Question #1:
x + 2y = 11
2x – y = 7
| Column A | Column B |
| x + y | x·y |
(a) The quantity in Column A is greater.
(b) The quantity in Column B is greater.
(c) The two quantities are equal.
(d) The relationship cannot be determined from the information given.
Solution:
x + 2y = 11
2x – y = 7
The system of equations can easily be solved if we multiply equation 2 by 2 and then add the equations.
5x = 25
x = 5
y = 2x – 7
y = 3
The quantity in Column A is x + y = 8.
The quantity in Column B is xy = 15.
Question #2: Functions g and f are defined by g(x) = | 1 – x2 |, f(x) = 32x.
| Column A | Column B |
| f(g(2)) | g(f(2)) |
(a) The quantity in Column A is greater.
(b) The quantity in Column B is greater.
(c) The two quantities are equal.
(d) The relationship cannot be determined from the information given.
Answer: g(2) = | 1 – 22 | = 3
f(g(2)) = 36.
f(2) = 34.
g(f(2)) = | 1 – 38 |
g(f(2)) = 38 – 1
As 38 – 1 > 36, (b) is the correct answer.
Question #3: For any x ≠ 0 and x ≠ 1, (x2 + 4x – 5)/(x2 – x) =
(a) x – 5
(b) 4 – 5/x
(c) x + 5/x
(d) 1 + 5/x
(e) 3x – 4
Answer:
(x2 + 4x – 5) = (x – 1)(x + 5)
(x2 – x) = (x – 1)x
(x2 + 4x – 5)/(x2 – x) = (x + 5)/x
(x2 + 4x – 5)/(x2 – x) = 1 + 5/x
Question #4: The two circles in the figure below intersect each other in M and N. Both circles have radii of 10 inches. AB = 4 inches where A and B are the points of intersection of segment O1 and O2 with the two circles.
Figure not drawn to scale.
| Column A | Column B |
| O1O2 | 4·AB |
(a) The quantity in Column A is greater.
(b) The quantity in Column B is greater.
(c) The two quantities are equal.
(d) The relationship cannot be determined from the information given.
Answer:
O1O2 = O1A + O2B + AB
O1A = r – AB
O2B = r – AB,
where r = 10 inches is the radius.
O1O2 = 2r – AB
O1O2 = 16 inches
4AB = 16 inches so the two quantities are equal.
Question #5: The rectangle solid in the figure below has w = 3 and l = 9.
| Column A | Column B |
| The area of rectangle ABCD | 30 |
(a) The quantity in Column A is greater.
(b) The quantity in Column B is greater.
(c) The two quantities are equal.
(d) The relationship cannot be determined from the information given.
Answer:
The area of rectangle ABCD is:
For different values of h, 
can be equal to 30, less than 30 or greater than 30.
(d) is the correct answer.
Question #6: How many committees of 4 students can be selected from a group of 7 students?
(a) 35
(b) 40
(c) 46
(d) 52
(e) 90
Answer: The numbers of different subsets of r objects that can be selected from n objects without regards to the order of selection is n!/[(n - r)!r!].
In our case, 7!/[4!3!] = 35 is the number of possible committees.
Question #7: If a and b are positive integers and a·b = 200, which of the following can be the sum a + b?
(a) 40
(b) 46
(c) 33
(d) 55
(e) 50
Answer: 200 = 2·2·2·5·5.
If a and b are positive integers, the 2 numbers and their sum can be:
2 + 100 = 102
4 + 50 = 54
5 + 40 = 45
8 + 25 = 33
10 + 20 = 30
(c) is the correct answer.
>Question #8: A year ago the annual subscriptions for newspaper A and newspaper B were both $50. Newspaper A increased the price of subscription by 20% while the price of the newspaper B’s subscription increased by 10%. What is the difference between the subscription costs of newspaper A and newspaper B today?
(a) $4
(b) $5
(c) $2
(d) $8
(e) $10
Answer: The difference between the subscription costs of newspaper A and newspaper B today is 1.2·$50 – 1.1·$50 = $5.
Question #9: Which of the following categories of employees is the highest cost for Company A ?
(a) Managers
(b) Engineers
(c) Technicians
(d) Operators
(e) Other employees
Answer: We need to multiply the average annual salary of each category by the numbers of employees in that category. We find that the operators are the highest cost for Company A.
Question #10: If a ◊ b = (a + b)2,
[a ◊ (1 - a)]◊(a2 – 1) =
(a) a4
(b) 4a2
(c) a5
(d) a3 – 1
(e) a
Answer:
a ◊ (1 – a) = (a + 1 – a)2.
a ◊ (1 – a) = 1.
[a ◊ (1 - a)]◊(a2 – 1) = 1 ◊ (a2 – 1).
[a ◊ (1 - a)]◊(a2 – 1) = (1 + a2 – 1)2.
[a ◊ (1 - a)]◊(a2 – 1) = a4.