Question
In a mixture, nitrogen and HCL are present in a 4:1
ratio. After removing 25% of the mixture and replacing it with 'x' liters of HCL, the ratio of nitrogen to HCL in the mixture becomes 1:4. Then, 'y' liters of nitrogen are added to the mixture, resulting in a nitrogen-to-HCL ratio of 3:2. What percentage of 'y' is 'x'?Solution
Let the quantity of the nitrogen initially = ‘4a’ litres And, the quantity of the HCL initially = ‘a’ litres Quantity of the nitrogen that is taken out = 4a × 25% = a litres Quantity of the HCL that is taken out = a × 25% = 0.25a litres According to the question, (4a – a):(a – 0.25a + X) = 1:4 12a = 0.75a + X X = 11.25a…………………….(1) Again, (4a – a + Y):(a – 0.25a + X) = 3:2 (3a + Y):(0.75a + 11.25a) = 3:2 (From equation 1) 2(3a + Y) = 3 × 12 3a + Y = 18a Y = 15x Required percentage = (11.25a/15a) × 100 = 75%
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