Fruit can be preserved through drying to remove excess moisture.
The water content of a certain fruit, by mass, is \(70\%\). Therefore, \(30\%\) of the fruit, by mass, is other material.
When left in the sun to dry, the fruit loses \(80\%\) of its water content, and the amount of other material remains the same.
Rounded to the nearest tenth, what percent of the dried fruit is water?
Solution 1
Let’s consider a piece of fruit that originally weighs \(100\) g. Since \(70\%\) of the mass is water, that means that \(70\) g is water and \(30\) g is other material.
When left in the sun to dry, the fruit loses \(80\%\) of its water mass. So it loses \(80\%\) of \(70\) g \(= 0.8 \times 70 = 56\) g of water, and \(70 - 56 = 14\) g of water remains.
The dried fruit still contains \(30\) g of other material. Therefore, the dried fruit consists of \(14\) g of water and \(30\) g other material, for a total of \(44\) g.
Therefore, the dried fruit is \(\dfrac{14}{14+30} \times 100 \% = \dfrac{14}{44} \times 100 \% \approx 31.8\%\) water.
Solution 2
Suppose the fruit originally weighs \(x\) g. Since \(70\%\) of the mass is water, that means that \(70\%\) of \(x = 0.7 \times x = 0.7x\) g is water and \(30\%\) of \(x\) = \(0.3x\) g is other material.
When left in the sun to dry, the fruit loses \(80\%\) of its water mass. So it loses \(80\%\) of \(0.7x = 0.8\times 0.7x = 0.56x\) g of water, and therefore \(0.7x-0.56x = 0.14x\) g of water remains.
The dried fruit still contains \(0.3x\) g of other material. Therefore, the dried fruit consists of \(0.14x\) g of water and \(0.3x\) g of other material, for a total of \(0.44x\) g.
Therefore, the dried fruit is \(\dfrac{0.14x}{0.14x+0.30x} \times 100 \% = \dfrac{0.14x}{0.44x} \times 100 \% \approx 31.8\%\) water.