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Mixture 1 - DS

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Mixture 1 - DS

by karthikpandian19 » Mon Dec 19, 2011 1:18 am
Solution Y is 30 percent liquid X and 70
percent water. If 2 kilograms of water
evaporate from 8 kilograms of solution
Y and 2 kilograms of solution Y are
added to the remaining 6 kilograms of
liquid, what percent of this new solution
is liquid X?
(A) 30%
(B) 33.33%
(C) 37.5%
(D) 40%
(E) 50%
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by chetansharma » Mon Dec 19, 2011 1:29 am
IMO C

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by knight247 » Mon Dec 19, 2011 2:02 am
Soln Y Contains
30% X and 70% Water

8 Litres of water contains 30*8/100 = 2.4 Litres of soln x ....(1)
and 8-2.4= 5.6 Litres of Water out of which two litres of water evaporate leaving
3.6 Litres of water...(2)

Now 2 litres of Y are added to the above 6 litres of solution y. The breakup on these two litres is as follows:
Litres of Solution X=30(2)/100=0.6 ....(3)
Litres of Water=1.4 ....(4)

From 1 and 3 the total amt of X in the final soln is 2.4+0.6= 3
From 2 and 4 The total volume of water is 3.6+1.4= 5
Total volume of mixture is 8. Hence percentage of soln x in the final mixture is (3/8) *100 = 37.5 Hence C


However, this problem is best solved using weighted averages.
(6/8)*(40)+(2/8)*(30)=37.5 Hence C
Last edited by knight247 on Mon Dec 19, 2011 2:28 am, edited 1 time in total.
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by chetansharma » Mon Dec 19, 2011 2:09 am
knight247 wrote:Soln Y Contains
30% X and 70% Water

If 2 out of 8 litres evaporate then there are 6 litres left.
Litres of Solution X =30(6)/100= 1.8....(1)
Litres of Water=6-1.8= 4.2....(2)

Now 2 litres of Y are added to the above 6 litres of solution y. The breakup on these two litres is as follows:
Litres of Solution X=30(2)/100=0.6 ....(3)
Litres of Water=1.4 ....(4)

From 1 and 3 the total amt of X in the final soln is 1.8+0.6= 2.4
The total volume of the mixture is 8. So we have (2.4/8)*100=[spoiler]30%[/spoiler] Hence A

However, this problem is best solved using weighted averages.
(6/8)*(30)+(2/8)*(30)=30 Hence A
Hi Knight247,

Your approach is correct and I have also followed the same way. But the catch in the question is that 2kgs of WATER is evaporated not the 2kgs of Y solution. Check the question again. U shd be able to get the correct soln after correcting the error.

Regards,
Chetan
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by shankar.ashwin » Mon Dec 19, 2011 2:09 am
Only water evaporates. Amount of solution X is the same.

Initial quantity of liquid X = 0.3*8 = 2.4

After water evaporates 2 litres of solution Y is added - Liquid X = 0.3 * 2 = 0.6

Total Liquid X = 3.

Percentage = 3/8 * 100 = 37.5
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by ankush123251 » Mon Dec 19, 2011 2:12 am
In 8 kg of Y,

X = .3 * 8 = 2.4 kg
Water(W) = .7 * 8 =5.6 kg.

Now if 2 kg of water is removed,

Water left = 3.6 kg.

and 2 kg of Y is added in which ,

X =.3 * 2 =.6 kg.

then in 8 kg Y we have,
Amount of X = 2.4 + .6 = 3 kg

So,
% of X =3/8 * 100 =37.5.
Option C.
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by knight247 » Mon Dec 19, 2011 2:29 am
Sorry about the oversight guys. Have edited my post. Cheers!
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by GMATGuruNY » Mon Dec 19, 2011 3:43 am
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by karthikpandian19 » Tue Dec 27, 2011 10:03 pm
OA is C
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