Hard math questions?

Question

These were from some math contests.

1. Juan rolls a fair regular octahedral die marked with the numbers 1 through 8. Then Amal rolls a fair six-sided die. What is the probability that the product of the two rolls is multiple of 3?

(I thought it is 2/8 x 2/6 = 7/12, but it wasn't the answer)

2.The area of a sector of a circle is 198 pi, and the arc length of this sector is 22pi. Find the degree measure of the central angle of this sector.

3. Find the real value of x for which: 5^(x+1)=3^(4x+3) x 7^(-2x+1).

4. A circle is inscribed in a square of side 8 inches. An equilateral triangle is inscribed in the circle. Find the area of this triangle.

Answer 1

possible multiple-of-3 products and the rolls that give them (the first number is the octahedral die; the second the 6-sided):

3 (1,3) (3,1,)

6 (1,6) (2,3) (3,2) (6,1)

9 (3,3)

12 (2,6),(3,4,) , (4,3) (6,2)

15 (3,5) (5,3)

18 (3,6), (6,3)

21 (7,3)

24 (4,6), (6,4) (8,3)

30 (5,6) (6,5)

36 (6,6)

42 (7,6)

48 (8,6)

total outcomes with multiples of 3:  24

total outcomes: 6 * 8 = 48

probability of a multiple of 3 as a product: 24 / 48 = 1/2

(oops... I had missed some... thanks!)

2.  let t be the central angle:

the ratio of the sector area to total area = t / 2pi

t / 2pi = 198 pi / (pi * r^2)

t = 396 pi / r^2

ratio of arc length / circumference = t / 2pi

t / 2pi = 22pi / (2 pi * r)

t = 22pi / r

t = 396 pi / r^2 = 22pi / r

cross-multiply:

396 pi * r = 22 pi * r^2

18 = r

radius = 18

t = 22pi / 18 = 11 pi / 9 = central angle (from arc length relation)

check in sector relation:

11pi / 9 = 396 pi / (18^2) = 396 pi / 324 = 198 pi / 172 = 99pi / 81 = 11pi / 9 check

t = central angle of 11pi / 9 in a circle of radius 18

3.  5^(x + 1) = 3^(4x + 3) * 7^(-2x + 1)

take log of both sides:

log (5^(x + 1) = log [3^(4x + 3) * 7^(-2x + 1)]

log (5^(x + 1) = log [3^(4x + 3)] + log [7^(-2x + 1)] (using property that log ab = log a + log b)

(x + 1) log 5 = (4x + 3) log 3 + (-2x + 1) log 7 (using property that log (a^r) = 4 log a)

x log 5 + log 5 = 4x log 3 + 3 log 3 - 2x log 7 + log 7 (distributing)

x log 5 - 4x log 3 + 2x log 7 = 3log 3 + log 7 - log 5 (bring all x terms to LHS)

x (log 5 - 4 log 3 + 2 log 7) = log 9 + log 7 - log 5

x (log 5 + log 49 - log 81) = log 63 - log 5

x log (5 * 49 / 81) = log (63 / 5)

x = log (63/5) / log (245 / 81)

caution: I haven't double-checked the arithmetic, but the procedure is sound:

take logs; bring exponents outside as factors; isolate x-terms; simplify logs using properties of logs:

log (ab) = log a + log b

log (a/b) = log a - log b

log (a^r) = r log a

4)  Area = 12 sqrt(3)

if circle is inscribed in this square, the radius is 4 "

The circumcenter of the triangle will be the center of the circle.

Divide the the triangle into 3 smaller congruent isosceles triangles with one vertex at the center of the circle, and the sides going to the vertices of the bigger triangle (with leg length 4, the radius of the circle).  The vertex angle of these three will be 120.  Now bisect each vertex angle to create 6 smaller 30-60-90 triangles, each with hypotenuse = 4, the radius.  The legs will be 2 and 2 * sqrt(3), and the area will be (2 * 2 sqrt(3)) / 2 = 2 sqrt(3).  Since there are six of these that make up the inscribed triangle, the area is 6 * 2 sqrt(3) = 12 sqrt(3)

(whew)

Answer 2

q1

P(product is multiple of 3)

= 1 - P(neither roll a multiple of 3)

= 1 - (4/6)*(6/8)

= 1 - 4/8

= 0.5

q2

198pi = x * pi * r^2

22pi = x * pi * 2r

r / 2 = 9

r = 2*9 = 18

x = 22 / (2*18) = 11/18 of 360 degrees

x = 220 degrees

q3

5^(x+1) = 3^(4x+3) * 7^(-2x+1)

5*5^x = 27*81^x * 7*(1/49^x)

= 189 * (81/49)^x

(49*5/81)^x = 189/5

(245/81)^x = 37.8

x = log 37.8 / (log 245 - log 81)

= 3.28178477

q4

x / (3*8/4) = 2 / √3

x = 2 * 6 / √3 = 4√3

area = 6x / 2 = 12√3

Answer 3

1. 2/8 + 2/6 -  (2/8 x 2/6) = 1/2

The way I did it is P(Rolling a 3 or 6 on the 8 sided die) + P(Rolling a 3 or 6 on the 6 sided die) - P(Rolling a 3 or 6 on the eight sided die and Rolling a 3 or 6 on the 6 sided die)

I believe you forgot that last part. This eliminates those dupicate values in the sample space ie (3,3), (6,6)

2. I didn't attempt but I believe notthejake probably got this one.

3. I get the same answer in the same way as notthejake

4. Circle inscribed in square implies circle has a diameter of 8. Inscribe the equilaterial triangle means from the center of the circle to the vertex of the triangle will create 3 smaller isosceles triangle. Each of the isosceles triangles has a two legs that are 4in in length (radius of the circle).

The angle between them must be 120°. Using law of cosines you can now figure out the legs of one of your equilaterial triangle. a² = b² + c² -2bc cosA. The result of one of the legs is √(48). Now you can figure out the height of the triangle, because if you draw the height in, it will be a 30-60-90 triangle with a hypotenuse = √(48), short leg = √(48)/2 and long leg (which is the height) = √(48)*√3/2

Using 1/2 b*h = (1/2)(√(48))(√(48)*√3/2) = 12√3