Student 4 must be paired with student 12. Student 5 is paired with student 11, so that leaves student 4. Student 11 can’t be paired with student 14. Student 14 is paired with student 2, so that leaves student 11. From there, work backwards from student to student eliminating possible choices until student 1’s pair is identified. Student 1 can be paired with student 15 (sum of 16) not student 8 (already paired with student 17) or student 3 (sum of 4). We have our fourth pair! Student 2 is paired with student 14! Student 2 can be paired with student 14 (sum of 16), not student 7 (already paired with student 18), not student 2, because he is student 2. Student 3 can be paired with student 13 (sum of 16) or student 6 (sum of 9) or student 1 (sum of 4). Student 4 can be paired with student 12 (sum of 16) or student 5 (sum of 9). Student 5 can be paired with student 11 (sum of 16) or student 4 (sum of 9). Student 6 can be paired with student 10 (sum of 16) or student 3 (sum of 9). Student 7 is already paired with student 18 (see above). Student 8 is already paired with student 17 (see above). Student 9 is already paired with student 16 (see above). Student 10 can be paired with student 15 (sum of 25) or student 6 (sum of 16). Student 11 can be paired with student 14 (sum of 25) or student 5 (sum of 16). Student 12 can be paired with student 13 (sum of 25) or student 4 (sum of 16). Student 13 can be paired with student 12 (sum of 25) or student 3 (sum of 16). Student 14 can be paired with student 11 (sum of 25) or student 2 (sum of 16). Student 15 can be paired with student 10 (sum of 25) or student 1 (sum of 16). Of the nine pairs, we have 3 pairs figured quickly! Student 16 has to be paired with student 9. Likewise, student 17 has to be paired with student 8. Pair student 18 with student 7, which is the only option for reaching a perfect square. Is there a key that unlocks this problem quickly? I don’t know.Įventually, I did figure that 1 had to be paired with 15, and this is how I did it… I knew that 1 had to be paired with 3 or 15 (I had known that since the first pass through.). That made student 17 pair with student 8. There was no way to pair student 17 with another student and get a number less than 17, so the only option was to add to 25. There wasn’t a student who could pair with 17 to reach the perfect square 36 (17 + 18 = 35). Student 17 had to be paired with another student to give a perfect square. I eliminated choices A and C, because 1 + 16 and 1 + 9 don’t give perfect squares.Ī little more deduction told me 17 had to be be paired with 8 to give the perfect square 25. I had about 4 minutes left at this point and decided to continue with the the brute force approach (another critical error?). A probability distribution (That’s new!). The dumb “how many quarters are there?” problem. I made my way through the rest of the test. Yes, almost all ACT Math items have a “quick way.” So, contrary to what I think most students should do at this point (which is to give it their best guess and move on), I skipped it. I started working and realized I couldn’t see the “quick way.” I made a mental list of all possible perfect square sums from students 1 through 18: 4, 9, 16, 25. (This item is owned by ACT Inc and printed in the first practice test in The Official ACT Practice Guide 2016-2017.) I noticed the problems were taking me a little more time than I wanted, but I only had a phone calculator and was only willing to use a small notepad to write something down if absolutely necessary. I was halfway through at the 19-minute mark and had no troubles. ![]() This time, I set the timer and began working. Recently, ACT Inc put out a new “Official ACT Prep Guide” and, as I always do, I sat down to take the tests (timed, of course) and to categorize the new items. I formulated the first title after I had to skip THE problem the first time. ![]() ![]() The original title of this post was supposed to be “The Math Problem I Had to Skip and Come Back To,” but it didn’t work out that way. THE Problem I Missed on the ACT Math Test
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