In what ratio must a mixture of 30% alcohol strength be mixed with that of 50% alcohol strength so as to get a mixture of 45% alcohol strength ?
Options:
A . 3 : 1
B . 1 : 2
C . 1 : 3
D . 2 : 1
Answer: Option C Answer: (c)Let x litres of first mixture is mixed with y litres of the second mixture. According to the question, ${x × 30 /100 + y × 50/100}/{x × 70/100 + y × 50/100} = 45/55$${0.3x + 0.5y}/{0.7x + 0.5y} = 9/11$6.3x + 4.5y = 3.3x + 5.5y6.3x – 3.3x = 5.5y – 4.5y3x = y ⇒ $x/y = 1 : 3$
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