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17 April, 11:02

Mr. Acosta works in the lab at a pharmaceutical company he needs to make 50 L of 31% acid solution to test a new product key supplier only ships the 27% and 32% solution Mr. Costa decides to make the 31% solution by mixing the 21% solution with the 32% solution how much of the 27% solution will Mr. Costa need to use?

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  1. 17 April, 11:05
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    10 litres of the 27% acid solution is needed

    Step-by-step explanation:

    In this question, we are tasked with calculating the amount in litres of the 27% solution that Mr. Costa will need

    we proceed as follows;

    let us suppose Acosta needs X lts of 27%solution and Y lts of 32% solution to make 50L of 31% acid solution.

    So total sum of x and y is 48

    x + y = 50 ... (i)

    amount of acid in 27% solution is 0.27Y

    amount of acid in 32% solution is 0.32Y

    Total amount of acid in 50 liters of 31% solution is 50 * 0.31 = 15.5

    So 0.27x + 0.32y = 15.5 ... (ii)

    From i, we can say y = 50 - x, we then substitute this into 2 to yield the following;

    0.27x + 0.32 (50 - x) = 15.5

    0.27x + 16 - 0.32x = 15.5

    ⇒ 0.32x - 0.27x = 16 - 15.5

    0.05x = 0.5

    x = 0.5/0.05

    x = 10

    Hence, 10 litres of the 27% acid solution is needed
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