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Fresh fruits contain 75% while dry fruits contain 20% water. If the weight of dry fruits is 300 kg what was its total weight when it was fresh?

Doubt: I really do not understand what to equate.

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    $\begingroup$ I would start by looking at what we have right now. We have a pile of fruits (incl. 20% water) which weighs 300 kg. We know that 80% of that is the "pure" fruits, so that's 0.8*300=240 kg. Now, let's look at the initial state. In the initial state, the weight of the "pure" fruits was still 240 kg but that's only 25% of the total mass. If 240 kg is 25% of the total mass, what is the total mass? $\endgroup$
    – Matti P.
    Aug 4, 2021 at 9:28

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Let $x_1 = 300$ (kg) and let $x_2$ be the total mass you are looking for.

From your first sentence, you can write $(1-0.75)x_2 = (1-0.20)x_1$. This is the ratio of pure fruits they both have. So from $x_1$, $80$% is pure fruit, which is $240$. This gives $0.25x_2 = 240$, which leads to $x_2 = $ total mass $ =960$kg

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