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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