] = [pic]
37. Let A[pic] = 1, A[pic] = 3, and A[pic] = A[pic] + A[pic], where n ≥ 3. If A[pic] is an odd number,
then n could be :
(A) 24 (B) 33 (C) 42 (D) 50 (E) 63
Use the information given to write terms in the sequence :
1. 3. 4. 7, 11, 18, 29, 47, 76, 123, 199, 322,…
Note : Every third number is even (3rd, 6th, 9th, etc.). All other numbers are odd.
Eliminate answers that are multiples of 3 (24, 33, 42, and 63). The only answer that
is not a multiple of 3 is 50. Answer : 50
38. At what bearing would the jet pilot head if he wants to fly due north at 325 mph when a 30 mph west wind
is blowing? (nearest minute)
(A) 354°44’ (B) 275°16’ (C) 185°43’ (D) 174°43’ (E) 5°16’
Make sketch using information in the problem.
Tan[pic] = [pic] ; [pic] = Tan[pic][pic] = 5°16’
360° - 5°16’ = 354°44’
40. Simplify : (sin[pic] + cos[pic])[pic] + (sin[pic] - cos[pic])[pic]
(A) [pic] - [pic] (B) [pic] (C) 2 (D) [pic] (E) [pic]
Solution : (sin[pic] + cos[pic])[pic] + (sin[pic] - cos[pic])[pic]
[pic] + [pic] + [pic] + [pic] - [pic] + [pic]
[pic] + [pic] = 2([pic] + [pic]) = 2(1) = 2
41. A horizontal tangent line to the curve y = [pic] - 4x can be found at __________.
(A) (0, - 4) (B) (2, - 1 [pic]) (C) (- 2, - 5 [pic]) (D) (2, 5 [pic]) (E) (- 2, 5 [pic])
To determine where a horizontal tangent line is located, find where y’ = 0.
Solution : y = [pic] - 4x
y = [pic] - 4x
y’ = x[pic] - 4
x[pic] - 4 = 0 ; (x + 2)(x - 2) = 0 ; x = 2 or x = - 2
f(2) = [pic] - 4(2) = [pic] - 8 = 2 [pic] - 8 = - 5 [pic]
f(- 2) = [pic] - 4(- 2) = - [pic] + 8 = - 2 [pic] + 8 = 5 [pic]
The horizontal tangent lines are located at (2, - 5 [pic]) and (- 2, 5 [pic]).
The answer is E.
44. If f”(x) = 8 and f’(x) = 7 and f(- 1) = 8, then f(0) = __________.
(A) 7 (B) 3 (C) 1 (D) - 1 (E) - 8
f’(x) = [pic] = [pic]
f’(x) = 8x + C ; f’(1) = 8(1) + C = 7 ; C = - 1
f’(x) = 8x - 1
f(x) = [pic] = [pic] = 4x[pic] - x + C
f(- 1) = 4(- 1)[pic] - (- 1) + C = 8 ; 5 + C = 8 ; C = 3
f(x) = 4x[pic] - x + 3 ; f(0) = 4(0)[pic] - 0 + 3 = 3
45. Box A contains ships number 1, 2, 3, and Box B contains ships numbered 1, 2, 3. One chip is
selected at random from each box. What is the probability that the average of the two randomly
selected ships is 1.5?
(A) [pic] (B) [pic] (C) [pic] (D) [pic] (E) [pic]
Think of all of the combinations of chips from Box A and Box B.
(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)
There are nine possible outcomes. For the average of two ships to be 1.5, then the sum of the
two chips must be 3. The only combination that adds up to 3 are (1, 2) and (2, 1). The probability
that the average of the two randomly selected chips is 1.5 is 2 out of 9 or [pic].
46. The equation of the tangent line to the curve y = 2x[pic] + 3 which is parallel to the line
y = 8x + 3 is y = __________.
(A) 8x - 5 (B) 4x + 3 (C) 4x - 2 (D) 8x + 11 (E) 8x
y = 2x[pic] + 3 ; y’ = 4x ; 4x = 8 ; x = 2
f(2) = 2(2[pic]) + 3 = 11 ; The line is tangent to the curve at (2, 1). Using the point (2, 1)
and slope of the tangent line, m = 3, write an equation.
m = [pic] ; 8 = [pic] ; 8x - 16 = y - 11 ; y = 8x - 5
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