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Menagerie

Logic puzzles require you to think. You will have to be logical in your reasoning.

 Puzzle ID: #32823 Fun: (2.32) Difficulty: (3.03) Category: Logic Submitted By: cnmne

Zachary has a private zoo. He has five groups of animals in his zoo: snakes, birds, mammals, insects, and spiders. Assume that, typically: animals have 1 head, snakes have 0 legs, birds have 2 legs, mammals have 4 legs, insects have 6 legs, and spiders have 8 legs. Zachary has some unusual animals in his zoo. He has: a snake with 3 heads, a bird with 2 heads, a mammal with 3 legs, an insect with 4 legs, and a spider with 7 legs. From the following information, determine how many of each group of animals that Zachary has in his menagerie.

1) There are a total of 100 heads and 376 legs.
2) Each group has a different quantity of animals.
3) The most populous group has 10 more members than the least populous group.
4) There are twice as many insect legs as there are bird legs.
5) There are as many snake heads as there are spider heads.

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 scallio Sep 03, 2006 Owww... my brain hurts. That took way too long! Great teaser! jazzmusician46 Sep 06, 2006 Wow! rachayl Sep 14, 2006 Wow, that was algebralicious Cindy650 Nov 07, 2006 5599277Wow javaguru Mar 20, 2009 The difficulty rating on this one seems kind of high. The problem solves much more easily and elegantly than the given solution. As pointed out, there are 97 animals, or an average of between 19 and 20 of each. Since the difference between the fewest and most is 10, the fewest is around 14 to 15 and the most is around 24 to 25. First give the unfortunate animals back their missing legs, making 380 legs. So where A = arachnids (spiders), I = insects, M = mammals and B = birds, the formula for the number of legs is: 8A + 6I + 4M + 2B = 380 4A + 3I + 2M + B = 190 So I and B must either both be even or both be odd. The relationship between birds and insects is: 4B = 6I - 2 B = (3I - 1)/2 This forces I to be odd, so I and B are both odd. Plugging in odd values in the range for I gives: I = 13, B = 19 I = 15, B = 22 I = 17, B = 25 I = 19, B = 28 Given these choices, it's probably I = 17, B = 25. It could be I = 13, B = 19, but that means that the most would be 23 and the other two need to total 97 - 55 = 42, constraining them to (19,23), (20,22) and (21,21), and since there are two more snakes than spiders, (19,23) won't work. So try I = 17, B = 25. 4A + 51 + 2M + 25 = 190 4A + 2M = 114 2A + M = 57 So M has to be odd, and the least has to be 15. Since there are two more snakes than spiders, there have to be either 15 mammals or 15 snakes. Try 15 for M first since you already know M has to be odd: 2A + 15 = 57 2A = 42 A = 21 S = A - 2 = 19 (snakes) Check it: A + I + M + B + S = 97 21 + 17 + 15 + 25 + 19 = 97 Took a couple minutes, but it solved easily with very little trial and no error. javaguru Mar 20, 2009 I got my spiders and snakes mixed up twice above...there are more spiders than snake, not snakes than spiders. grilled_cheese3 Jun 04, 2009 that was the easiest teaser ive solved. by the way i never got a lower than an a on my report card and got a letter from the president saying im the smartest kid in my school district. elementary school that is. javaguru Jun 06, 2009 What does being smart have to do with this being "easiest". You must not have done many of the teasers here as there are some really stupid simple teasers here. Oh, and I'd really be surprised if the school district sent a letter proclaiming someone as the smartest kid in the district.