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# A chemical supply company has 60 liters of a 40% HNO3 solution.

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Manager
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A chemical supply company has 60 liters of a 40% HNO3 solution.  [#permalink]

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01 Oct 2016, 11:07
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A chemical supply company has 60 liters of a 40% HNO3 solution. How many liters of pure undiluted HNO3 must the chemists add so that the resultant solution is a 50% solution?

A) 12
B) 15
C) 20
D) 24
E) 30
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Re: A chemical supply company has 60 liters of a 40% HNO3 solution.  [#permalink]

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02 Oct 2016, 01:31
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lalania1 wrote:
A chemical supply company has 60 liters of a 40% HNO3 solution. How many liters of pure undiluted HNO3 must the chemists add so that the resultant solution is a 50% solution?

A) 12
B) 15
C) 20
D) 24
E) 30

Another simple one.

60 liters of a 40% HNO3 solution means HNO3 = 24 liters in 60 liters of the solution.

Now, let x be the pure HNO3 added.

As per question,

24 + x = 50% of (60 + x)

or x =12. Hence, A
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A chemical supply company has 60 liters of a 40% HNO3 solution.  [#permalink]

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02 Oct 2016, 16:27
lalania1 wrote:
A chemical supply company has 60 liters of a 40% HNO3 solution. How many liters of pure undiluted HNO3 must the chemists add so that the resultant solution is a 50% solution?

A) 12
B) 15
C) 20
D) 24
E) 30

Use weighted Avg:

$$\frac{W1}{W2}=\frac{(A2-Avg)}{(Avg-A1)}$$
Here W1 = 60; W2 is what we need to find.
Avg = 50%; A1 = 40%; A2 = 100%
$$\frac{60}{W2}=\frac{(100-50)}{(50-40)}=\frac{5}{1}$$
W2 = 12
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Re: A chemical supply company has 60 liters of a 40% HNO3 solution.  [#permalink]

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05 Apr 2018, 17:13
lalania1 wrote:
A chemical supply company has 60 liters of a 40% HNO3 solution. How many liters of pure undiluted HNO3 must the chemists add so that the resultant solution is a 50% solution?

A) 12
B) 15
C) 20
D) 24
E) 30

We can lert n = the amount of pure undiluted HNO3 to be added and create the equation:

(0.4 x 60 + n)/(60 + n) = 1/2

2(24 + n) = 60 + n

48 + 2n = 60 + n

n = 12

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Re: A chemical supply company has 60 liters of a 40% HNO3 solution.  [#permalink]

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13 Nov 2018, 12:31
Hi swerve,

We're told that a chemical supply company has 60 liters of a 40% HNO3 solution. We're asked for the number of liters of pure undiluted (meaning 100%) HNO3 the chemists must add so that the resultant solution is a 50% solution. This question is essentially a 'Weighted Average' question, but there's a math 'shortcut' that you can use to simplify the calculations involved. We can TEST THE ANSWERS to take advantage of the shortcut.

To start, since we're mixing in a pure (100%) solution, it won't take much of that solution - relatively speaking - to raise a 40% solution to an overall 50% solution. Thus, the correct answer is likely one of the smaller answers. Both Answer A (12) and Answer B (15) are factors of 60, so those numbers would form a simple ratio (1:5 and 1:4, respectively) with the 60 liters that are already there. This means that we can calculate the average using the ratio - and not the larger numbers involved.

Answer A: 12 liters --> forms a ratio of 1:5 with the original 60 liter mixture

[(1)(100%) + (5)(40%)]/(1+5) = 300%/6 = 50%
This is an exact match for what we were told, so this must be the answer

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Re: A chemical supply company has 60 liters of a 40% HNO3 solution.  [#permalink]

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13 Nov 2018, 13:21
lalania1 wrote:
A chemical supply company has 60 liters of a 40% HNO3 solution. How many liters of pure undiluted HNO3 must the chemists add so that the resultant solution is a 50% solution?

A) 12
B) 15
C) 20
D) 24
E) 30

Perfect opportunity for the Alligation and Bruce Lee (when we make numerators equal)!

$$? = x$$

$$\frac{{60}}{{60 + x}} = \frac{{100 - 50}}{{100 - 40}} = \frac{{5 \cdot \boxed{12}}}{{6 \cdot \boxed{12}}}\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,60 + x = 72\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,? = x = 12$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: A chemical supply company has 60 liters of a 40% HNO3 solution.  [#permalink]

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28 Dec 2018, 12:50
Does anyone have a log of questions similar to this one? Please relay!

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Re: A chemical supply company has 60 liters of a 40% HNO3 solution.  [#permalink]

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03 Jan 2019, 15:04
colorblind wrote:
lalania1 wrote:
A chemical supply company has 60 liters of a 40% HNO3 solution. How many liters of pure undiluted HNO3 must the chemists add so that the resultant solution is a 50% solution?

A) 12
B) 15
C) 20
D) 24
E) 30

Use weighted Avg:

$$\frac{W1}{W2}=\frac{(A2-Avg)}{(Avg-A1)}$$
Here W1 = 60; W2 is what we need to find.
Avg = 50%; A1 = 40%; A2 = 100%
$$\frac{60}{W2}=\frac{(100-50)}{(50-40)}=\frac{5}{1}$$
W2 = 12

Hello!

Could someone please explain to me why I can't solve it by the attached image?

Thank you so much!
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A chemical supply company has 60 liters of a 40% HNO3 solution..jpg [ 796.03 KiB | Viewed 589 times ]

Re: A chemical supply company has 60 liters of a 40% HNO3 solution.   [#permalink] 03 Jan 2019, 15:04
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