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The 20kg mixture comprised of water and sand is mixed evenly so that t

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New post 22 Feb 2017, 01:42
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The 20kg mixture comprised of water and sand is mixed evenly so that the ratio of water to sand is 8 to 2. If part of the mixture is replaced by sand to make the ratio of water to sand as 6 to 4, what is the amount of mixture replaced by sand, in kg?

A. 4.0      
B. 5.0      
C. 6.8      
D. 7.6      
E. 8.4
[Reveal] Spoiler: OA

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Re: The 20kg mixture comprised of water and sand is mixed evenly so that t [#permalink]

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New post 22 Feb 2017, 01:50
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AustinKL wrote:
The 20kg mixture comprised of water and sand is mixed evenly so that the ratio of water to sand is 8 to 2. If part of the mixture is replaced by sand to make the ratio of water to sand as 6 to 4, what is the amount of mixture replaced by sand, in kg?

A. 4.0      
B. 5.0      
C. 6.8      
D. 7.6      
E. 8.4


W/S = 8/2 = 4/1 i.e. water is 80% of Mixture

New Ratio of W/S = 6/4 = 3/2 i.e. water is 60% of Mixture

Using Weighted Average Method

Let, x is the Weight of the mixture replaced. So the amount of water that will decrease will be 80% of x

i.e. Water Remaining after Replacement = (80/100)20 - (80/100)*x = (60/100)*20

i.e. 1600 - 80x = 1200

i.e. 80x = 400
i.e. x = 5 kg

Answer: Option B
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The 20kg mixture comprised of water and sand is mixed evenly so that t [#permalink]

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New post 22 Feb 2017, 12:27
AustinKL wrote:
The 20kg mixture comprised of water and sand is mixed evenly so that the ratio of water to sand is 8 to 2. If part of the mixture is replaced by sand to make the ratio of water to sand as 6 to 4, what is the amount of mixture replaced by sand, in kg?

A. 4.0      
B. 5.0      
C. 6.8      
D. 7.6      
E. 8.4


let x=amount of mixture replaced by sand
.2*20-.2x+x=.4*20
x=5 kg
B

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Re: The 20kg mixture comprised of water and sand is mixed evenly so that t [#permalink]

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New post 11 Mar 2017, 13:27
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Maybe another way to approach this problem is to think of how much water do you need to rpelace with sand

before and after rations are: 16/4 and 12/8
So we need to replace 4 kg of water (16-12). As we take the whole mixture we would use proportion 16/4 or 4/1 hence 4 kg of water + 1 kg of sand=5 kg of mixture total
Answer B

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Re: The 20kg mixture comprised of water and sand is mixed evenly so that t [#permalink]

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New post 27 Aug 2017, 02:00
Let the amount of mixture replaced be one of the option. Now the best option to use is B as option B when divided in ratio of 8:2 will give a good integer to work with.
Using option B:
Out of total of 20 kg, 16 kg is water and 4 kg is sand
If 5 kg of mixture is out, in the same ratio of 8:2 (4:1) 1kg sand will be out and 4 kg water will be out

Remaining quantity of sand = 3 (4-1) but 5 kg (as per option) of sand is added to the mixture
So final quantity of sand in new mixture = 8 (3+5)
Final quantity of water = 12 (16-4)
Ratio of water to sand = 12:8 = 6:4
Thus correct answer is Option B
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Re: The 20kg mixture comprised of water and sand is mixed evenly so that t [#permalink]

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New post 27 Aug 2017, 04:18
We can approach this question in two ways by considering quantity of water or quantity of sand .
I will take quantity of sand
Note that the amount of total mixture is same i.e 20 kg
Suppose x kg of mixture is taken out and replaced with with x kg of sand
So we have Initial 4 kg of sand
then we remove x kg of mixture then it will remove x/5 kg of sand also
Therefore the equation can be set up as
4-x/5+x=8
x=5 kg
hope it helps
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Re: The 20kg mixture comprised of water and sand is mixed evenly so that t   [#permalink] 27 Aug 2017, 04:18
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