I carried 1000,kg of watermelon in summer by | vedclass

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(D) Initially,the total weight of the watermelon is $1000\,kg$. Since the water content is $99\%$,the weight of the solid part (non-water) is $1\% 	e

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$500$

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(D) Initially,the total weight of the watermelon is $1000\,kg$. Since the water content is $99\%$,the weight of the solid part (non-water) is $1\% 	ext{ of } 1000\,kg = 0.01 	imes 1000 = 10\,kg$. Let the new total weight of the watermelon at the destination be $x\,kg$. At the destination,the water content is $98\%$,which means the solid part is $2\% 	ext{ of } x$. Since the weight of the solid part remains constant,we have: $0.02 	imes x = 10\,kg$. $x = \frac{10}{0.02} = 500\,kg$. The reduction in weight is $1000\,kg - 500\,kg = 500\,kg$.

$I$ carried $1000\,kg$ of watermelon in summer by train. In the beginning,the water content was $99\%$. By the time $I$ reached the destination,the water content had dropped to $98\%$. The reduction in the weight of the watermelon was $.......\,kg$.

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