Acar radiator needs a 60% antifreeze solution. the radiator now holds 35 l of a 50% solution. how many liters of this should be drained and replaced with 100% antifreeze to get the desired strength?

7 L should be drained off and replaced with 7L 100% antifreeze.

Step-by-step explanation:

The first step is to figure out the number of Liters of antifreeze are present.

50% of 35L is antifreeze.

50% * 35 L = 50/100 * 35L = 17.5 L are antifreeze.

Now you have to figure out how many L of antifreeze you need for 60% antifreeze 60*35 = 21.00 L of antifreeze are needed for a 60% solution.

The difference is 21 - 17.5 = 3.5 L of antifreeze are needed.

Since you are using 100% antifreeze, all you need do is drain off 3.5 L of what is in there and replace it with 3.5 L of 100% antifreeze. The problem is not quite that simple because the 3.5 L of antifreeze that you drained off contains 1.75 L of antifreeze. So you have to refine your answer.

Let the number of Liters you have to drain off = x

Let the number of 100% antifreeze you need to put in of 100% also be x.

50% (35 - x) + 100%*x = 60%35L     Remove the brackets on the left.

17.5 - 50%x + 100%x = 60%*35 L     Expand the right side.

17.5 - 50%x + 100%x = (60/100)*35L Find the number of Liters on the right

17.5 - 50%x + 100%x  = 2100/100 = 21L of  antifreeze in a 60% solution. Now combine the percents on the left.

17.5 + 50%x = 21L     Subtract 17.5 from both sides.

17.5 - 17.5 + 50%x = 21L - 17.5

50%x = 3.5L                Translate 50% to a decimal

(50/100)x = 3.5L       Multiply both sides by (100/50)

(50/100)x * (100/50) = 3.5 * (100/50)

x = 7 L

What you do is take out 7 L of 50% antifreeze and put in 7L of 100% antifreeze in the 7Ls place.

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i hope this answer !

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geometry: triangles

solvers

lessons

answers archive

call the larger triangle abc

then the smaller triangle cde is similar to abc (all angles are equal, aaa).  

ab is the side that is 36, and we wish to find de, the side corresponding to ab

area_cde = area_abc - area_cde --> area_abc / area_cde = 2

let x=length of de

ratio of areas = (ratio of sides)^2

+2+=+%2836%2fx%29%5e2+

+2+=+1296%2fx%5e2+

++x%5e2+=+1296%2f2+=+648+

25.45584