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# Consider three mixtures - the first having water and liquid A in the r

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Re: Consider three mixtures - the first having water and liquid A in the r [#permalink]
Bunuel - can you please post the solution for this question? The one posted above is a little confusing. Can we solve this question without having to assume a value for k?
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Re: Consider three mixtures - the first having water and liquid A in the r [#permalink]
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Took longer than the allocated 2 mins so I hope someone could provide a more elegant way of answering.

Water in Mixture A - 1/3
Water in Mixture B - 1/4 (25% water)
Water in Mixture C - 1/5

ratio of mixed in the proportion 4 : 3 : 2
Water in Liquid A - (1/3)(4/9) = 4/27
Water in Liquid B - (1/4)(3/9) = 3/36
Water in Liquid C - (1/5)(2/9) = 2/45

Sum of all 3 -> (4/27+3/36+2/45) * 100% = approx 28%

which is > water in Mixture B
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Re: Consider three mixtures - the first having water and liquid A in the r [#permalink]
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Aim here is to arrive at some common parameters.
Step 1 W:A = 1:2 , W:B = 1:3, W:C = 1:4

Step 2 Take LCM of 3, 4, & 5 = 60 ....which becomes our qty of soln

Step 3 Now convert ratio into absolute terms ie (Soln A : W 20, A 40) , (Soln B :W 15, B 45) , (Soln C: W 12, C 48)

Step 4 Multiply the absolute amount with desired ratio ie 4:3:2 So Water W (80+45+24), Soln B (45*3 = 135) and hence option C

I hope this is helpful IMO!!
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Re: Consider three mixtures - the first having water and liquid A in the r [#permalink]
Since water is a common element amongst all the three old mixtures and is also the an important element in the new mixtures, let's pick water to make estimations.

Mixture....... Water component....... Mixture share in new proportion.......Final Share of water, per mixture
$$A...............\frac{1}{3}............................\frac{4}{9}....................\frac{1}{3} * \frac{4}{9} = \frac{4}{27}$$
$$B...............\frac{1}{4}............................\frac{3}{9}....................\frac{1}{4} * \frac{3}{9} = \frac{3}{36}$$
$$C...............\frac{1}{5}............................\frac{2}{9}....................\frac{1}{5} * \frac{2}{9} = \frac{2}{45}$$

Now, let's equalize the denominators by taking LCM.
LCM (27, 36, 45) = 540

Thus, final share of water, per mixture is as follows:

$$Mixture.........Final Share of water, per mixture\\ A..................\frac{80}{540}\\ B..................\frac{45}{540}\\ C..................\frac{24}{540}$$

Total Water = $$(\frac{80}{540}+\frac{45}{540}+\frac{24}{540}) * 540 = 149$$

Now, we need to find the share of LIQUID in each of the mixtures, as each mixture is comprised of a liquid and water.

Thus,
Share of liquid from Mixture A = $$\frac{4}{9} * 540 * \frac{2}{3} = 160$$
Share of liquid from Mixture B = $$\frac{3}{9} * 540 * \frac{3}{4} = 135$$
Share of liquid from Mixture C = $$\frac{2}{9} * 540 * \frac{4}{5} = 96$$

Re: Consider three mixtures - the first having water and liquid A in the r [#permalink]
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