+76 1.)Roy prepared for the opening of school. He started buying padpapers and ballpens at $17.00 and $10.00 respectively. He bought a total number of 9 items. Had he bought 1 padpaper and less than 1 more pen, he would have spent $125. How many pens did he buy?
2.) A boy has a liter of 95% octane gas. He wanted to increase the concentration to 96% octane so that his motorbike wouuld run better. How much of pure octane gas should he add?
3.) Solution XRD is one half sulfuric acid while solution QXJ is 3/4 sulfuric acid. How many liters of solution QXJ should be added to a liter of solution XRD to obtain a solution that is 3/5 sulfuric acid?
4.) How much water must be evaporated from a 20 liter of 32% salt solutionif the desired concentration is 50%?
5.) There are 2 vats of milk, one containing 25% protein and the other 40% protein. These two are to be mixed to get 12 liters of a 35% protein content milk. How many liters of each must be used?
6.) In the chemistry lab, Efren mixed salt and water to produce 4 liters of a 50% solution. He wanted to know how many liters he will have to evaporate to obtain 75% solution?
7.) A 200mL shampoo with 80% cleansing power is to be diluted ith water. This is done by drawing out some amount and replacing it with water. Agnes wnats a mild shampoo of 60% cleansing power. How much must she draw off ad replace?
8.) A radiator that holds 16 quarts is full of 10% antirust solution. How much of this solution must be drawn off and be replaced with pure antirust so that the contents will be 35% antirust?
+128521 2) Note that pure octane = 100% = 1.00
.95(1) + 1.00 (x) = .96(1 + x) simplify
.95 + x = .96 + .96x subtract .96x, .95 from both sides
.04x = .01 divide both sides by .04
x = 1/4 L
+128521 3) Note.... 1/2 = .50 3/4 = .75 3/5 = .60 ....so we have
%purity * known amount of XRD + %purity * unknown amount of QXJ = final %purity [ known + unknown amts. ]
.50(1) + .75x = .60 (1 + x) simplify
.50 + .75x = .60 + .60x subtract .60x, 50 from both sides
.15x = .10 divide both sides by .15
x = [.10 / .15 ] L = [10/15 ] L = [2/3]L
+128521 4).....How much water = x .....it's easier if we express everything in terms of % water....... note that 32% salt solution = 68% water solution and pure water = 100% = 1 and a 50% solution = .50 water
.68(20) - 1x = .50 [ 20 - x] simplify
13.6 - 1x = 10 - .50x subtract 10 from both sides.....add 1x to both sides
3.6 = .50x divide both sides by .50
x = 3.6 / .50 = 7.2 L
+128521 5) All of these work the same, pretty much....
% of 25% protein milk * unkown amt + % of 40% protein milk * [ 12 - unknown amt ] = 40% protein milk * 12 L
.25x + .40(12 - x) = .35(12) simplify
.25x + 4.8 - .40x = 4.2
-.15x + 4.8 = 4.2 subtract 4.8 from both sides
-.15x = -0.6 divide both sides by -.15
x = 4 L of 25% and 12 - 4 = 8L of 40%
+128521 6) I'm assuming that we want to end up with a solution that is 75% salt.....this is like 4.....and 75% salt = 25% water....so we have
50% solution of water* 4 L - 100% water * unknown amt evaporated = 25% water [ 4L - unknown amt]
.50(4) - 1x = .25 [ 4 - x] simplify
2 - 1x = 1 - .25x add 1x to both sides, subtract 1 from both sides
1 = ..75x = (3/4)x multiply both sides by 4/3
[4/3 ] L = x this is the amt to be evaporated
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+128521 7) 80% cleansing powder = 20% water = .20 60% cleansing powder = 40% water = .40 pure water = 100% = 1
This one is a little trickier......I hope I've done it correctly....!!!
Let's call the amount we draw off = x And this is = to the amount of pure water we replace it with....so we have....
.20water * [ starting amount - amt drawn off] plus "x" amt of pure water = 40% water * 200L
.20(200-x) + 1x = .40 [ 200] simplify
40 - .20x + 1x = 80
40 + .80x = 80 subtract 40 from each side
.80x = 40 divide both sides by ..80
x = 40/.80 = 50 mL = amt drawn off
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