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Re: If Adrian mixes 15 liters of saline solution A with 5 liters of a 40% [#permalink]
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QZ wrote:
If Adrian mixes 15 liters of saline solution A with 5 liters of a 40% saline solution to get a 25% saline solution, what percent saline is solution A ?

a. 5
b. 10
c. 15
d. 20
e. 25



Ratio in which Solution A and Other solution was mixed = 15: 5 = 3: 1

Saline Concentration of Solution A = x

Saline Concentration of Other Solution = 40%

Overall Concentration = 25%

Using Alligation we can write -

(40-25): ( 25 - x) = 3: 1

15 = 75 - 3x

3x = 60

x = 20%

Correct Answer is Option D.


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Re: If Adrian mixes 15 liters of saline solution A with 5 liters of a 40% [#permalink]
QZ wrote:
If Adrian mixes 15 liters of saline solution A with 5 liters of a 40% saline solution to get a 25% saline solution, what percent saline is solution A ?

a. 5
b. 10
c. 15
d. 20
e. 25


let x=saline % of A
15*x+5*.4=20*.25
x=1/5=20%
D
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Re: If Adrian mixes 15 liters of saline solution A with 5 liters of a 40% [#permalink]
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AkshdeepS wrote:
If Adrian mixes 15 liters of saline solution A with 5 liters of a 40% saline solution to get a 25% saline solution, what percent saline is solution A ?

a. 5
b. 10
c. 15
d. 20
e. 25


Let solution A be an x% saline solution; we can create the equation:

15(x/100) + 5(0.4) = 0.25(15 + 5)

15x/100 + 2 = 5

15x/100 = 3

15x = 300

x = 20

Answer: D
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Re: If Adrian mixes 15 liters of saline solution A with 5 liters of a 40% [#permalink]
VeritasKarishma
If I solve this with your method, I'm not getting the correct answer -

w1/w2 = a2 - Avg / Avg - a1

15/5 = 40-25 / 25-x

3 = 20/25-x

75 - 3x = 20

x = 55/3 which is 18 point something but not 20. Can you please explain where I'm going wrong?
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Re: If Adrian mixes 15 liters of saline solution A with 5 liters of a 40% [#permalink]
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pearljiandani wrote:
VeritasKarishma
If I solve this with your method, I'm not getting the correct answer -

w1/w2 = a2 - Avg / Avg - a1

15/5 = 40-25 / 25-x

3 = 20/25-x

75 - 3x = 20

x = 55/3 which is 18 point something but not 20. Can you please explain where I'm going wrong?


Calculation error.

40 - 25 = 15
Correct it to get x = 20
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Re: If Adrian mixes 15 liters of saline solution A with 5 liters of a 40% [#permalink]
Top Contributor
Let A be a y% saline solution.
Thus 15(y/100) + 5(0.4) = 0.25(15+5)
=>15y/100 = 3
=> y =20 (D)

Hope this helps. Keep working :)
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If Adrian mixes 15 liters of saline solution A with 5 liters of a 40% [#permalink]
I would use the mixture/allegation concept here.

20---------40 (Already given)
-----25-------(Avg given)
15----------5 (both nos given)

The subtraction in mixture/allegation should always subtract the higher value from the smaller value to give the resultant. We must get a positive ratio down because the items in the third line represent the actual weights. So we need to ensure that we get a positive value in the 3rd line. The 1st line must always be populated by "per cents" "per kg" value etc.

Hence, 25-5=20.
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Re: If Adrian mixes 15 liters of saline solution A with 5 liters of a 40% [#permalink]
We can solve this using the mixture table. Since we know the final solution contains 5 lts of saline (25% of 20 lts), we can figure out the absolute value of saline in soln. B and then the remainder saline, would be coming from sol. A.
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Re: If Adrian mixes 15 liters of saline solution A with 5 liters of a 40% [#permalink]
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