A farmer has 10 tons of watermelons stored in a barn. The
watermelons contain 99% water, by weight. During storage, the
melons dry out so that their water content is decreased to 98% of their
new weight.
What is the weight of the watermelons now?
A farmer has 10 tons of watermelons stored in a barn. The
watermelons contain 99% water, by weight. During storage, the
melons dry out so that their water content is decreased to 98% of their
new weight.
What is the weight of the watermelons now?
This is pretty much just arithmetic. There are 10 tons of watermelons and the are 99% water. That means they are (10)(.99)= 9.9 tons water, 0.1 tons "non-water". The melons "dry out", losing only water so that their water weight is 98% of their new weight. Let the weight of water lost, in tons, be "w". Then their total weight is 10- w and their water weight is 9.9- w. That is a ratio of $\displaystyle \frac{9.9- w}{10- w}= .98$. Solve for w. The new weight of the water melons is 10- w tons.
Hello, SuperMaruKid!
A farmer has 10 tons of watermelons stored in a barn.
The watermelons contain 99% water, by weight.
During storage, the melons dry out, so that their water content
is decreased to 98% of their new weight.
What is the weight of the watermelons now?
Originally, the watermelons are 99% water and 1% solids.
The farmer has 20,000 pounds of watermelons
The amount of solids is: $\displaystyle 1\% \times 20,\!000 \:=\:200\text{ pounds.}$
When the melons are reduced to $\displaystyle x$ pounds, they will be 2% solid.
. . $\displaystyle 2\% \times x \:=\:0.02x\text{ pounds}$
The amount of solids remains constant: .$\displaystyle 0.02x \,=\,200$
Therefore: .$\displaystyle 0.02x \:=\:200 \quad\Rightarrow\quad x \:=\:10,\!000\text{ pounds} \:=\:5\text{ tons}$