Quantity A | Quantity B |
The daily rent of a man who pays $975 rent per month |
The daily rent of a man who pays $11,650 rent per year |
-
Solution
(A) Don’t be misled by the word daily into finding the actual daily rent of the man in column A, by dividing his monthly rent by the number of days in a month. You may waste a lot of time deciding whether to divide by 31, 30, or 30.25 to obtain a messy figure you don’t need. Since you know there are 12 months in a year, multiply 975 by 12 to obtain $11,700 as the annual rent in (A), which is greater than quantity (B), $11,650. A man who pays more rent per year will also pay more rent per day.
Quantity A | Quantity B |
2x + 3y | 3x + 4y |
-
Solution
(D) Quantity (B) “seems greater” by exactly 1x + 1y, because (3x + 4y) – (2x + 3y) = 1x + 1y. Though (B) appears greater, x and y may be negative numbers, so (D) is correct. For example, if x = y = 1, then (A) = 5 < (B) = 7, but if x = y = –1, then (A) = –5 > (B) = –7.
Quantity A | Quantity B |
The area of a square with perimeter equal to 64 |
The area a circle with diameter equal to 18 |
-
Solution
In the xy-plane, P is the point of intersection of the two lines given by the following equations:
y = 2x – 6 ; y = (1/2)x + 6
Quantity A | Quantity B |
The x-coordinate of point P | The y-coordinate of point P |
-
Solution
Quantity A | Quantity B |
5x | 100 |
-
Solution
(C) When lines intersect, the vertical angles, those directly across from each other, are equal. This means that we can label the three angles within the triangle 2x°, 3x°, and 4x°. By the triangle sum theorem, 2x° + 3x° + 4x° = 180°. Then 9x = 180, and x = 20. So 5x = 100, (C).
Quantity A | Quantity B |
The area of a quadrilateral with perimeter equal to P |
The area of a quadrilateral with perimeter equal to 2P |
-
Solution
Quantity A | Quantity B |
The number of meters in 10 kilometers |
The number of square centimeters in 1 square meter |
-
Solution
(m + 5)(m – 5) = 0
(k + 4)(k – 4) = 0
Quantity A | Quantity B |
m 2 | k 2 |
-
Solution
(A) Expand (m + 5)(m – 5) as m 2– 5m + 5m – 25 = m 2– 25. If this equals 0, then m 2= 25.
Similarly, (k + 4)(k – 4) = k 2 – 4k + 4k – 16 = k 2 – 16 = 0. Add 16 to both sides, and k 2= 16, (A) is greater.
Quantity A | Quantity B |
y | z |
-
Solution
(A) In a triangle, the larger sides are across from the larger angles, and the smaller sides are across from the smaller angles. Since the side length x + .001 > x, the angle opposite x + .001 is greater than the angle opposite x. So y > z, (A).
Lines L1 and L2 are parallel.
Quantity A | Quantity B |
x | 91 |
-
Solution
m = 3n
n = 5p
Quantity A | Quantity B |
m | p |
-
Solution
(D) By substitution, m = 3(5p) = 15p. This means that m is 15 times p. It looks as if m, Quantity (A), is greater, but if the variables are negative, p is greater, so (D) is correct.
Quantity A | Quantity B |
The greatest possible area of a triangle with side lengths 4, 5, and x |
11 |
-
Solution
4n = 256
Quantity A | Quantity B |
3n | (3/2)n3 |
-
Solution
(B) Solve 4n = 256 by multiplying fours on your calculator until you reach 256 as a product. Note that 4 × 4 × 4 × 4 = 16 × 16 = 256, meaning that the exponent n equals 4. Then (A) is 34 = 3 × 3 × 3 × 3 = 9 × 9 = 81. Quantity (B) is greater, because (3/2) × 4 3 = (3/2) × 64 = 96, and 96 > 81.