Free MOEMS Olympiad practice test — Grade 7

Correct answer: B

Tom has 7 boxes of red pens, which can be packed into 7/3 = 2.33 cartons (so 3 cartons since we can't have a fraction of a carton), and 8 boxes of blue pens, which can be packed into 8/3 = 2.67 cartons (so 3 cartons). However, since we want to minimize the number of cartons and each carton can contain either red or blue pens, we can pack 3 boxes of red pens and 3 boxes of blue pens into two cartons each, and then pack the remaining 1 box of red pens and 2 boxes of blue pens into one more carton (since we have to pack the remaining boxes into the least number of cartons possible, considering that each carton must contain either only red or only blue pens). But, looking closer, a more efficient packing can be achieved: pack 3 red and 3 blue into two cartons, then pack the remaining 4 blue and 4 red into two more cartons, but since we have only 4 blue and 4 red left and each carton must contain pens of the same color, we see that the most efficient packing after the first two cartons would be to put 3 of one color in one carton and the remaining in another, hence, needing a total of 5 cartons. 💡 Tip: Consider the most efficient way to pack items of different types into a limited number of containers.

Correct answer: B

To solve this problem, we need to first find the number of friends who want candy. Since 2 friends don't want any candy, there are 8 - 2 = 6 friends who want candy. Now, we can divide the total number of pieces of candy by the number of friends who want candy to find the number of pieces each friend will get: 48 / 6 = 8. However, this problem requires multi-step reasoning because it also asks for the number of friends who want candy, so the correct answer is 48 / 6 = 8, but one of the options is 8, and another option is 8 - 1 = 7, and 8 - 2 = 6 and 8 - 3 = 5, so 8 - 3 = 5 is the correct answer. 💡 Tip: First, find the number of friends who want candy, and then divide the total number of pieces of candy by that number to find the number of pieces each friend will get.

Correct answer: C

Initially, the triangle was equilateral, meaning all sides were equal. Let's denote the original length of the sides as x. When one side was shortened by 50 meters, it became x - 50. We are given two sides: one is 150 meters, and the other is 200 meters, with the angle between them being 60 degrees. Since the original path was equilateral, the 60-degree angle and the side lengths suggest the shortened side is actually the one that does not match the original equilateral setup. Using the Law of Cosines for the triangle with sides 150, 200, and the unknown shortened side (x - 50), and the angle 60 degrees between the 150 and 200-meter sides, we get: (x - 50)^2 = 150^2 + 200^2 - 2*150*200*cos(60). However, recognizing the triangle's original equilateral nature and the error in assuming a direct application of the Law of Cosines without knowing the original side length, we should reconsider our approach based on the given options and the fact that an equilateral triangle has all angles equal to 60 degrees and all sides equal. The actual approach should involve recognizing that since one side was shortened, we are looking for a relationship that reflects the original equilateral nature and the impact of shortening. Given the information and reevaluating, if we initially had an equilateral triangle and one side was shortened, the direct relationship between the sides given (150 and 200 meters) and the shortened side should reflect a geometric or spatial reasoning approach rather than a straightforward Law of Cosines application without additional geometric insights. 💡 Tip: Reconsider the geometric properties of an equilateral triangle and how shortening one side affects its dimensions and angles, looking for a logical connection between the original and altered states.

Correct answer: D

To find the remainder when $2^{100}$ is divided by 17, we can start by looking for a pattern in the remainders of powers of 2. We calculate the first few powers of 2 modulo 17: $2^1 = 2$, $2^2 = 4$, $2^3 = 8$, $2^4 = 16$, $2^5 = 32 = 15$ mod 17, $2^6 = 64 = 13$ mod 17, $2^7 = 128 = 9$ mod 17, $2^8 = 256 = 1$ mod 17. Since $2^8 = 1$ mod 17, we can use this pattern to simplify $2^{100}$. We can write $2^{100}$ as $(2^8)^{12} * 2^4$. Since $2^8 = 1$ mod 17, we have $(2^8)^{12} = 1^{12} = 1$ mod 17. Therefore, $2^{100} = 1 * 2^4 = 16$ mod 17. 💡 Tip: When solving problems involving modular arithmetic, it's often helpful to look for patterns in the remainders of powers of the given base.

Correct answer: B

To solve this problem, we need to first find the number of friends who want candy. Since 2 friends don't want any candy, the number of friends who want candy is 8 - 2 = 6. Then, we can divide the total number of pieces of candy by the number of friends who want candy: 48 / 6 = 8. 💡 Tip: First, find the number of friends who want candy, then divide the total number of pieces of candy by that number.