Suppose the vessel initially contains 8 litres of liquid.

Let x litres of this liquid be replaced with water.

Quantity of water in new mixture = \(\left(3-\frac{3x}{8}+x\right) liters\)

Quantity of syrup in new mixture = \(\left(5-\frac{5x}{8}\right) liters\)

So,  \(\left(3-\frac{3x}{8}+x\right)=\left(5-\frac{5x}{8}\right)\)

5x + 24 = 40 - 5x

 10x = 16

x =  \(\frac{8}{5}\)

So, part of the mixture replaced = \(\left(\frac{8}{5}\times\frac{1}{8}\right)=\frac{1}{5}\)

Since first and second varieties are mixed in equal proportions.

So, their average price = Rs.     = Rs.130.50

So, the mixture is formed by mixing two varieties, one at Rs. 130.50 per kg and the other at say, Rs. x per kg in the ratio 2 : 2, i.e., 1 : 1. We have to find x.

By the rule of alligation, we have:

\(\frac{x-153}{22.50}=1\) 

x - 153 = 22.50

x = 175.50

Suppose the can initially contains 7x and 5x of mixtures A and B respectively

Quantity of A in mixture left = \(\left(7x-\frac{7}{12}\times9\right) liters=\left(7x-\frac{21}{4}\right) liters\)

Quantity of B in mixture left = \(\left(5-\frac{5}{12}\times9\right) liters=\left(5x-\frac{15}{4}\right) liters\)

So,  \(\frac{\left(7x-\frac{21}{4}\right)}{\left(5x-\frac{15}{4}\right)+9}=\frac{7}{9}\)

\(\frac{28x-21}{20x+21}=\frac{7}{9}\)

 252x - 189 = 140x + 147

 112x = 336

 x = 3.

So, the can contained 21 litres of A.

Let the cost of 1 litre milk be Re. 1

Milk in 1 litre mix. in 1^st can = \(\frac{3}{4}\)  litre, C.P. of 1 litre mix. in 1^st can Re   \(\frac{3}{4}\)

Milk in 1 litre mix. in 2^nd can = \(\frac{1}{2}\) litre, C.P. of 1 litre mix. in 2^nd can Re   \(\frac{1}{2}\)

Milk in 1 litre of final mix.= \(\frac{5}{8}\)    litre, Mean price = Re    \(\frac{5}{8}\)

By the rule of alligation, we have:

C.P. of 1 litre mixture in 1^st can  =  \(\frac{3}{4}\)

C.P. of mixture in 2^nd can  =  \(\frac{1}{2}\)

Main price  =  \(\frac{5}{8}\)

So, Ratio of two mixtures = \(\frac{1}{8}:\frac{1}{8}=1:1\)

So, quantity of mixture taken from each can = \(\left(\frac{1}{2}\times12\right)= 6 litres\)

By the rule of alligation:

Cost of 1 kg pulses of 1^st kind = Rs.15

Cost of 1 kg pulses of 2^nd kind Rs.20

Main price = Rs. 16.50

So,Required rate = 3.50 : 1.50 = 7 : 3.

By the rule of alligation:

Cost of 1 kg pulses of 1^st kind = Rs.15

Cost of 1 kg pulses of 2^nd kind Rs.20

Main price = Rs. 16.50

So,Required rate = 3.50 : 1.50 = 7 : 3.

Let C.P. of 1 litre milk be Re. 1

Then, S.P. of 1 litre of mixture = Re. 1, Gain = 25%.

C.P. of 1 litre mixture = Re. \(\left(\frac{100}{125}\times1\right)=\frac{4}{5}\)

By the rule of alligation, we have:

C.P. of 1 litre of milk = Rs.1

C.P. of 1 litre of water = 0

So, Ratio of milk to water = \(\frac{4}{5}:\frac{1}{5}=4:1.\)

Hence, percentage of water in the mixture =  \(\left(\frac{1}{5}\times100\right)\)  % = 20%

S.P. of 1 kg of mixture = Rs. 9.24, Gain 10%.

So, C.P. of 1 kg of mixture = Rs. \(\left(\frac{100}{110}\times9.24\right)=Rs. 8.40\)

By the rule of allilation, we have:

C.P. of 1 kg sugar of 1^st kind = Rs.9

Cost of 1 kg sugar of 2^nd kind = Rs.7

Main Price = Rs. 8.40

So, Ratio of quantities of 1^st and 2^nd kind = 14 : 6 = 7 : 3.

Let x kg of sugar of 1^st be mixed with 27 kg of 2^nd kind.

Then, 7 : 3 = x : 27

x= \(\left(\frac{7\times27}{3}\right)= 63 kg.\)

Amount of milk left after 3 operations = \(\left[40\left(1-\frac{4}{140}\right)^{3}\right]litres\)

=  \(\left(40\times\frac{9}{10}\times\frac{9}{10}\times\frac{9}{10}\right) =29.16 litres.\)

By the rule of alligation, we have:

Strength of first jar = 40%

Strength of 2^nd jar = 19%

Main Strength = 26%

So, ratio of 1^st and 2^nd quantities = 7 : 14 = 1 : 2

So, Required quantity replaced = \(\frac{2}{3}\)

To solve this problem, let us first define some variables.
Let’s say,
Q = Quantity of whisky in the jar
%A = Percentage of alcohol in the jar
%B = Percentage of alcohol in the whisky replaced

We are given the following information:
Q = 1
%A = 40 %
%B = 19 %

Now, the task is to find the quantity of whisky replaced.

Now, using the formula for finding the percentage, we can write:

%A = (Quantity of whisky with 40% alcohol/Total Quantity of whisky)*100

%B = (Quantity of whisky with 19% alcohol/Total Quantity of whisky)*100

From the given information, we can write:

40% = (Q/Q)*100
19% = (Q2/Q)*100

Where Q2 is the quantity of whisky replaced with 19% alcohol.

Now, we need to find the value of Q2.

To do this, let us rearrange the equation to get:
Q2 = (19/40)*Q

Now, we need to find the value of Q.

To do this, let us rearrange the equation to get:
Q = (40/19)*Q2

Substituting the value of Q2 in the equation, we get:
Q = (40/19)*(19/40)*Q

Simplifying the equation, we get:
Q = (40/19)*Q

Dividing both the sides by Q, we get:
(40/19)*Q/Q = 1

Simplifying the equation, we get:
40/19 = 1

Now, we need to find the value of Q2.

Substituting the value of Q in the equation, we get:
Q2 = (19/40)*(40/19)*Q

Simplifying the equation, we get:
Q2 = Q

Therefore, the quantity of whisky replaced is equal to the total quantity of whisky in the jar.

Hence, the answer is Option B 2/3

If you think the solution is wrong then please provide your own solution below in the comments section .