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

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

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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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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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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]

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hazelnut 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



We can let the number of kilograms of water = x and the number of kilograms of sand = y. Thus, x + y = 20.

We also see that:

x/y = 8/2

x/y = 4/1

x = 4y

Thus:

4y + y = 20

5y = 20

y = 4, so x = 16

We are given that part of the mixture is replaced by sand to make the ratio of water to sand 6 to 4. We can let n represent the number of kilograms of mixture that is replaced; thus, the number of kilograms of sand that is added to the mixture is also n.

When we replace n kilograms of mixture, (1/5)n kilograms will be sand and (4/5)n kilograms will be water, since the ratio of water to sand is 4 : 1. So, we are removing (4/5)n kilograms of water from the existing 16 kilograms of water, and we are removing (1/5)n kilograms of sand from the existing 4 kilograms of sand.

However, we are also adding back n kilograms of sand. Thus:

water/sand = 6/4

(16 - (4/5)n)/(4 - (1/5)n + n) = 6/4

(16 - (4/5)n)/(4 + (4/5)n) = 3/2

2(16 - (4/5)n) = 3(4 + (4/5)n)

32 - (8/5)n = 12 + (12/5)n

Multiplying the whole equation by 5, we have:

160 - 8n = 60 + 12n

100 = 20n

n = 5

Alternate solution:

The original water:sand ratio of 8:2 indicates that sand comprises 20% of the original 20 kg mixture. We will remove x kilograms of the original mixture which was 20% sand, and we will replace what was removed with pure sand (100% sand). The result will be 20 kilograms of a mixture which is now 40% sand. We can express this algebraically as:

20(0.2) – x(0.2) + x(1.0) = 20(0.4)

4 - 0.2x + x = 8

0.8 x = 4

x = 5

Thus we replace 5 kg of the original mixture with pure sand to create the new mixture.

Answer: B
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Re: The 20kg mixture comprised of water and sand is mixed evenly so that t   [#permalink] 13 Dec 2017, 13:32
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