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

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

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New post 02 Mar 2019, 09:52
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Question Stats:

47% (02:04) correct 53% (01:47) wrong based on 75 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
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Re: A large watermelon weighs 20 kg with 96% of its weight being water. I  [#permalink]

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New post 02 Mar 2019, 10:27
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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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A large watermelon weighs 20 kg with 96% of its weight being water. I  [#permalink]

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New post 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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Re: A large watermelon weighs 20 kg with 96% of its weight being water. I  [#permalink]

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New post 21 Aug 2019, 09:23
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


initial melon: 4%
final melon: 5%
initial weight: 20kg
water that evaporated: x
final weight: 20-x

5%=4%(initial/[final=initial-x])… 5%=4%(20/20-x)… 20(1%)=5%x… 20(1%)/5%=x… x=20(1)/5=4
final weight: 20-(4)=16

Answer (A)
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Re: A large watermelon weighs 20 kg with 96% of its weight being water. I  [#permalink]

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New post 21 Aug 2019, 09:32
1
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


Given:
1. A large watermelon weighs 20 kg with 96% of its weight being water.
2. 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.

Asked: Its reduced weight will be?

Let the reduced weight be x kg
20*4% = .8 kg = Non-water portion of watermelon
5% x =.8 kg
x = .8/5% = 16 kg

IMO A
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Re: A large watermelon weighs 20 kg with 96% of its weight being water. I   [#permalink] 21 Aug 2019, 09:32
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