A bucket contains a mixture of two liquids A and B in the proportion 7 : 5. If 9 litres of mixture is replaced by 9 liters of liquid B, then the ratio of the two liquids becomes 7 : 9. How much of the liquid A was there in the bucket ?
A. 21 liters
B. 23 liters
C. 25 liters
D. 27 liters
Answer: Option A
Solution(By Examveda Team)
Suppose the can initially contains 7x and 5x litres of mixtures A and B respectively. When 9 litres of mixture are drawn off, quantity of A in mixture left:$$\eqalign{ & = \left[ {7x - {\frac{7}{{12}}} \times 9} \right]\, \text{litres} \cr & = \left[ {7x - {\frac{{21}}{4}} } \right]\,{\text{litres}} \cr & {\text{Similarly quantity of B in mixture left}}, \cr & = \left[ {5x - {\frac{5}{{12}}} \times 9} \right]\, \text{litres} \cr & = \left[ {5x - {\frac{{15}}{4}} } \right]\,{\text{litres}} \cr & \therefore \,{\text{ratio becomes}}, \cr & \frac{{ {7x - {\frac{{21}}{4}} } }}{{ {\left(5x - {\frac{{15}}{4}}\right)+9 } }} = \frac{7}{9} \cr & \Rightarrow \frac{{ {28x - 21} }}{{ {20x + 21} }} = \frac{7}{9} \cr & \Rightarrow {252x - 189} = 140x + 147 \cr & \Rightarrow 112x = 336 \cr & \Rightarrow x = 3 \cr & {\text{So the can contained}}, \cr & = 7 \times x \cr & = 7 \times 3 \cr & = 21\,{\text{litres of A initially}}{\text{.}} \cr} $$
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You didn't add 9l of liquid B 🙄🤔
you didn't add 9L of liquid B. you took 9L in the ratio7:5. but didn't add 9L
How can you write
[7x −(21/4)]/ [5x - (15/4)] = 7/9
=>(28x -21)/(20x+21)=7/9
Wonderful
5x - 5/12*9 + 9ltr = 5x - 15/4 + 9 = 20x-15+36/4 = 20x+21/4
(28x-21/4)/(20x+21/4) = 7/9
(28x -21)/(20x +21) = 7/9
112x = 336
X = 3
add 9litres liquid with B . Then u wil get
Can anyone explain how
[5x −(15/4)] = (20x+21)