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# A large watermelon weighs 20 kg with 96% of its weight being water. I

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Manager
Joined: 29 Nov 2018
Posts: 138
Concentration: Marketing, Strategy
A large watermelon weighs 20 kg with 96% of its weight being water. I  [#permalink]

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02 Mar 2019, 09:52
3
00:00

Difficulty:

55% (hard)

Question Stats:

56% (02:03) correct 44% (02:02) wrong based on 36 sessions

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A large watermelon weighs 20 kg with 96% of its weight being water. It is allowed to stand in the sun for a while and some of the water evaporates so that now only 95% of it is water. Its
reduced weight will be?

1. 16 kg
2. 19.2 kg
3. 18 kg
4. 19.8 kg
5. 19 kg
Senior Manager
Joined: 04 Aug 2010
Posts: 389
Schools: Dartmouth College
Re: A large watermelon weighs 20 kg with 96% of its weight being water. I  [#permalink]

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02 Mar 2019, 10:27
2
cfc198 wrote:
A large watermelon weighs 20 kg with 96% of its weight being water. It is allowed to stand in the sun for a while and some of the water evaporates so that now only 95% of it is water. Its
reduced weight will be?

1. 16 kg
2. 19.2 kg
3. 18 kg
4. 19.8 kg
5. 19 kg

Since 96% of the 20 kg watermelon is water, 4% of the 20 kg is non-water:
$$(0.04)(20) = 0.8$$ kg

Since 95% of the post-evaporation watermelon is water, the remaining 5% must be composed of the 0.8 kg of non-water:
$$0.05x = 0.8$$
$$x = \frac{0.8}{0.05} = \frac{80}{5} = 16$$ kg

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Joined: 21 Feb 2019
Posts: 36
A large watermelon weighs 20 kg with 96% of its weight being water. I  [#permalink]

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02 Mar 2019, 12:00
Water: $$\frac{96}{100}(20) = 19,2$$

Loss of weight is due to loss of water. Then we can set:
$$19,2 - x = \frac{95}{100}(20-x)$$
$$19,2 - x = \frac{19}{20}(20-x)$$
$$19,2 - x = 19 - \frac{19}{20}x$$
$$0,2 = \frac{1}{20}x$$
$$x = 4$$

New weight: $$20 - 4 = 16$$

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A large watermelon weighs 20 kg with 96% of its weight being water. I   [#permalink] 02 Mar 2019, 12:00
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