# Mr Farmer loaded a cart with fruit and set off to market. When he started his trip, 90%of the weight of the fruit in the cart was water.Mr Farmer loaded a cart with fruit and set off to market....

Mr Farmer loaded a cart with fruit and set off to market. When he started his trip, 90%of the weight of the fruit in the cart was water.

Mr Farmer loaded a cart with fruit and set off to market. When he started his trip, 90% of the weight of the fruit in the cart was water. The day was hot, by the time Mr Farmer reached the market, only 60% of the weight of the fruit in the cart was water. When he weighed his load, he found that it was 15 kg lighter than at the beginning of the day. What was the weight of the fruit in Mr Farmer loaded on the cart?

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Let 'x' be the weight of fruit when Mr Farmer loaded the cart.

The water content in the fruit at loading the cart = 90% = 0.9x

Weight of fruit without water = x-0.9x = 0.1x

Let 'y' be the weight of fruit when Mr Farmer reached the market.

The water content in the fruit at loading the cart = 60% = 0.6y

Weight of fruit without water = y-0.6y = 0.4y

As the weight of fruit without water in both cases is equal, therefore:

0.1x = 0.4y or y=x/4

Loss in weight of fruit = x-y = 15Kg

substituting the value of y in the above equation, we get:

x-x/4 = 15kg

multiplying both sides by 4

4x-x = 60 Kg

3x = 60kg

x = 20Kg

**The weight of the fruit Mr Farmer loaded on the cart was 20Kg**