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100 kgs of an alloy of tin and lead in the ratio 1:3 is mixed with x kgs of an alloy of tin and lead in the ratio 3:2. If the overall alloy should contain between 40% and 50% tin, what is the range of values x can take?

A. 100 kgs ≤ x ≤ 200 kgs
B. 110 kgs ≤ x ≤ 240 kgs
C. 105 kgs ≤ x ≤ 220 kgs
D. 75 kgs ≤ x ≤ 250 kgs
E. 60 kgs ≤ x ≤ 150 kgs

Are You Up For the Challenge: 700 Level Questions

    The ratio of tin and lead in 100 kg alloy = 1: 3
      Amount of tin = \(\frac{100}{4}*1 = 25\) kg
      Amount of lead = \(\frac{100}{4}*3 = 75\) kgs
    The ratio of tin and lead in x kg = 3: 2
      Amount of tin = \(\frac{x}{5}*3 =\frac{3x}{5}\) kg
      Amount of lead = \(\frac{x}{5}*2 = \frac{2x}{5}\) kg
    Total amount of alloy = \(100 + x\)
      Overall amount of tin =\(25 + \frac{3x}{5}\)
    Now, If the overall tin is 40%
      Then, \(\frac{40}{100}*(100+x) = 25 + \frac{3x}{5}\)
      \(40 + \frac{2x}{5} = 25 + \frac{3x}{5}\)
      \(40 – 25 = \frac{3x}{5} – \frac{2x}{5}\)
      \(15*5 = x\)
      \(x = 75\) kg
    If the overall tin = 50%
      Then,\( \frac{50}{100}*(100+x)=25 + \frac{3x}{5}\)
      \(50 + \frac{x}{2} = 25 +\frac{3x}{5}\)
      \(50 – 25 = \frac{6x}{10} – \frac{5x}{10}\)
      \(25*10 = x\)
      \(x = 250\)
    Thus, \(75 kg ≤ x≤ 250kg\)
Answer: Option D.
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100 kgs of an alloy of tin and lead in the ratio 1:3 is mixed with x kgs of an alloy of tin and lead in the ratio 3:2. If the overall alloy should contain between 40% and 50% tin, what is the range of values x can take?

Case1: if the overall alloy should contain 40% tin:

25%-----40%----------60% (x kgs)
(100 kgs)
--> \(\frac{x}{100}= \frac{(40-25)}{(60-40)}= \frac{15}{20}= \frac{3}{4}\)
\(x= \frac{300}{4}= 75\)

Case2: if the overall alloy should contain 50% tin:

25%-------------50%------60% (x kgs)
(100 kgs)

--> \(\frac{x}{100}= \frac{(50-25)}{(60-50)}= \frac{25}{10}= \frac{5}{2}\)
\(x= \frac{100*5}{2}= 250\)

The range of values of x
\(75 < x < 250\) kgs

Answer (D).
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