Question

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) 45 g40 g×100
(2) 40 g45 g×100
(3) (45-40)g40g×100
(4) (45-40)g45 g×100

Answer 1

The percentage of water absorbed would be $(\frac{(45-40)g}{40g})\times 100$.

Explanation:

1. From given data

 Mass of dry raisins = 40g

 mass of wet raisins = 45g

2. Now

   Amount of water absorbed by raisins = increase in mass of raisins = (45-40)g

3. Now percentage of water absorbed by raisins = $(\frac{increase in mass of raisins}{Initial mass of raisins})\times 100$

4. So

Percentage of water absorbed by raisins = $(\frac{(45-40)g}{40g})\times 100$

Answer 2

The answer is Option (C)

Explanation:

Mass of dry ones = 40 g  

Mass of wet ones = 45 g

Water absorbed = 5 g (As 45-40 = 5)

Percentage = $\frac {increased\ or\ decreased}{original} \times 100$

                   = $\frac {45 g - 40 g}{40} \times 100$  

 So Correct is Option C