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Percentage problem

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Percentage problem

by mathewmithun » Mon Nov 14, 2011 3:12 am
Fresh watemelons contain 90% water by weight whereas dry watermelons contains 20% water by weight. what is the weight of dry watermelon obtained from 20kgs of fresh watermelon?

a: 2.4KG b: 2KG c: 2.5KG d: Cannot be determined
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by shankar.ashwin » Mon Nov 14, 2011 3:20 am
I am assuming they ask for weight of pulp here.

20kg fresh has 18 kg water and 2 kg pulp.

2kg pulp in a dry water melon (80% pulp) translates = 2/0.8 = 2.5 kgs of dry water melon C IMO
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by mathewmithun » Mon Nov 14, 2011 3:24 am
shankar-you are correct...
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by neelgandham » Mon Nov 14, 2011 3:30 am
mathewmithun wrote:Fresh watemelons contain 90% water by weight whereas dry watermelons contains 20% water by weight. what is the weight of dry watermelon obtained from 20kgs of fresh watermelon?
a: 2.4KG b: 2KG c: 2.5KG d: Cannot be determined
Quantity of non-water component in 20 kgs of fresh watermelon = 20*(10/100) = 2 kgs
The question can be rephrased to
'What amount of dry watermelon contains 2 kgs of non-water component'

100 kgs of dry watermelon contains 80 kgs of non-water component
? kgs of dry watermelon contains 2 kgs of non-water component

100/? = 80/2 => ? = 200/80 = 2.5 kgs

IMO C
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by tpr-becky » Mon Nov 14, 2011 2:29 pm
the trick here is to figure out how much fruit (without water) there is and then work from there.

20kg of fresh watermelon is 10% fruit so there is 2kg of fruit.

The fruit will not change, only the ratio of water to fruit. So in a dry watermelon there will be 2kg of fruit, which is 80% of the weight of a dry watermelon. So if 2kg is 80% of the weight that means that 2kg = 80/100 (dry weight) and thus it calculates to 2.5kg of dry weight so C is the answer.
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