A mixture containing 18% of methanol is mixed with another mixture containing b% of methanol in the ratio 4:5. If the percentage of methanol in the final mixture is 16%, what is the percentage of b?
Show answer & explanation
Correct answer: B
Explanation
Given:
Methanol in the first mixture = 18%
Methanol in the second mixture = b%
Mixture ratio = 4 : 5
Final mixture = 16%
Let
First mixture = 4 units
Second mixture = 5 units
(4 × 18 + 5 × b) / (4 + 5) = 16
(72 + 5b) / 9 = 16
72 + 5b = 144
5b = 72
b = 14.4%