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An experiment was set up to determine the percentage of water absorbed by raisins. If the mass of dry raisins was 40 g, and the mass of wet raisins was 45 g, then the percentage of water absorbed would be

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An experiment was set up to determine the percentage of water absorbed by raisins. If the mass of dry raisins was 40 g, and the mass of wet raisins was 45 g, then the percentage of water absorbed would be

(1) \(\frac{45g}{40g}\times 100\)

(2) \(\frac{40g}{45g}\times 100\)

(3) \(\frac{(45 - 40)g}{40g}\times 100\)

(4) \(\frac{(45 - 40)g}{45g}\times 100\)

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Correct answer: (3) \(\frac{(45 - 40)g}{40g}\times 100\)

Explanation:

Percentage of water absorbed by raisins

= \(\frac{W_2 - W_1}{W_1} \times 100\)

The percentage of water absorbed in the above case would be

= \(\frac{(45 - 40)g}{40g}\times 100\)

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