+136 In a store window, there was a box of berries having a total weight of 200 kg. The berries were 99% moisture, by weight. After two days in the sun, the moisture content of the berries was only 98%, by weight. What was the total weight of the berries after two days, in kg?
1% of the moisture weight is 198*0.01 = 1.98 kg. Subtracting that from total moisture 198 - 1.98 = 196.02 kg.
Add back the berry content 2kg. Answer = 196.02 + 2 = 198.02 kg.
+2511 The above solution and solution method is wrong.
...a box of berries having a total weight of 200 kg. The berries were 99% moisture, by weight. After two days in the sun, the moisture content of the berries was only 98%, by weight. What was the total weight of the berries after two days, in kg?
Solution:
\(\text {200 Kg of berries are 99% water. The weight of the water is then}\\ \left(0.99\cdot 200\right) = 198 ~ kg\\ \text {Let x be the weight of the water lost after exposure to the sun. }\\ \left(0.99 \cdot 200-0.98(200-x)\right)=x\\ \begin{aligned} 198-(196-0.98x)&=x\\ 198-196+0.98x&=x\\ 2+0.98x&=x\end{aligned}\\ \begin{aligned}1+0.98x-0.98x&=x-0.98x\\ 2&=0.02x\\ x&=100\\ \end{aligned}\\ \text { }\\ 200-x=200-100=100 ~ kg \leftarrow \text {The total weight of the berries after 2 days}\\ \)
GA
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