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lalania1
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

Answer: A
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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

Final Answer:

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lalania1
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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Does anyone have a log of questions similar to this one? Please relay!

Bunuel
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lalania1
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
A chemical supply company has 60 liters of a 40% HNO3 solution..jpg [ 796.03 KiB | Viewed 8042 times ]

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