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There are two piles of coal with a total of 19 tons. If 40% of the coal is transported from the first pile and three tons are transported from the second pile, then the weight is equal. How many tons of coal is there in the first pile
1÷(1-40%)=1÷60%=5/3
(19-3) / (5 / 3 + 1) = 16 / 8 / 3 = 6 tons
The second pile was 6 + 3 = 9 tons
The first pile was 19-9 = 10 tons
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The total weight of two pairs of coal is 27 tons. If two fifths of the first ton is transported and three tons of the second pile is transported, the weight of the two piles of coal is equal
I can answer it,
(27-3) / [(3 / 5) + 1] = 15 tons (a)
B: 27-15 = 12 tons
There are 78.5 tons of coal in the two piles. After the first pile carries four fifths of the coal and the second pile carries three fourths of the coal, the remaining weight of the two piles of coal is exactly the same
Less tons?
Suppose the first pile of coal has x tons, then the second one has (78.5-x) tons
（1-0.8）x=（1-0.75）（78.5-x）
x=43.61
A: the first batch of coal was about 43.61 tons
I sold two fifths of the total yesterday and two thirds today. How many kilos did I sell in two days
14 out of 15 of the total. As for how many kilos, you can multiply by the total!
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