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Mixture Q

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by engg.manik » Tue Oct 13, 2009 6:52 am
A jug is ful of 20% solution of Alcohol, 5 litre are drawn off & replaced by water. The strength of solution is now found to be 18%. What is the capacity of Jug.
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by 2010gmat » Tue Oct 13, 2009 10:24 am
X = solution
X/5 = Alcohal

(X/5-1)/X = 18/100

X = 50 lts
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by smartm0ve » Tue Oct 13, 2009 7:17 pm
250 ?
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by ssuarezo » Tue Oct 13, 2009 9:02 pm
2010gmat wrote:X = solution
X/5 = Alcohal

(X/5-1)/X = 18/100

X = 50 lts
I agree with 2010gmat. In 5 liters, there's one liter alcohol, so 20% minus one liter should be the 18% we need.
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by vivek.kapoor83 » Tue Oct 13, 2009 10:27 pm
i think
let x be the capacity
alcohal - x/5
water - 4x/5
after taking 5 liter off
water - (4x/5)+5 ........(1)
alchoal - 18x/100 (18% given)........(2)
adding 1 n 2
x= 250
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Mixture Q explain pls

by ausgmat2008 » Tue Oct 13, 2009 11:27 pm
Dear Vivek
Are you please able to explain your solution?
thanks in advance
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by sanjana » Wed Oct 14, 2009 4:14 am
IMO : 50

Let there be X litres of the solution

Alcohol : x/5
Water : 4x/5

a:w = 1 : 4

After removing 5l

Alcohol : (x-5)/5
Water : 4(x-5)/5

Now 5 l of water added :
Water : (4(x-5)/5) + 5

Now,in the resulting solution
Water 18/100x = 9x/50
Alcohol : 41x/50

Hence ,
(4(x-5)/5) + 5 = 41x/50
Solving, x=50.

Can you please provide the OA?
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