A can contains a mixture of two liquids A and B in the ratio 7:5 when

Using the values here might be the most straightforward way for this question.

As A:B::7:5 ---> only option C is a multiple of 7 and hence it is a good place to start. Also A:B::7:5 means that , A = (712)*Total and B = (5/12)*Total

If A = 21 , B = 15 ---> remove 9 litres ---> you remove (7/12)*9 of A ---> A remaining = 21-(7/12)*9 = 63/4

Similarly, for B, you remove (5/12)*9 ---> B remaining = 15 - (5/12)*9 = 45/4 and then add 9 more litres of B ---> 9+45/4 = 81/4

Thus A/B (final ratio) = (45/4)/(81/4) = 7:9 , the same as the final ratio mentioned in the question.

Hence E is the correct answer.

FYI, the algebraic equations will become:

A/B = 7/9 = (7x-(7/12)*9)/ (5x-(5/12)*9+9), where 7x and 5x are initial quantities of A and B respectively.

Thus, 7/9 = (7x-(7/12)*9)/ (5x-(5/12)*9+9) ---> giving you x=3. Thus A (original) = 7*3 = 21.

Hope this helps.

let x=total liters in can
5x/12-(9)(5/12)+9=9x/16
x=36
(7/12)(36)=21 liters of A in can initially

7x/12 - 9*7/12 = x*7/16
Solving, x=36
A = 7x/12 = 7*36/12 = 21

Method 2 ROA
5/12 - - - - 1
----9/17----
3 : 1
1 =9
4=36
A = 36*7/12 = 21

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To solve removal and replacement questions like the one above, we can use the following equation.

\(\frac{Final}{Initial} = (1 - \frac{b}{a})^n\)


Final = The Final Quantity of that component who's concentration is decreasing.

Initial = The Initial Quantity of that component who's concentration is decreasing.

b = Amount of liquid replaced

a = Final Volume in the container after the replacement of quantity b.

n = number of times the operation is done


Let the initial volume = x

b = 9

Since, we are removing and replacing the same amounts i.e 9 liters, therefore a = x

Initial quantity = 7x/12

Final Quantity = 7x/16

n = 1


Substituting in the equation above, we get \(\frac{\frac{7x}{16}}{\frac{7x}{12}} = (1 - \frac{9}{x})^1\)

\(\frac{12}{16} = 1 - \frac{9}{x}\)

\(\frac{9}{x} = 1 - \frac{3}{4} = \frac{1}{4}\)

x = 36


Therefore Initial amount = 7x/12 = (7 * 36) / 12 = 21 L


Option E

Arun Kumar

Using the formula will help us in arriving the answer C1* V1 = C2 * V2

Initial Volume = V1
9lts removed Volume now = V1-9
Concentration of A is 7/12

Now, B is added Volume is V1 again.

Hence
7/12 (V1-9) = 7/16 *V1

V1=36
Liquid A = 7/12*36 = 21

Solution:

We can let 7k and 5k be the number of liters of liquids A and B, respectively. When 9 liters of the mixture is drawn, 7/12 x 9 = 7/4 x 3 = 21/4 liters are liquid A and 5/12 x 9 = 5/4 x 3 = 15/4 liters are liquid B. We can create the following equation for the ratio of the two liquids after the removal:

(7k - 21/4) / (5k - 15/4 + 9) = 7/9

(28k - 21) / (20k - 15 + 36) = 7/9

7(20k + 21) = 9(28k - 21)

140k + 147 = 252k - 189

336 = 112k

3 = k

Initially, there were 7(3) = 21 liters of liquid A.

Answer: E

Conclusion= Solution(A&B) & B are getting mixed.

Let x be the total volume of the solution
A in solution = 7/12 x ----------Refer 1
9 L of mixture is drawn off so in 9 L how much is A ?
A= 7/12 * 9 = 63/12

till now the equation is 7/12 x - 63/12

now the same quantity is filled up with B in the solution. What is the quantity of A in pure B ? It's 0.
So now the equation becomes 7/12 x - 63/12 + 0
How much is A in the resultant mixture ?
7/16 x ( remember the total quantity is x)

Therefore, 7/12 x - 63/12 + 0 = 7/16 x
x= 36
How much is A in 36 ?
7/12 x -----Refer (1)
7/12 * 36
= 21 L
Option E

Hi,

let x be volume of the mixture. We know that A=(7/12)x and B=(5/12)x. Now according to the text, what happens is:

((7/12)x-(7/12)*9) / ((5/12)x-(5/12)*9+9)=A/B

Solve for x to get A=21L.

\(Initial \ Ratio: \frac{A}{B} = \frac{7}{5}\)

\(Initial \ Quantity \ of \ A = \frac{7}{12}\)

\(Final \ Ratio: \frac{A}{B} = \frac{7}{9}\)

\(Final \ Quantity \ of \ A = \frac{7}{16}\)

\(Formula: \ F = I*(1-\frac{y}{t})^n\)

F = Final Quantity
I = Initial Quantity
y = Amount being replaced
t = Total Quantity
n = No. of times it is repeated

F = \(\frac{7}{16}\)
I = \(\frac{7}{12}\)
y = 9 litres
t = 12x
n = 1

\(\frac{7}{16} = \frac{7}{12}*(1-\frac{9}{12x})^1\)

\(\frac{3}{4} = \frac{12x-9}{12x}\)

\(12x - 9 = 9x\)
\(3x = 9\)
\(x = 3\)

\(Thus, \ initial \ quantity \ of \ liquid \ A \ was = 7x = 7*3 = 21 (E)\)

Let 7x and 5x be number of litres of liquid of A and B respectively:

A removed 7/12*9 = 21/4
B removed 5/12*9 = 154
B added - 9

(7x - 21/4) / (5x - 15/4 + 9) = 7/9

(28x - 21) / (20x - 15 + 36) = 7/9

7(20x + 21) = 9(28x - 21)

140x + 147 = 252x - 189

336 = 112x

3 = x

So is 7*3 = 21 (E)

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