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chemicals solution % ..Please see attachmnt

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by 4GMAT_Mumbai » Mon Aug 23, 2010 8:19 pm
Hi,

Let x liters of first solution and y liters of second solution be mixed

What do we have?

(0.2x + 0.3y) / (x + y) = 22 / 100

20x + 30y = 22x + 22y

8y = 2x

4y = x

Hence, x / (x+y) = 4y / (4y + y) = 4y / 5y = 0.8

Hence, 80%.

Sanity check: the final solution has 22% of A. X has 20% of A and Y has 30% of A. One needs to use (lot) more of solution X so that the balance shifts to 22%. So, 80% is more like it - rather than 44% (guess that is what has been marked)

Hope this helps. Thanks.
Naveenan Ramachandran
4GMAT, Dadar(W) & Ghatkopar(W), Mumbai
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by ymach3 » Mon Aug 23, 2010 9:22 pm
I tried solving by picking numbers....

X=0.2A + 0.8B--- Say X = 20ltr ---- A=4 ltrs and B=16ltrs

Y=0.3A + 0.7B---- Say Y= 20 ltr--- B= 6 ltrs and B=14 ltrs

When X+Y , A=10 Ltrs and B=30 Ltrs.... i.e A %=25% and B=75%..... and X=50% of (X+Y)...

But we want A=22%, so X should be less than 50%.... 44% in options.....B

I am not sure whether my approach is correct...

Request you guys to share your thoughts on this approach...


Shout at me if i made a blunder...
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