What was the weight of the fruit in Mr Farmer loaded on the cart?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...

What was the weight of the fruit in Mr Farmer loaded on the cart?

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.

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

Let the weight of the fruit loaded by Mr. Farmer when he set off to the market be X. Here, 90% of the weight of the fruit is water. When he reached the market 60% of the weight of the fruit was water and the fruit weighed 15 kg less than what its weight was at the beginning of the day.

The weight of water in the fruit when he set off was 0.9*X, when he reached the market it was 0.6*X.

0.9*X - 0.6*X = 15

=> 0.3*X = 15

=> X

=> X = 15/0.3

=> X = 50

The weight of the fruit loaded into the cart by the farmer was 50 kg.

najm1947's profile pic

najm1947 | Elementary School Teacher | (Level 1) Valedictorian

Posted on

The above answer has a verification problem as under:

Weight of fruit loaded in cart = 50kg

Weight of water = 90% = 0.90*50 = 45Kg

Weight of fruit without water = 50-45 = 5kg

Weight of fruit in market after drying = 50-15 = 35kg

weight of water = 60% of fruit = 0.60*35 = 21kg

weight of fruit without water = 34-21 = 13kg

This weight 13kg does not match the weight of fruit without water at start i.e. 5k hence this answer is not correct.

The correct answer is given at link:

http://www.enotes.com/math/q-and-a/mr-farmer-loaded-cart-with-fruit-set-off-market-358847 and the same is being reproduced below:

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

The original weight without water = 20-0.9*20 = 2kg

The wieght without water after drying = (20-15)*(1-0.6) = 2kg

Both the weights are equal hence the answer is verified to be correct.

 

 

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