How do you do this?

<p>Type of Chocolate Bar: Milk, Dark, Bittersweet
Percent of Cocoa by weight: 35, 50, 70</p>

<p>A website sells three types of chocolate bars of equal weight. If Serena orders two chocolate bars at random from the website, then melts them together, what is the probability that the resulting mix contains at least 50 cocoa percent by weight?</p>

<p>(A) 1/9
(B) 1/3
(C) 4/9
(D) 1/2
(E) 2/3</p>

<p>I'm making the assumption here that the question means that a milk chocolate bar is always 35% cocoa, dark is 50%, bittersweet is 70%. (Not that milk chocolate is available in 35, 50, or 70% and etc.)</p>

<p>This means that there are 9 possible combinations of two chocolate bars (3^2, or 3x3) for Serena to buy. In order to create a chocolate mix with over 50% chocolate, the average of the %s from the two original bars must be over 50.</p>

<p>So the possible choices are:
- milk and bittersweet (avg = 52.5)
- bittersweet and milk (this is a different choice from above!)
- dark and dark (avg = 50)
- dark and bittersweet (avg = 60)
- bittersweet and dark (again, this is a different choice)
- bittersweet and bittersweet (avg = 70)</p>

<p>With 6 choices yielding at least 50% cocoa out of 9 possible combinations, that's 6/9, or 2/3.</p>

<p>So (E) is the correct answer.</p>

<p>This is how i did the problem
Basically you have 3 candy bars, A, B and C.
A is 35% coco therefore 65% other
B is 50% coco therefore 50% other
C is 70% coco therefore 30% other
The key to this question is that you simply divide the weight of coco by the total weight (coco and other) and then u compare ur percentages with the different combinations.
When I did this problem i came up with 3 combinations, AB, BC, and AC. To figure out percentage of coco of AB and added their coco weights(35+50) then i divided it by total weight which includes "coco" weights and "other" weights (35+50+65+50). If you look carefully there's no need to even use a calculator. You know that in order for it to be at least 50%, the "other" weight must be equal or less than the coco weight. For AB, the percentage is
80/(80+"other" weight) If the other weight is more than 80, you know that number is less than .50. So I put in all the other numbers for BC and AC and found that only AB was less than 50%. (80/195). Therefore 2 out of the 3 combinations worked or 2/3 or E.</p>

<p>That's what I thought but it's actually D. I can't even understand their explanation.</p>

<p>can't be: </p>

<p>Make a punnett square for this and you will see:</p>


<p>it's from princeton 11 practice tests. I just double checked and I swear it's D.</p>

<p>that is odd, b/c i got E ask well...i'll work on it some more, but...</p>

<p>r3n-_-, why do you care about the SAT still? I thought you've already achieved your 2400 (with private 1 on 1 tutoring), according to some rumors. Are they true? It certainly appears that they are you must have been a liar as far as I'm concerned.</p>

<p>It's not as if these books are without their occasional typos..</p>

<p>Just hope that the actual SAT's answer key has none.</p>

<p>I have the PR's 11 book and the answer is C. I got E too but when I checked the answer it said that the probability of choosing either a dark or bittersweet is 2/3, so the probability that they both are dark or bittersweet is (2/3)X (2/3) = 4/9.</p>

<p>i think the ans is wrong becoz by doing 2/3 x 2/3, it eliminated the possiblity of having a milk+bittersweet combination. (35+70)/200 >50%</p>


<p>I got E as well. I think it might be a typo.</p>

<p>I also got E... Let w be the weight of each bar. Thus 2w will be the weight of the resulting mix. The cocoa percent for each is .35w, .5w and .7w. The possible sums are </p>

<p>.35w + .35w=.7w
.35w + .5w= .85w
.5w+.5W= 1W
.5w+.7w = 1.2w
.35w+.7w = 1.05w
.7w+.7w = 1.4w</p>

<p>The former 2 are less than 50%. If it said more than 50%, then D would be correct...</p>