Download Answers to 2014 Math Day Competition

Answers to 2014 Math Day Competition
1.
Answer is 29/15
2.
Answer is 43
3.
Answer is -7
4.
Answer is 15
5.
Answer is 4. Reflecting each vertex in the opposite side creates another equilateral
triangle, one whose altitudes are twice those of the original triangle. If the scale factor is
1:2, then the ratio of areas is 1:4.
6.
Answer is 1232. Let x be the number of students on opening day. Then
(x+48)(0.95) = 1216 gives x = 1232.
7.
Answer is 2013. Use a property of logarithms to rewrite the sum of the logs: logb3 +
logb11 + logb61 = logb(3 • 11 • 61) = logb(2013). Thus, b = 2013.
8.
Answer is 336
9.
Answer is 1/8. There are 26 = 64 sequences of heads and tails. We count the number of
sequences that have exactly 6 or 5 or 4 heads in a row. There is only 1 sequence with 6
heads in a row: HHHHHH. There are 2 sequences with exactly 5 heads in a row:
THHHHH and HHHHHT. There are 5 sequences with exactly 4 heads in a row:
XTHHHH or THHHHT or HHHHTX, where X can be H or T. So the desired probability
is (1 + 2 + 5)/64 = 1/8.
10.
Answer is 72 in.2. Each face is missing a 1 × 1 in. square, so the area of the remaining
portion is 9 – 1 = 8 in.2. Removing the cube from a face adds 4 in.2 of area. Therefore, for
each face of the original cube, the area is 12 in.2. There are 6 such faces, so the total
surface area is 72 in.2.
11.
Answer is 11/2
12.
Answer is 64. If the product of the elements is divisible by 5, then the subset must contain
5, 10, or both 5 and 10. If 5 is in the subset and 10 is not, 8C2 = 28 subsets are possible.
Similarly, if 10 is in the subset and 5 is not, 8C2 subsets are possible. When both 5 and 10
are in the subset, there are 8 possible choices for the third element. Thus, 28 + 28 + 8 =
64 subsets.
13.
Answer is 46 nickels. Since every friend brought in the same amount and the total was
$5.29, we need to consider the divisors of 529 = 23 • 23. The options are 1 student, 23
students, or 529 students. The only reasonable possibility is 23. There are many ways to
form 23¢ but only one way that would consist of 6 coins: 1 dime, 2 nickels, and 3
pennies. When 23 friends each gave Teresa 2 nickels, she had 46 nickels.
14
Answer is 10/3
15.
Answer is 32/45 . 0.71. Let W be the event that a witch cast the spell, let A be the event
that an apprentice cast the spell, and let M be the event that an impostor cast the spell. Let
E be the event that a spell iseffective. We need the conditional probability that a witch
(
)
cast a spell given that the spell was effective: P W E =
P(W ∩ E )
. We know that
P( E )
P(W 1 E) = (1/3)(0.8). The probability that the spell was effective is the sum of three
probabilities:
P(E) = P(W 1 E) + P(A 1 E) + P(M 1 E) = (1/3)(0.8) + (1/6)(0.5) + (1/2)(0.05) = 3/8.
Finally, (4/15)/(3/8) = 32/45, or approximately 0.71.
16.
Answer is 7. The only possible values for f and f – 3 must be divisors of 70 (1, 2, 5, 7,
10, 14, 35, and 70). The only two numbers in the set that differ by 3 are 7 and 10.
17.
Answer is 4. Let b, s, and v represent, respectively, the number of students who play
basketball only, soccer only, or volleyball only. Let m, n, and p represent, respectively,
the number of students who play basketball and soccer only, soccer and volleyball only,
and volleyball and basketball only. Finally, let x be the number of students who play all
three sports. Then we have the following:
b + m + p + x = 21
s + m + n + x = 18
v + n + p + x = 15
Adding all these equations gives
3x + 2(m + n + p) + (b + s + v) = 54
or
2x + (m + n + p) + (x + m + n + p + b + s + v) = 54. So 2x + 6 + 40 = 54, 2x = 8,
and x = 4.
18.
Answer is 9009. The largest two-digit number has 9801 as its square. Thus, the
palindrome that we seek is at most 9779, and the next several (smaller) possibilities have
the form 9xx9. All four-digit palindromes are divisible by 11, so at least one of the twodigit factors of the number must be a multiple of 11. Examine the two-digit factor pairs,
beginning with 11 • 9 = 99 as one of the factors: 99(98), 99(97), 99(96), . . . , 99(90). Of
all these products, only one has 9 in the units place: 99(91). The product is 9009, the
largest palindrome that is the product of two two-digit numbers.
19.
Answer is 15/4. The area of a triangle is the product of a side times the altitude to that
side, so we can say that in any given triangle a side and its corresponding altitude are
inversely proportional. If we name the shortest two sides a and b and the longest side c,
we know by the triangle inequality theorem that a + b > c. Thus, 1/6 + 1/10 > 1/h, so
1/h < 8/30 = 4/15, and h > 15/4.
20.
Answer is 859 and 883. We can begin by dividing 1742 in half to get 871, which is
divisible by 13. The primes surrounding 871 are 863 and 877. However, 1742 – 863 =
879, which is not prime, and 1742 – 877 = 865, which is also not prime. Thus, we must
look at the next possible pair of primes—namely, 859 and 881. The partner of 881 is 861,
which is not prime, but the partner of 859 is 883, which is prime.