A completely filled car radiator with capacity 8L contains a mixture of 40% antifreeze (by volume). If the radiator is partly drained and refilled with pure antifreeze, how many liters should be drained from the radiator so as to have a mixture of 70% antifreeze?
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`40% times 8L=3.2L` (antifreeze) and `60% times 8L = 4.8 L` (water)
We need 70% antifreeze.
`therefore 70% times 8 L = 5.6 L` (antifreeze) and `30% times 8L = 2.4 L` (water)
Let x = 100% antifreeze (which is the same quantity as the amount we will drain)to obtain the required consistency. Therefore we know that (8-x) will drain the radiator sufficiently.
We also know that we need 5.6 L antifreeze (once we have made the relevant adjustments) to produce a 70% solution. So, we can deduce that:
`100% times x + 40% times (8-x) = 5.6 L` in order to meet the requirements as we will still have some of the 40% consistency left after draining the required amount.
Calculate `x` from the equation:
`(100x)/100 +(40/100)(8-x) = 5.6`
`x + 3.2 - 0.4x = 5.6`
`therefore 0.6x= 5.6-3.2`
therefore x=4 L
So if 4liters are drained, there will be 4 liters left. Of that 4 liters there is a 40% solution of antifreeze: `40% times 4L=1.6 L` . Add this to the 4liters that will be poured in = 5.6L which is the required amount to create a 70% mixture.
Ans: 4 liters of the solution should be drained from the radiator so as to create a 70% mixture when more antifreeze is added.
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