Idaho Society of Professional Engineers
Friday Update - 06/11/04
UPCOMING EVENTS:
● IBPEPLS Board Meeting, June 11 & 12, 2004
● NSPE 2004 Convention and Expo, July 8 -
10, 2004, Honolulu, Hawaii
● NSPE Western and Pacific Regional
Meeting, September 17-18, 2004, Coeur d'Alene, ID
Share your thoughts on the future of NSPE—NSPE
has created a Future Directions Task Force (FDTF) to assess, modify, or even
possibly reinvent the Society. The FDTF is seeking your input (members and
non-members, PEs and non-PEs) on important issues such as the targeted
membership, the program/product/services mix, and current strengths and
weaknesses, among others. Please take a few moments to
fill out our survey!
MATHCOUNTS PROBLEM OF THE WEEK
Can you solve this MATHCOUNTS problem? The answer will appear in next week's
edition of the Friday Update!
Wrapping Up The School Year
Final exams often accompany the final days of school. Sherry knew that for her
four quarter grades in Science, she had 87%, 89%, 92% and 91%. If her final exam
is 1/3 of her final grade, and each quarter grade is 1/6 of the final grade,
what percent does Sherry need to score on her final exam to have a final average
of at least 92% for the year? Express your answer as a decimal to the nearest
tenth.
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After taking her Biology exam, Sherry was relieved that she only had one exam
left. It was her Algebra I exam, which she felt very prepared for since she had
been on the MATHCOUNTS team all year! The last question on the test read: “If
the following equation has two positive integer solutions, what is the
difference between the greatest possible coefficient of x and the least possible
coefficient of x?” There was then a quadratic equation, but the coefficient of
the x-term was missing: x2 + ?x + 20 = 0. What is the correct answer to this
problem on Sherry’s exam?
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Once Sherry finished her Algebra I exam, the last thing she had to do was turn
in her padlock for her locker. Unfortunately, it had been a while since she used
her locker and she was not sure that she remembered what her three-number
combination was. The numbers on the dial were the whole numbers through 45. She
remembered that the three numbers formed an arithmetic sequence, the first
number was 6, and the numbers increased in value. How many different
combinations satisfy these conditions?
Answer to last week's problem:
There are four pictures with a combined width of 1.25 × 4 = 5 inches. The page
is 8.5 inches across, which leaves 3.5 inches of empty space. Notice that if S =
empty space and P = picture, the arrangement is SPSPSPSPS. This shows there are
five empty regions for a row of four pictures, and each of these five regions
must all be the same size. Dividing 3.5 by 5, we see that there is .7 or 7/10 of
an inch between two adjacent pictures in a row.
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Let’s just consider the girls. The tallest girl goes in the middle of the back
row. There is no option there. The two shortest girls are placed on the two
ends, but we’re not told which end the absolute shortest girl must be on. So
between the two shortest girls, either one could have gone on the left. So there
are 2 options for the left-most back position, but once that’s filled, there’s
only 1 option for the right-most back position. The second and fourth positions
can then be filled in either order by the two remaining girls. So there are 2
options for the second position, and the girl left over goes to the fourth
position. In total, then, there are 2 × 2 × 1 × 1 × 1 = 4 ways they could be
arranged. (Notice that once the first and second positions are filled, the other
girls have no options as to where to stand… it has already been determined.) The
same is really true for the boys in the front row. They just don’t have a
tallest boy in the middle, but as we saw above, that didn’t impact the number of
arrangements of the girls. There are then 2 × 2 × 1 × 1 = 4 ways the boys could
be arranged. Each of the 4 girl-arrangements can be paired with any of the 4
boy-arrangements, for a total of 16 possible arrangements.
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There are ten multiples of seven in the book (ranging from 7 × 1 = 7 to 7 × 10 =
70). There are seven multiples of ten in the book (ranging from 10 × 1 = 10 to
10 × 7 = 70). Notice, though, that page 70 was counted as a multiple of seven
and as a multiple of ten. Therefore, there were 10 + 7 – 1 = 16 pages on which
Talia’s friend left a message or signature.
If you want to see last week's problem again, click on
http://www.mathcounts.org/Queries/POW_Archive.taf?_function=detail&Q_A_uid1=497&_UserReference=4EA2CDF0CDDF822140C74384
Idaho Society of Professional Engineers
PO Box 170239
Boise, ID 83717-0239
208-426-0636
Fax: 208-426-0639
E-Mail: ispe@rmci.net
Web Site: www.Idahospe.org
Idaho
Society of Professional Engineers
