Math Should Never Be Boring! More Math Brain-Teasers (Free ... - Free Printable

1) Fill the 5-gallon container. Pour water from it into the 3-gallon container until the 3-gallon is full, leaving 2 gallons in the 5-gallon container. Empty the 3-gallon container. Pour the remaining 2 gallons from the 5-gallon container into the 3-gallon container. Fill the 5-gallon container again. Pour water from the 5-gallon container into the 3-gallon container (which already has 2 gallons) until it's full. This requires 1 gallon, leaving exactly 4 gallons in the 5-gallon container. Empty the 3-gallon container. Pour the 4 gallons from the 5-gallon container into the 3-gallon container, filling it with 3 gallons and leaving 1 gallon in the 5-gallon container.

2) 5 socks. In the worst case, you pull out one sock of each color first (black, blue, brown, red). The fifth sock you pull out must match one of the first four, guaranteeing a pair.

3) Put it in front of a mirror. The reflection will show another dollar, effectively doubling it visually.

4) All months have at least 28 days.

5) 7 cartons. Let L be the number of large boxes and S be the number of small boxes. We have 8L + 10S = 96 and L > S. Solving for integer solutions: if L=7, then 8*7=56, so 10S=40, S=4. Since 7>4, this satisfies the condition. Total cartons = L + S = 7 + 4 = 11? Wait, re-read: "how many cartons did he ship?" Each carton holds either 8 large or 10 small. So total cartons = L/8 + S/10? No, that doesn't make sense. Re-interpret: He shipped some cartons containing large boxes and some containing small boxes. Let C_L be cartons of large boxes, C_S be cartons of small boxes. Then 8*C_L + 10*C_S = 96, and 8*C_L > 10*C_S. Also, C_L and C_S are integers. Try C_L=7: 8*7=56, then 10*C_S=40, C_S=4. Check 56>40? Yes. Total cartons = 7+4=11. But the answer might be different. Try C_L=2: 16, then 10*C_S=80, C_S=8. 16>80? No. C_L=4: 32, 10*C_S=64, C_S=6.4, not integer. C_L=5: 40, 10*C_S=56, C_S=5.6, no. C_L=6: 48, 10*C_S=48, C_S=4.8, no. C_L=7: 56, C_S=4, as above. C_L=8: 64, 10*C_S=32, C_S=3.2, no. C_L=9: 72, 10*C_S=24, C_S=2.4, no. C_L=10: 80, 10*C_S=16, C_S=1.6, no. C_L=11: 88, 10*C_S=8, C_S=0.8, no. C_L=12: 96, C_S=0. 96>0? Yes, but are there more large than small? 96 large, 0 small, so yes. Total cartons=12. But the problem says "more large boxes than small boxes", which is true for both (56>40 and 96>0). But perhaps they mean strictly more than zero small boxes? The problem doesn't specify. However, in the context, probably they expect the solution with both types. But let's see the answer. Perhaps I misread. "he sent a total of 96 boxes" – meaning 96 individual boxes. And "more large boxes than small boxes". So 8C_L + 10C_S = 96, 8C_L > 10C_S. With C_L=7, C_S=4: 56 large, 40 small, 56>40, total cartons=11. With C_L=12, C_S=0: 96 large, 0 small, 96>0, total cartons=12. But perhaps the problem implies he used both types? Or maybe not. But 11 is a common answer. Let me check if there are other solutions. C_L=2, C_S=8: 16 large, 80 small, 16>80? No. C_L=4, C_S=6.4, invalid. Only integer solutions are (C_L,C_S) = (7,4), (12,0), and also (2,8) but 16<80, (0,9.6) invalid, (1,8.8) invalid, (3,7.2) invalid, (5,5.6) invalid, (6,4.8) invalid, (8,3.2) invalid, (9,2.4) invalid, (10,1.6) invalid, (11,0.8) invalid. So only (7,4) and (12,0). Now, "more large boxes than small boxes" – for (12,0), it's true, but perhaps the problem intends for there to be small boxes. However, the question is "how many cartons did he ship?" For (7,4), it's 11. For (12,0), it's 12. But let's see the context. Perhaps I need to find which one is intended. Maybe "more large boxes than small boxes" means the number of large boxes is greater than the number of small boxes, which is satisfied by both, but perhaps they want the minimum cartons or something. But the problem doesn't specify. However, in many such problems, they expect the solution with both types. But let's calculate: 8*7 + 10*4 = 56+40=96, and 56>40, so it works. Total cartons 11. I think 11 is the intended answer.

6) 12 jumps. The hole is 14.5 feet deep. Each jump gains 2 feet net (jumps 3, slides back 1). After 11 jumps, he is at 22 feet? No, after n jumps, he is at 2n feet. After 11 jumps, he is at 22 feet? But the hole is only 14.5 feet, so he would be out before 11 jumps. Let's think: after first jump, he is at 3 feet, then slides to 2 feet. After second jump, from 2 to 5, slides to 4. After third, to 7, slides to 6. After fourth, to 9, slides to 8. After fifth, to 11, slides to 10. After sixth, to 13, slides to 12. After seventh, he jumps from 12 to 15, which is above 14.5, so he gets out without sliding back. So 7 jumps. Let me verify: start at 0. Jump 1: to 3, slide to 2. Jump 2: to 5, slide to 4. Jump 3: to 7, slide to 6. Jump 4: to 9, slide to 8. Jump 5: to 11, slide to 10. Jump 6: to 13, slide to 12. Jump 7: to 15, which is greater than 14.5, so he escapes. So 7 jumps.

7) Let B be boy's current age, S sister's, F father's. B = 2S, B = F/2, so F = 2B = 4S. In 50 years, sister will be S+50, father will be F+50 = 4S+50. Given that S+50 = (1/2)(4S+50). So S+50 = 2S + 25. Then 50 - 25 = 2S - S, so S=25. Then B=2S=50. So the boy is 50 years old now.

8) Both have two faces (sides).

9) The letter 'e'. It appears once in "minute", twice in "week", and once in "year".