+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.
+2236 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:
200 Kg of berries are 99% water. The weight of the water is then(0.99⋅200)=198 kgLet x be the weight of the water lost after exposure to the sun. (0.99⋅200−0.98(200−x))=x198−(196−0.98x)=x198−196+0.98x=x2+0.98x=x1+0.98x−0.98x=x−0.98x2=0.02xx=100 200−x=200−100=100 kg←The total weight of the berries after 2 days
GA
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