7th Graders: Follow this link for review questions before Thursday's test on Signed Numbers https://sites.google.com/site/lassmath/my-forms 7th Graders - Here's a link to our online curriculum: https://im.kendallhunt.com/MS/students/2/index.html Here's a link to our online glossary: https://im.kendallhunt.com/MS/students/2/glossary.html Problem of the Week Counterfeit Coins
Five counterfeit coins are mixed with nine authentic coins. If two coins are drawn at random, find the probability that one coin is authentic and one is counterfeit.
Solution 45/91. If the counterfeit coin is chosen first, the probability is 5/14 • 9/13. If the authentic coin is chosen first, the probability is 9/14 • 5/13. The total probability is 5/14 • 9/13 + 9/14 • 5/13 = 45/91. Tires tires tires
A three-wheeled vehicle travels 100 km. Two spare tires are available. Each of the five tires is used for the same distance during the trip. For how many kilometers is each tire used?
Solution 60. Since only three of the five tires are in use at any time, the total distance traveled by all the tires is 3(100) = 300 km. However, each of the five tires is used for the same distance during the trip. Thus, each tire is used for 300 ÷ 5 = 60 km. Workshop
Toni makes stools and tables in her workshop. At the end of a long day, she has made 35 legs. If each stool has 3 legs and each table has 4 legs, and the legs are all identical, what are the possible combinations of stools and tables that can she now assemble if she uses all 35 legs? Solution 1 stool and 8 tables; or 5 stools and 5 tables; or 9 stools and 2 tables. Since each stool requires 3 legs and each table requires 4 legs, the scenario can be described as all possible integral solutions for 3x+ 4y= 35. This has 3 possible solutions: (1, 8), (5, 5), (9, 2). Note that in terms of legs, 4 stools are equivalent to 3 tables.
Palindromic Speeding
The odometer of a family car shows 15,951 miles. The driver noticed that this number is palindromic: it reads the same backward as forward. Surprised, the driver saw his third palindromic odometer reading (not counting 15,951) exactly five hours later. How many miles per hour was the car traveling in those 5 hours (assuming speed was constant)?
Solution 62. Realistically, the first digit of 15,951 could not change in 5 hours. Therefore, 1 is the first and last digit of the new number. Clearly, the second and fourth digits changed to 6. So the next three resulting palindromes are 16,061, 16,161, and 16,261. Thus, the car traveled 310 miles in 5 hours and must have been traveling 62 miles per hour.
Equal Row Sums
Using the diagram, place the numbers 1 to 10 in the circles so that the sums in the rows of three circles are the same and the sums in the rows of four circles are the same.  Solution One possible arrangement is shown. The sum of the rows of four is 23, and the sum of the rows of three is 16.  Engine Failure
An aircraft is equipped with three engines that operate independently. The probability of an engine failure is .01. What is the probability of a successful flight if only one engine is needed for the successful operation of the aircraft?
Solution 0.999999. Let P(S) be the probability of a successful flight, P(S') the probability of an unsuccessful flight, and P(Fn) the probability of nengines failing. Since the flight is unsuccessful only when all three engines fail, then the probability of unsuccessful flight is P(S') = P(F1 ∩ (F2 ∩ F3)) =(.01)(.01)(.01) =(.01)3. But P(S) = 1 − P(S') = 1 − (.01)3 = 1 − .000001 = 0.999999 Milk or Bread or Both?
A customer enters a supermarket. The probability that the customer buys bread is 0.6, the probability that he buys milk is 0.5, and that he buys both bread and milk is 0.3. What is the probability that the customer would buy either bread or milk or both?
Solution .8, or 4/5. Let Brepresent the event that the customer buys bread, Mthe event that the customer buys milk. Then, according to the rule of addition, we have P(B∪M) = P(B) + P(M) − P(B∩ M) = .60 + .50 − .30 = .80
Alternate solution.A Venn diagram offers a visual approach. If we first show the probability that the customer buys mile andbread—P(M∩ B)—then we can complete the diagram by subtraction: P(M∪B) = .2 + .3 + .3 = .8 Rectangular Solids
How many rectangular solids are possible with a volume of 100 cubic meters and sides of only whole numbers? Solution 8. The solutions by dimension are (1, 1, 100), (1, 2, 50), (1, 4, 25), (1, 5, 20), (1, 10, 10), (2, 2, 25), (2, 5, 10), and (4, 5, 5).
10 Digit Number
Write a ten-digit number so that the first digit indicates how many 0s are in the number, the second digit indicates how many 1s are in the number, the third digit indicates the number of 2s, etc. Solution 6,210,001,000
Flat Tire
After a cyclist has gone 2/3 of his route, he gets a flat tire. Finishing on foot, he spends twice as long walking as he did riding. If his walking and riding rates are both constant, how much faster does he ride than walk? Solution 4 times as fast. He walks one-third of the way, or half as far as he rides, but it takes him twice as long. Therefore, he rides four times as fast as he walks.
Pluto Math
Pluto's inhabitants use the same mathematical operators that we do (+, −, etc.). They also use an operator, @, that we do not know. The following are true for any real numbers x and y. x@ 0 = x x@ y= y@ x x @ y = ((y–1) @ x) + (x+1) = (y-2) @ x + 2(x+1) = ((y–3) @ x) + 3(x+1)…. What is the value of 12 @ 5? Solution 77. We have 12 @ 5 = 5 @ 12 = (4 @ 12) + 13 = (3 @ 12) + 13 + 13 = (3 @ 12) + 26. Continuing, (3 @ 12) + 26 = (2 @ 12 + 39) = (1 @ 12) + 52 = (0 @ 12) + 65 = (12 @ 0) + 65 = 77 Path Distance
Jeremy walks along a spiral path (as shown). If the path is 2 meters wide, how far does he walk?  Solution 97 meters. By superimposing a grid of 2X2 squares, one can see that the path Jeremy travels can be broken into 13 sections of lengths 13, 12, 12, 10, 10, 8, 8, 6, 6, 4, 4, 2, and 2 meters. Chicken Nuggets
Chicken nuggets come in packets of 6, 9, or 20. What is the largest number of nuggets that you cannot buy when combining various packets?
Solution 43. Any multiple of 3 that is greater than 3 can be obtained from packets of 6 and 9 nuggets. Since 36 = 9 + 9 + 9 + 9, 38 = 20 + 9 + 9, and 40 = 20 + 20, any even number ≥ 36 can be achieved by adding 6s to each of these. Similarly, by adding another 9, any odd number ≥ 45 can be achieved. But 43 is not yet guaranteed, so we need to examine the possibility of combinations that yield 43 nuggets. Since 43 is not a multiple of 3, likewise, 43 – 20 = 23 is not a multiple of 3, and 43 – 2 • 20 = 3 is too small to achieve. Consequently, 43 cannot be obtained.
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