An alloy contains a mixture of two metals X and Y in the ratio of 2 : 3. The second alloy contains a mixture of the same metals, X and Y, in the ratio 7 : 3. In what ratio should the first and the second alloys be mixed so as to make a new alloy containing 50% of metal X?
Answer & explanation
Correct answer: option 4
Quantity of X in 1st mixture = \(\frac{2x}{5}\)
Quantity of X in 2nd mixture = \(\frac{7y}{10}\)
Quantity of X in Resultant mixture = \(\frac{1}{2}\) × (x + y)
Then, \(\frac{2x}{5}\) + \(\frac{7y}{10}\) = \(\frac{1}{2}\) × (x + y)
⇒ \(\frac{2x}{5}\) - \(\frac{1}{2}\)x = \(\frac{1}{2}\)y - \(\frac{7y}{10}\)
⇒ \(\frac{4 - 5}{10}\)x = \(\frac{5 - 7}{10}\)
⇒ -\(\frac{x}{10}\) = -\(\frac{2y}{10}\)
⇒ \(\frac{x}{y}\) = \(\frac{2}{1}\)
Then, x : y = 2 : 1.