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AMC 10 Test Review Mock-Version A

AMC 10 Test Review Mock-Version A. The Fibonacci sequence 1, 1, 2, 3, 5, 8, 13, 21, … starts with two 1s, and each term afterwards is the sum of its two predecessors. Which one of the ten digits is the last to appear in the units position of a number in the Fibonacci sequence?.

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AMC 10 Test Review Mock-Version A

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  1. AMC 10 Test ReviewMock-Version A

  2. The Fibonacci sequence 1, 1, 2, 3, 5, 8, 13, 21, … starts with two 1s, and each term afterwards is the sum of its two predecessors. Which one of the ten digits is the last to appear in the units position of a number in the Fibonacci sequence? The sequence of units digits is: 1, 1, 2, 3, 5, 8, 3, 1, 4, 5, 9, 4, 3, 7, 0, 7, 7, 4, 1, 5, 6, . . . . 6 is the last digit to appear in the units position. Answer: 6 (C)

  3. 2. DP, DB trisects ADC, we have ADP = 30, ADB = 60 ADP = 30  AP = 1/3 AD = 1/3 = 3/3 ADB = 60  AB =3 AD = 3; PB = AB – AP = 23/3 DPB = ABD = 30  DP = PB = 23/3 ABD = 30  DB = 2AD = 2 Perimeter = 2 + 23/3 + 23/3 = 2 + 43/3 (B)

  4. Let f - # of freshmen; s -- # of sophomore 2/5 f = 4/5 s  f = 2s Answer: (D)

  5. By triangle inequality: 6 – 4 < X < 6 + 4  2 < X < 10 6 – 4 < Y < 6 + 4  2 < Y < 10 The largest distance of any 2 numbers between 2 and 10 is less than 8 The smallest distance of any 2 numbers between 2 and 10 is 0. Answer: (D)

  6. Note that: 111111111 X 111111111-----------------------------------------------111111111 111111111 …… 111111111 ------------------------------------------------ 12345678987654321 Answer: (1+2+3+…+8)*2 + 9 = 81 (E)

  7. Fat in 2% milk = 2% V (where V is volume) Fat in 2% milk = 60% (fat in whole milk) Fat in whole milk = 2% V * 100/60 = 1/30 V Fat % in whole milk = 1/30 V / V = 1/30 = 100/30 % = 10/3 % Answer: (C) 10/3

  8. Let P be the price at beginning of Jan. Price(end of March) = 1.2 * 0.8 * 1.25 * P = 1.2 P Price(end of April) = P Decrease amount = 1.2 P – P = 0.2 P Decrease percentage = 0.2P/1.2P * 100% = 100/6 %  16.7 % Answer: (B) 17

  9. Find the millets in each day, & observe the formula Day 1: ¼ Day 2: ¼ + ¾ * ¼ Day 3: ¼ + ¼ * ¾ + (¾)2 * ¼ = Day 4: ¼ + ¼ * ¾ + (¾)2 * ¼ + (¾)3 * ¼ Day N: ¼ + ¼ * ¾ + (¾)2 * ¼ + … + (¾)n-1 * ¼ = ¼ ((1 – ¾ )n – 1 ) /(1 – ¾) = 1 – ( ¾ )n To make 1 – ( ¾ )n > ¾  ( ¾ )n < ¼  n = 5 Answer: (D) Friday

  10. Since ABC is equilateral BAC = BCA = 60 OAC = OCA = 30 From ACO  AOC = 120 By symmetry, AOB = BOC = 60 BO = 2 * CO = 2 * DO BD = BO – DO = 2 DO – DO = DO BD /BO = DO/ (2 DO) = 1/2 Answer: (B) 1/2

  11. 8 6 From Pythagorean theorem: AC = 10  AM = 5 By AME and ABC similar right triangle AM / AB = 5/ 8 Area(AME) / Area(ABC) = (5/8) (5/8) = 25/64 Area(ABC) = ½ * 6 * 8 = 24 Area(AME) = 25/64 * 24 = 75/8 Answer: (D)

  12. AD = 3, AB = 4  BD = 5 Observe that EAB  ABD EB / AB = DB / AD EB = AB * DB/AD = (4 * 5) / 3 = 20/3 CBF  ABD  BF / BC = DB / AB FB = BC * DB/AB = (3 * 5) / 4 = 15/4 EF = EB + EF = 20/3 + 15/4 = 125/12 Answer: (C) 125/12

  13. Let Va and Vl be speed of Andrea and Lauren Va = 3Vl and Va + Vl = 60 (km/hr) Va = 45, Vl = 15 After 5 min, distance D = 20 – 5 = 15 Time for Lauren = D/ Vl = 15/15 = 1 (hr) Answer: (D) 65 min

  14. Radius of the circle = 1/3 > 1/2 Half of the diagonal of the square is (12 + 12)/2 = 1/2 > Radius (e.g. 1/2) Thus the corner of the square is outside the circle & the circle is not totally inside the square (shown) Let a be length of XB as shown in the picture. a2 + ( ½ ) 2 = (1/3) 2 a = 3/6 = ½ (1/3) = ½ OB = ½ OA ABC is equilateral  Area(ABC) = 3/12 Area(lune) = 1/6 Area(circle) – Area(ABC) =  /18 - 3/12 Answer: 4 ( /18 - 3/12 ) = 2/9 - 3/3 (C)

  15. R2 = 156  R = 156 Since OA = 43 < 156  A in circle. Let S be side-length of ABC Let X be intersection of OA and BC  OA  BC Since ABC is equilateral, BX = S/2; AX = S3/2 From right OCX: 156 = (S/2)2 + (S3)/2 + 43)2 156 = 1/4S2 + 3/4S2 + 12S + 48  S2 +12S – 108 = 0 (S-6)(S+18) = 0  S = 6; Answer: (B) 6

  16. Let X = value of the expression, we get: X2 = (9 – 62) + (9 + 6 2) + 2 (92 – 62 * 2) = 18 + 2 (81 – 72) = 18 + 2 9 = 18 + 2*3 = 24 Thus X = 24 = 26 Answer: (B) 26

  17. Let P2 be population in 1991; Q2 + 9 be population in 2001, and R2 be population in 2011 P2 + 150 = Q2 + 9 Q2 – P2 = 141 = 3 * 47 = 1 * 141 Q + P = 47; Q – P = 3  Q = 25; P = 22 ---- (1) Q + P = 141; Q – P = 1  Q = 72; P = 71 ---- (2) From (1), R2 = P2 + 300 = 784 = 262; a perfect square From (2), R2 = P2 + 300 = 5341; not a perfect square Thus (1) is the only choice for P & Q. Answer = 300/P2 = 300/22262% (E)

  18. A C B For any fixed point A on the circle, clockwise chords from A intersects with clockwise chords from B only if distance between A & B is less than r. Similarly, clockwise chords from A intersects with clockwise chords from C only if distance between A & C is less than r. (see picture) Note that the arc between BC is 1/3 of the circle Answer: 1/3 (D)

  19. From I: (X + Y + Z) /2 =?= (X+Y)/2 + (X+Z)/2 (X + Y + Z) /2 =X= (2X+Y+Z)/2 From II: X + (Y+Z) /2 = (2X + Y + Z) /2 From III: X @ (Y + Z) /2 =?= (X + Y)/2 @ (Y + Z)/2 (2X + Y + Z) /2 = (2X + Y + Z) /2 Answer: II & III are true. (E)

  20. Since AB//CD  AMD = CDM Since AMD = CMD CDM is isosceles  CM = CD = 6 Since BCM is right triangle & BC = ½ CM BCM is 30-60 right triangle  BMC = 30 CMD + AMD = 180 – 30 = 150 By CMD = AMD  AMD = 150/2 = 75 Answer: (E) 75

  21. Square on both sides  5|X| + 8 = X2 - 16 5|X| = X2 – 24 1) If X > 0: 5X = X2 – 24 X2 – 5X – 24 = 0  (X-8)(X+3) = 0  X = 8 2) X < 0: - 5X = X2 – 24 X2 + 5X – 24 = 0  (X+8)(X-3) = 0  X = -8 Answer: 8 * (-8) = - 64 (A)

  22. Draw the shaded region R as shown,where EFBC, HIAB, with E, I bisecting BC and AB respectively. Since BG(i.e.BD)  AC & ABC = 120 ABG = CBG = 60  ACB = CAB = 30 CEFBEF; BIHAIH  BEFBFGBGHBHI Area(BEF) = ½ BE * (3/3 BE) = 3/6 BE*BE = 3/6 Area(R) = 4 * Area (BEF) = 23/3 Answer: (C) 23/3

  23. Total distribution: 37 To find the over-count: # of distribution with no red: 27 # of distribution with no blue: 27 # of distribution with no red & no red: 1 # of valid distribution = 37 - 27 - 27 + 1 = 1932 Answer: (C) 1932

  24. 1 b a c d 4,5,6 e f 9 Observe that 1 can only go in the upper left corner, 9 can only go in the lower right corner. Also observe that the middle box M can only be 4, 5, or 6. 1) M = 4: (2,3) -> (a,c),(c,a), symmetric; 5  b or e, symmetric; each 5 leads to two choices of 6, which then leads to one or two choices of 7 & 8. Total choices: 2 x 2 x (1 x 2 + 1 x 1) = 12 2) M = 5: 2  (a,c) (symmetric); let 2(a), (b,c) = (3,4),(4,3) or (c,e) = (3,4); (b,c) = (3,4),(4,3)  two choices for 6, which leads to one or two choices for 7 & 8 (c,e) = (3,4)  two choices for 6, which leads to one or two choices for 7 & 8 (c,e) = (3,4)  two choices for 6, which leads to one or two choices for 7 & 8 Total choices: 2 x (2 x (1 x 2 + 1 x 1) + 1 x ( 1 x 1 + 1 x 2)) = 18 3) M = 6: (2,3)  (a,b),(c,e) or (2,3)  (a,c), (c,a); total will be: 2x2 + 2x2x2 = 12, since: (2,3)  (a,b): (4,5)  (c,e); (2,3)  (c,e): (4,5)  (a,b), and each leads to two choices for 7 & 8 (2,3)  (a,c),(c,a): (4,5)  (b,e) or (eb), and each leads to two choices of 7 & 8. Answer: 1) + 2) + 3) = 12 + 18 + 12 = 42(D)

  25. 4 3 5 4 3 2 3 2 1 Alternatively, the question can be answered by finding the number of Young-tableaux associated with the 3x3 Young diagram. Given the hook-lengths in the diagram: 5, 4, 4, 3, 3, 3, 2, 2, 1 According to hook-length theorem: # of possible Young-tableaux = 9! / (5x4x4x3x3x3x2x2x1) = 42 Answer: 42(D)

  26. Let scores as (a, aN, aN2, aN3) and (a, a+M, a+2M, a+3M) A + aN+aN2+aN3 = 4a+6M+1 a(N4 – 1)/(N-1) = 4a + 6M + 1 Note that N < 5, otherwise N3 >= 125 > 100 Also N > 1 since a, aN, … is an increasing sequence Listing possible answers 1) N = 2  a = 1, 5  M = x, 9 2) N = 3  a = 1, 2  M = x, x 3) N = 4  a = 1;  M = x Answer: 5 + 5*2 + 5 + 5+9 = 34(E)

  27. By P(1) = a: P(1) – a = 0, this means that 1 is a root of P(x) –a Similarly, we get 3, 5, 6 are roots of P(x) - a Let P(x) - a = (x-1)(x-3)(x-5)(x-7)*Q(x) for some Q(x) That is: P(x) = (x-1)(x-3)(x-5)(x-7)Q(x) + a ------- (1) From (1), use x = 2, 4, 6, and 8, we get: P(2) = (2-1)(2-3)(2-5)(2-7)Q(2) + a = -a P(4) = (4-1)(4-3)(4-5)(4-7)Q(4) + a = -a P(6) = (6-1)(6-3)(6-5)(6-7)Q(6) + a = -a P(8) = (8-1)(8-3)(8-5)(8-7)Q(8) + a = -a

  28. -15 Q(2) + a = -a  -15 Q(2) = -2a 9 Q(4) + a = -a  9 Q(4) = -2a -15 Q(6) + a = -a  -15 Q(6) = -2a 105 Q(8) + a = -a  105Q(8) = -2a Thus -2a = -15Q(2) = 9Q(4) = -15Q(6) = 105Q(8) Thus a must be a multiple of 15, 9, 105 Hence the smallest a is LCM(15, 9, 105) = 315 (B)

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