Math Counts Test Review 2011 State

Area(8-in diameter) = 42 * π 3. Kara has three non-overlapping circles. The area of the circle 8 inches in diameter is how much greater than the combined areas of the circle 6 inches in diameter and the circle 2 inches in diameter? Express your answer in terms of π. Area(8-in diameter) = 42 * π Area (6-in) + Area(2-in) = 32 * π + 12 * π Difference = 16 π – 10 π = 6 π

Starting with the largest quantity! 12. Javier needs to exchange his dollar bill for coins. The cashier has 2 quarters, 10 dimes and 10 nickels. Assuming the cashier gives Javier the correct amount and at least one quarter, how many possible combinations of coins could Javier receive?. Starting with the largest quantity! With 2 Q: # of possible dimes - 5, 4, 3, 2 , 1, 0 With 1 Q: # of possible dimes - 7, 6, 5, 4 Total: 11

Area(ADE) / Area(ABCE) = 1/5 Area(ADE) / Area(ABCD) = 1/6 17. In a rectangle ABCD, point E is on side CD. The area of triangle ADE is one-fifth of the area of quadrilateral ABCE. What is the ratio of the length of segment DE to the length of segment DC? Express your answer as a common fraction. Area(ADE) / Area(ABCE) = 1/5 Area(ADE) / Area(ABCD) = 1/6 Let H be height, X be DE, Y be DC ½(H*X) / (H * Y) = 1/6 ½ ( X / Y ) = 1/6  X/Y = 1/3

18. For a certain sequence of numbers, the sum of the first n numbers in the sequence is given by n3 + 4n for all positive integers n. What is the tenth number in the sequence? (N1 + N2 + … + Nn ) = N3 + 4 * N N1 + N2 + … + N10 = 103 + 4 * 10 ----- (1) N1 + N2 + … + N9 = 93 + 4 * 9 ----- (2) (1) - (2): N10 = 103 – 93 + 4*10 – 4*9 = 271 + 4*(10 – 9) = 275

Note ABC and ADC are right triangle Assume the circle has radius R 25. Quadrilateral ABCD is inscribed in a circle with segment AC a diameter of the circle. If ∠DAC = 30° and ∠BAC = 45°, the ratio of the area of ABCD to the area of the circle can be expressed as a common fraction in simplest radical form in terms of π as a + √b / ( c * π) . What is the value of a + b + c? Note ABC and ADC are right triangle Assume the circle has radius R We got: DC = R, AD = √3R; AB = BC = √2 R Area(ADC) = ½ √3R2; Area(ABC) = ½ (√2 R)2 Ratio = ½ (√3+2)R2/(R2π) = (2+√3)/(2π); Ans: 7

Let H – prob. of head, T – prob. of tail. 27. In three flips of an unfair coin the probability of getting three heads is the same as the probability of getting exactly two tails. What is the ratio of the probability of flipping a tail to the probability of flipping a head? Express your answer as a common fraction in simplest radical form. Let H – prob. of head, T – prob. of tail. H * H * H = H * T * T + T * H * T + T * T * H H * H = T * T + T * T + T * T = 3 * T * T T2 / H2 = 1/3 T / H = 1/√3 = √3/3

29. A bag contains red balls and white balls 29. A bag contains red balls and white balls. If five balls are to be pulled from the bag, with replacement, the probability of getting exactly three red balls is 32 times the probability of getting exactly one red ball. What percent of the balls originally in the bag are red? Assume R red balls and W white balls in bag Prob(3 R) = (R /(R+W) )3 * (W/(R+W)2 * (5*4*3)/(3*2*1) = 10 * (R3 W2)/(R+W)5 Prob(1 R) = (R /(R+W) ) * (W/(R+W)4 * 5 = 5 * (R W4)/(R+W)5 10 * (R3 W2/(R+W)5 = 5 * (R W4)/(R+W)5 * 32 R2 = 16 W2  R2/W2 = 16  R/W = 4; R/(R+W) = 4/5 = 80%

Hint: use (x + a) (x – a) = x2 – a2 30. What is the value of 52,683 × 52,683 – 52,660 × 52,706? Hint: use (x + a) (x – a) = x2 – a2 52660 = 52683 - 23; 52706 = 52683 + 23 52660 * 52706 = (52683 – 23)(52683 + 23) = 52683 * 52683 – 23*23 Answer = 23*23 = 529

Observe any sum is multiple of 3 The smallest is 3+6+9 = 18 G5. How many distinct sums can be obtained by adding three different numbers from the set {3, 6, 9. …, 27, 30}? Observe any sum is multiple of 3 The smallest is 3+6+9 = 18 The largest is 24 + 27 + 30 = 81 From: (81 – 18)/3 = 63/3 = 21 We observe that there are 21 + 1 = 22 different multiple of 3’s in between 18 and 81. Answer: 22

Hans can win in 1st, 2nd and 3rd rounds To win in 1st rd: 60% * 40% G7. Hans and Franz are in a shooting competition. The object of the match is to be the first to hit the bull’s-eye of a target 100 feet away. The two opponents alternate turns shooting, and each opponent has a 40% chance of hitting the bull’s-eye on a given shot. If Hans graciously allows Franz to shoot first, what is the probability that Hans will win the competition and take no more than three shots? Express your answer as a decimal to the nearest hundredth. Hans can win in 1st, 2nd and 3rd rounds To win in 1st rd: 60% * 40% To win in 2st rd: (60% * 60%) * (60% * 40%) To win in 3st rd: (60% * 60%) * (60% * 60%) * (60% * 40%) Sum = 60%*40%(1+(60%)2+(60%)4) = 0.3575  0.36

Split the object into a vertical prism & 4 pyramids G8. The solid figure shown has six faces that are squares and eight faces that are equilateral triangles. Each of the 24 edges has length 2 cm. The volume of the solid can be expressed in cubic centimeters in the form (a*b)/c where c has no perfect square factor except 1 and where a and b are relatively prime. What is the value of a + b + c? Split the object into a vertical prism & 4 pyramids prism: base: 2x2; height: 22 pyramid: base: 2 x 22; slant: 2; height: 1 (note the formation of 45-45-90 right triangles) Volume of rectangular: 2x2x22= 82 Volume of pyramid: 2x22x1x 1/3=42/3 Volume = 82 + 4*(42/3) = 402/3 a + b + c = 40 + 2 + 3 = 45

M4. A train travels at a constant rate of 55 miles per hour through a tunnel. Forty-five seconds after the front of the train enters the tunnel the front of the train exits the tunnel. How many feet long is the tunnel? Note that 1 mile = 5280 feet Speed = 55 * 5280 / (60 * 60) = 11 * 22 / 3 Distance = speed * time Answer = 11 * 22 / 3 * 45 = 11*22*15 = 3630

M5. The L-shaped piece shown will be placed on the grid so that it covers exactly three unit squares of the grid. The sum of the numbers in the grid’s covered three unit squares will be S. If rotating the L-shaped piece is permitted, what is the sum of all the values of S for all possible placements on this grid of the L-shaped piece? For 1-2-4-5, we can get: 124,125,145,245 For 2-3-5-6, we can get: 235,236,256,356 For 4-5-7-8, we can get: 457,458,478,578 For 5-6-8-9, we can get: 568,569,589,689 Sum = 3 (1+2+4+5 + 2+3+5+6 + 4+5+7+8 + 5+6+8+9) = 3 * 80 = 240

M8. Let M be the midpoint of the segment FG M8. Let M be the midpoint of the segment FG. Let A and B be points coplanar to points F and G. Points A and B are positioned on the same side of the line containing segment FG such that FMA and MGB are equilateral. The lines FB and GA intersect at point K. What is the measure of GKB? FMA =GMB = 60  AMB = 60 With MA = FM = MG = MB   AMB is equilateral Thus AB//FG  ABGM is rhombus  AG  FB  ABMF is rhombus  FB  AM From MGB  AGB = AGM = 60/2 = 30 From ABM  ABF = MBF = 60/2 = 30 From GFB: GKB = 180 - 30 – (30+60) = 60

Prob. of 2 red: 5/10 * 5/10 Prob. of 2 blue: 3/10 * 3/10 M9. A bag contains five red marbles, three blue marbles and two green marbles. Six marbles are to be drawn from the bag, replacing each one after it is drawn. What is the probability that two marbles of each color will be drawn? Express your answer as a common fraction. Prob. of 2 red: 5/10 * 5/10 Prob. of 2 blue: 3/10 * 3/10 Prob. of 2 green: 2/10 * 2/10 Prob. of 2 each: ¼ * 9/100 * 4/100 = 9/10000 Possible ways to get 2 each: 6!/(2!2!2!) = 90 (i.e. possible arrangment of 2 each) Ans: 90 * 9/10000 = 81/1000

Let the sequence be: a, a+k, a+2k, … a+10k M10. The geometric mean of two positive numbers a and b is √a√b. The third term of an arithmetic sequence of positive numbers, in which the difference between the terms is not zero, is the geometric mean of the first and eleventh terms. What is the ratio of the second term to the first term of the sequence? Express your answer as a common fraction. Let the sequence be: a, a+k, a+2k, … a+10k We have: a + 2k = √(a(a+10k))  (a+2k)2 = a2 + 10 ak a2 + 4ak + k2 = a2 + 10 ak  4ak + k2 =10 ak a2 + 4ak + k2 = a2 + 10 ak  4ak + k2 =10 ak k = 3/2 a  (a+k)/a = (a+3/2 a)/a = 5/2