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weighted average problems

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Intern
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Joined: 22 Oct 2014
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Re: weighted average problems [#permalink]

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New post 16 Aug 2015, 10:29
Compiling the steps in method 3:

3.
The difference between 10% and 3% is 7%; and, the WA is 8%
Here, x is (10-8)/7 = 2/7 away from 8%, to right in number line (if I see visually)
and y is (3-8)/7 = -5/7 away from 8%, to the left in number line.
As the differentials are distance from WA, you can omit negative sign for calculation and make them equal *

The fractions are:
2/7 distance from x = \(\frac{2}{7}x\)
5/7 distance from y = \(\frac{5}{7}y\)

As both distances come and meet at WA 8%* so,
\(\frac{2}{7}x = \frac{5}{7}y\)
\(2x = 5y\) #Multiply by 7 (Not taking reciprocals)
\(x = \frac{5y}{2}\)
\(\frac{x}{y} = \frac{5}{2}\)
So the ratio from the fraction is \(x:y = 5:2\)

Is this correct?

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Re: weighted average problems [#permalink]

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New post 16 Aug 2015, 10:31
VeritasPrepDennis wrote:
Here is a better problem that ilustrates the concept.

Say you have unlimited quantities of 15% HCL (hydro-chloric acid) and 50% HCL. How many gallons of the 15% HCL will you need to use to make 7 gallons of a 30% HCL solution?
A) 2
B) 3
C) 4
D) 5
E) 5 2/3


As of your new problem,
Let x = gal of 15% HCL, we need the value of x.

Method 2: Using normal WA
15 (x) + 50 (y) = 30 (x+y) #we know x+y = 7, but should we use that value here?
15x + 50y = 30x + 30y
20y = 15x
x/y = 20/15
x/y = 4/3
x : y = 4 : 3

Method 1: Using differentials:
x (+15) + y (-20) = 0
15x - 20y =0
15 * 4 - 20 * 3 = 0
x=4
y=3

Answer : C) 4
Correct?

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Re: weighted average problems [#permalink]

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New post 17 Aug 2015, 06:59
appleid wrote:
VeritasPrepDennis wrote:
Here is a better problem that ilustrates the concept.

Say you have unlimited quantities of 15% HCL (hydro-chloric acid) and 50% HCL. How many gallons of the 15% HCL will you need to use to make 7 gallons of a 30% HCL solution?
A) 2
B) 3
C) 4
D) 5
E) 5 2/3


As of your new problem,
Let x = gal of 15% HCL, we need the value of x.

Method 2: Using normal WA
15 (x) + 50 (y) = 30 (x+y) #we know x+y = 7, but should we use that value here?
15x + 50y = 30x + 30y
20y = 15x
x/y = 20/15
x/y = 4/3
x : y = 4 : 3

Method 1: Using differentials:
x (+15) + y (-20) = 0
15x - 20y =0
15 * 4 - 20 * 3 = 0
x=4
y=3

Answer : C) 4
Correct?



appleid -

Yes, C is the answer. Notice how this question utilizes the relationship of the numbers.

In method #2, you asked if you needed to use the x + y = 7. Yes, in this case, you would use the series of equations (.15)x + (.50)y = (.30)(x+y) and x +y=7 to solve: two varialbles requires two equations.

For me, it is simpler to use the number line and ratio concept (more of a conceptual or visual thinker) than the abstract, formula approach. But either works!
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Re: weighted average problems [#permalink]

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New post 20 Aug 2016, 20:25
Hello from the GMAT Club BumpBot!

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Re: weighted average problems [#permalink]

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New post 24 Aug 2017, 10:37
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: weighted average problems   [#permalink] 24 Aug 2017, 10:37

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