Park Forest Math Team

Meet #1

Arith1netie

Self-study Packet

Problem Categories for this Meet:

l. Mystery: Problem solving

2. Geometry: Angle measures in plane figures including supplements and complements

3. Number Theory: Divisibility rules, factors, primes, composites

4. Arithmetic: Order of operations; mean, median, mode; rounding; statistics

5. Algebra: Simplifying and evaluating expressions; solving equations with 1 unknown including identities

2017-18

Category 4 Arithmetic Meet #1 - October, 2015 1) Using the standard order of operations, find the value of 2) The arithmetic mean (average) of four different whole numbers is 65. The largest of the numbers is 80. What is the smallest possible value of any of the four numbers? 3) The average of a set of seven different positive integers is 26. The median is 15. What is the largest possible value of any of the seven integers? In October, 1959, The "Twilight Zone" first premiered on television.

Answers

1)

2)

3)

Solutions to Category 4 Arithmetic Meet #1 - October, 2015 1) Answers = 1) 553 = 2) 23 = 3) 128 = = = = 2) 80 + 79 + 78 + X = 4(65) 237 + X = 260 X = 23 3) The smallest three numbers must be below 15 and the largest three must be greater than 15. If the average of the seven numbers is 26, then their sum is 7(26), or 182. To get the largest of the seven numbers, let the three numbers below 15 be the smallest possible different values, namely 1, 2, and 3. Let the first two numbers that are greater than 15 be 16 and 17. The sum of the first six numbers is 1 + 2 + 3 + 15 + 16 + 17 = 54. Therefore, the seventh and final number is 182 - 54, or 128.

1)

2)

3)

Answers

Category 4ArithmeticMeet #1 - October, 2013

1) If a number is added to the following set, N, there would be two modes. What is that number?

N = 11, 7, 6, 3, 7, 12, 8, 3, 4, 10, 7, 9

2) A number, W, is added to the set of twelve numbers in the set N from problem #1 to create the set Ω. If the mean of the numbers in set Ω is 7.6, rounded to the nearest tenth, then what is the largest possible whole-number value of W ?

3) A = 6 x 10 ÷ 2 x 3 B = 15 - 32 + 23 + 1

C = A

B

What is the value of C ?

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Solutions to Category 4ArithmeticMeet #1 - October, 2013

Answers1) The current mode is 7, as there are more

1) 3 sevens in set N than any other number. If there were another 3 in the set, then

2) 12 there would be as many threes as sevens.

3) 6

2) The total of the twelve numbers in set W is 87. If the mean is about 7.6, then the total sum of the thirteen numbers, including W, is about (13)(7.6), or 98.8. If 13 is added to set W, then the new sum would be 100. To test, divide 100 by 13. The quotient is 7.692... which does NOT round to 7.6. Therefore, try adding 12 instead of 13 to set W. That would bring the thirteen-number total to 99. To test, divide 99 by 13. The quotient is 7.615... which meets all conditions of the problem.

3) A = 6 x 10 ÷ 2 x 3 Multiply and divide from left to right. = 60 ÷ 2 x 3 = 30 x 3 = 90 B = 15 - 32 + 23 + 1 Evaluate powers first = 15 - 9 + 8 + 1 Now add and subtract, in order, from left to

= 6 + 8 + 1 right. = 14 + 1 = 15

C = 90

15 = 6

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Meet #1 October 2011

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Category 4 – Arithmetic

1. Find the value of the expression:

Give your answer as a decimal rounded to the nearest hundredth.

2. The following table lists the scores distribution on a recent test:

Score Number of students who got

that score

7 6

What is the positive difference between the mean score and the median score?

3. Jeff wrote down a list of natural numbers, all different from each other.

He noticed the following properties of the list:

The mean value was .

The median was one of the numbers in the list.

The sum of all numbers in the list was times the median.

What is the largest possible number of numbers in the list if

the median is a two-digit number? Answers

1. _______________

2. _______________

3. _______________

Meet #1 October 2011

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Solutions to Category 4 – Arithmetic

1.

2. According to the table, there are a total of students/scores.

students scored or below, and scored or above, so the median is

. To calculate the mean score:

( )

The difference between the median and the mean then is .

3. Let N be the number of numbers in the list, M be the value of the median, and S

be the sum of the numbers in the list. Since the mean is 20, S must be 20N.

If N is even, then M could not be in the list because M would be halfway

between the two middle numbers, and not be either one. So, N is odd.

Also, from the problem, S=4M. So S=20N=4M or 5N=M. If N=5, then M=25,

so one list could be 1,9,25,30,35, so we know 5 is a possible number of

numbers in the list. If N=7, then S=140 and M=35. Suppose the list in

increasing order starts with 1,2,3,35, which are the smallest possible. The last 3

numbers in the list are greater than 35, but they must add to 99, which is

impossible. If the first 3 numbers were larger, then the sum left over is even

smaller than 99, so it is always impossible to have N=7. It gets worse for larger

lists, so the largest possible N is 5.

Note: The original solution had 19 as the answer, but only 5 was accepted as correct.

Answers

1.

2.

3.

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Category 4 - Arithmetic

Meet #1, October 2009

1. Find the value of the expression below:

5 ∙ 42 + 4 2 + 45 ÷ 15 ∙ 3

2. The table below lists individual points scored by players of the Orlando Magic in

game 3 of this year’s NBA finals.

What is the positive difference between the mean individual score and the median

individual score?

3. Express

2∙3∙5∙6

6∙66+6

66

as a decimal rounded to the nearest hundredth.

H. TURKOGLU

18

R. LEWIS 21

D. HOWARD 21

C. LEE 4

R. ALSTON 20

M. PIETRUS 18

T. BATTIE 4

J. NELSON 2

M. GORTAT 0

Answers

1. _______________

2. _______________

3. _______________

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Solutions to Category 4 - Arithmetic

Meet #1, October 2009

1. 5 ∙ 42 + 4 2 + 45 ÷ 15 ∙ 3 = 5 ∙ 16 + 4 2 + 3 ∙ 3 =

5 ∙ 202 + 9 = 5 ∙ 400 + 9 = 2,009

2. If we list the scores from least to greatest 0, 2, 4, 4, 18, 18, 20, 21, 21, we can

see that the median score is 18.

Added up, there were 108 points scored, so the mean is 108

9= 12.

The difference is 18 − 12 = 6.

3. 2∙3∙5∙6

6∙66+6

66

=180

3612

1

= 5

12= 0.416 ≅ 0.42

Answers

1. 2,009

2. 6

3. 0.42

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Category 4 Arithmetic Meet #1, October 2007

1. Find the value of this expression :

+⋅++⋅

.

2. The stem and leaf plot below shows the number of hits each player for the Boston Red Sox had at the end of the season. What is the mean number of hits for the Red Sox? Red Sox 20 3 19 2 5 18 2 4 6 17 6 16 3 15 7

3. In a list of 5 positive integers, the mean is 17, the median is 18, and the mode is unique. What is the largest possible number in the list? **"The mode is unique" means that there is just one mode. **

Answers 1. _______________ 2. _______________ 3. _______________

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==++=

++⋅=

++⋅=

+⋅++⋅

Solutions to Category 4 Arithmetic Meet #1, October 2007 1. 4

2. The Red Sox mean = (203+195+192+186+184+182+176+163+157) 9 = 182 ** You can add numbers in a stem and leaf plot without actually adding them!!! You can add the leaves 3+5+2+6+4+2+6+3+7=38 and add the stems(once each per leaf) 20+19+19+18+18+18+17+16+15=160 separately. Since the stems were actually the tens place, multiply that sum by 10 : 160x10 = 1600 and then add the leafs which came from the ones place 1600+38 = 1638. Now you can divide by 9 to get 182! 3. In a list of 5 positive integers, the mean is 17, the median is 18, and the mode is unique. What is the largest possible number in the list? If the mean is 17, then the sum of the numbers must be 5x17 = 85. The median is 18, so the middle number will be 18. If we want the biggest possible number we want the rest to be small, so we should make the first two as small as possible(both 1). The 4th number must be greater than 18 since if it were 18 there would be two modes(1 & 18), so the 4th number is 19. Adding up the 4 we have so far gives us 39. The last number would then be 85 – 39 = 46. If you make the smaller two numbers 1 and 2 you can make the 4th number 18, but those first four numbers would still have a sum of 39 and the biggest number would still be 46.

Answers 1. 4 2. 182 3. 46

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Category 4 Arithmetic Meet #1, October 2005 1. Find the value of the expression below.

10 ⋅ 32 − 52 −117

+ 18 ÷ 3 ⋅ 2

2. Each X in the lineplot below shows the score of a mathlete on a certain middle school math team. Find the median and the mode of these scores?

X X X X

X X X X X X X X X X X X X X X X X X 0 2 4 6 8 10 12 14 16 18

3. So far this quarter, Gil has taken 6 quizzes and 2 tests, and there is one more test to take. His scores on the quizzes are 85, 83, 98, 93, 86, and 87. His scores on the tests are 91 and 87. In computing an average, Gil’s teacher counts each test as two quizzes. He will also round an average of 89.5 up to a 90, which is an A-. What is the lowest score that Gil can get on the final test of the quarter and still have an average of 89.5?

Answers 1. _______________ 2. Median ________ Mode _________ 3. _______________

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Solutions to Category 4 Arithmetic Meet #1, October 2005

1. Following the order of operations, we get

10 ⋅ 32 − 52 −117

+ 18 ÷ 3 ⋅ 2

= 10 ⋅ 9 − 25 −117

+ 6 ⋅ 2

= 90 − 147

+ 12

= 90 − 2 + 12= 100

2. The median score is the middle score when the scores are arranged from least to greatest. The middle of these 22 scores is the average of the eleventh and twelfth scores. Since these are both 10, the median is 10 points. The mode is the most frequent score, which is 8 points, made obvious by the tallest bar of X’s. (Both answers must be correct for the 2 points to be awarded.)

3. The three tests that Gil will take in the quarter count as much as six quizzes, since each one counts double. If Gil is to have an average of 89.5, he must have a total of 89.5 × 12 = 1074 points. So far he has 85 + 83 + 98 + 93 + 86 + 87 = 532 points in quizzes and 2 × 91 + 2 × 87 = 182 + 174 = 356 points in tests, for a total of 888 points. He needs 1074 – 888 = 186 points. Dividing this by 2, we see that Gil must get at least a 93 on his last test to earn an A-.

Answers 1. 100 2. Median 10 Mode 8 3. 93