You will need an understanding of physics and calculus to solve this science challenge.
Diagram of the Spinning Stick.
A uniform stick with length 2 metres rotates at a speed of 1 rotation per minute on a flat plane. It comes into contact with a stationary ball. The stick is kept spinning about it’s origin at the exact same speed by a motor. The ball is only affected by the stick’s rotation, and does not experience friction or roll over the stick. The ball will eventually escape the stick from centrifugal force. The challenge is to find the velocity of the ball when it is flung off of the stick.
We wanted to find a way to factor out the 1 from √x+1 to find it’s derivative. It is impossible to factor out the 1. The answer is thus simply √x+1 / 2x+2.
This challenge covers the topics of calculus and algebra.
In calculus we often find the derivatives of things. The derivitave of 1, π, of any other constant c, is 0. The derivative of x is 1, and that of any constant multiplied by x is that constant. And then the derivative of x2 is 2. All common knowledge. But what about the derivative of √x? That would be √x / 2x, We multiply by the exponent and then decrease it by 1. This does work with x2. But what about the derivative of √x+1? Try to factor the 1 out of the radical. That is today’s challenge. Good luck!
We wanted to solve the Machiavellian Matrix. We can began by dividing the top row by xyz to simplify it to ( 1 1 1 1 ). Then we subtract row 1 from row 3 to fin that x = -1/3. This makes x2 = 1/9. Then we can get a system of equations for y and z, that y + z = 4/3, and y2 + z2 = 8/9. If y + z = 4/3 then z = 4/3 – y, and we can substitute to get the quadratic equation. Solve to get (y – 2/3) * (y – 2/3) = 0, so y = 2/3. Our answer is thus that x = -1/3, y = 2/3, and z = 2/3.
\left(\begin{matrix}
1 & 1 & 1 & 1 \\
x & y & z & 1 \\
4 & 1 & 1 & 0
\end{matrix}\right)
\overrightarrow{R_3-R_1→R_1}
\left(\begin{matrix}
3 & 0 & 0 & -1 \\
x & y & z & 1 \\
4 & 1 & 1 & 0
\end{matrix}\right) \\[10pt]
y^2+\left(\frac43-y\right)^2-\frac89=0
You will need serious matrix solving skills to solve today’s challenge.
Todays challenge is strait forward: solve the Machiavellian Matrix. Machiavellian means crafty and duplicitous, like a trickster. Sometimes this means breaking the rules or finding loopholes. With matrices, machiavellian often means horrifying. Variables are x, y, and z, sum on the right.
\left(\begin{matrix}
xyz & xyz & xyz & xyz \\
x & y & z & 1 \\
4 & 1 & 1 & 0
\end{matrix}\right)
We wanted to find what portion of the population owns a dog. First, we know that 60% of the population own at least one pet. Since all pets are randomly distributed, people with 1 pet must have a three fifths chance of owning a second pet, and so on. This makes the average number of pets per person equal to 3/5 + (3/5)2 + (3/5)3 + … . The general formula for a sequence like this is shown below. In this case, we get 1.5. Now, we make an equation. Let the ratio of dogs to cats to fish to people be x:y:z:1. We know that 3x = 2y, x+y+z = 1.5, and 6y / (1-y) = 5z / (1-z). We can easily substitute y to get x+1.5x+z = 1.5 and 9x / (1-1.5x) = 5z / (1-z). Substituting z is impossible. We must use a matrix. In this case it is long and difficult and we will not show it. The positive solution is x = 1/3, z = 2/3, and so y = 1/2. Since the ratio of dogs to people is 1:3, 1/(3+1) of the population must own dogs to make 1/(3+1-1) = 1/3. And so our answer is that 1/4 of the population owns at least 1 dog.
\sum_{k=1}^\infin\left(\frac ab \right)^k=\frac a{b-a}
You will need matrices and combinatorics to solve this challenge.
In a city with infinite people, there are only three types of pets. They are dogs, cats, and fish. 60% of people have a pet. The ratio of dogs to cats is 2:3, and the ratio of poeple with cats to those with fish is 5:6. All pets are randomly distributed. The challenge is to find what fraction of the people own dogs.
We wanted to find the probability that the row started by a pear ends with one. Since no conditions limit the number of fruits, there are infinite solutions. Our answer is thus one third.
You will need an understanding of algebra and combinatorics to solve today’s challenge.
Today’s challenge begins with 3 rows of fruits. They begin with an apple, and orange, and a pear. The fruits behind the front are unknown. Directly behind any apple, there is an equal chance of a plum, a banana, or an orange. There is no other fruit directly behind an apple, but one apple is at the end of a row. Behind any plum is the same fruit that was in front of it. Directly behind an orange, there is a one third probability there are grapes. There is an equal probality of pears and blueberries, and 1 orange leads to a banana. For any banana there is an equal chance of blueberries or a plum. 1 banana is followed by grapes, and another ends a row. Behind blueberries, there is a one third chance of a pear, otherwise grapes. Directly behind grapes, there is a 50% chance of more grapes, a 25% chance of a plum, and otherwise a pear. Finally, directly behind a pear there is the fruit of the colour in the rainbow following that of the fruit in front of the pair. The exception is the 1 pear that ends a row. The apples in this question are red, which is considered to follow purple, the colour of the plums. The fruit following the pair at the front is an apple. The challenge is to use this information to find the probability that the row started by a pear ends with one.
We wanted to find the probability, x, that Bob wins a game of Stupid Spin at Scam Casino. The probability Bob gets a point is a, otherwise Scam Casino gets one. The first to b points wins. First, the maximum spins in a round is 2b – 1, where 1 player wins by just 1 point. If we always play this many spins, you will always win at b points. We can simplify the problem to whomever has more points after 2b – 1 spins. Bob wins if the Casino has 0 to b-1 points. This gives us the first part of the equation. The probability of any given score for the Casino is equal to the number of ways that score can happen times the probability of it happening. The number of ways is easily done with the choose function. The probability of that is a to the power of the number of points scored by Bob, part three, multiplied by 1-a to the power of the number of points scored by the casino, the final part. We combine these 4 parts to get our answer, the equation below.
