AMC 10 Test ReviewMock-Version A

1. 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)

DP, DB trisects ADC, we have ADP = 30, ADB = 60

ADP = 30 AP = 1/3 AD = 1/3 = 3/3

2.

ADB = 60 AB =3 AD = 3; PB = AB – AP = 23/3DPB = ABD = 30 DP = PB = 23/3

Perimeter = 2 + 23/3 + 23/3 = 2 + 43/3 (B)ABD = 30 DB = 2AD = 2

Let f - # of freshmen; s -- # of sophomore2/5 f = 4/5 s f = 2s

Answer: (D)

By triangle inequality: 6 – 4 < X < 6 + 4 2 < X < 10

Answer: (D)

6 – 4 < Y < 6 + 4 2 < Y < 10The largest distance of any 2 numbers between 2 and 10 is less than 8The smallest distance of any 2 numbers between 2 and 10 is 0.

Note that: 111111111 X 111111111----------------------------------------------- 111111111 111111111 …… 111111111------------------------------------------------ 12345678987654321

Answer: (1+2+3+…+8)*2 + 9 = 81 (E)

Fat in 2% milk = 60% (fat in whole milk)

Answer: (C) 10/3

Fat in 2% milk = 2% V (where V is volume)

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 %

Price(end of March) = 1.2 * 0.8 * 1.25 * P = 1.2 P

Answer: (B) 17

Let P be the price at beginning of Jan.

Price(end of April) = P

Decrease amount = 1.2 P – P = 0.2 PDecrease percentage = 0.2P/1.2P * 100% = 100/6 % 16.7 %

Day 1: ¼

Answer: (D) Friday

Find the millets in each day, & observe the formula

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: (B) 1/2

Since ABC is equilateral BAC = BCA = 60OAC = OCA = 30From ACO AOC = 120By symmetry, AOB = BOC = 60BO = 2 * CO = 2 * DOBD = BO – DO = 2 DO – DO = DOBD /BO = DO/ (2 DO) = 1/2

Answer: (D)

From Pythagorean theorem:AC = 10 AM = 5

8

6

By AME and ABC similar right triangle AM / AB = 5/ 8Area(AME) / Area(ABC) = (5/8) (5/8) = 25/64 Area(ABC) = ½ * 6 * 8 = 24Area(AME) = 25/64 * 24 = 75/8

Answer: (C) 125/12

AD = 3, AB = 4 BD = 5Observe that EAB ABDEB / AB = DB / ADEB = AB * DB/AD = (4 * 5) / 3 = 20/3CBF ABD BF / BC = DB / ABFB = BC * DB/AB = (3 * 5) / 4 = 15/4EF = EB + EF = 20/3 + 15/4 = 125/12

Let Va and Vl be speed of Andrea and Lauren

Answer: (D) 65 min

Va = 3Vl and Va + Vl = 60 (km/hr)Va = 45, Vl = 15After 5 min, distance D = 20 – 5 = 15Time for Lauren = D/ Vl = 15/15 = 1 (hr)

Area(lune) = 1/6 Area(circle) – Area(ABC) = /18 - 3/12

Radius of the circle = 1/3 > 1/2Half 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 OAABC is equilateral Area(ABC) = 3/12

Answer: 4 ( /18 - 3/12 ) = 2/9 - 3/3 (C)

(S-6)(S+18) = 0 S = 6; Answer: (B) 6

R2 = 156 R = 156Since OA = 43 < 156 A in circle.Let S be side-length of ABCLet X be intersection of OA and BC OA BCSince ABC is equilateral, BX = S/2; AX = S3/2From right OCX: 156 = (S/2)2 + (S3)/2 + 43)2

156 = 1/4S2 + 3/4S2 + 12S + 48 S2 +12S – 108 = 0

X2 = (9 – 62) + (9 + 6 2) + 2 (92 – 62 * 2)

Answer: (B) 26

Let X = value of the expression, we get:

= 18 + 2 (81 – 72) = 18 + 2 9 = 18 + 2*3 = 24Thus X = 24 = 26

P2 + 150 = Q2 + 9

Answer = 300/P2 = 300/222 62% (E)

Let P2 be population in 1991; Q2 + 9 be population in 2001, and R2 be population in 2011

Q2 – P2 = 141 = 3 * 47 = 1 * 141Q + 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 squareFrom (2), R2 = P2 + 300 = 5341; not a perfect square Thus (1) is the only choice for P & Q.

Similarly, clockwise chords from A intersects with clockwise chords from C only if distance between A & C is less than r. (see picture)

Answer: 1/3 (D)

For any fixed point A on the circle,

BC

clockwise chords from A intersects with clockwise chords from B only if distance between A & B is less than r.

Note that the arc between BC is 1/3 of the circle

A

From I: (X + Y + Z) /2 =?= (X+Y)/2 + (X+Z)/2

Answer: II & III are true. (E)

(X + Y + Z) /2 =X= (2X+Y+Z)/2From 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

Since BCM is right triangle & BC = ½ CM

Answer: (E) 75

Since AB//CD AMD = CDMSince AMD = CMDCDM is isosceles CM = CD = 6

BCM is 30-60 right triangle BMC = 30 CMD + AMD = 180 – 30 = 150By CMD = AMD AMD = 150/2 = 75

5|X| = X2 – 24

Answer: 8 * (-8) = - 64 (A)

Square on both sides 5|X| + 8 = X2 - 16

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: (C) 23/3

Draw the shaded region R as shown,where EFBC, HIAB, with E, I bisecting BC and AB respectively.Since BG(i.e.BD) AC & ABC = 120ABG = CBG = 60 ACB = CAB = 30 CEFBEF; BIHAIH BEFBFGBGHBHIArea(BEF) = ½ BE * (3/3 BE) = 3/6 BE*BE = 3/6Area(R) = 4 * Area (BEF) = 23/3

To find the over-count:

Answer: (C) 1932

Total distribution: 37

# 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: 1) + 2) + 3) = 12 + 18 + 12 = 42 (D)

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.

1

9

4,5,6

ac

b

e

d

f

2) M = 5: 2 (a,c) (symmetric); let 2(a), (b,c) = (3,4),(4,3) or (c,e) = (3,4); Total choices: 2 x 2 x (1 x 2 + 1 x 1) = 12

Total choices: 2 x (2 x (1 x 2 + 1 x 1) + 1 x ( 1 x 1 + 1 x 2)) = 183) M = 6: (2,3) (a,b),(c,e) or (2,3) (a,c), (c,a); total will be: 2x2 + 2x2x2 = 12, since:

(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

(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: 42 (D)

Alternatively, the question can be answered by finding the number of Young-tableaux associated with the 3x3 Young diagram.

5

Given the hook-lengths in the diagram: 5, 4, 4, 3, 3, 3, 2, 2, 1According to hook-length theorem:# of possible Young-tableaux = 9! / (5x4x4x3x3x3x2x2x1) = 42

44 3

3

3 2

2

1

Answer: 5 + 5*2 + 5 + 5+9 = 34 (E)

Let scores as (a, aN, aN2, aN3) and (a, a+M, a+2M, a+3M) A + aN+aN2+aN3 = 4a+6M+1a(N4 – 1)/(N-1) = 4a + 6M + 1Note that N < 5, otherwise N3 >= 125 > 100Also N > 1 since a, aN, … is an increasing sequenceListing 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

By P(1) = a: P(1) – a = 0, this means that 1 is a root of P(x) –aSimilarly, we get 3, 5, 6 are roots of P(x) - aLet 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)

P(2) = (2-1)(2-3)(2-5)(2-7)Q(2) + a = -aFrom (1), use x = 2, 4, 6, and 8, we get:

P(4) = (4-1)(4-3)(4-5)(4-7)Q(4) + a = -aP(6) = (6-1)(6-3)(6-5)(6-7)Q(6) + a = -aP(8) = (8-1)(8-3)(8-5)(8-7)Q(8) + a = -a

-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) = -2a105 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, 105Hence the smallest a is LCM(15, 9, 105) = 315 (B)