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    Topics || Problems

    An alloy contains 20% silver and 30% lead. How much silver and how much lead should be added to 100 pounds of alloy in order to obtain an alloy containing 25% silver and \(33\frac{1}{3}\%\) lead? Solution:
    S1S2S3Mixture
    Silver20%100%025%
    Lead30%0100%\(33\frac{1}{3}\%\)
    Total100xy100+x+y

    \(0.2(100) + x +0y = 0.25 (100+x+y)\)

    \(20+x = 25 +0.25x +0.25y\)

    \(0.75x = 5+ 0.25y\)

    \(x = \frac{20}{3}+ \frac{y}{3}\)

    \(0.3(100) + 0x +y = \frac{1}{3} (100+x+y)\)

    \(30 + y = \frac{100}{3} + \frac{x}{3} + \frac{y}{3}\)

    \(-\frac{10}{3} +\frac{2y}{3} = \frac{x}{3} \)

    \(x = -10+2y \)


    \(\frac{20}{3}+ \frac{y}{3} = -10+2y\)

    \(\frac{50}{3}= \frac{5y}{3}\)

    \(y = 10\)

    \(x = 10\)

    To obtain the mixture a 10 lbs silver and a 10 lbs lead are needed.