
A can contains a mixture of two liquids A and B in ratio 7:5 when 9 litres of mixture are drawn off and the cane is filled with B, the ratio of A and B becomes 7:9. How many litres of liquid A was contained by the cane initially?
( a ) 10
( b ) 20
( c ) 21
( d ) 25
Answer
483.6k+ views
Hint: to solve this question what we can do is first we will find the volume of liquids A and B in removed 9 litres of liquid and then we will find the ratio of liquid B when 9 litres of liquid B was added and then finally we will substitute the volumes of liquid A and in ratio equals to 7:9 and then we will solve to find the initial volume of liquid A.
Complete step by step answer:
Now, in question it is given that a cane has a mixture of two liquids A and B which are in ratio of 7:5.
And, on removal of 9 litres of mixture and filling cane with liquid B makes the ratio of liquids A and B to 7:9. Then we have to calculate the initial volume of liquid A.
Now, let initial volume of A = 7x and initial volume of B = 5x.
So, the total volume in the cane will be equal to the sum of volumes of A and B liquid together.
So, Total volume of Cane = 7x + 5x = 12x
Now, it is said that 9 litres of liquid is removed from the cane. So, here the key point is 9 litres of liquid is a mixture of A and B liquids.
Remained volume of liquid A $=7x-\left( \dfrac{7x}{12x}\times 9 \right)$, where $\left( \dfrac{7x}{12x}\times 9 \right)$ represents volume of liquid in 9 litre of removed volume.
On solving we get, Remained Volume of liquid A $=7x-\dfrac{21}{4}$
Same, Remained volume of liquid B $=5x-\left( \dfrac{5x}{12x}\times 9 \right)$, where $\left( \dfrac{5x}{12x}\times 9 \right)$ represents volume of liquid in 9 litre of removed volume.
On solving we get, Remained Volume of liquid B $=5x-\dfrac{15}{4}$
Now, it is given that the ratio of volume of liquids A and B after removing 9 litres from a cane and then adding 9 litres of liquid B in the cane is equal to 7:9.
Now, total Volume of liquid B in cane $=\left( 5x-\dfrac{15}{4} \right)+9$
So, $\dfrac{7x-\dfrac{21}{4}}{\left( 5x-\dfrac{15}{4} \right)+9}=\dfrac{7}{9}$
On solving by cross multiplication method, we get
$63x-\dfrac{189}{4}=35x-\dfrac{105}{4}+63$
$63x-35x=\dfrac{189-105}{4}+63$
$28x=84$
Obtaining value of x, we get
$x=3$
Initial volume of liquid A = 7x
Putting value of x, we get
Initial volume of liquid A $\text{= 7 }\!\!\times\!\!\text{ 3 = 21 litres}$
Hence, Initial volume of liquid A in cane is 21 litres.
Note: Expression a:b represents the ratio of two numbers a and b where there can be some constant common in both numbers. Understanding the Language of question is important in questions of ratio and proportion. Unit must be mentioned in the final answer which helps in identifying the answer from the solution.
Complete step by step answer:
Now, in question it is given that a cane has a mixture of two liquids A and B which are in ratio of 7:5.
And, on removal of 9 litres of mixture and filling cane with liquid B makes the ratio of liquids A and B to 7:9. Then we have to calculate the initial volume of liquid A.
Now, let initial volume of A = 7x and initial volume of B = 5x.
So, the total volume in the cane will be equal to the sum of volumes of A and B liquid together.
So, Total volume of Cane = 7x + 5x = 12x
Now, it is said that 9 litres of liquid is removed from the cane. So, here the key point is 9 litres of liquid is a mixture of A and B liquids.
Remained volume of liquid A $=7x-\left( \dfrac{7x}{12x}\times 9 \right)$, where $\left( \dfrac{7x}{12x}\times 9 \right)$ represents volume of liquid in 9 litre of removed volume.
On solving we get, Remained Volume of liquid A $=7x-\dfrac{21}{4}$
Same, Remained volume of liquid B $=5x-\left( \dfrac{5x}{12x}\times 9 \right)$, where $\left( \dfrac{5x}{12x}\times 9 \right)$ represents volume of liquid in 9 litre of removed volume.
On solving we get, Remained Volume of liquid B $=5x-\dfrac{15}{4}$
Now, it is given that the ratio of volume of liquids A and B after removing 9 litres from a cane and then adding 9 litres of liquid B in the cane is equal to 7:9.
Now, total Volume of liquid B in cane $=\left( 5x-\dfrac{15}{4} \right)+9$
So, $\dfrac{7x-\dfrac{21}{4}}{\left( 5x-\dfrac{15}{4} \right)+9}=\dfrac{7}{9}$
On solving by cross multiplication method, we get
$63x-\dfrac{189}{4}=35x-\dfrac{105}{4}+63$
$63x-35x=\dfrac{189-105}{4}+63$
$28x=84$
Obtaining value of x, we get
$x=3$
Initial volume of liquid A = 7x
Putting value of x, we get
Initial volume of liquid A $\text{= 7 }\!\!\times\!\!\text{ 3 = 21 litres}$
Hence, Initial volume of liquid A in cane is 21 litres.
Note: Expression a:b represents the ratio of two numbers a and b where there can be some constant common in both numbers. Understanding the Language of question is important in questions of ratio and proportion. Unit must be mentioned in the final answer which helps in identifying the answer from the solution.
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