<p>Hi, guys hopefully you can help me sort this out.</p>
<p>There was a problem today that I though had no possible solution.</p>
<p>The question read something like this (I have some numbers from the calculator)</p>
<p>At a convenience store, 18 people purchase milk, 23 purchased eggs and 26 purchased cheese. The ones who bought milk did not purchase either eggs or cheese. Of the 60 customers, how many bought both eggs and cheese?</p>
<p>)2
)7
)10
)12
)13</p>
<p>If I was not mistaken, that is, if I did not misread the question (I was exhausted), I believe there is no possible answer choice that works fro this problem. Also, if one were to solve the problem, the answer would come out to 3.5, which is unlikely. Please help me sort his out. Thanks guys. BTW I remember this being question 8 or 9 of the final math section.</p>
<p>60-18 gives you 42 who did not buy milk, and therefore had to buy cheese/eggs. the sum of the total cheese/egg customers is 23+26=59, therefore there are 59-42=17 who should have bought both, unless i just did this completely wrong</p>
<p>**49… the answer is 7</p>
<p>@peanut32</p>
<p>First I did 60-18= 42 customers who purchased eggs, cheese or both.</p>
<p>23+26= 49 purchases of eggs and cheese.</p>
<p>So 7 customers purchased both eggs and cheese (49-42= 7)</p>
<p>the key was to realize that you had to subtract the milk customers from 60 then find the difference between the sum of the remaining numbers and the total customers excluding those who bought milk</p>
<p>wouldn’t you have to multiply the 7 by two to compensate for both groups?</p>
<p>For that one, I did 60 (total) - 18 (milk only) and got 42. Then I added the 23 (eggs) and 26 (cheese) to get 49. 49-42 = 7, or at least that’s how I figured it lol.</p>
<p>And with the prime factor one, they had like 12p and which could be p. So multiply all the answer choices by 12. The first 3 were divisible by 5, which is greater than 3. The fourth was divisible by 3 only I think, while the last was divisible by 7. The only greatest prime number not over 3 was the fourth one, 16*12 (192).</p>
<p>but i think that was experimental because i didn’t have it and i definitely had experimental writing, unless they give out different tests, which sounds pretty reasonable</p>
<p>i had experimental writing, and got the prime factor question.</p>
<p>i think they gave out different tests.</p>
<p>did you guys get a significant amount of whole number answers for the grid in? (9, 1200, ect.)</p>
<p>I just realized what I did wrong.
I don’t know why I did but I added 23 and 26 and got 49 then I multiplied the seven by 2 and subtracted that from 49. I got 35, which I compared to the (60-18) and thought it was wrong. Can someone explain to me why this method is wrong?</p>
<p>what was your line of thought for multiplying 7x2 ? I’ll see if I can help you figure it out</p>
<p>What’s the problem for 9:10? And is 9:10 correct?</p>
<p>
</p>
<p>I remember the answer to number 9 on the grid-in was 9. 
(The question dealing the coloring books and crayons).</p>
<p>I remember on a past SAT that I took that number 10 or number 11 on the grid-in was equal to it’s numerical counterpart as well.</p>
<p>was the computer lab answer 124? was that answer choice E or D? i think it was D</p>
<p>ah i see why i was wrong
well, that’s one known one wrong so far, i probably got more wrong as it is
what did you guys get for the one that asked which product would be positive for the inequality x^2-x-6 > 0?</p>
<p>anyone remember the question whose answer is 3/5 grid in
like graph of students participating in something…</p>
<p>how about the question maximum different studetnts lab, is it 140-8*2=124?</p>
<p>
</a></p>
<p>Umm, 48’s the desired number. The answer is 4 turns:</p>
<p>You start with 3 and turn once to get to 6, turn another time to get to 12, turn another time to get to 24, turn one last time to get to 48. 4 turns.</p>
<p>anyone can answer my question: which one is the one with 3/5 grid in, about probability to choose student or something from a chart</p>
<p>err hahahahah yea I meant 4. Heres an updated list gotten through pm/reworking, might be slightly outta order, like I said before my memory isn’t too hot.</p>
<ol>
<li>12 (find x, x=2g, 3=g/2)??</li>
<li>77th in line</li>
<li>12(averages problem find x)</li>
<li>4 cranks</li>
<li>II only (pythagorean theorem one)</li>
<li>75(what was this one)?</li>
<li>9(what was this one)?</li>
<li>2.40(auto industry problem)?</li>
<li>AP<AC (point p anywhere in rectangle)</li>
<li>-9<Y<1</li>
<li>Y=3X(just plugged in 2,6 to each one)</li>
<li>(X-3)(X+2)(had to factor each to find answer)</li>
<li>4:9 (ratio of something to something, cant remember)</li>
<li>(X+5)^2 - X^2 = 125 (word problem, rearranged to solve for x)</li>
<li>I and III (Cant remember this one, but I think i remember answer this)</li>
<li>72pi (semicircle one)</li>
<li>|H-68|<6 (pilot one)(option d?)</li>
<li>f(-1)=f(1) is False</li>
<li>12</li>
<li>1/4 tomato/potato question</li>
</ol>
<p>If any are wrong please correct me. Do you remember if the answer for 17 was option D?</p>