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8 liters of solution is removed from 20% milk solution and 8 liters of

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8 liters of solution is removed from 20% milk solution and 8 liters of  [#permalink]

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New post 25 Jun 2019, 03:08
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Re: 8 liters of solution is removed from 20% milk solution and 8 liters of  [#permalink]

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New post 25 Jun 2019, 03:33
2
8 litres of mixture taken out also had 20% milk => milk taken out was 8 x 0.2=1.6 litres

milk/total mixture =0.2=> milk=0.2 total qty

as the total qty remains the same, substitute the above eq as shown below

=>(milk-1.6)/total qty=0.16
=>0.2 total qty - 0.16 total qty= 1.6
=>total qty= 1.6/0.04=40
IMO option B
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8 liters of solution is removed from 20% milk solution and 8 liters of  [#permalink]

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New post 25 Jun 2019, 04:03
Bunuel wrote:
8 liters of solution is removed from 20% milk solution and 8 liters of water is added to the solution. The resulting solution has 16% milk in it. What was the initial quantity of the 20% milk solution?

A. 38 liters
B. 40 liters
C. 42 liters
D. 44 liters
E. 45 liters


Milk/total solution = 20% initially then 16%;
M=0.2T

16%=[M-(8*20%)]T;
16%=(0.2T-1.6)/T
0.04T=1.6; T=40 IMO B
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Re: 8 liters of solution is removed from 20% milk solution and 8 liters of  [#permalink]

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New post 25 Jun 2019, 08:25
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Bunuel wrote:
8 liters of solution is removed from 20% milk solution and 8 liters of water is added to the solution. The resulting solution has 16% milk in it. What was the initial quantity of the 20% milk solution?

A. 38 liters
B. 40 liters
C. 42 liters
D. 44 liters
E. 45 liters


let x=initial quantity of 20% milk solution
.8x-8(.8)+8=.84x→
x=40 liters
B
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Re: 8 liters of solution is removed from 20% milk solution and 8 liters of  [#permalink]

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New post 26 Jun 2019, 03:41
Bunuel wrote:
8 liters of solution is removed from 20% milk solution and 8 liters of water is added to the solution. The resulting solution has 16% milk in it. What was the initial quantity of the 20% milk solution?

A. 38 liters
B. 40 liters
C. 42 liters
D. 44 liters
E. 45 liters


total qty of milk ; .2*8 ; 16
milk /total = 20%
milk-1.6/total = 16%
milk = .2*total
.2*total-1.6/total = .16
solve for total = 40
IMO B
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Re: 8 liters of solution is removed from 20% milk solution and 8 liters of  [#permalink]

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New post 01 Jul 2019, 02:12
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Bunuel wrote:
8 liters of solution is removed from 20% milk solution and 8 liters of water is added to the solution. The resulting solution has 16% milk in it. What was the initial quantity of the 20% milk solution?

A. 38 liters
B. 40 liters
C. 42 liters
D. 44 liters
E. 45 liters


Let the initial quantity of the 20% milk solution be X, of which 0.2X would be milk and the remaining (0.8X) be water

Now since 8 litres of this solution is removed and is dilated by adding 8 litres of water, amount of milk reduced would be = 20% of 8 litres, i.e. 1.6 litres

The remaining portion of milk in the solution = 0.2X - 1.6 litres

Now, it is given that the concentration of the resulting solution (or to simply put, % of milk in the resulting solution) = 16%

We can write this as (0.2X - 1.6)/X *100 = 16, solving which we get 4X = 160, or X = 40 (D)

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Re: 8 liters of solution is removed from 20% milk solution and 8 liters of  [#permalink]

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New post 01 Jul 2019, 17:40
1
Bunuel wrote:
8 liters of solution is removed from 20% milk solution and 8 liters of water is added to the solution. The resulting solution has 16% milk in it. What was the initial quantity of the 20% milk solution?

A. 38 liters
B. 40 liters
C. 42 liters
D. 44 liters
E. 45 liters


When 8 liters of the solution is removed, the 8 liters consists of 0.2 x 8 = 1.6 L of milk, and 6.4 L or water. If we let s = the initial amount, in liters, of the solution, so originally there are 0.2s liters of milk and we can create the equation:

(0.2s - 1.6 + 0)/(s - 8 + 8) = 16/100

(0.2s - 1.6)/s = 4/25

25(0.2s - 1.6) = 4s

5s - 40 = 4s

s = 40

Alternate Solution:

We start with x liters of 20% milk. We remove 8 liters of this solution, so we are removing 8 liters of 20% milk. Then we add back 8 liters of water (which is 0% milk), and the result is x liters of 16% milk. We can summarize this in an equation as:

x(0.20) - 8(0.20) + 8(0) = x(0.16)

0.20x - 1.6 + 0 = 0.16x

0.04x = 1.6

4x = 160

x = 40

Answer: B
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Re: 8 liters of solution is removed from 20% milk solution and 8 liters of  [#permalink]

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New post 02 Jul 2019, 09:36
2
Bunuel wrote:
8 liters of solution is removed from 20% milk solution and 8 liters of water is added to the solution. The resulting solution has 16% milk in it. What was the initial quantity of the 20% milk solution?

A. 38 liters
B. 40 liters
C. 42 liters
D. 44 liters
E. 45 liters


we can do witha allegations:

we see we have 80%water solution, 8 litre is removed and is replayed by 100% water solution to make a 84% water solution.
so

80 100
84
16:4
=4:1
So, 1/5 of the total solution was repplaced with 8 litres.
so, total solution is 40
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Re: 8 liters of solution is removed from 20% milk solution and 8 liters of  [#permalink]

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New post 11 Jul 2019, 02:59
can we do it this way?

20/100 * x = 8 ltrs.
upon solving, we get x = 40 ltrs. ?

(percentage method?)
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Re: 8 liters of solution is removed from 20% milk solution and 8 liters of   [#permalink] 11 Jul 2019, 02:59
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