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A and B plant trees. It takes 13 more time for a to plant this batch of trees alone than B. if they work together, B will plant 36 more trees than a when they finish the task. How many trees are there in all?
If a takes 13 more time than B, the time ratio is 4:3, the efficiency ratio is 3:4, 36 (43 + 4-33 + 4), = 36 (47-37), = 36 (17), = 36 × 7, = 252 trees; a: there are 252 trees in total
A unit plans to plant trees on both sides of the two roads leading to the gymnasium. Now a batch of seedlings are transported back. It is known that the length of one road is more than 6000 meters longer than the other. If one tree is planted every 4 meters, it will be 2754 less. If one tree is planted every 5 meters, it will be 396 more. How many seedlings are there?
According to the analysis, four rows of trees are planted on both sides of the two roads, and the number of seedlings in each row is equal to the interval plus one, that is to say, the number of seedlings on one road is equal to two times of the interval plus two, so the number of seedlings on two roads is equal to four times of the interval plus four
I can see it clearly. There is a little doubt. After setting the total distance as s, because the two roads are disjoint, the number of trees on each road is 1 / 4 plus 1, or 1 / 5 plus 1. Trees should be planted at both ends of the two roads. If the combined distance is s, it seems that one tree is missing. Can you think like this? It's a headache, I don't know if I can think like this. If I'm wrong, please pay,
Suppose that the shorter road is s meter, the longer road is 2S + 6000, and the number of saplings is x2 * s / 4 + 2 + 2 * (2S + 6000) / 4 + 2 = x + 2754.12 * s / 5 + 2 + 2 * (2S + 6000) / 5 + 2 = x-396.2, then we get x = 13000 trees, x + 2754 = s / 4 * 4 + (4), x-396 = s / 5 * 4 + (4), and the 4 in brackets in the analytical expression of x = 13000 is the top of the two roads
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