Nigel Car Radiator

Let x=volume of solution being drained or added (Since they will be the same volume)

(25%)(16-x) + (100%)(x) = (40%)(16)

Using simple arithmetic skills, x = 3.2 quarts

Since the first case (where it's 25% antifreeze solution) states that that there is a 16-quart radiator filled with 25% antifreeze solution, we get that there are 4 quarts of pure antifreeze, mixed with 12 quarts of other parts in that radiator.

Let x=the quantity of solution he must drain; quantity of pure antifreeze to be added

Since there are x quantities of the solution to be drained, and the one to be drained is also 25% of pure antifreeze in that solution, then there would be $4-0.25x$ quarts of pure antifreeze left in the solution

Now, we would add a pure antifreeze [i.e. 100% or 1] that would once again fill the 16-quart capacity of Nigel's radiator, and when we regained that quantity, it's said that the solution would become 40% antifreeze solution. (The ratio of the quantity of pure antifreeze to the quantity of the whole solution is 0.4 or 40%) .

Hence, we get an equation of: $\frac { (4-0.25x)+x }{ 16 } =0.4$

Solving further... $\Rightarrow \quad 4+0.75x=6.4\\ \Rightarrow \quad 0.75x=2.4\\ \Rightarrow \quad \boxed{ x=3.2}$

The initial solution has 4 quart antifreeze and the aim is to make it 6.4 quart. So 2.4 quart must be drained. But at the same time 0.6(25% of 2.4) quart antifreeze also will get drained. So this amount(0.6 quart) of solution must be drained again. But here also 0.15 quart of antifreeze gets drained and to balance that another 0.15 quart solution must be drained. And this sequence goes on.

So the total amount of solution to be drained = 2.4 + 0.6 + 0.15 + .......................... (An infinite G.P.)

Hence the final answer is $\frac{2.4}{1-\frac{1}{4}} = 3.2$

1. Currently there are 4 quarts of antifreeze in the solution (given 25% concentration). The final output should have 6.4 quarts (given 40% concentration).

2. Hence, the final solution needs 6.4 - 4 = 2.4 additional quarts of antifreeze.

3. Removing and adding any amount of solution and replacing it with antifreeze will effectively add antifreeze at 75% of the solution amount (since the solution removed already has 25% antifreeze).

4. Hence, we need to find, of what amount is 2.4, 75%. 2.4 * (4/3) = 3.2 Final answer = 3.2 quarts

(16-x) 0.25 + x = 6.4, antifreeze balance

X = 3.2

1. Overall amount of anti-freezer increases , (40% of 16) - (25% of 16) = 2.4 quarts
2. On the other hand: a) first, you have to drained off X quarts of "mixer" . and then, b) fill the tanker with the equal X quarts of "anti-freezer" --> a) reduced anifreezer from X is 0.25X [since it is 25% of mixer] --> b) added anifreezer is X [now, you fill with only anitifreezer]
3. Now, euation of total amount of increased anit-freezer:
   -.25X +X = 2.4 =>x=3.2
   

Let x be the Volume drained and added since they both are the same Antifreeze left in the Radiator=16-x Pure Antifreeze left = 25(16-x)/100 Pure Antifreeze after x is added=(16-x)/4+x Percentage=((16-x)/4+x)/16*100=40 On Solving We Get x as 3.2

0.25(16-x) + 100/100x = 0.4(16) => removing x amount of 25% from 16 quart and add 100% x amount of antifreeze so that the new mixture of 166 quart is 40% antifreeze. X=3.2