78 practice 2012 a

2010 TOURNAMENTGrade 8 and below

Table of ContentsIndividual Event: Problems .................................................................................................................... 1

Team Event: Problems .......................................................................................................................... 2

Individual Event: Model Solutions .................................................................................................. 3-4

Team Event: Model Solutions .............................................................................................................5-6

Tiebreakers: Problems and Model Solutions .......................................................................................... 7

PRACTICE PROBLEMS FOR 2012

Copyright 2011 by Mathematical Olympiads for Elementary and Middle Schools, Inc. GRADE 8 AND BELOW 11111

2010 Individual EventNo calculators are permitted during this tournament.

Time Limit: 30 minutes.

1. Rearrange the six letters in the word REDDER so that 1 letter separates the two Rs, 2 letters separatethe two Es, and 3 letters separate the two Ds. There are two such arrangements. Write either one of them.

2. John buys 4 action figures and has $24 left. If instead he buys 7 action figures, he would have no moneyleft. What is the cost of 1 action figure?

3. What is the least number of hours in exactly D days, if D has more than 2000 hours?

4. What number is 40% of the way from –19 to – 4?

5. The measure of one angle of a triangle is 55°. A second angle contains as many degrees as the thirdangle. How many degrees are contained in the largest angle?

6. The 16 digits of a credit card number are written in the boxes below. The product of any four consecutivedigits is 84. What is the sum of A and B?

7. The area of a circular pond is 225π square meters. Bordering the pond is a pavedcircular path whose area is 175π square meters. How wide is the paved path alone, inmeters?

8. A number N is 1 more than a multiple of 4, 3 less than a multiple of 6, and leaves aremainder of 6 when divided by 9. What is the least value of N, if N is also greater than700?

9. Express the extended fraction at the right as a simple fraction in lowest terms.

10. At the right is rectangle ABCD. Point P lies in side DC. ∠APD and ∠BPC arecomplementary angles. AB = 5 and PA = 4. Find the area of rectangle ABCD.

4 1 A B

132

546

−−

A

C

B

D

5

P

4

POND

PATH

?

23


2010 Team EventNo calculators are permitted during this tournament.

Time limit: 20 minutes

11. What is the value of (487 × –591) + (487 × 591)?

12. A hexagon is a polygon of 6 sides. How many diagonals does it have?

13. The product of a 3-digit number and a 1-digit number is 3714. What is the 3-digit number?

14. Sara and Josh each threw 5 darts at the target shown. Sara scored 19 points by landing 2darts in A and 3 in B. Josh scored 17 points by landing 1 dart in A and 4 in B. How manypoints are given for a dart landing in A?

15. On Monday Emily buys a fashion doll for $25. On Tuesday she sells it for $18. On Wednesdayshe buys it back for $22. On Thursday she sells it for $26. On Friday she buys it back for$20. What must she then sell it for so that she makes a total of $5 on all these transactions?

16. The figure shown is formed using 6 line segments. In all, how many triangles are shown in thisfigure?

17. The circle graph illustrates the number of students in each grade of a middleschool. If 37.5% of all the students in the school are in grade 8, what is thedegree-measure of central angle AOB?

18. The 6-digit number 9_51_6 is a multiple of 36. Two of the digits are missing and all six digitsare different. Write the complete 6-digit number.

19. Given three positive numbers A, B, and C. If A is divided by B, the simplified result is 25 . If

B is divided by C, the simplified result is 34 . But if A is multiplied by C, the result is 120.

What is the value of A?

20. Each box above the bottom row contains the sum of the two numbers right belowit. For example, the entry in box M is the sum of the entries in boxes A and B. Thesum of all the entries in Row 1 is 57. The sum of all the entries in Row 2 is 85.What is the entry in box X?

A

B

Row 4 X Row 3 Row 2 MRow 1 A B

A

O

B

Studentdistributionby grade forgrades 5-8

6

8 7

5


1 more than a multiple of 4

701, 705, 709,

713, 7173 more than a multiple of 6

705, 711, 717

6 more than a multiple of 9 708, 717

INDIVIDUAL EVENT SOLUTIONS, 2010

ANSWERS: 1) DRERDE or EDRERD 2) 8 3) 2016 4) –13 5) 756) 10 7) 5 8) 717 9) 10) 12

1. Work backwards. The Ds are separated by 3 letters, so either the first or last letter of the six-letter word is D.The word is D _ _ _ D _ or _ D _ _ _ D. The Es are separated by 2 letters, so only the following placementsare possible: D _ E _ D E or E D _ E _ D. When the Rs fill in the remaining positions, both possible six-letterwords are completed: DRERDE or EDRERD.

2. METHOD 1: The additional 3 action figures would cost the remaining $24. Thus each action figure costs $8.

METHOD 2: By algebra, if a is the cost in dollars of one action figure, the equation is 7a = 4a + 24. Solving,3a = 24 and a = 8.

3. 2000 ÷ 24 = 83 days, 8 hours. Add 16 hours to complete the 84th day. The total is 2016 hours.

4. From –19 to –4 is an increase of 15. 40% of 15 = × 15 = 6. Adding 6 to –19 produces –13.

5. The sum of the measures of the angles of a triangle is 180°. The sum of the measures of the second and thirdangles is 125. For ease of computation, represent the third angle by 3x and the second angle by 2x, respectively.Then 5x = 125, x = 25 and the two angles are 50° and 75°. The largest angle is 75°.

6. The second sentence tells us that 84 = 4×1×A×B = 1×A×B×4 = A×B×4×1 = B×4×1×A. Then the 16-digitcredit card number is 41AB 41AB 41AB 41AB. Thus, 4×1×A×B = 84 and A×B = 21. Then either A = 3and B = 7 or A = 7 and B = 3. In either case, A+B = 10.

7. The radius of the circular pond whose area is 225π is 15 m. The path and pond combine to form a larger circlewhose area is 400π sq m and whose radius is thus 20 m. Then the path alone is 20 – 15 = 5 m wide.

8. METHOD 1: Note that each remainder is 3 less than the related divisor. If we add 3 to N, we will get anothermultiple of that divisor. Then the LCM of 4, 6, and 9 is 36 and N is 3 less than the first multiple of 36 that isgreater than 700. The first such multiple is 720. The value of N is 717.

METHOD 2: List those numbers greater than 700 that satisfy eachcondition. Use the tests of divisibility for 4, 6 (that is 3 and 2), and 9,respectively, to determine that 700 is a multiple of 4, 702 is a multiple of6, and 702 is a multiple of 9. Add the 1, 3, and 6, respectively, to get thetable at the right. The least number to appear on all three lists, 717, isthe correct solution.

1920

25


9. First simplify 5

6

3

4 − = 1

6

3

3 = 19

6

3 = 3 ÷ 19

6 = 31 × 6

19 = 1819 .

Then simplify 18

19

1

2 − = 20

19

1 = 1 ÷ 20

19 = 19

20.

10. The sum of complementary angles APD and BPC is 90º. Then m∠APB = 90, ΔAPB isa right triangle, and we may use the Pythagorean Theorem (figure 1): since42 + (PB)2 = 52, PB = 3. The area of triangle APB = 1

2 × 4 × 3 = 6. Draw PQ(figure 2) to divide ABCD into two rectangles.

METHOD 1: Look at rectangle AQPD; AP divides it into two triangles of equal area.Likewise, look at rectangle QBCP; BP divides it into two triangles of equal area. Thenthe area of rectangle ABCD is twice the area of triangle APB, which is 6. The area ofrectangle ABCD is 12.

METHOD 2: Look at right triangle APB. To find its area, use the altitude PQ to sideAB: 1

2 × 5 × (PQ) = 6; solving, PQ = 125

. Since PQ = AD = 125

, the dimensions ofrectangle ABCD are 5 by 12

5 and the area is 12.

(Method 2 suggests a faster way to find the length of an altitude to the hypotenuse ofa right triangle. If the legs are a, b, the hypotenuse is c, and the altitude to that hypotenuseis h, then ab = ch.)

A B5

P

4

figure 1

132

546

−−

5 A

C

B

D P

4

Q

figure 2


TEAM EVENT SOLUTIONS, 2010

ANSWERS: 11) 0 12) 9 13) 619 14) 5 15) 2816) 12 17) 135 18) 985,176 19) 6 20) 113

11. (487 × –591) + (487 × 591) = 487 × (–591 + 591) = 487 × 0 = 0.

12. METHOD 1: From each of the 6 vertices a diagonal can be drawn to 3 of the other vertices.But each of these 18 diagonals is counted twice, so the actual number of diagonals is 9.

METHOD 2: Draw the figure. Draw and count its diagonals. There are 9 diagonals.

13. Search for the one-digit factor of 3714. Since 3714 is even, 2 is a factor: 3714 = 2 × 1857. All factors of 1857 areodd: so test 1857 for divisibility by 3, 5, 7 and 9. Since the sum of the digits of 1857 is 21, 3 is also a factor of3714. Then 3714 = (619 × 3) × 2 = 619 × 6. Because 619 is not divisible by 3, 5, 7, or 9, the one-digit factor is6 and the three-digit factor is 619 and no other possibilities exist.

14. METHOD 1: Assume the first four darts for each person landed the same way: 1 dart in A and 3 darts in B.Then Sara’s 5th dart landed in A and Josh’s 5th dart in B. Since she scored 2 points more than he did, a dart inA scores 2 points more than a dart in B. Suppose Sara’s 3 darts that landed in B had landed in A instead. Thenshe would have scored 6 more points, for a total of 25 points. With five darts in A, each dart in A scores 5 points.To check, each dart in B scores 3 points and the totals are Sara: 2×5 + 3×3 = 19; Josh: 1×5 + 4×3 = 17.

METHOD 2: Let a and b represent the points for one dart in regions A and B, respectively. Then:(1) Sara’s point-score is 2a + 3b = 19 and(2) Josh’s point-score is 1a + 4b = 17.

Multiply equation (1) by 4 and equation (2) by 3. The resulting equations are:(3) Sara: 8a + 12b = 76 and(4) Josh: 3a + 12b = 51

Subtract (4) from (3) to get:(5) 5a = 25(6) Then a = 5. A dart landing in A scores 5 points. Be sure to check.

15. METHOD 1: In all Emily paid $25 + $22 + $20 = $67. In all she received $18 + $26 = $44. To make an overallprofit of $5, she must sell it for $67 – $44 + $5 = $28.

METHOD 2: Think of signed numbers. Represent each amount paid out as negative and each amount receivedas positive, and let x represent the amount she gets from the sixth transaction: –25 + 18 –22 + 26 –20 + x = 5.Solving, x = 28. She must sell for $28 to show a $5 profit.


16. Organized counting works best. Label each region as shown.In the table at the right, the triangles are counted accordingto the number of regions each includes. A total of 12 trianglesof varying sizes can be formed.

17. METHOD 1: 37.5% = 38 of all students are in grade 8. The sum

of all four central angles is 360°. Then angle AOB contains 38 of 360°

= 135°

METHOD 2: 37.5% compares to 100% in the same way that n° compares to 360°.Eliminating labels and cross-multiplying, we get n × 100 = 37.5 × 360. Then 100n = 13500and n = 135. There are 135°.

METHOD 3: The double of 37.5% is 75%. 75% of a circle is 270°. Since half of 75% is37.5%, half of 270° is the central angle. The central angle contains 135°.

18. Any multiple of 36 is also a multiple of 4 and 9. Since the last two digits of any multiple of 4 is also a multipleof 4, the last two digits are 16, 36, 56, 76, or 96. Since the sum of the digits of any multiple of 9 is also a multipleof 9, there are only five possible 6-digit numbers: 955116, 935136, 915156, 985176, and 965196. Of these, only985176 has all six digits different.

19. METHOD 1: Another way of writing A ÷ B is AB . Then A

B = 25 and B

C = 34 . Notice that if the left sides of both

equations are multiplied, the result is AB × B

C = AC . Notice also that A

C × AC = A². Thus when the left sides of

all three equations are multiplied, AB × B

C × AC = A². When the right sides of all three equations are multiplied,

the simplified result is 25 × 3

4 × 120 = 36 . Since A² = 36, then A = 6.

METHOD 2: Since AB = 2

5 , then 2B = 5A and B = 52A . Since B

C = 34 , then 4B = 3C and B = 3

4C . Then 5

2A = 3

4C

since each equals B, and 20A = 6C, which simplifies to C = 103A . Replacing C in the equation AC = 120 produces

(A)(103A ) = 120. Thus,

2103A = 120, A2 = 36, and A = 6.

20. Represent the numbers in the four boxes in row 1 by a, b, c, and d, respectively and then add to get the numbersin the other 6 boxes.The results appear in each box in rows 2, 3, and 4.

Row 1 contains a + b + c + d. Their sum is 57.Row 2 contains (a + b) + (b + c) + (c + d) = (a + b + c + d) + (b + c).

Then b+c = 85 – 57 = 28.Row 3 contains (a + 2b + c) + (b + 2c + d) = (a + b + c + d) + (2b + 2c).Row 4 contains (a + 3b + 3c + d) = (a + b + c + d) + (2b + 2c).

Since (2b + 2c) = 2(b + c), X = 57 + 2(28) = 113.

Alternately, row 2 can be thought of as (a + 2b + 2c + d) = 85.Row 4 contains (a + 2b + c) + (b + 2c + d)

= (a + 2b + 2c + d) + (b + c) = ( 85 ) + ( 28 ) = 113.

z

a cyb

A

O

BStudent

distributionby grade forgrades 5-8

6

8 7

5

Row 4 XRow 3 a+2b+c b+2c+d

Row 2 a+b b+c c+dRow 1 a b c d

x

# OF REGIONS

REGIONS IN TRIANGLE

# OF TRIANGLES

1 a, b, c 32 ax, by, cz, ab, bc 5

3 abc 14 abxy, bcyz 26 abcxyz 1

TOTAL 12

77777

2010 TIEBREAKERSQuestions are given one at a time. Winning places are awarded in the order that correct answers aresubmitted. Incorrect answers result in elimination. No calculators are permitted during this tournament.Time limit: 5 minutes per question.

T1. Suppose 2 grapefruits weigh the same as 6 pears and 4 tangerines weigh the same as 3 pears. Then 5grapefruits weigh the same as how many tangerines? (Assume all of each type of fruit weigh the same.)

T2. What is the value of 2 × 1860 + 1860 × 3 + 10 × 930?

T3. In this addition, different letters represent different digits. What is the value of ME?

T4. The average of five numbers is 45. Three of the numbers are 25, 30 and 40. What is theaverage of the other two numbers?

T5. What number between 101 and 139 is divisible by each of the following: 1, 2, 3, 4, and 5?

2010 TIEBREAKER SOLUTIONSANSWERS: T1) 20 T2) 18,600 T3) 85 T4) 65 T5) 120

T1. Since 2 grapefruits weigh the same as 6 pears, 1 grapefruit weighs the same as 3 pears. But 4 tangerines alsoweigh the same as 3 pears, so 1 grapefruit weighs the same as 4 tangerines. Then 5 grapefruits weigh the sameas 20 tangerines. [An algebraic solution mirrors this approach exactly.]

T2. By the associative property, 10 × 930 = (5 × 2) × 930 = 5 × (2 × 930) = 5 × 1860. Then by the commutativeand distributive properties, 2 × 1860 + 3 × 1860 + 5 × 1860 = (2 + 3 + 5) × 1860 = 10 × 1860 = 18,600.

T3. METHOD 1: Look at the ones place: E can only be 0 or 5. Suppose E = 0. Then 3 × M = WE;but WE would end in a zero. This is false for all nonzero values of M. Thus E = 5. Now look at thetens place: 3 × M + 1 ends in a 5, so 3 × M ends in a 4 and M = 8. Finally, ME = 85. Checking,85 + 85 + 85 = 255, which matchs all the conditions of the original problem.

METHOD 2: Look at the ones place: E can only be 0 or 5. The greatest sum for any three 2-digit numbers isless than 300, so W = 1 or W = 2. (The leading digit cannot be 0.) Thus the sum, WEE, is 100, 200, 155, or 255.Only 255 is divisible by 3, so ME = 255 ÷ 3 = 85.

T4. The sum of the five numbers is 45 × 5 = 225. The sum of the three given numbers is 95. Then 225 – 95 = 130is the sum of the other two numbers and the average of those two numbers is 65.

T5. METHOD 1: Any multiple of 1, 2, 3, 4, and 5 is also a multiple of their LCM, 60. The desired multiple is 120.

METHOD 2: The numbers divisible by 5 between 101 and 139 are 105, 110, 115, 120, 125, 130, and 135.Ofthese, only 120 is divisible by 4. Checking, 120 is also divisible by 3, 2, and 1.

M EM E

+ M EW E E

M EM E

+ M EW E E

78 practice 2012 a

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