In what ratio should milk and water be mixed so that after selling the mixture at the cost price a profit of 50/3 % is made?

Let us assume the number of boys = B and number of girls = G.
According to question,
B + G = 30
Lets us assume total weight of boys = W1 and total weight of girls = W2
average weight of boys = total weight of boys/number of boys
total weight of boys/number of boys = 20
W1/B = 20
W1 = 20B

average weight of girls = total weight of girls/number of girls
25 = W2/G
W2 = 25G

Data is not sufficient to solve the equation.
since we do not know either the average weight of the whole class or the ratio of no. of boys to girls.

Let us assume the ratio factor is x.
According to question,
Quantity of milk = 4x liters and Quantity of Water = x liters.
According to question,
4x + x = 45
? 5x = 45
? x = 9
Quantity of milk = 4x = 9 x 4 = 36 liters
Quantity of water = x = 9 liters
Let y liters of water be added to make the ratio 3:2
Then,
36/(9 + y) = 3/2
? 72 = 27 + 3y
? y = 15 liters
If you add 15 liters of water, the ratio will become 3:2.

The difference counts only due to 40% of the profit which was distributed according to their investments.

Let total profit = R.
40% of R is distributed in the ratio, 125000 : 85000 = 25 : 17

Share of 1st partner = 40% of R x25/(25 + 17)
= 40% of 25R/42 = (40/100) x (25R/42) = 5R/21

Share of 2nd partner
= 40% of 17R/42 = (40/100) x (17R/42) = 17R/105

Now according to the question
5R/21 - 17R/105 = 600
? R(25 - 17)/105 = 600
? R = (600 x 105)/8
= ? 7875

A's share : B's share : C's share
= [16000 x 3 + (16000 - 5000) x 9] : [12000 x 3 + (12000 + 5000) x 9] : (21000 x 6)
= (16 x 3 + 11 x 9) : (12 x 3 + 17 x 9) : (21 x 6)
= 147 : 189 : 126
= 7 : 9 : 6

Hence, B's share exceeds C's share by = {13200/(7 + 9 + 6)} x (9 - 6)
= (13200 x 3)/22
= ? 1800

Let investment of B be a for b months.
Then, A's investment = 3a for 2b months
? A : B = (3a x 2b) : (a x b) = 6ab : ab = 6 : 1

Let the total profit = m.
Then. m x 1/7 = 6000
? m = 6000 x 7 = ? 42000
? 20% of 42000 = ? 8400

Method 1 to solve the equation.
Let us assume the number of liters of the 90% purity solution = A
and the number of liters of the 97% purity solution = B.
According to question,
Since there are 21 liters of the solution,
A + B = 21 ...................... (1)
Since after mixing the two solutions the new mixture has 94% purity,
Concentrate of A + Concentrate of B = Concentrate of (A + B)
A x 90% + B x 94% = (A+ B) x 97%
? A x 90/100 + B x 97/100 = (A + B) x 94/100
? 90A + 97B = (A + B) x 94
? 90A + 97B = 94A + 94B
? 94A + 94B - 90A- 97B = 0
? 4A - 3B = 0 ........................(2)
Multiply the 3 with Equation (1) and add with Equation (2),
3A + 3B + 4A - 3B = 63 + 0
? 7A = 63
? A = 63/7 = 9
Put the value of A in Equation (1) , we will get
9 + B = 21
B = 21 - 9
B = 12
The first solution would be A = 9 liters.

Method 2 to solve the equation.
Hit and trail method.
94% is closer to 97% but barely meaning the mixtures will not be equal parts but will be slightly more of the higher purity. Quickly eliminate A and B. Out of the others 9 is the easy choice. If the other choices were closer to half this wouldn't work.

Using Alligation rule,
(Quantity of cheaper tea) / (Quantity of dearer tea) = (d - m) / (m - c) = 7/3
Therefore, they must be mixed in the ratio of 7 : 3.

Cost price of the mixture = 15 × (100 / 180) = Rs. 25/3 per kg
(Quantity of rice @ Rs. 8 per kg) / (Quantity of rice @ Rs.10 per kg) = (5 / 3) / (1/3) = 1/5
Quantity of rice @ Rs. 10 per kg = 25 × (1/ 5) = 5 kgs.

He should mix 30% and 50% in the ratio 5 : 15 or 1 : 3.
( 30% Solution ) / (50% Solution) = 1 / 3 or 1 : 3

According to figure we can find that the ration would be 1 : 7.
Quantity sold at 10% profit = 1 / (1 + 7)× 160 = 20 kgs.
Quantity sold at 6% loss = (160 ? 20) = 140 kgs.