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# If Adrian mixes 15 liters of saline solution A with 5 liters of a 40%

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Senior Manager
Joined: 13 Apr 2013
Posts: 259
Location: India
Schools: ISB '19
GMAT 1: 480 Q38 V22
GPA: 3.01
WE: Engineering (Consulting)
If Adrian mixes 15 liters of saline solution A with 5 liters of a 40% [#permalink]

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29 Dec 2017, 08:58
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25% (medium)

Question Stats:

71% (01:06) correct 29% (01:11) wrong based on 35 sessions

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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
[Reveal] Spoiler: OA

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Manager
Joined: 23 May 2017
Posts: 196
Concentration: Finance, Accounting
WE: Programming (Energy and Utilities)
Re: If Adrian mixes 15 liters of saline solution A with 5 liters of a 40% [#permalink]

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29 Dec 2017, 09:03
$$\frac{x}{100}$$* 15 + $$\frac{40}{100}$$* 5 = $$\frac{25}{100}$$ * 20

$$\frac{x}{100}$$ * 15 + 2 = 5

$$\frac{x}{100}$$ = $$\frac{3}{15}$$

x = 20 %
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If Adrian mixes 15 liters of saline solution A with 5 liters of a 40% [#permalink]

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29 Dec 2017, 10:32
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/100 = unknown percent saline in solution A
% = percent saline in each solution

(% A)(Vol A) + (% B)(Vol B) = (% A+B)(Vol A+B)

$$\frac{x}{100}(15) + .40(5) = .25(15 + 5)$$

$$\frac{15}{100}x + 2 = 5$$

$$\frac{15}{100}x= 3$$

$$x = (3 * \frac{100}{15}) = 20$$

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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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31 Dec 2017, 10:38
Expert's post
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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%

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Joined: 07 Dec 2014
Posts: 885
Re: If Adrian mixes 15 liters of saline solution A with 5 liters of a 40% [#permalink]

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31 Dec 2017, 12:03
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
Re: If Adrian mixes 15 liters of saline solution A with 5 liters of a 40%   [#permalink] 31 Dec 2017, 12:03
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