____________________________________________________________ There are three cookie bakers, Alice, Bob, and Cindy, in the village. Alice works twice as fast as Bob does, and Bob works twice as fast as Cindy does. During the past 3 hours, the three cookie bakers together made 560 cookies. How many of those cookies were made by Bob?
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m is the smallest integer such that 2016m is a perfect square. n is the smallest integer such that 2016/n is a perfect square. What is m/n?
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An alloy consists of three metals X, Y, and Z. We know the density of X,Y and Z are 3 g/cm3 , 6 g/cm3 , and 8 g/cm3 , respectively. The density of the alloy, which consists of 6 g of X, 18 g of Y, and t g of Z, is 6.4 g/cm3 . What is t? (Assume mixing the metals will not change their combined volume)
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Compute 1 − 3 + 5 − 7 + 9 − 11 + · · · + 97 − 99 + 101
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Suppose that ABC4 + 200 = ABC9, where A, B, and C are valid digits in base 4 and 9. What is the largest possible value of A + B + C in base 10?
There are three cookie bakers, Alice, Bob, and Cindy, in the village. Alice works twice as fast as Bob does, and Bob works twice as fast as Cindy does. During the past 3 hours, the three cookie bakers together made 560 cookies. How many of those cookies were made by Bob?
Call the number of cookies that Cindy can make in one hour = N
Call the number that Bob can make in one hour = 2N
And let Alice make 2 (2N) = 4N
In therr hours they can make
3 ( N + 2N + 4N) = 560
3( 7N) = 560
21N = 560 divide both sides by 21
N = 80/3
So in 3 hours Bob can make 3(2N) = 3(2* 80/3) = 2 * 80 = 160
m is the smallest integer such that 2016m is a perfect square. n is the smallest integer such that 2016/n is a perfect square. What is m/n?
Factor 2016 = 2^5 * 3^2 * 7
We need to have m = 2 * 7 = 14 for 2106m to be a perfect square
And we need n to be same for 2016/n to be a perfect square
So.....m/n = 14/14 = 1
Compute 1 − 3 + 5 − 7 + 9 − 11 + · · · + 97 − 99 + 101
We will have ( 101 + 1)/ 2 = 51 terms
We can write this in a slightly different manner
(101 + 1) + (-99 - 3) + ( 97 + 5) + (-95 - 7) .......
We will have 50 pairings which will sum to 0
The middle term will be unpaired with any other term in this summation
And this term = 51
So.....the sum = 51
Suppose that ABC4 + 200 = ABC9, where A, B, and C are valid digits in base 4 and 9. What is the largest possible value of A + B + C in base 10?
We have
A(4)^2 + B(4) + C + 200 = A(9)^2 + B(9) + C
A( (9^2 - 4^2) + B(9-4) - 200 = 0
65A + 5B - 200 = 0
13A + B - 40 = 0
13A + B = 40
A = 3 B = 1
As large as C can be in base 4 = 3
A + B + C = 3 + 1 + 3 = 7