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Re: A company produces a cleaning solution that has 60% concentration of C [#permalink]
yogi02 wrote:
which batch to make 60...ambiguity in question


in the premise, one batch is mentioned,
and that batch is formed of adding two different concentrations: 200L of 30% K and 150L of 40% K

Conc. 1:
K:Water:total
30:70:100 (ratio)
60:140:200 (L)

Conc. 2:
K:Water:total
40:60:100 (ratio)
60:90:150 (L)

so by adding, K:Water:total = 120:230:350 (L) --> which is the final form of the current batch

to convert it to 60% by adding x (L) of K,
\(\frac{120+x}{350+x}\) = \(\frac{60}{100}\),
and x = 225L
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Re: A company produces a cleaning solution that has 60% concentration of C [#permalink]
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200*30%=60L chemical K
150*40%=60L chemical K
200+150=350L (concentration)

—> 60+60=120L —chemical K
350–120=230L —water

If you add the pure chemical K, the amount of total water cannot be changed in the concentration.
—> according to the question, the amount of water should be 40%:

230–>40%
x—> 100%

—> x= 100*230/40= 23*25=575
575–350=225L (pure chemical K)

The answer choice is D.

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Re: A company produces a cleaning solution that has 60% concentration of C [#permalink]
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cecconis wrote:
A company produces a cleaning solution that has 60% concentration of Chemical K. The current production batch contains 200L of an ingredient with a 30% concentration of Chemical K and 150L of an ingredient with a 40% concentration of Chemical K. How much pure chemical K must be added to the batch to attain the correct concentration of Chemical K?

A. 150L
B. 175L
C. 200L
D. 225L
E. 250L


We can create the equation:

(200 x 0.3 + 150 x 0.4 + K)/(350 + K) = 0.6

(60 + 60 + K)/(350 + K) = 3/5

5(120 + K) = 1050 + 3K

600 + 5K = 1050 + 3K

2K = 450

K = 225

Alternate Solution:

We have 200 liters of 30% solution and 150 liters of 40% solution, and we will add x liters of 100% solution, thus obtaining (200 + 150 + x) liters of a 60% solution. We can summarize this in the following equation:

200(0.30) + 150(0.40) + x(1.00) = (200 + 150 + x)(0.60)

60 + 60 + x = 120 + 90 + 0.6x

120 + x = 210 + 0.6x

0.4x = 90

x = 225 liters

Answer: D
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Re: A company produces a cleaning solution that has 60% concentration of C [#permalink]
cecconis wrote:
A company produces a cleaning solution that has 60% concentration of Chemical K. The current production batch contains 200L of an ingredient with a 30% concentration of Chemical K and 150L of an ingredient with a 40% concentration of Chemical K. How much pure chemical K must be added to the batch to attain the correct concentration of Chemical K?

A. 150L
B. 175L
C. 200L
D. 225L
E. 250L


200L of an ingredient with a 30% concentration of Chemical K = 30%of 200 = 60L
and 150L of an ingredient with a 40% concentration of Chemical K = 40% of 150 = 60L
_______
Total of Chemical K = 120L

And Total solution is now = 200L + 150L =350L
Now let X amount of Chemical K is added to above 120L of Chemical K ,which is required to be 60% of total solution

120+X = 60% of (350 + X)
120 + X = 3(350+X)/5
600 + 5X =1050+3X
2X= 450
X= 225L

Ans D
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Re: A company produces a cleaning solution that has 60% concentration of C [#permalink]
Is it only me who got confused by the wordings?
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Re: A company produces a cleaning solution that has 60% concentration of C [#permalink]
A company produces a cleaning solution that has 60% concentration of Chemical K. The current production batch contains 200L of an ingredient with a 30% concentration of Chemical K and 150L of an ingredient with a 40% concentration of Chemical K. How much pure chemical K must be added to the batch to attain the correct concentration of Chemical K?

A. 150L
B. 175L
C. 200L
D. 225L
E. 250L

Weighted Average formula for Mixtures :\( A2 - Aavg/Aavg - A1\)
A1 = 200 * \(30/100\) (stem: first part of current batch) + \(40/100\) * 150. (2nd part of current batch)
Percentage of Chemical K in current batch = \(120/350\) * 100 =2\(40/7\) = A1 (Approx 34)
W1 = Weight of current batch = 200 + 150 = 350
A2 = % of chemical k in pure chemical k = 100
W2 = ?
Therefore substituting values.
\(350/W2 \)= \(100 - 60/60 - 34\)
\(350/W2\) = \(40/26\)
4550 = 20W2
Therefore W2 = 225
D
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A company produces a cleaning solution that has 60% concentration of C [#permalink]
cecconis wrote:
A company produces a cleaning solution that has 60% concentration of Chemical K. The current production batch contains 200L of an ingredient with a 30% concentration of Chemical K and 150L of an ingredient with a 40% concentration of Chemical K. How much pure chemical K must be added to the batch to attain the correct concentration of Chemical K?

A. 150L
B. 175L
C. 200L
D. 225L
E. 250L


Can someone pls tell why this method is wrong and is there a concept error
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Re: A company produces a cleaning solution that has 60% concentration of C [#permalink]
A2D2 wrote:
cecconis wrote:
A company produces a cleaning solution that has 60% concentration of Chemical K. The current production batch contains 200L of an ingredient with a 30% concentration of Chemical K and 150L of an ingredient with a 40% concentration of Chemical K. How much pure chemical K must be added to the batch to attain the correct concentration of Chemical K?

A. 150L
B. 175L
C. 200L
D. 225L
E. 250L


Can someone pls tell why this method is wrong and is there a concept error


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Re: A company produces a cleaning solution that has 60% concentration of C [#permalink]
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Re: A company produces a cleaning solution that has 60% concentration of C [#permalink]
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