Challenges cover a variety of topics
Today’s challenge requires the concept of Algebra.
5 glasses contain amounts of various liquids. The glasses are numbered 1 to 5. Non-prime glasses contain liquid A, and prime glasses contain liquid B. Odd glasses contain 750mL, and even ones contain 500mL. When you mix 2 glasses they then contain the same quantity of liquid in the same proportions. The challenge is to find the minimum number of mixes to ensure that each glass contains the same quantity of liquid A.
We wanted to find the minimum number of points of each colour for a set. The most important thing to notice is that if one type of point connects to another type at a different frequency than vice versa, than the ratio of their sizes is the inverse of that difference. Put simply, if red points connect to 2 orange points and each orange point connects to 1 red point, there are twice as many orange points as red points. Repeat this to find a collective ratio of 1:2:4:3:1:1. Since red points connect to 1 other red point, we can thus find our answer. There is a minimum of 2 red points, 4 orange points, 8 yellow points, 6 green points, 2 blue points, and lastly, 2 purple points.
Graph theory and algebra are required to solve this challenge.
Our challenge begins with a set of points. There are points of red, orange, yellow, green, blue, and purple. These points are all interconnected as follows. Every red point connects to 1 red point, 2 orange points, and 3 green points. Each orange point connects to 1 red point, 2 orange points, 2 yellow points, and 1 blue point. All yellow points connects to 1 orange point, 1 other yellow point, 3 green points, and 1 purple point. Green points all connect to 1 red point, 4 yellow points, and 1 blue point. All of the blue points connect to 2 orange points, 3 green points, and 1 purple point. And finally, purple points connect to 4 yellow points, 1 blue point, and 1 purple point. The question is this: what is the minimum number of points for each colour?
In this challenge, we wanted to find the antiderivative of f(x), given the above, and that the derivative of h(x) is 0.5. First, lets find f(x) in terms of h(x).
f(x)=-1/h(x)Then we can find the derivative of f(x) to be -2. The antiderivative of this, f(x), is -2x+c. The antiderivative of f(x), therefore, is -x^2+cx+c
-x^2+cx+cCalculus is required to solve today’s challenge.
let:f(x)=1/g(x),h(x)=-g(x)Today’s challenge is to find the antiderivative of f(x), given that the derivative of h(x) is 0.5. Remember that both functions are related to g(x).
We wanted to know how many kilograms of hydrogen Joanne the scientist has. We know that she has 9kg of pure H₂O, no isotopes. The atomic weight of oxygen is 16, and the atomic weight of hydrogen is 1. Since there are 2 hydrogen atoms in H₂O, the ratio of hydrogen to oxygen weight is 2:16, or 1:8. This makes the hydrogen portion of total weight 1/9. Since Joanne has 9kg of water, she has exactly 1kg of hydrogen!
This challenge challenges your algebra skills.
Joanne is doing a science experiment. She is trying to purify water so that there are no isotopes. This means that there will be only normal H₂O molecules. If Joanne has 9kg of water after all of this, how many kilograms of hydrogen atoms does she have?
We wanted to find the length of the side which is comprised of the non-shared sides. The first thing we need to note is that, as stated in the problem, the angle opposite 30° is also 30°. We will later be able to use the law of sines to solve for the lower length. As for the angle opposite the 80°, it has to be exactly 100°. We know this because there are exactly 2 possible triangles, as mentioned in the problem, and we have both. And we can use the sine law again to find the length of the side opposite the newly-proven 100° angle. Doing this will allow us to get the answer, approximately 47.27cm.
Solving this challenge requires an understanding of geometry, trigonometry, and angle theorems.
In today’s challenge, we face two triangles who are not congruent. This is surprising because they appear identical, and we are told they share two sides. We can find that they also share and angle. However, because the angle isn’t between the sides, it’s only SSA. SSA is not a valid congruency theorem, because in some cases it gives two possible solutions. In this case, we have both of them, directly connected. The question is this: what is the length of the line that is cut unevenly? That is, the one whose length equals the sum of the sides that the triangles do not share.
In this challenge, we wanted to find the maximum number of connections that we could add within the given conditions. Those conditions were only connecting on dotted lines, and exactly on path from A to B that goes through all other points. Note that there may be other paths from A to B, but those may not go through every point. The first step to solving this is to consider the points on the right. If we end with them, then we must come from the upper one. Then we can go to the lower. That’s because the upper one cannot reach B. If we do this, we can have an extra connection between the lower points. That one connection is needed to reach the maximum. The rest follows quickly. We are coming through the middle, so the left points must connect. We have to include the right points, so the others can all connect to B. At this point, we must remove as few outer connections as possible on the left. That would be 1, from the upper left to the centre. Now we have our final answer: 13 connections!