<p>The answers from 3-8 are correct.</p>
<p>For #2 I keep getting the answer 2 quarts.
Here’s how I did it:
Total volume will remain the same i.e. 12 quarts. Now, In a 30% solution of the total volume the volume of antifreeze will be 0.3<em>12=3.6 quarts
So, somehow you need to obtain 3.6 quarts after all the mixing. Let’s assume that we replace x quarts of solution leaving behind (12-x) quarts of 20% solution and we add x quarts of 80% solution.
So, the equation becomes 0.2</em>(12-x)+0.8*x=3.6
Solving, you get x=2 quarts.</p>
<p>So, I think the answer should be 2 quarts of the solution must be replaced.</p>
<p>For #1, I think the question would make a lot more sense if it read “How many…4% butterfat?” instead of "How many…4% milk?</p>
<p>@ccprofile: I’m pretty sure the total volume changes, because you are adding another solution to the 12 quarts.</p>
<p>yo, hold it a sec. for number 4, assuming they each got the same distance when the hiker and ranger caught each other, and D = RT, shudnt it be:
80 minutes * x mph = 40 minutes * (x + 6) mph,
80x = 40x + 240,
x = 6 for the hiker, and 12 mph for the ranger?</p>
<p>@GammaGrozza:But the question says “replaced.” So, I believe that instead of adding more antifreeze we have to replace some volume of the 20% solution with 80 % solution, leaving the total volume unchanged.</p>
<p>The ranger started down the mountain trail after 80 minutes and then took additional 40 minutes to catch up. Hence, it should be 120 minutes * x mph = 40 minutes * (x + 6) mph</p>
<p>^^^ well well well, arent we smart? haha, thanks.</p>
<p>4% milk is milk with 4% fat. It’s like when you go to the store and get 2% milk, you’re getting reduced fat milk with only 2% fat. That’s the naming convention the distributors use. </p>
<p>So for the question:</p>
<p>You have 100 quarts of 3% milk, meaning 3% of that 100 quarts if fat. You want to add some milk that is 36% fat (cream is just milk with higher fat content) to get a final solution thats 4% fat. </p>
<p>So let x = volume in quarts of 36% fat cream you’re adding. </p>
<p>Then the amount of fat you’re adding from that cream is .36x. </p>
<p>The total amount of fat in the final solution then is .36x + .03(100).</p>
<p>The percent fat is just fat over total volume, and the total volume is 100 + x.</p>
<p>So,</p>
<p>.36x + 3 = .04
100+x</p>
<p>Or,</p>
<p>.36x + 3 = .04(100+x)</p>
<p>And then you solve.</p>
<p>Edit: which is 2.778, the answer you got so you were correct. Just saw the discussion about it and thought I’d do it out.</p>
<p>I agree with ccprofile about number 2. Since the question says replace, it implies that you’re taking out some 20% and adding an equal amount of 80% leaving the total volume unchanged. </p>
<p>Assuming this:</p>
<p>(12-x).2 + .8x = .3(12)</p>
<p>and x = 2</p>
<p>Everything else is correct.</p>
<p>I posted the 1st problem on the aops forum. No one seems to be able to agree on an answer!
Also, thanks for checking my answers grozza, ccprofile, and balaylay =DD</p>
<p>@guys who corrected me.</p>
<p>yeah, my bad. I guess I’m just so used to chem diluting questions where you get a larger total volume. This question format is dumb, imo. It is more likely to trip up test-takers who have taken a high school chem class and remembered the material (which should be rewarded).</p>
<p>
Hm, sounds like something you’d see on the SAT. I guess other math books model themselves after the test.</p>
<p>Ray has $435 in fives, tens, and twenties. He has 2 more $10 bills than $5 bills. How many of each type of bill does he have?</p>
<p>Does this problem even give enough information to solve it?</p>
<p>I think it needs to include information about how many twenties he has.</p>
<p>First you get the general equation: </p>
<p>435 = 20(y) + 10(x + 2) + 5(x)</p>
<p>Where “x” and “y” are the number of bills.</p>
<p>Then we get 435 - 20y = 10x + 20 + 5x</p>
<p>Giving us: 415 - 20y = 15x </p>
<p>Now “x” must be a integer: So 415 - 20y must be (congruent) to 0 (Mod 15)</p>
<p>We know the following are multiples of 15: (15, 30, 45, 60, 75, 90, 105, 120, 135, 150, 165, 180, 205…) we should now see the pattern…</p>
<p>Also “y” must be an integer: and we need 415 - 20y to give us a number that is divisible by 15. We know that every 60 or 3y we we get a number divisible by 15… for example y = 2 and x = 25 is a solution… because 435 = 20(2) + 10(25 + 2) + 5(25)</p>
<p>However, y = 5 and x = 21 is also a solution… because 435 = 20(5) + 10(21 + 2) + 5(21)</p>
<p>So there are multiple solutions.</p>