# Math Winter Packet - 5th grade - 2012

by Ruby
(Upper Marlboro, MD)

page 1

WINTER BREAK MATHEMATICS
HOMEWORK PRACTICE
December 24, 2012 – January 1, 2013

Help Lee figure out her score in some games she played with friends.

 Dec 31, 2012 Math Winter Packet by: Staff Answer Part I Hi Ruby, Some of the questions in this winter packet are similar to the questions you submitted one year ago. (You can download the Math Standards for grade 5 in the State of Maryland from: http://mdk12.org/share/frameworks/CCSC_Math_gr5.pdf.) Page 3 of winter packet 1-3. Help Lee figure out her score in some games she played with friends. Her score is always in between the lowest score and the highest score shown. Use only the cards below to determine her scores. *Remember you can only use each card once. As before, the easiest way to compare the scores on page 3 of the winter packet is to CONVERT every NUMBER TO its DECIMAL EQUIVALENT, and then compare the numbers. You can convert any fraction into its decimal equivalent by dividing the numerator by the denominator. For example, the score listed on page 3 of the winter packet is 4 ½. To convert 4 ½ to its decimal equivalent: 4 ½ = 4 + ½ = 4 + (1 ÷ 2) = 4 + 0.5 4 ½ = 4.5 I have converted all of the scores which are listed as fractions on page 3 to their decimal equivalents. Now it’s just a matter of fitting Lee’s scores in between the highest and lowest scores for each row. Every score listed will only fit with the range in one row. The average of the high and low scores for every row is computed below: Row 1: average of low and high scores Average of 4.5 and 5.0 = (4.5 + 5.0)/2 = 4.75 Lee’s 2nd score belongs in this row. It is also the average of the high and low scores: 4.75 (or 4 ¾)

 Dec 31, 2012 Math Winter Packet by: Staff ………………………………………………………. Part II Row 2: average of low and high scores Average of 0.325 and 0.35 = (0.325 + 0.35)/2 = 0.3375 Lee’s 3rd score belongs in this row. It is above average: 0.349 Row 3: average of low and high scores Average of 0.35 and 0.375 = (0.35 + 0.375)/2 = 0.3625 Lee’s 1st score belongs in this row. It is above average: 0.365 Row 4: average of low and high scores Average of 0.2 and 0.225 = (0.2 + 0.225)/2 = 0.2125 Lee’s 4th score belongs in this row. It is above average: 0.221 Row 5: average of low and high scores Average of 2.75 and 3.5 = (2.75 + 3.5)/2 = 3.125 Lee’s 6th score belongs in this row. It is far below average: 2.751 The final answers are: *Bonus: Lee is missing a score. What could it be? Lee’s missing score: 3.521 Which score did you not give to Lee? 3.521 What could be the lowest and highest score for that number? Lowest: any number less than 3.521 Highest: any number greater than 3.521 3.521 does not fall between any of the high/low scores listed in the table of values. • Place Lee’s 0.375 and 3.521 on the number line below. Make sure to label your number line.

 Dec 31, 2012 Math Winter Packet by: Staff ……………………………………………………….Part III 4. The stem-and-leaf plot below shows the results of a student survey about the amount of time spent playing outside. Stem: the left-hand column. This contains the whole number of minutes in multiples of 10. 2 = 20 minutes, 3 = 30 minutes, etc.Leaf: the right-hand column. This column contains the number of minutes in the one’s place. 0 = 0 minutes, 5 = 5 minutes, etc.So, the first row contains the following number of minutes:20 minutes, 25 minutes, 25 minutesThe second row contains the following number of minutes:30 minutes, 35 minutes, 37 minutes, 39 minutesThe third row contains the following number of minutes:40 minutes, 45 minutes, 45 minutes, 45 minutes, 46 minutes

 Dec 31, 2012 Math Winter Packet by: Staff ………………………………………………………. Part IV The fourth row contains the following number of minutes: nothing The fifth row contains the following number of minutes: 60 minutes, 65 minutes, 66 minutes, 68 minutes The sixth row contains the following number of minutes: 70 minutes, 75 minutes Based on the plot, how many students spent less than an hour playing outside? Explain why your answer is correct. To answer this question, count the number of data points which are less than 60 minutes. The first row: 3 students 20 minutes, 25 minutes, 25 minutes The second row: 4 students 30 minutes, 35 minutes, 37 minutes, 39 minutes The third row: 5 students 40 minutes, 45 minutes, 45 minutes, 45 minutes, 46 minutes Total = 3 + 4 + 5 = 12 Answer: 12 students spent less than an hour playing outside. --------------------------------------------------- Use the digits below and the rules for divisibility to write four 4-digit numbers that are divisible by 2, 3, 5 and 9. Digits that can be used: 0, 0, 0, 0, 1, 2, 2, 3, 4, 4, 5, 5, 6, 7, 7, 8 You can only use a digit more than once if it is shown more than once. *Remember: You have to make four numbers. This particular problem looks like it could be difficult. However, it is easy to solve using the following rules: Divisibility by 2 ________________________________________ Any even number is divisible by 2. Even numbers end in 0, 2, 4, 6, or 8. The last digit of each of the four digit numbers will be 2, 2, 4, 6, or 8 Divisibility by 3 Any number is divisible by 3 if the sum of its digits is divisible by 3. Divisibility by 5 Every number that is divisible by 5 ends in 0 or 5. Since each of the four digit numbers must also be divisible by 2, the last digit of all four numbers must be 0. Divisibility by 9 Any number is divisible by 9 if the sum of the digits is divisible by 9. If a number is divisible by 9, it will also be divisible by 3. The last digit of the 4-digit numbers must be 0. The sum of the first three digits must by a multiple of 9. Here are four answers: 1440 8730 5220 5670 You can rearrange the digits to find other answers: 1440 or 4140 or 4410 8730 or 8370 or 7830 or 7380 or 3870 or 3780 5220 or 2520 or 2250 5670 or 5760 or 6570 or 6750 or 7560 or 7650

 Dec 31, 2012 Math Winter Packet by: Staff ………………………………………………………. Part V 9. You are the “Advertising Chief” for a video game company. You have a budget of \$200,000.00 to spend each week on television advertising time. Advertising time is \$5,000 per minute for prime time and \$3,000 per minute for off time. Prime time is considered to be from 8:00 p.m. to 11:00 p.m. everyday. Make a schedule for your weekly budget. Include the days, times, and number of minutes each ad will run. If you don’t want your budget cut, spend as close as you can to \$200,000.00. Weekly Budget Prime Time Budget + Off Time Budget ≤ 200000 x = prime time minutes y = off time minutes 5000*x + 3000*y ≤ 200000 5000x + 3000y ≤ 200000 • When this inequality is plotted, the area inside the boundaries forms the Bounded Feasible Region: Bounded Feasible Region (shaded yellow area) Any combination of prime time minutes (x) and off time minutes (y) that falls within the shaded yellow area or along the red line will be within budget. To spend the entire \$200,000 each week, use a combination of x and y which falls on the red line. 10. Jack has 100 football cards that he wants to put in a photo album. Each page in the album will hold eight cards. a. How many pages in the album will have cards on them? Pages = 100 / 8 = 12.5 13 pages of the album will have football cards. b. Using the same information, how many pages will Jack completely fill with his cards? 12 pages will be completely filled. c. Jack spent \$100 on eight packs of cards. How much did each pack cost? Cost = 100 / 8 = \$12.50 per pack d. Compare the three problems that you completed for #5. How are they the same? How are they different? There is no #5 listed! Thanks for writing. Staff www.solving-math-problems.com