Two vessels A and B contain alcohol and water in the ratios 5:7 and 3:5…
2019
Two vessels A and B contain alcohol and water in the ratios 5:7 and 3:5 respectively. Their contents are mixed in the ratio 3:2. From this mixture, 100 mL of solution is taken. To make the alcohol:water ratio in this 100 mL equal to 1:2, how much water must be added?
Answer: A. 20 mL — Key idea: find the alcohol fraction in the mixed solution, compute amounts in 100 mL, then determine how much water to add so water becomes twice the alcohol.…
- A.
20 mL
- B.
40 mL
- C.
30 mL
- D.
25 mL
Attempted by 85 students.
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Correct answer: A
Key idea: find the alcohol fraction in the mixed solution, compute amounts in 100 mL, then determine how much water to add so water becomes twice the alcohol.
Alcohol fraction in A = 5/(5+7) = 5/12; in B = 3/(3+5) = 3/8.
Mixing A and B in ratio 3:2 gives alcohol fraction = [3*(5/12) + 2*(3/8)] / (3+2) = (15/12 + 6/8)/5 = (1.25 + 0.75)/5 = 2/5.
In 100 mL of this mixture: alcohol = 100*(2/5) = 40 mL, so water = 100 − 40 = 60 mL.
To get alcohol:water = 1:2, water must be 2×alcohol = 2×40 = 80 mL. Therefore add 80 − 60 = 20 mL of water.
Answer: 20 mL