Method 1 Suppose the big monk eats three, and the little monk eats 1 ÷ 3 = 1/3. Assuming that each young monk is also three, then 100 monks need to eat * * *: 3x100 = 300 (steamed bread) and the actual ratio, one * * * eats too much: 300- 100 = 200 (one steamed bread); A young monk ate too much: 3- 1/3 = 8/3 (a steamed bread); Number of young monks: 200 ÷ 8/3 = 75 (one person); And the number of big monks. 1 A big monk and three little monks eat four steamed buns, which means that every four steamed buns are just given to 1 a big monk and three little monks. We might as well divide every four 100 steamed buns into one group, * * can be divided into: 100÷4=25 (group), and 100 monks are just divided into 25 groups. Each group must have 1 big monk and 3 little monks, so the big monk * * has: 1×25=25 (one), and the little monk * * has: 3×25=75 (one), while this spider has 8 legs, the dragonfly has 6 legs and 2 pairs of wings, and the cicada has 6 legs. Now these three bugs * * 60. How many are there in each? Scheme 1: Let Spider A be 8a+6 (18-a) =118, and get a=5. Then, let dragonfly and cicada *** 18-5= 13 only set dragonfly B+ 13-b = Method 2: If you haven't learned to solve the equation, you can solve the hypothesis method like this. Let's assume that all bugs have only six feet, and the number of feet is 18× 6 = 108. The extra feet are spiders' (118-108) ÷ (8-6) = Similarly, suppose that dragonflies and cicadas only have a pair of wings (1× (18-5) =/kloc. The number of dragonflies is 7 cicadas (18). Give a website/view/0b3aca650b1c59eef8c7b418.html, and hope to adopt it. O(∩_∩)O Thank you for asking: add me qq22 1732474 1.