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One fish tank has 3% salt mixed in. Another tank has 10% salt. A certain type of fish requires a 100gallons tank which is 6% salinity. How much from each tank should be used to acquire the correctmixture for this species of fish?Explain

1 Answer

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Given:

First tank: 3% salt

Second tank : 10% salt

The fish requires 6% salt with a volume of 100 gallons

Let the required volume from the first tank be x.

Hence the required volume from the second tank would be:


=\text{ 100-x}

The amount of salt from the first tank plus the amount of salt from the second tank should be equal to the amount in the mixture:


\begin{gathered} x\text{ }*\text{ 0.03 + \lparen100-x\rparen }*\text{ 0.1 = 0.06 }*\text{ 100} \\ 0.03x\text{ + 10 - 0.1x = 6} \end{gathered}

Collect like terms:


\begin{gathered} 0.03x\text{ -0.1x = 6 - 10} \\ -0.07x\text{ = -4} \\ Divde\text{ both sides by -0.07} \\ x\text{ = 57.14 gallons} \end{gathered}

Hence the required volume of the first tank is 57.14 gallons. The volume of the second tank is:


\begin{gathered} =100\text{ - 57.14} \\ =\text{ 42.86 gallons} \end{gathered}

Answer:

We would required 57.14 gallons from the first tank and 42.86 gallons from the second tank

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