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4 September, 01:44

A science teacher has a supply of 20% saline solution and a supply of 50% saline solution. How much of each solution should the teacher mix together to get 60 ML of 28% saline solution for an experiment

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  1. 4 September, 04:30
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    Let f = amount of 50% sugar solution in mL

    Let e = amount of 80% sugar solution in mL

    You want 150 mL of total solution, which is a combination of the two existing solutions in unknown amounts.

    f + e = 150

    The next is the sugar content. We know that 50% sugar and 80% sugar in each solution, so the sugar content itself can be represented as 0.5f and 0.8e respectively.

    Adding them together, we want a total of a 60% solution from both amounts. So now we have:

    0.5f + 0.8e = 0.6 (f + e)

    Now we have a system of two equations and two unknowns. First, I'll simplify that second equation a little.

    0.5f + 0.8e = 0.6f + 0.6e

    0.2e = 0.1f

    From there, I'll take the first equation and make it one variable in terms of the other:

    f + e = 150

    f = 150 - e

    Now we can substitute the expression (150 - e) for f in the simplified second equation.

    0.2e = 0.1f

    0.2e = 0.1 (150 - e)

    Now we can solve for e.

    0.2e = 15 - 0.1e

    0.3e = 15

    e = 15 / 0.3

    e = 50

    Now that we know that, we can solve for f

    f = 150 - e

    f = 150 - 50

    f = 100

    100 mL of 50% and 50 mL of 80% z

    If you look at your multiply choice, none of the options add up to 150 mL, so I'm thinking that maybe the problem is supposed to be to 105. If that's the case, it's still a system of equations using a different equation 1 (e + f = 105, instead of e + f = 150).
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