Question

A farmer has a truckload of watermelons to take to market. At the beginning of the trip, the watermelons weigh 1000 kg. 99% of this weight is made up of water. At the end of the trip. due to dehydration, 98% of the weight of the watermelons is made up of water. How much does the truckload weigh at the end of the trip?

Answer 1

Answer:

The weight of watermelon at the end of the trip is 500 kg.

Step-by-step explanation:

Given:

The weight of watermelons is 1000kg.

At the beginning of the trip watermelon is 99% of the weight:

99% of 1000kg

$\frac{99}{100}\times 1000$

$=0.99\times 1000$

$= 990$ kg

Now, at the end of the trip it is 98% of weight. So, let $x$ kg of water has been evaporated. So, remaining weight of the watermelon is $1000-x$.

If $x$ kg of water is removed, then the amount of water left is $990-x$.

As per question, water now weighs 98% of the total weight. So,

98% of ($1000-x$) = $990-x$

$\frac{98}{100}\times (1000-x)=990-x$

$980-0.98x=990-x$

$-0.98x+x=990-980$ (Bringing like terms on one side)

$0.02x=10\\x=\frac{10}{0.02}=500\ kg$

Therefore, the weight of watermelon at the end of the trip is given as:

Weight = $1000 - x$ = 1000 - 500 = 500 kg

Therefore, the watermelon weighs 500 kg at the end of trip.

Answer 2

Answer:

500 kg

Step-by-step explanation:

At the beginning of the trip, the watermelons weigh 1000 kg.

99% of this weight is made up of water.

So, in 1000 kg of watermelons there are 990 kg of water and 10 kg of solid.

If after the trip due to dehydration x kg of the weight reduces, then in (1000 - x) kg of watermelons there are 10 kg of solid and (990 - x) kg of water.

Now, given that the % of the water in the load is 98%.

So, we can write that $\frac{990 - x}{1000 - x} = \frac{98}{100}$

⇒ 99000 - 100x = 98000 - 98x

⇒ 2x = 1000

⇒ x = 500

Therefore, the track load at the end of the trip will be (1000 - x) = 500 kg. (Answer)