
A jar contains a mixture of two liquids A and B in the ratio 4: 1. When 10 litres of the mixture is replaced with liquid B, the ratio now becomes 2 : 3. What was the volume of liquid A present in the jar earlier?
(a) 20 litres
(b) 10 litres
(c) 16 litres
(d) 15 litres
Answer
592.8k+ views
- Hint: Assume variable for amount of liquid A and liquid B. Write the equations relating them using the ratio initially and using the ratio after 10 litres of the mixture is replaced with liquid B. Solve the two equations in two unknowns to find the volume of liquid A present initially.
Complete step-by-step solution -
It is given that a jar contains a mixture of liquids A and B. Let the volume of liquid A be x litres and the volume of liquid B be y litres.
It is given that the volume of liquids A and B are in the ratio 4: 1, then we have:
\[\dfrac{x}{y} = \dfrac{4}{1}\]
Cross-multiplying the terms, we have:
\[x = 4y.............(1)\]
It is given that 10 litres of the mixture is replaced with liquid B. We know that in 10 litres of the liquid, we have a 4 : 1 ratio of liquids A and B.
Volume of liquid A in 10 litres = \[\dfrac{4}{{1 + 4}} \times 10\]
Volume of liquid A in 10 litres = \[\dfrac{4}{5} \times 10\]
Volume of liquid A in 10 litres = \[8litres\]
Volume of liquid B in 10 litres = 10 – 8
Volume of liquid B in 10 litres = 2 litres
Hence, the 8 litres and 2 litres of liquid A and liquid B respectively are replaced by liquid B.
Hence, the volume of liquid A becomes x – 8, and the volume of liquid B becomes y – 2 +10 which is y + 8 litres.
The new ratio is now 2 : 3, hence, we have:
\[\dfrac{{x - 8}}{{y + 8}} = \dfrac{2}{3}\]
Using equation (1), we have:
\[\dfrac{{4y - 8}}{{y + 8}} = \dfrac{2}{3}\]
Cross-multiplying, we have:
\[3(4y - 8) = 2(y + 8)\]
Simplifying, we get:
\[12y - 24 = 2y + 16\]
\[12y - 2y = 24 + 16\]
\[10y = 40\]
Solving for y, we have:
\[y = \dfrac{{40}}{{10}}\]
\[y = 4litres...........(2)\]
Using equation (2) in equation (1), we have:
\[x = 4(4)\]
\[x = 16litres\]
Hence, the correct answer is option (c).
Note: Note that 10 litres of the mixture is replaced by liquid B, you may wrongly assume that 10 litres of liquid A is replaced with liquid B in which case you will get option (a) as the correct answer, which is wrong.
Complete step-by-step solution -
It is given that a jar contains a mixture of liquids A and B. Let the volume of liquid A be x litres and the volume of liquid B be y litres.
It is given that the volume of liquids A and B are in the ratio 4: 1, then we have:
\[\dfrac{x}{y} = \dfrac{4}{1}\]
Cross-multiplying the terms, we have:
\[x = 4y.............(1)\]
It is given that 10 litres of the mixture is replaced with liquid B. We know that in 10 litres of the liquid, we have a 4 : 1 ratio of liquids A and B.
Volume of liquid A in 10 litres = \[\dfrac{4}{{1 + 4}} \times 10\]
Volume of liquid A in 10 litres = \[\dfrac{4}{5} \times 10\]
Volume of liquid A in 10 litres = \[8litres\]
Volume of liquid B in 10 litres = 10 – 8
Volume of liquid B in 10 litres = 2 litres
Hence, the 8 litres and 2 litres of liquid A and liquid B respectively are replaced by liquid B.
Hence, the volume of liquid A becomes x – 8, and the volume of liquid B becomes y – 2 +10 which is y + 8 litres.
The new ratio is now 2 : 3, hence, we have:
\[\dfrac{{x - 8}}{{y + 8}} = \dfrac{2}{3}\]
Using equation (1), we have:
\[\dfrac{{4y - 8}}{{y + 8}} = \dfrac{2}{3}\]
Cross-multiplying, we have:
\[3(4y - 8) = 2(y + 8)\]
Simplifying, we get:
\[12y - 24 = 2y + 16\]
\[12y - 2y = 24 + 16\]
\[10y = 40\]
Solving for y, we have:
\[y = \dfrac{{40}}{{10}}\]
\[y = 4litres...........(2)\]
Using equation (2) in equation (1), we have:
\[x = 4(4)\]
\[x = 16litres\]
Hence, the correct answer is option (c).
Note: Note that 10 litres of the mixture is replaced by liquid B, you may wrongly assume that 10 litres of liquid A is replaced with liquid B in which case you will get option (a) as the correct answer, which is wrong.
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