<p>The student band consists of tubas, kazoos, and harps. Overall, 9 students can play the tuba, 13 students can play the kazoo, and 20 students can play the harp. If six of the students each can play all three instruments and two of the students each can play two of the instruments, how many students are in the student band?</p>
<p>A) 32</p>
<p>B) 33</p>
<p>C) 36</p>
<p>D) 41</p>
<p>E) 42</p>
<p>What? Did I happen to post an extremely easy question that none cares to answer or something else?</p>
<p>the answer is A. [Student</a> Band | Hard SAT Math Questions](<a href=“http://satmathquestions.com/?p=136]Student”>http://satmathquestions.com/?p=136) lol.</p>
<p>That solution is wrong. They put 2 in the center of the diagram, it should be 6. (problem says 6 of the students each play all three instruments. Only 2 play 2). They have the 2 and the 6 switched in the solution. The answer should be 28.</p>
<p>^ ahhh, i got 28 too but I thought I was wrong cause it wasn’t a choice. So i googled it and i didn’t read the solution. My bad.</p>
<p>Anyone who can post the solution too? only the answer is not going to do much good to a dabbler like me…</p>
<p>This question is worded very poorly. In any case, drawing a Venn diagram is a good way to solve it (the solution above is wrong because the 2 and 6 are in the wrong positions). Here’s an alternate solution without a diagram:</p>
<p>If you add the number of students that play tuba, the number that play kazoo and the number that play harp you get 9+13+20=42. But you may have counted some students twice. For example, if a student plays both tuba and kazoo, then you’ve counted him once for tuba, and once for kazoo. So you need to subtract each of the intersections taken two at a time: 42 - 2 - 6 - 6 - 6= 22. Finally, you’ve subtracted 6 off once too many, so add it back once: 22 + 6 = 28.</p>
<p>I essentially applied the following formula:</p>
<p>(A or B or C) = A + B + C - (A and B) - (A and C) - (B and C) + (A and B and C).</p>
<p>Specific,
Your math question is very good , can you recheck the wording again, something is not right,
The formula (A or B or C) = A + B + C - (A and B) - (A and C) - (B and C) + (A and B and C) = 9 +13 +20 - 2 -2 -2 +6= 42 creates a nonsense result</p>
<p>If you don’t want memorize that formula, you can draw a Venn diagram. I find that the easiest way to solve these</p>
<p>I did it without a formula. If u have 20, 13, and 9. Yet 6 kids play all 3. U cud drop18 and add 6. In effect, turn ur 20 to an 8. ( where the kids are doesnt change anything) now u have a similar but not exact issue. 2 kids are beong counted double. (for ease of mental math i chose 13) 13-4=9 9+2=11. Now what do u have? 11+9+8=28. </p>
<p>Once broken into sub problems it is easier,in my eyes, to solve it.</p>
<p>what level problem is this?</p>
<p>I don’t believe that this is actually an SAT problem - it is worded poorly. It would have to be level 5 if something similar were to appear.</p>
<p>DrSteve,
Use the formular:(A or B or C) = A + B + C - (A and B) - (A and C) - (B and C) + (A and B and C).
A + B + C = 42
(A and B)= 2 (A and C) =2 (B and C) =2 and (A and B and C)= 6
so (A or B or C)= 42-(2+2+2) +6 =42
Please show me how can you said 42 - 2 - 6 - 6 - 6= 22 +6 =28. I do not understand this. I do not want to miss this question, saw similar couple times in the SAT test. Can you be kind to re-write the question or make a different question and solution. Many thanks.</p>
<p>Here is a slightly different way to go:</p>
<p>If you start by adding up the three types of players, you get 42. Now what’s wrong with that? Well, to start with, you counted some people twice – how many? The 2 who play two instruments. So to adjust our sum, take 42 -2 = 40. But wait: thre are also 6 people who play all three instruments. Our initial sum counted them THREE times. To adjust, we have to subtract out 6 and then 6 again (or 6x2=12). So from 40 - 12 we get 28.</p>
<p>If you wanted to do it all in one line: 9 + 13 + 20 - 2 - 6 - 6 = 28.</p>
<p>I think it would help you more if you understand why we add or subtract rather than developing and memorizing a formula to apply. The basic idea in these overlapping category problems is that if you are not paying attention, you end up counting the “overlapping” people more times than you need to.</p>
<p>pckeller
Great solution and understanding. Well done brother.</p>
<p>Pc has a nice solution there, but just for completeness I’ll try to clarify what I did. Let’s do a simpler problem first:</p>
<p>The student band consists of tubas and kazoos. Overall, 9 students can play the tuba, and 13 students can play the kazoo. If six of the students can play both instruments, how many students are in the student band?</p>
<p>Here, if you add the number of students that play tuba and the number of students that play kazoo you get 9 + 13 = 21. But you’ve counted the ones that play both twice, so you need to subtract them off once: 21 - 6 = 15.</p>
<p>The above solution leads to the following simple formula:</p>
<p>(A or B) = A + B - (A and B)</p>
<p>I suggest you also draw a Venn diagram of the above situation to increase your understanding.</p>
<p>Now, the generalization of this formula to 3 things is what I used above:</p>
<p>(A or B or C) = A + B + C - (A and B) - (A and C) - (B and C) + (A and B and C)</p>
<p>Clearly A = 9, B = 13, C = 20, and (A and B and C) = 6. </p>
<p>The other 3 are not set in stone, because those extra 2 can be anywhere (the problem isn’t specific abut which 2 instruments are played). So let’s just assume that the extra 2 are both tuba and kazoo.</p>
<p>Then (A and B) = 8, (A and C) = 6, and (B and C) = 6.</p>
<p>Thus we have: </p>
<p>(A or B or C) = A + B + C - (A and B) - (A and C) - (B and C) + (A and B and C)
= 9 + 13 + 20 - 8 - 6 - 6 + 6 = 28.</p>
<p>I don’t think I need to explain this further - Pc 's explanation of what is going on conceptually is very good. But again, it would be a good idea for you to draw a Venn diagram to get a deeper understanding of the situation, and to see why this formula works.</p>
<p>By the way, a problem of this level is very rare on the SAT, and if it appears it is a Level 5 problem.</p>