A factory cleans its machinery with a solution that is 87% water and

Weighted Average

In mixture problems that involve percent water and percent "something else," we can use a straightforward weighted average.
Track on only one concentration, percent water or percent vinegar. Either works.

(The formula below works for any two liquids whose ratio we must find. I mention "water" because in a mixture problem, if pure water is added or subtracted, its concentration of the other thing such as vinegar is 0%. If we track on water, water is 100% water.)

We have a solution that is 40% vinegar.
The original is thus 60% water
We need a resultant solution that is 87% water

Let A = original solution with 60% water

Let w = amount of water to be added
w = 100% water, in decimal form = 1

A + w = Total volume of resultant mixture

% = concentration / percent water

If A and B are the two liquids to be mixed, one weighted average formula is

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

Original 60% water solution = A
Water to be added = B = w

\(.60(A) + (1)w = .87(A + w)\)

\(.60(260) + 1w = .87(260 + w)\)

\(156 + 1w = (.87)(260) + .87w\)

\(1w - 0.87w = 226.2 - 156\)

\(.13w = 70.2\)

\(w=\frac{70.2}{.13}=540\)

\(w\) = amt of water to be added = \(540\)

Answer D

let w=water to be added
.6*260+w=.87(260+w)
w=540 liters
D

Can someone help me with this question? I am unable to understand the solutions provided. Thanks
p.s: Is this a mixture problem?

From allegation- (40-13):(13-0)=X:260 on solving, X=540

Sent from my XT1562 using GMAT Club Forum mobile app

We need the vinegar solution to be 13% vinegar.

We have 260 litres that contain 0.4(260) = 104 litres of vinegar.

We can let x = the number of litres of water to be added and create the following equation:

104/(260 + x) = 13/100

10,400 = 13(260 + x)


10,400 = 3,380 + 13x

7,020 = 13x

540 = x

Answer: D

Another way to look at the problem:

Initial Solution:

Volume of water = 260 *60/100 = 156 litres ---(1)

The addition of water:

Let the volume of water to be added = x litres ---(2)

Final Solution: (Given in the problem)

Volume of water in the final solution = 87% = 87/100

We can represent the percentage of water in the final solutions as:

(156 + x)/(260+x) = 87/100

15600 + 100x = 87*260 +87x
15600 - 22620 = 87x - 100x
- 7020 = -13x
x = 540

Alligation approach:

[quote="Blackbox"]Can someone help me with this question? I am unable to understand the solutions provided. Thanks
p.s: Is this a mixture problem?

hi. let me try to explain. to have optimal proportion of mixture with two liquids: water must be 87% of the total solution, while vinegar must be 13% of the total solution. we have total solution which is 260 liters. 40 % of 260 = 104 is vinegar. the rest is water (260-104 = 156) now we are asked how much more water do we need to have 13 % of vinegar and 87 % of water. now we can use simple proportional method 13/87 = 104/x cross multiply and x = 696. since 156 liters of water is already there we need to subtract 696-156 = 540 YAY ! here is answer!!

is it clear ?

have a great day

Can you please explain how did you form this equation?
104/(260 + x) = 13/100
Thanks

VeritasKarishma am I on the right track?? I just learned your wow-some weighted avg. method 2days ago and now I'm applying it over here-


We have an existing 60% solution of water(40% is vinegar; so, 60% must be water) & 100% of water solution which we need to add to make the desired 87%(water) solution.

Using scale method we have,

0.6____________0.87______1

Look the difference between the weighted avg.(0.87) and (1)water is less so the water pulls the avg. towards its side and must be in greater qty.

Therefore the ratio of the final solution = (1-0.87)/(0.87-0.60) = 13/27 = vinegar/water or, water/vinegar = 27/13 ;

Now, Let the amount of water added be 'w' lit.

Therefore, w/(260+w) = 27/(27+13)
or, w/(260+w) = 27/40
or, w = 540.

Hence, Answer is choice D ;

Happy Folks?!

Amount of vinegar = 40/100 * 260 = 104
Let Total solution be x
so 13x/100 = 104
x = 800
so the amount of water required = 800-260 =540
Hence D

The first half is correct but the second half needs to be clarified.

w1/w2 = (A2 - Aavg)/(Aavg - A1) = (1-0.87)/(0.87-0.60) = 13/27

Note what w1 and w2 are - they are the amounts of solution 1 (60% water) and solution 2 (100% water)

So in 13 parts of solution 1, you add 27 parts of pure water to get the 87% solution.
The actual amount of solution 1 is 260 lts so the multiplier is 20. The actual amount of pure water added will be 27*20 = 540 lts

Given: 1. A factory cleans its machinery with a solution that is 87% water and the remainder vinegar.
2. The factory manager starts with 260 Litres of a 40% vinegar solution

Asked: how much water must the manager add to the existing solution to attain the proper vinegar concentration for cleaning the machinery?

Vinegar in 260 litres of 40% vinegar solution = .4 * 260 = 104 litres
Water in 260 litres of 40% vinegar solution = 260 - 104 = 156 litres

The ratio of vinegar to water required = 13%/87% =13/87

Let vinegar be 13x and water be 87x
But the vinegar = 104 litres
13x = 104 => x=8
Water = 87x = 696
But water already available l= 156 litres
Water to be added = 696 - 156 = 540 litres.

IMO D

Instead of solving, use approximation.

water is 60% in 260 litre solution => 156 liter

Now we have to add water to make it 87% water (close to 90%)

156+x / 260 +x

What value can x take? Use approximation....

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