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A cleaner is sold in low and high concentrations so that customers can combine amounts from each in order to obtain a desired quantity and concentration. The low concentration is 3% pure cleaner and the high concentration is 18% pure cleaner. How many liters of the low and high concentrations must be combined to obtain 10 liters that is 8% pure cleaner? Choose 1 answer: A 6 liters of low and 4 liters of high (B) 6(1)/(3) liters of low and 3(2)/(3) liters of high (C) 6(2)/(3) liters of low and 3(1)/(3) liters of high (D) 7 liters of low and 3 liters of high

A cleaner is sold in low and high concentrations so that customers can combine amounts from each in order to obtain a desired quantity and concentration. The low concentration is 3% 3 \% pure cleaner and the high concentration is 18% 18 \% pure cleaner. How many liters of the low and high concentrations must be combined to obtain 1010 liters that is 8% 8 \% pure cleaner?\newlineChoose 11 answer:\newline(A) 66 liters of low and 44 liters of high\newline(B) 613 6 \frac{1}{3} liters of low and 323 3 \frac{2}{3} liters of high\newline(C) 623 6 \frac{2}{3} liters of low and 313 3 \frac{1}{3} liters of high\newline(D) 77 liters of low and 33 liters of high

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Q. A cleaner is sold in low and high concentrations so that customers can combine amounts from each in order to obtain a desired quantity and concentration. The low concentration is 3% 3 \% pure cleaner and the high concentration is 18% 18 \% pure cleaner. How many liters of the low and high concentrations must be combined to obtain 1010 liters that is 8% 8 \% pure cleaner?\newlineChoose 11 answer:\newline(A) 66 liters of low and 44 liters of high\newline(B) 613 6 \frac{1}{3} liters of low and 323 3 \frac{2}{3} liters of high\newline(C) 623 6 \frac{2}{3} liters of low and 313 3 \frac{1}{3} liters of high\newline(D) 77 liters of low and 33 liters of high
  1. Define Variables: Let's denote the amount of low concentration cleaner as LL liters and the amount of high concentration cleaner as HH liters. We know that the total amount of cleaner we want is 1010 liters, so we can write the first equation:\newlineL+H=10L + H = 10
  2. Set Up Equations: We also know that the final mixture should be 8%8\% pure cleaner. We can write a second equation representing the total amount of pure cleaner in the final mixture:\newline0.03L+0.18H=0.08×100.03L + 0.18H = 0.08 \times 10
  3. Solve System: Now we have a system of two equations with two variables:\newline11) L+H=10L + H = 10\newline22) 0.03L+0.18H=0.80.03L + 0.18H = 0.8\newlineWe can solve this system using substitution or elimination. Let's use the substitution method. From the first equation, we can express LL in terms of HH:\newlineL=10−HL = 10 - H
  4. Substitute and Expand: Substitute L=10−HL = 10 - H into the second equation:\newline0.03(10−H)+0.18H=0.80.03(10 - H) + 0.18H = 0.8\newlineExpand the equation:\newline0.3−0.03H+0.18H=0.80.3 - 0.03H + 0.18H = 0.8\newlineCombine like terms:\newline0.15H=0.50.15H = 0.5
  5. Find High Concentration: Now, solve for HH: \newlineH=0.50.15H = \frac{0.5}{0.15}\newlineH=103H = \frac{10}{3}\newlineH=3(13)H = 3\left(\frac{1}{3}\right) liters
  6. Find Low Concentration: Now that we have HH, we can find LL using the first equation:\newlineL=10−HL = 10 - H\newlineL=10−3(13)L = 10 - 3(\frac{1}{3})\newlineL=10−103L = 10 - \frac{10}{3}\newlineL=303−103L = \frac{30}{3} - \frac{10}{3}\newlineL=203L = \frac{20}{3}\newlineL=6(23)L = 6(\frac{2}{3}) liters
  7. Final Answer: We have found the amounts of low and high concentration cleaners needed:\newlineL = 6(23)6\left(\frac{2}{3}\right) liters of low concentration\newlineH = 3(13)3\left(\frac{1}{3}\right) liters of high concentration\newlineThis corresponds to answer choice (C)(C).

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