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,

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.

A.  120

B.  600

C. 1,200

D. 1,320

E. 1,600

F.  1

G.  5

H.  7

J.  15

K. –7

A. 0.05125

B. 0.1825

C. 0.22

D. 0.375

E. 0.5125

F. 12

G. 16

H. 20

J.  24

K. 32

A. 18

B. 24

C. 57

D. 78

E. 96

F. 7

G. 

H. 

J.  14

K. 

A.  5

B. 10

C. 20

D. 
E. 

F.  60

G.  90

H. 150

J.  180

K. 210

d	depth in feet	0	33	66	99	132
V	volume in liters	1

A. V =

B. V =

C. V = – + 1

D. V =

E. V =

F. 

G. 

H. 

J.  

K. 

A. –2

B. –1

C.  

D.  1

E.  2

F.  6

G.  2

H.  1

J.  –1

K.  –6

Set 5: Mathematics 

1. The correct answer is D.

60% of 2,200 = 0.6(2,200) = 1,320

A. 120 is 100 = of 2,200, not 60%.

B. 600 is 100 = of 2,200, not 60%.

C. 1,200 is 100 = of 2,200, not 60%.

E. 1,600 is 100 = of 2,200, not 60%.

2. The correct answer is F.

| 3 – 7 | – | 4 – 1 | = | –4 | – | 3 | = 4 – 3 = 1

No other answer can be correct because 3 – 7 is uniquely –4, | –4 | is uniquely 4, 4 – 1 is uniquely 3, | 3 | is uniquely 3, and 4 – 3 is uniquely 1.

3. The correct answer is C.

The average of and 0.065 =  =  =  = 0.22

A. The average of and 0.065 is 0.05125, since  =  =  = 0.05125, so 0.05125 can't be the average of and 0.065

B. The average of and 0 is 0.1875, since  =  = 0.1875, so 0.1875 can't be the average of and 0.065

D. The average of and is 0.375, since  =  =  = 0.375, so 0.375 can't be the average of and 0.065

E. The average of and 0.65 is 0.5125, since  =  =  = 0.5125, so 0.5125 can't be the average of and 0.065

4. The correct answer is K.

The top and bottom are 2 by 2 so each has area 4, so 2(4) = 8.
The front and back are 2 by 3 so each has area 6, so 2(6) = 12.
The sides are 2 by 3 so each has area 6, so 2(6) = 12.
The total surface area is 8 + 12 + 12 = 32.

F. 12 is the surface area of only the front and back, not of the whole box.

G. 16 is the surface area of only the bottom, front, and one side, not the whole box.

H. 20 is the surface area of only the top, bottom, front, and back, not the whole box.

J. 24 is the surface area of only the front, back, and sides, not the whole box.

5. The correct answer is C.

The area of the shaded region is the area of the 8-by-12 rectangle minus the area of the right triangle with base 8 and height 9 minus the area of the right triangle with base 2 and height 12 – 9 = 3.
So A = 8(12) – (8)(9) – (2)(3) = 96 – 36 – 3 = 57.

Check: The shaded region is a right triangle with base 8 and height 9 plus a trapezoid with parallel sides 8 and
8 – 2 = 6 and height 12 – 9 = 3.
So A = (8)(9) + (8 + 6)(3) = 36 + 21 = 57.

6. The correct answer is G.

Since the perimeter is 28 inches, each side is  = 7 inches. Then in ABC, AB = 7, BC = 7, and AC =  =  =  = .

F. 7 is the length of each leg of ABC so can't be the length of the hypotenuse .

H. is the length of the diagonal of a rectangle that is 7 by , since  =  = , not the length of the diagonal of ABCD.

J. 14 is the length of the diagonal of a square with sides inches, since  =  =  = 14, but the perimeter of such a square would be , not 28.

K. is the length of the diagonal of a square with sides inches, since  =  = , but the perimeter of such a square would be , not 28.

7. The correct answer is E.

ABC is a 30°–60°–90° triangle with sides in the ratio , so  = and AB = .

A. can't be 5 units long because if AB = 5, then
AB < BC, since 5 < 10. This implies that
m(C) < m(A), which is impossible since m(C) = 60° and m(A) = 30°.

B. can't be 10 units long because if AB = 10, then ABC is isosceles and m(C) = m(A). But then the angle sum for ABC would be 30° + 30° + 90° = 150° 180°.

C. can't be 20 units long because if AB = 20, then AC =  = and the ratio of the sides would be 10:20:, which is not the same as for 30°–60°–90° triangles.

D. can't be units long because  =  = < 10, so AB < BC. This implies that m(C) < m(A), which is impossible since m(C) = 60° and m(A) = 30°.

8. The correct answer is F.

On rainy days, Zina walks 120 + 90 or 210 yards. On sunny days, she walks  =  =  = 150 yards. The difference is 210 – 150 = 60 yards.

G–K. The route on the sidewalks is clearly 120 + 90 = 210 yards. Since the shortcut is the hypotenuse of a right triangle, its length, s, must exceed 120 yards, since the hypotenuse of a right triangle must be the longest side. So, s > 120, –s < –120, 210 – s < 210 – 120,
210 – s < 90. Therefore, the difference between the long way and the short way to school must be less than 90 yards and none of G–K can be correct.

9. The correct answer is A.

If n = 0, 1, 2, . . . , when d = 33n, V = . Therefore, if d = 33n, n = , so V =  =  = .

Check:
  if d = 0, V =  = 1
  if d = 33, V =  =  =
  if d = 66, V =  =  =
  if d = 99, V =  =  =
  if d = 132, V =  =  =

B. is not correct because if d = 0, V =  = –1, not 1.

C. is not correct because if d = 66, V = – + 1 = 0, not .

D. is not correct because V is undefined when d = 0.

E. is not correct because if d = 0, V =  =  = –33, not 1.

10. The correct answer is F.

H and K can be eliminated because, with a very large number in the numerator and 5 in the denominator, they are very large. F, G, and J all have the same numerator, so the smallest of the three will be the one with the largest denominator. Since x – 1 < x < x + 1, x + 1 is the largest denominator and is the smallest fraction.

11. The correct answer is B.

Since the figure is a square, the angle at A is a right angle, so is perpendicular to the side adjacent to . This side goes through (0,2) and (–2,0), so its slope is = 1. Therefore, the slope of is the negative reciprocal of 1, which is –1.

A. If the slope of were –2, then, since (0,2) is on , (1,0) would be on , since = –2. But then AB = = . The side with endpoints (0,2) and (–2,0) has length = . The figure can't be a square if the slope of is –2.

C, D, E. slants toward the negative x-axis, so it has a negative slope. C–E cannot be correct because they are all positive.

12. The correct answer is J.

x = 6 – x², so x² + x – 6 = 0, and (x + 3)(x – 2) = 0. Then x + 3 = 0, so x = –3 or x – 2 = 0, so x = 2. When x = –3, 6 – x² = 6 – (–3)² = 6 – 9 = –3, so x = 6 – x²; and when x = 2, 6 – x² = 6 – 2² = 6 – 4 = 2, so x = 6 – x². Thus –3 and 2 are both solutions, and their sum is –3 + 2 = –1.

F, G, H, K. A quadratic equation has at most 2 solutions; –3 and 2 have been shown to be solutions, so no other solutions exist. The sum of –3 and 2 is uniquely –1, so no other number can be the sum of the 2 real solutions of x = 6 – x².