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

Question

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

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

GMATinsight · Accepted Answer

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

gracie · Answer

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

SeregaP · Answer

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

HolaMaven · Answer

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

arvind910619 · Answer

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

ScottTargetTestPrep · Answer

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

hadimadi · Answer

Hi,

cool question!

Let x denote the total initial weight of the mixture in kg, hence x=20. We know that currently the ratio W:S is  / . What we do is remove y units of the mixture and add y units of sand. For every y unit of mixture removed, we lose 8/10 W, and 2/10 of S, and we add a whole unit of sand, until we get the ratio 6:4. In other terms:

/ =6/4
-> / =6/4, plugging in x=20
->  / =6/4
-> 64-(32/10)*y=24+(48/10)*y
-> 40=8y
-> y=5

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