TRY THIS 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
METHOD 1-LONG CUT
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
METHOD 2-SHOT CUT
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.
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