2011-2012 Mathcounts workout 9

#261: The answer is 241.3












#262: Let a students wear blue jerseys and b students wear red jerseys.
There are two cases to consider: one is 2 plays wear blue and 1 wears red or 2 wear red and 1 wear blue.
bC2 x aC1 + bC1 x aC2 = [b x (b - 1)]/(2!) times a  + b times [a(a-1)]/2!  = 25
Both sides times 2 and you have ab² - ab + ba² - ab = 50 ,  ab (b-1) + ab (a -1) = 50
ab (b - 1 + a -1) = 50,  ab(a + b -2) = 50 , Factor "50" and you see ab = 10 and a + b -2 = 5
works if the two numbers are 2 and 5. The answer is 2 + 5 = 7

#263: If 13 years ago the price is y, after 13 years, with the ticket price increases 4% each year, it would
be y x ( 1.04)¹³.
y/y x ( 1.04)¹³ = 0.60057 The answer is 60%

#264: Let the three sides be x, xr, and xr² and you want to find what
is x/xr², or 1/r²
Using Pythagorean theory, you have x² + (xr )² = (xr²)²
x² + x²r² = x²r⁴ Divide both
sides by x² and the equation becomes 1 + r² = r⁴
r⁴ - r² -1 = 0, r² = 1 + √ 5  / 2
1/r² = 1/ (1 + √ 5  / 2) =  0.62111 The answer is 0.62

#265:This one is very straightforward. The area of the rectangle is (3 + 3) x 3 =18
Minus the 2 quarter circle: 18 - 1/2 of (3² Π) = 3.8628... The answer is 3.9

#266: A number ends in 2 or 7 if when divided by 5, leaves a remainder of 2.
So check, 102, 107, 112, 117. etc.. to see if it leaves a remainder of 1 when divided by 3.
The divisible rule for 3 is the sum of all the digits is divisible by 3, so in this case you want to see if
the sum will leave a remainder of 1 when divisible by 3.
The number 112 fits but it doesn't leave a remainder of 3 when divided by 7.
Use 112 and add ( 3 x 5 ) each time to see if the next number fits the criteria.
112 + 15 = 127 (no)
127 + 15 = 142 (no)
142 + 15 = 157 -- yes so the answer is 157

#267: See the solution. The answer is 2.5 units.















#268: 21 x 5 = 105 (total sum)  Since the numbers are all different positive integers, the smallest median
is 3. The five numbers look like this  1 , 2, 3, ___, ___
The largest median is 1, 2, 33, 34, 35 [Use (105 - 1 - 2) divided by 3 to get the median of the larger 3 numbers. If not evenly divided, then you need to do some trial and error tests for that.].
From 3 to 33 inclusive there are 33-3 + 1 = 31 numbers.

#269: Since the time stays constant no matter what length unit you choose, we can choose the candle is 1 unit long. Let t represent the time, in hours, until the slower burning candle is twice the height of the faster burning candle.The faster candle burns completely in 6 hours so in 1 hour, 1/6 of the candle will burn. In t
hours, t times 1/6 or t/6 candle will burn leaving the length 1 - (t/6) long.
Similarly, the slower candle burns completely in 9 hours so in 1 hour, 1/9 of the candle will burn. In t hours, 1/9 of t or t/9 will burn, leaving only 1- (t/9) of the candle.
Now you can set up the equation and solve for t:
1 - (t/9) = 2 [ 1- (t/6) ] . Multiply both sides by 9 and you have 9 - t = 18 - 3t, 2t = 9 so t = 4.5 hours.

#270: Let x be the number of books Ms. Meany originally has, x + n would the total books she has later on.
(x + n) C 2 - xC2 = 99
(x + n) (x + n - 1)/ 2! - x (x -1)/2! = 99
Multiply both sides by 2 and you have (x + n) (x + n-1) - x (x -1) = 198
Consolidate and you have (x + n)² - (x + n) - x² -x = 198
Further consolidate and take out the factor and you have n (n + 2x -1) = 198 = 1 x 198 = 2 x 99 = 3 x 66
= 6 x 33 = 9 x 22 = 11 x 18
Add up all the n (those numbers in red) and you have 1 + 2 + 3+ 6 + 9 + 11 = 32