Chapter 19
Army Selection Instrument for Flight Training (SIFT) Answers/Explanations
Ready to find out how you did on the Army Selection Instrument for Flight Training (SIFT) practice test in Chapter 18? In this chapter, we present both more-detailed explanations for the test questions and a basic answer key. We suggest checking out the explanations so that you get a better understanding of why your answer is right or wrong and get a little extra review in the process.
Subtest 1: Simple Drawings
1. A.
2. E.
3. C.
4. B.
5. B.
6. A.
7. C.
8. C.
9. E.
10. D.
11. A.
12. B.
13. C.
14. E.
15. E.
16. B.
17. A.
18. D.
19. B.
20. A.
21. E.
22. B.
23. E.
24. E.
25. A.
Subtest 2: Hidden Figures
26. E.
Illustration by Wiley, Composition Services Graphics
27. C.
Illustration by Wiley, Composition Services Graphics
28. D.
Illustration by Wiley, Composition Services Graphics
29. A.
Illustration by Wiley, Composition Services Graphics
30. B.
Illustration by Wiley, Composition Services Graphics
31. D.
Illustration by Wiley, Composition Services Graphics
32. E.
Illustration by Wiley, Composition Services Graphics
33. C.
Illustration by Wiley, Composition Services Graphics
34. B.
Illustration by Wiley, Composition Services Graphics
35. A.
Illustration by Wiley, Composition Services Graphics
36. A.
Illustration by Wiley, Composition Services Graphics
37. D.
Illustration by Wiley, Composition Services Graphics
38. C.
Illustration by Wiley, Composition Services Graphics
39. B.
Illustration by Wiley, Composition Services Graphics
40. E.
Illustration by Wiley, Composition Services Graphics
Subtest 3: Army Aviation Information Test
41. C. The rotor mast is rigged to the left side.
42. A. The four forces that act on an aircraft are lift, weight, thrust, and drag.
43. A. Cyclic control changes the pitch all at the same time.
44. B. The velocity of air flowing over the advancing versus retreating rotor is unequal lift, called dissymmetry of lift.
45. D. The main purpose of the tail rotor is to counteract the torque effect of the main rotor blades.
46. B. Coning of the main rotor blades is the upward bending from the combination of lift and centrifugal forces.
47. D. When at a hover, the helicopter tends to move, or drift, in the direction of the tail rotor thrust; this propensity is called translating tendency.
48. D. Wind velocity is the only listed factor that doesn’t affect density altitude.
49. D. During an engine failure, you must immediately lower the collective pitch to maintain rotor rpm and enter into an autorotative state.
50. E. A weight-and-balance envelope is a short distance fore and aft of the main rotor blade for center of gravity.
51. A. The foot, or anti-torque, pedals allow you to control torque effect of the main rotor blades.
52. C. Gyroscopic precession is when a force applied to a rotating disc has the effect occur 90 degrees later and in the plane of rotation.
53. D. When the helicopter enters into undisturbed air, additional lift, called translational lift, is achieved. Translational lift is the initial buffeting you feel upon takeoff and shortly before landing.
54. C. Ground effect can result in becoming airborne before the desired takeoff speed.
55. D. Retreating blade stall is where the retreating blade airspeed decreases with an increased forward aircraft speed. After you reduce the retreating rotor airspeed to a wing-stalling condition, retreating blade stall will occur (and that is bad).
56. B. The most favorable conditions are low density altitude, low gross weight, and moderate to strong winds.
57. D. The cyclic controls the tilt of the main rotor disc.
58. D. Pitch is the movement around the aircraft’s lateral axis.
59. B. The collective pitch controls the pitch of all main rotor blades equally and simultaneously.
60. A. Bernoulli’s basic principle applies here: Lift is developed when airflow passes around an airfoil (wing).
Subtest 4: Spatial Apperception Test
61. A.
62. C.
63. C.
64. E.
65. B.
66. A.
67. B.
68. D.
69. B.
70. C.
71. E.
72. C.
73. B.
74. D.
75. A.
76. B.
77. A.
78. A.
79. B.
80. C.
81. C.
82. E.
83. B.
84. B.
85. E.
Subtest 5: Reading Comprehension Test
86. B.
87. B.
88. C.
89. C.
90. D.
91. C.
92. B.
93. D.
94. B.
95. A.
96. B.
97. C.
98. D.
99. A.
100. B.
101. A.
102. D.
103. B.
104. C.
105. C.
Subtest 6: Math Skills Test
106. C. (5 feet ÷ 10 feet) × 60 feet = 30 feet.
107. A. Seven days of rations for 6 people equals a total of 42 ration portions. Next, add the 4 insurgents to equal a total of 10 people to feed and divide the total number of rations by the number of people. 42 ÷ 10 = 4.2 days.
108. B. 100 percent of the initial price minus 25 percent = 75 percent. Multiply this result by 0.75 (to figure in an additional 25 percent markdown) to get a total price of 56.25 percent of the initial price. This answer means your savings is 100 percent – 56.25 percent = 43.75 percent off.
109. A. Subtract 5/16 from 1, and you have a total of 11/16 of the pizza leftover. Convert this result to a decimal by dividing 11 by 16 to get 0.6875, and divide this amount by the three people sharing the pizza to get a total per-person amount of 0.229 of the pizza.
110. C. One inch equals 60 miles on the map scale. You know that 1 inch equals 2.54 centimeters. Therefore, 2.54 centimeters equals 60 miles. Divide 6 centimeters by 2.54 centimeters and then multiply that result by 60 miles to get 141.6 miles. (6 centimeters ÷ 2.54 centimeters) × 60 miles = 2.36 × 60 miles = 141.6 miles.
111. D. You first have to get the total weight in ounces. 16 ounces = 1 pound, so 10,000 pounds × 16 ounces/pound = 160,000 ounces. Divide the number of ounces by the amount going into each bag (80 ounces) to get the answer. 160,000 ounces ÷ 80 ounces per bag = 2,000 bags shipped.
112. B. Find the total amount each group receives: (3 × $675 = $2,025), (2 × $600 = $1,200), (4 × $595 = $2,380). Add the total amounts to get the total amount paid out each week: $2,025 + $1,200 + $2,380 = $5,605. Divide that result by the nine employees to get the $622.78 average weekly pay, and divide this number by 40 hours per week to get $15.569 or $15.57 in average hourly pay per employee.
113. A. Split the total donation into units and then use the ratio to find how much goes to each recipient: $225,000 ÷ (4 + 3) = $225,000 ÷ 7 = $32,142.86. Take this amount and multiply by 4 to get university A’s amount of $128,571.44. To double check, you can find university B’s amount by multiplying $32,142.86 by 3 to get $96,428.58 and then add both results to ensure they equal the total donation: $128,571.44 + $96,428.58 = $225,000.02
114. D. The sum of the average of the five tests to receive an A must be 450 (90 average × 5 tests). Subtract from 450 the score of each of the four previous tests, and you have 90 left over, so that’s what the student must score to average 90 and make an A in the class. You can also set the missing score as the variable x: (88 + 84 + 92 + 96 + x) ÷ 5 = 90.
115. B. If a = 5b, you can figure that 2a =10b, so you can substitute 2a for 10b in the second equation to get 2a = 4c. Divide both sides by 2 to get a = 2c.
116. B. 106 is the same as a 1 followed by 6 zeros, or 1,000,000.
117. C. First, calculate the distance required until the aircraft can see each other by taking the total distance and subtracting the distance at which they can acquire one another (900 miles – 10 miles = 890 miles). Divide this distance by the combined airspeeds at which the aircraft are converging (680 miles per hour + 630 miles per hour = 1,310 miles per hour). 890 miles ÷ 1,310 miles per hour = .679 hours. But wait; the answer choices are in minutes! Never fear; just multiply .679 hours × 60 minutes to get 40.74 or 41 minutes.
118. A. Seven hours times the average of 61 miles per hour gives you 427 miles. Divide the total miles by the gas mileage of 27 miles per gallon. 427 ÷ 27 = 15.8 gallons used.
119. D. First, figure out the separate rates of flow. To fill the 600-gallon tank, hose A = 600 ÷ 30 minutes = 20 gallons per minute or 1,200 gallons per hour. To empty the tank, hose B = 600 ÷ 60 minutes = 10 gallons per minute or 600 gallons per hour. If you open both, you have water coming in at the rate of 1,200 gallons per hour and leaving at 600 gallons per hour for a net filling rate of 600 gallons per hour. The tank will fill in 1 hour.
120. B. Use the formula a2 + b2 = c2, or 82 + 122 = c2 to find the length of the hypotenuse. This setup gives you 64 + 144 = c2. Solving this equation gives you 208 = c2. Therefore, 
, which equals 14.42 feet or 14 feet, 5 inches (to get the 5 inches you multiply 0.42 feet × 12 inches per foot = 5 inches). To get the triangle’s perimeter, add up the three side lengths: 8 feet + 12 feet + 14 feet, 5 inches = 34 feet, 5 inches.
121. D. Divide 2 acres by 5 bags to get a per-acre amount of 2.5 bags of salt. Multiply this number times 61⁄3 acres to get the total amount of 15.83 bags of salt.
122. A. Take 16 ounces × 0.4 to get 6.4 ounces of pure juice. Add 2 ounces to both the juice and the total amount. Divide the new juice amount (6.4 ounces + 2 ounces = 8.4 ounces) by the new total amount (16 ounces + 2 ounces = 18 ounces) to get 8.4 ounces. Divide that by 18 ounces = 46.7 percent = 47 percent
123. B. 75 percent of the straws are blue, so you have a 75-percent chance, or a three out of four chance, of grabbing a blue straw.
124. C. You know that in a right triangle, one angle is 90 degrees, so that’s the first part of the answer. You also know that the sum of the angles in a right triangle is 180 degrees. Therefore, you can subtract the known angles from that total to find the remaining angle. 180 – 90 – 25 = 65 degrees.
125. C. The reciprocal of 5 is the same as 1/5 = 0.2 or 20 percent
126. A. The ladder leaning against the wall forms a right triangle, so you can use the Pythagorean theorem (the ladder is the hypotenuse): a2 + 162 = 222. a2 + 256 = 484. a2 = 484 – 256 = 228. 
127. B. The total perimeter is 150 feet, which means that the total of one of the long sides plus one of the short sides equals 75 feet. You know that the short side is half the length of the long side, so one side is 25 feet and the other is 50 feet. Another way of looking at it is x + x + 2x + 2x = 150; 6x = 150; x = 25. The short side is 25 feet, and the long side is 2 × 25 or 50 feet. Now you can find the area by just multiplying the base × height: 25 feet × 50 feet = 1,250 square feet.
128. D. First, you must figure out the radius. Divide the diameter by 2 to get the radius of 50 centimeters. Next convert the height of 2.5 meters to 250 centimeters so that everything uses the same scale. Use the formula for volume of a cylinder: V = hπr2. V = 250 cm × π × (50 cm2) = 250 cm × π × 2,500 cm2 = π × 625,000 cm3 = 1,963,495 cm3. (If you use 22/7 for π, you get 1,964,287 cm3.) Now take 1,963,495 cm3 and divide it by 3,781.4 cm3 per gallon to get 519 gallons.
129. B. Divide the difference by the price of last year’s bag to get the percentage change of this year’s price from last year’s price. $25.11 – $24.75 = $0.36. Divide $0.36 by $24.75 to get 0.0145 or 0.015, which equals 1.5 percent.
130. C. The train travels 550 miles in 6 hours or 360 minutes. Divide 550 miles by 360 minutes to get 1.53 miles per minute. Multiply this result by 22 minutes: 1.53 × 22 = 33.61 miles.
131. A. First, convert the yards to feet. 4 yards = 12 feet and 7 yards = 21 feet. To get the square footage, multiply the length times width, or 12 feet × 21 feet = 252 square feet. Multiply this amount by $3.99 per square foot to get the total price of $1,005.48.
132. D. On the first day, the travelers completed 2,900 × 1/5 = 580 miles, and on the second they completed 2,900 × 1/7 = 414 miles. Subtract both of these distances from the total trip distance to get 1,906 miles left in the trip. Divide this result by the four days that they have left, and the average comes to 476.5 miles per day.
133. B. Multiply the available time of 2.7 hours by 8 and then divide by 5 (the number of aircraft available) to get the new amount of search time of 4.32 hours.
134. D. Subtract John’s regular work week of 40 hours from his actual work week of 53 hours to see how much time he earns overtime for. 53 – 40 = 13 hours. Next, multiply each amount of time by its appropriate salary. $21.50 × 40 = $860, and $21.50 × 1.5 × 13 hours = $419.25. Add the two to get $1,279.25.
135. A. This one is a simple chain of addition and subtraction: $360 + $50 – $27 – $250 = $133.
Subtest 7: Mechanical Comprehension Test
136. B. The formula used for determining how an inclined plane reduces effort is length of ramp ÷ height of ramp = weight of object ÷ force. In this case, x ÷ 3 feet = 300 pounds ÷ 150 pounds. (The question states that the amount of force must be reduced by half, so that’s why you divide 300 by 150.) x ÷ 3 = 300 ÷ 150, so x = 2 × 3 = 6 feet.
137. C. To carry the load evenly, you must place the load directly in the middle of the board. Because the board is 12 feet long, you must place the load at the 6-foot mark.
138. A. Because wheel B has a smaller circumference, it covers a shorter linear distance than wheel A when turning at the same rate. Therefore, wheel A covers a distance faster than wheel B. You can figure the circumference with the formula of πd or 2πr. The circumference of wheel A is 22π = 69 feet, and the circumference of wheel B is 11π = 34.5 feet. To cover the same amount of distance, wheel B must rotate exactly twice the amount of time as wheel A at the same number of revolutions per minute.
139. B. A stationary single pulley allows you to change the direction of force but doesn’t result in increased mechanical advantage; therefore, the correct answer is Choice (B).
140. C. Gear 1 will turn clockwise, and gear 2 will turn counterclockwise; because gear 2 is the same size as gear 1, both gears will turn at the same speed. Gear 3 will turn clockwise but is half the diameter of gear 2, so gear 3 will rotate at twice the speed of gears 1 and 2. Gear 4 will rotate counterclockwise and is the same size as gear 3, so it will also rotate twice as fast as gears 2 and 1.
141. D. The weight of A equals exactly the weight of B. An object will always remain balanced at its center of gravity, by definition.
142. B. Gear A has 14 teeth and makes 4 revolutions, so it rotates a total of 56 teeth (4 × 14 = 56). Gear B has a total of 20 teeth, so you divide the total teeth movement of 56 by 20 to get 2.8 revolutions.
143. A. A second-class lever is a lever with force or lifting action applied to one end (handles), a fulcrum at the opposite end, and the load to be carried in the middle.
144. C. The number of times a rope moves through the pulley determines the mechanical advantage. In this pulley system, the advantage is 2. Therefore, by applying a 50-pound force on the end of the rope, you can lift a weight that is 100 pounds.
145. D. Any of these choices can promote acceleration, so Choice (D) is the best answer.
146. B. 2π radians equal 360 degrees. Therefore, 6.28 radians = 360 degrees. Divide both sides of that equation by 6.28 to find that 1 radian equals 57.32 degrees. Finally, take 115 degrees ÷ 57.32 = 2 radians to get the final answer.
147. A. Specific gravity is the measurement of a fluid in relation to the weight of water. The higher specific gravity of the fluid will cause the float to remain higher, so the answer is Choice (A).
148. C. Force equals mass (m) × acceleration (A). Therefore, because the force is equal and opposite, m × A = –(m × A1). This setup gives you the following equation: 100 kilograms × 5 meters per second = –(150 kilograms × A). Simplified, you have 500 kilograms meters per second divided by 150 kilograms, or 3.33 meters per second. Because the raft is going in the opposite direction, the correct answer is –3.33 meters per second.
149. D. The linear distance traveled is greater on the outside but the angular distance remains constant.
150. B. Under the law of conservation of mass, m1V1 + m2V2 = (m1 + m2) × V. Therefore, 5 kilograms(10 meters per second) + 10 kilograms(–5 meters per second) = (5 kilograms + 10 kilograms) × V. This math works out to 0 = (15 kilograms) × V, so V = 0.
151. A. Nothing will happen because in order for any movement to occur, anvil B would have to be placed more than 1.5 meters to the right of the fulcrum.
152. A. Pressure equals force divided by area, or P = F ÷ A; you can rearrange this equation to get P × A = F. Therefore, if you plug in the numbers from the problem, 14.7 pounds per square inch × 145 square inches = 2,132 pounds.
153. C. The pulley function on the bottom is just to change direction, so you have a mechanical advantage of 2 on top.
154. B. Gear 1 rotates clockwise, and gear 2 rotates counterclockwise and turns both gear 3 and gear 5 clockwise. Gear 3 turns gear 4 counterclockwise.
155. D. First, you must get the total volume of the tank. Volume is length × width × height, or 5 feet × 2 feet × 3 feet = 30 cubic feet. You know that the weight per cubic foot is 62.5 pounds, so the total weight acting on the bottom of the tank is 1,875 pounds. Now you must figure out the area of only the bottom in square inches. Multiply (12 inches × 5 feet) × (12 inches × 2 feet), which equals 60 inches × 24 inches, or 1,440 square inches. Now simply divide the total weight of 1,875 pounds by 1,440 square inches to get the answer of 1.3 pounds per square inch.
156. C. Wheel B has a smaller circumference, so given the same number of revolutions, it will cover a smaller distance. Therefore, wheel A will reach the bottom first.
157. A. The formula to use is length of ramp ÷ height of ramp = weight of object ÷ effort. Using the numbers from the problem, you get 10 meters ÷ 3 meters = 100 kilograms ÷ effort. Therefore, cross-multiplying gives you 100 kilograms divided by 3.33 = 30 kilograms.
158. B. Because valve A is required to be open for water to flow in, you know that one must be open. Now you need to figure out how full the tank has to be to get to 75 percent. The tank holds a total of 20 inches, so take 0.75 × 20; the answer is 15 inches. If valve B is open, the water will only fill to the 2-inch mark, so that one must remain closed. If valve C is open, water will be able to fill to the 15-inch mark, so that valve must be opened. Valve D must also be open for the water to continue to flow out. The correct answer is Choice (B), valves A, C, and D.
159. C. Because gear 1’s radius is twice that of gear 2, the gear 2 will revolve twice as fast (10 revolutions) in the opposite direction of gear 1. Gear 3 will rotate opposite of gear 2 and the same as gear 1; it will rotate at 5/6 the speed of gear 2 (10 revolutions × 5/6 = 8.33 revolutions).
160. A. The vertical component of the momentum is 0 at this point, and the horizontal components remain unchanged.
161. D. The volume of water flowing past point B is the same as for points A and C. Due to the restriction of approximately half the width, the velocity approximately doubles as it passes point B. This is the law of conservation of mass, the fundamental principle that results in the Bernoulli equation.
162. D. Pulley D is rotating most quickly because it has the smallest circumference.
163. C. You must figure out which brace covers the most area. Because you have a right triangle, you can use the formula A = 1/2 base × height. The largest area is 4 × 4 ÷ 2 = 8, which is Choice (C).
164. A. The shorter pendulum length B completes the swing the fastest, and the longer pendulum length A will take the longest. Interestingly, the mass of the pendulum doesn’t matter!
165. B. The resistance in B is the sum of the two resistors and therefore twice the resistance in A and C.
Answers at a Glance
Subtest 1: Simple Drawings
1. A
2. E
3. C
4. B
5. B
6. A
7. C
8. C
9. E
10. D
11. A
12. B
13. C
14. E
15. E
16. B
17. A
18. D
19. B
20. A
21. E
22. B
23. E
24. E
25. A
Subtest 2: Hidden Figures
26. E
27. C
28. D
29. A
30. B
31. D
32. E
33. C
34. B
35. A
36. A
37. D
38. C
39. B
40. E
Subtest 3: Army Aviation Information Test
41. C
42. A
43. A
44. B
45. D
46. B
47. D
48. D
49. D
50. E
51. A
52. C
53. D
54. C
55. D
56. B
57. D
58. D
59. B
60. A
Subtest 4: Spatial Apperception Test
61. A
62. C
63. C
64. E
65. B
66. A
67. B
68. D
69. B
70. C
71. E
72. C
73. B
74. D
75. A
76. B
77. A
78. A
79. B
80. C
81. C
82. E
83. B
84. B
85. E
Subtest 5: Reading Comprehension Test
86. B
87. B
88. C
89. C
90. D
91. C
92. B
93. D
94. B
95. A
96. B
97. C
98. D
99. A
100. B
101. A
102. D
103. B
104. C
105. C
Subtest 6: Math Skills Test
106. C
107. A
108. B
109. A
110. C
111. D
112. B
113. A
114. D
115. B
116. B
117. C
118. A
119. D
120. B
121. D
122. A
123. B
124. C
125. C
126. A
127. B
128. D
129. B
130. C
131. A
132. D
133. B
134. D
135. A
Subtest 7: Mechanical Comprehension Test
136. B
137. C
138. A
139. B
140. C
141. D
142. B
143. A
144. C
145. D
146. B
147. A
148. C
149. D
150. B
151. A
152. A
153. C
154. B
155. D
156. C
157. A
158. B
159. C
160. A
161. D
162. D
163. C
164. A
165. B