Free MOEMS Olympiad practice test — Grade 8

Correct answer: B

First, Tom will fill the 6 larger cartons. Since each can hold 8 boxes of pens, 6 * 8 = 48 boxes can be packed into these. However, Tom only has 12 boxes, which is less than 48, so he won't even fill one large carton completely. Since he must use all the larger cartons first but can't because there aren't enough boxes, we need to rethink the approach based on the requirement to use all larger cartons first. Given Tom has 12 boxes and 6 large cartons can hold 8 boxes each, but he can't fill them all, let's correct the approach: If Tom fills the large cartons first, he can fill 1 large carton (since 12 boxes / 8 boxes per carton = 1.5, but he can't fill a fraction of a carton with whole boxes), with 4 boxes left over (12 total boxes - 8 boxes in one large carton). The remaining 4 boxes can then be placed into small cartons, requiring 1 small carton since each can hold 4 boxes. 💡 Tip: Carefully analyze the constraints and requirements of the problem to avoid misinterpreting the conditions.

Correct answer: C

This problem requires considering the different combinations of items (tents and sleeping bags) that can be brought by the friends. Since each friend can bring one or both items, but the total must be exactly 5 items, we have to consider various cases of how many friends bring one item and how many bring two, while also considering the limitation that each item can only be brought once. This involves both casework and logical deduction to ensure all possibilities are counted. 💡 Tip: Break down the problem into cases based on the number of friends bringing two items and those bringing one item, ensuring the total is exactly 5 items.

Correct answer: B

Let's denote the width of the garden as w. Since the length is 5 meters more than the width, the length is w + 5. The perimeter of a rectangle is given by P = 2l + 2w, where l is the length and w is the width. So, 2(w + 5) + 2w = 30. Simplifying, we get 2w + 10 + 2w = 30, which gives us 4w + 10 = 30. Subtracting 10 from both sides, we have 4w = 20. Dividing both sides by 4, we get w = 5. Therefore, the length is 5 + 5 = 10 meters. The area of the garden is length * width = 10 * 5 = 50 square meters. However, considering the given options and reevaluating the calculation: 4w = 20, so w = 5, and thus the length is 10. The area is indeed 50 square meters, but this matches an option, so it must be correct as per the given options. 💡 Tip: Use algebra to represent the unknowns and set up an equation based on the given information.

Correct answer: C

To find the smallest number that leaves a remainder of 1 when divided by 3, 4, and 5, we look for the least common multiple (LCM) of 3, 4, and 5, which is 60, and then add 1 to it. However, 60 itself is divisible by all three, so the number right after it, which is 61, would leave a remainder of 1 when divided by each of these numbers. But the question specifically asks for the smallest such number, and since 60 is the LCM, adding 1 gives us the smallest number that meets the criteria, but we must verify if 60+1 indeed satisfies the condition given, which it does. 💡 Tip: Understand the concept of the least common multiple (LCM) and how it applies to problems involving remainders.

Correct answer: B

First, find the rate at which each pipe fills the tank per hour. Pipe A fills 1/4 of the tank per hour, and Pipe B fills 1/6 of the tank per hour. When both pipes are open, their combined rate is 1/4 + 1/6 = (3+2)/12 = 5/12 of the tank per hour. In 2 hours, they fill 2 * (5/12) = 10/12 = 5/6 of the tank. This means 1/6 of the tank remains to be filled by Pipe B. Since Pipe B fills 1/6 of the tank per hour, it will take 1 hour to fill 1/6 of the tank, but considering it's filling the remaining 1/6 of the tank, the actual time needed for Pipe B to fill the rest after Pipe A is closed is based on its rate, so it's 1/6 tank / (1/6 tank per hour) = 1 hour for the remaining part, but given the setup, this explanation initially misleads by implying a direct calculation error; correctly, after filling 5/6 of the tank, 1/6 remains, and at Pipe B's rate of 1/6 tank per hour, it indeed takes 1 hour to fill this remaining part, aligning with option calculations considering combined work rates and remaining work. 💡 Tip: Calculate the combined work rate of both pipes and then determine the remaining work for Pipe B after Pipe A is closed.