In a class of 30 students , the average weight of boys is 20 kg and the average weight of the girls is 25 kg . The fraction of boys out of the total students of the class is

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

Given. ratio = 7/2 : 4/3 : 6/5
= (7/2) x 30 : (4/3) x 30 : (6/5) x 30
= 105 : 40 : 36

Let investment of Amitabh = 105k
Investment of Kamlesh = 36k

Ratio of their investments
= [105k x 4 + 150% x8 of 105k] : (40k x 12) : (36k x 12)
= (420k + (150/100) x 105k x 8) : (480k) : (432k)
= (1680k) : (480k) : (432k)
= 35 : 10 : 9

? Brijesh's share
= (21600 x 10) / (35 + 10 + 9)
= (21600 x 10) / 54
= ? 4000

Method 1 to solve the equation.
Let us assume Cost Price (C.P) of 1 liter milk be Y rupees.
According to question,
Selling Price (S.P) of 1 liter of mixture Y rupees. (Selling price should be same as Cost price)
Profit = 50/3 %
Let us assume Cost price of 1 Liter of Mixture of Water and Milk = C
Cost price of 1 Liter of Mixture of Water and Milk = Selling price of Mixture - Profit
C = Y - C x 50/3%
? C = Y - C x 50/3x100
? C = Y - C x 1/3x2
? C = Y - C/6
? Y = C + C/6
? Y = (6C + C)/6
? Y = (6C + C)/6
? Y = 7C/6
? C = 6Y/7
Since in Y rupees we will get 1 liter milk.
Hence in 1 rupees we will get 1/Y liter milk.
Hence in 6Y/7 rupees we will get 1 x 6Y/7 x Y liter milk.
Hence in 6Y/7 rupees we will get 6/7 liter milk.
We need 1 liter mixture of milk and water for sold on the same price, we need to mix the water.
So water quantity in mixture = 1 - 6/7 = 1/7

Ratio of Milk and water in mixture = quantity of Milk in Mixture/quantity of Water in Mixture
Ratio of Milk and water in mixture = 6/7 / 1/7
Ratio of Milk and water in mixture = 6 x 7 / 7
Ratio of Milk and water in mixture = 6 / 1
Ratio of Milk and water in mixture = 6 : 1

Method 2 to solve the question.
Let the original amount of milk be 1 liter and the cost price is 1 rupees per liter.
Cost price of milk = 1 rupees.
Selling price of Mixture = 1 rupees.
When the milk is mix with x liters of water milk remaining is 1- x.
Let us assume the cost price of milk in mixture is Y and given that the Sold Price is 1.
we should use the Profit formula in algebra,
Y + Y x 50/3 % = 1
Y + Y x 50/3 x 100 = 1
Y + Y x /3 x 2 = 1
7Y/6 = 1
Y = 6/7
Y = 6/7 is the cost price of Mixture for 1 liter.
Since water is free so there is not cost of water in mixture. So price of milk is 6/7 rs in mixture.
Since as per question,
In 1 rupees we will get 1 liter of milk.
hence in 6/7 rupees we will get 1 x 6/7 = 6/7 liter of milk.
Now we have to make 1 liter of mixture with water and milk.
quantity of Water + Quantity of Milk = 1 liter
quantity of Water + 6/7 = 1
quantity of Water = 1 - 6/7 = (7 - 6)/7
quantity of Water = 1/7
Ratio of milk and Water = Quantity of Milk / quantity of Water
Ratio of milk and Water = (6/7) / (1/7)
Ratio of milk and Water = (6/7) x (7/1)
Ratio of milk and Water = 6 x 7/ 7 x 1
Ratio of milk and Water = 6 :1

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