warm-up 1 - lexingtonjfrost/mc/2006 - 2007... · 54 mathcounts 2006-2007 answers standard...

MATHCOUNTS 2006-200718

Answers

1._ 13

(C, M)

2. 96 (C, F, G)

3._ 3 2 (F, M, T)

4. 2 (G, S, M, P, T)

5. 12 (C, F)

6. 2.5 (C, F)

7. 55 (C, F, M, P, T)

8. 4 (C, G, T)

9. 6 (C, M, P, T)

10. 80 (C, M)

Warm-Up 1

Standard Chapter/Unit Notes

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Solution/Representation - Problem #3 We have isosceles triangle ABC with angle C as its vertex angle. Because the measure of base angle A is 45º, we also know the measure of base angle B is 45º, and angle C must be a 90º angle to bring our angle sum for the triangle up to 180º. This makes triangle ABC a 45-45-90 triangle. Additionally, because CD is a perpendicular bisector coming down from the vertex angle of an isosceles triangle, it also is the angle bisector of that vertex angle. Therefore, we have two congruent, smaller 45-45-90 triangles within a larger 45-45-90 triangle. With all of these isosceles triangles, we can see that if AD = 3 meters, we also can conclude CD = 3 meters, BD = 3 meters and AB = 6 meters. Considering right triangle ACD with hypotenuse AC, we can set up the following equation with the Pythagorean Theorem: AC2 = 32_+_32. This leads to AC2 = 9 + 9 = 18, so AC = 18 = 3 2 meters.

The fact that this is a 45-45-90 triangle means that we can skip the calculations of the Pythagorean Theorem. In every 45-45-90 triangle, the length of the hypotenuse is equal to the product of the measure of one of the legs and 2 . In reverse, we can say that the length of either leg of a 45-45-90 triangle is equal to the length of the hypotenuse divided by 2 . Notice in triangle ACD, leg AD is 3 meters, so hypotenuse AC = 3 2 . Looking instead at triangle ABC, hypotenuse AB is 6 meters, so leg AC = 6/ 2 = 6 2 /2 = 3 2 . Memorizing the relationship between the leg and hypotenuse of a 45-45-90 triangle will be extremely useful!

A

C

BD45 45

45 45

3 3

3


AnswersWarm-Up 2

1. -2 (C, T)

2. 25 (C, F, M, P, S)

3. 4 (C, F M, S)

4. 5 (E, T)

5. 10 (E, F, G)

6. 8 (F, G, M, T)

7. 0 (C, M, P, T)

8. 13 (C, F, M, P, S, T)

9. 1* (E, G, S, T)

10. 57 (C, F, M, T)

* The plural form of the units will always be provided in the answer blank, even if the answer appears to require the singular form of units.


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Solution/Representation - Problem #9 Let there be q quarters, d dimes and n nickels. We know q ≥ 1, d ≥ 1, n ≥ 1 and 25q + 10d_+_5n = 60. We can see that since the value of the coins is greater than 50 cents, but less than 75, we can have only one or two quarters. Let’s look at each case. If q = 1, we then have 25 + 10d_+_5n = 60, which leads to 10d_+_5n = 35. From here there are a number of options. We could have 3 dimes and 1 nickel; 2 dimes and 3 nickels; or 1 dime and 5 nickels. What’s important, though, is that it’s possible to meet all of the requirements with 1 quarter. Looking at the second case with q = 2, we have 25(2) + 10d_+_5n = 60, which leads to 10d_+_5n = 10. The only way to make 10 cents without pennies is to have 1 dime and 0 nickels or 0 dimes and 2 nickels. We were told that we have at least one of each type of coin, so neither of these options work, and we must not have 2 quarters. Our answer is then 1 quarter.

Similarly, we could see that since we must have one of each type of coin in the jar, let’s take out one quarter, one dime and one nickel. Those three coins are worth 40 cents, leaving us coins in the jar with a total value of 20 cents. Clearly we can make 20 cents with dimes and nickels, but there can’t be another quarter in the jar. Therefore, there was just 1 quarter in the jar when we started.


AnswersWarm-Up 3

1._ 2425

(C)

2. 9 (C, M)

3. 66 (C, E)

4._ 12

(C, F, M, P, S)

5. 262 (C)

6. 6 (C, M,T)

7. 8 (C, F)

8. 492 (C)

9. 16 (C, F, M, P)

10. 110 (C, F)


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Solution/Representation - Problem #9 Each square in our figure has an area of 4 square units. However, after the first square, there is an overlap, and so not all 4 square units of each square can be counted. The first square does give us an area of 4 square units. A quarter of the second square - or 1 square unit - is “lost” behind the first square when it is added, so by adding the second square, we are adding only 3 square units, for a total of 7 square units. Similarly, when the third square is added, a quarter of it is lost behind the second square, so we gain only another 3 square units for a total of 10 square units. According to this pattern, adding the fourth and fifth squares will add an additional 3 + 3 = 6 square units, for a total of 16 square units. (Extension: Using this pattern, what is a formula we could write for the total area in Figure n, or the figure with n squares? A = 4 + 3(n - 1))

The solution above is based on the idea of adding 3 square units each time there is a square added. We could approach this differently. Consider Figure 3. There are three 2 by 2 square regions and two 1 by 1 square regions of overlap that have been counted twice. If we were to draw Figure 4, there would be four 2 by 2 square regions and three 1 by 1 regions of overlap that have been counted twice. So for Figure 5, there will be five of the 2 by 2 squares drawn, which amounts to 5 × 4 = 20 square units. However, there will be four 1 by 1 square regions of overlap that have been counted twice, which amounts to 4 × 1 = 4 square inches. The total area is then 20 - 4 = 16 square inches. (Extension: Using this pattern, what is a formula we could write for the total area in Figure n, or the figure with n squares? A = 4n - 1(n - 1) Can you show that this equation is consistent with the one at the end of the paragraph above?)


AnswersWarm-Up 4

1. 141,000 (C, F)

2. 359 (C, G, M, P, S)

3._14 (C, T)

4. 3 (C, F, M)

5. 2 (C, F, P)

6. 25 (C)

7. 17 (C, F, M)

8. 5 (M, P)

9. 14 (C, F, T)

10. 6π (C, F)

Solution/Representation - Problem #7 We have been asked to find the distance between the points (3, 5) and (-5, 20). The distance formula is certainly one way to attack the problem.

( ) ( ) ( ( )) ( ) ( ) ( )2 2 2 2 2 21 2 1 2d x x y y 3 5 5 20 8 15 64 225 289 17= − + − = − − + − = + − = + = =

Let’s take a look at the graph of these points in the coordinate plane. We can see that the segment connecting the two points has a negative slope and can be turned into the hypotenuse of a right triangle by extending horizontal and vertical segments, as shown. Using the x-coordinates, we see the horizontal leg of the right triangle is 3 - (-5) = 8 units long. Using the y-coordinates, we see the vertical leg of the right triangle is 20 - 5 = 15 units long. Now we can set up the Pythagorean Theorem and see AB2 = 82_+_152 = 64 + 225 = 289. Taking the square root, we have AB = 17 units.

Notice that all three of the side lengths of the right triangle we formed above are integer values. When this happens for a right triangle, we say that the three lengths form a Pythagorean Triple. The most common triples are 3-4-5, 5-12-13, 7-24-25 and 8-15-17, as well as the multiples of these triples. If any two side lengths of a right triangle are part of a triple, then we automatically know the length of the third side (as long as we check that the longest of the three lengths is on the hypotenuse). As soon as we see legs of 8 and 15, we know the hypotenuse is 17 without doing any calculations.

(3, 5)(-5, 5)

(-5, 20)

15

8


Answers


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1. (3.4,4.8) (C,F,M)

2. 25 (C,G,M,T)

3. 204 (C,F,M,T)

4._ 2:3_ (C,_F)

Warm-Up 5

5._ -6_ (C,_F,_G,_S,_T)

6._ 198_ (C,_F,_G,_T)

7. 13 (C,F,M)

8._ 3140

_ (C,_F,_S)

9. 5400 (C,F)

10. 18 (C,S,T)

Solution/Representation - Problem #1_ The_coordinates_of_the_points_(3,_4)_and_(3.8,_5.6)_are_not_nice_integers,_but_the_midpoint_formulawillworkjustfine.Themidpointofthepoints(x1,_y1)_and_(x2,_y2)_is_the_point__

( )++ + + = = = 1 21 2 y yx x 3 3.8 4 5.6 6.8 9.6, , , 3.4, 4.8

2 2 2 2 2 2.

_

Whatifwedidn’tknowthemidpointformula?Aftergraphingourtwopoints,weseethatmovingfrom(3,4)to(3.8,5.6)canbedonealongthesegmentconnectingthetwopoints,orthemovecanbedonewithacombinationofmoving0.8unitstotherightandthenup1.6units.Itmakes_sense_that_making_half_of_that_trip_would_put_us_at_the_midpoint_and_wouldentailmoving0.8÷2=0.4unitstotherightandthenup 1.6÷2=0.8units.Movingthesetwodistancesfrom(3,4)putsusat(3+0.4,4+0.8)=(3.4, 4.8).

(3,4)

(3.8, 5.6)

(3.8,4)0.8

1.6

3 4

4

5


Answers


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Warm-Up 6

1. 5 (C,F,M,P,T)

2. 24 (C,F,M,P,S,T)

3. 7 (C,F,M)

4._16 _ (C,_S)

5._ 8_ (C)

6. 0.5 (C,F)

7. 10 (C,F,G)

8._ 48_ (C,_F,_G,_S,_T)

9. 9 (G,M,P,T)

10. 7 3 (C,F,M)

Solution/Representation - Problem #5 Inthisproblemweseehowcheckingouranswersisveryhelpful!WhenMayasubstitutesher8in_for_x,shefindsthattherightsideoftheequationhasavalueof2(8)+20=36andtheleftsideoftheequationhasavalueof4(8)-_4_=_28.__The_expression_on_the_left_of_her_original_equation_is_then_36_-_28_=_8_greater_than_the_expression_on_the_right_side.

Whensolvingtheequation2x+20=4x_-_4,_we_can_remember_that_thisisthesameasfindingthepointofintersectionforthelinesy_=_2x+20andy_=_4x_-_4.__When_we_graph_the_two_lines,_we_see_that_they_intersect_at_the_point_(12,_44),_which_means_the_solution_to__2x+20=4x_-_4_is_12,_and_when_using_this_correct_answer,_both_the_rightsideandtheleftsideoftheoriginalequationwillhaveavalueof44.Inthisgraphweseethatwhenx_=_8_on_line_1,_the_y-valueis36and_when_x_=_8_on_line_2,_the_y-valueis28.Onthegraph,thedistancebetween_(8,_36)_and_(8,_28)_is_8units.Thisisobviouslynotthemostefficientsolutiontothisproblem,butithelpstofurtherclarifywhatouranswermeanswhensolvingsystemsofequations.

(12,44)

(8,28)

(8,36)

line_1

line_2


Answers


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1._ 2319 _ (C,_S)

2. 60 (C,F,M)

3. 2 (C,F,M,T)

4._1

12 (C,F,M,T)

Warm-Up 7

5._ 34_ (C,_E,_G,_T)

6._ 2_ (C,_F,_S)

7. 36 (C,F,M)

8._12 _ (C,_G)

9._ 28_ (C,_G,_S)

10. 3 (C,G,M,P)

Solution/Representation - Problem #1_ Whensimplifyinganexpressioninvolvingmixednumbers,ourfirststepisoftentogeteverythingintosimplefractions.Whenwedothiswiththisexpression,wehave:( )( ) ( )( )5 14 7 141 2 1 2

2 3 3 3 2 3 3 32 4 2 4 1 1+ + + = + + + .Fromherewehaveafewmoresteps:(I)multiplythefirstpairoffractions;(II)geteverythingwithacommondenominator;(III)addthenumerators;(IV)turnthisbackintoamixednumber;andfinally(V)simplifythefractionportionofthemixednumber.Eachofthesefivestepscanbeseenhere: 70 7 14 70 14 28 6 118 4 2

6 3 3 6 6 6 6 6 6 31 19 19+ + + = + + + = = = _.

_ Rather_than_turning_all_of_our_mixed_numbers_into_simple_fractions_and_then_changing_our_fractionattheendbackintoamixednumber,wecandoonlyonesimpleconversionattheveryend.__Think_of_the_mixed_numbers_as_a_type_of_binomial.__Consider_ ( ) = + +1 2 1 2

2 3 2 32 (4 ) (2 )(4 ) .Usingthedistributiveproperty,wehave + + = + + + = + + + = +4 51 2 2 1 1 2 1

2 3 3 2 2 3 3 3 3(2 )(4 ) (2)(4) (2)( ) ( )(4) ( )( ) 8 2 10 __Substitutingthisintotheoriginalexpression,wecannowfinishtherestofthesimplification.Noticetheonlyconversionbetweensimplefractionsandmixednumberscomesinthefinalstep:

+ + + + + + = + = + + = + =5 81 2 2 2 23 3 3 3 3 3 310 2 4 1 17 17 (2 ) 19 19 _.


Answers


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Warm-Up 8

1. 1.9×1019_ (C)

2._ 7 (C,F,M)

3. 300 (C,F,M,P)

4._16 (C,G,M,T)

5. 13,320 (F,T)

6. 62 (F,M,T)

7. 13 (P,S,T)

8. 12 (M)

9._ 18_ (C,_G)

10. 144 (C,F)

Solution/Representation - Problem #4_ Let_the_area_of_a_rectangle_represent_the_boys_and_girls_in_the_algebra_class,aswehavedoneinthefirstfigure.Weseethattheclassisdividedintohalfboysandhalfgirls,astheproblemstates.Now,halfoftheboysarewearing_glasses,_which_means_half_of_half_of_the_class,_or_(1/2)(1/2)_=_1/4_of_the_class_is_“boys_wearing_glasses.”__We_need_a_third_of_the_class_to_be_wearing_glasses,_so_1/3_-_1/4_=_4/12_-_3/12_=_1/12_of_the_class_must_be_“girls_wearing_glasses.”Ifwedivideourclassintotwelfths,weseethatwegetathird(or4/12)oftheclasswearingglassesifwehave3/12=1/4oftheclassbeingthe“boys_wearing_glasses”_and_1/12_of_the_class_being_the_“girls_wearing_glasses.”__Fromthefigure,wecanseethat1/6_of_the_girls_are_“girls_wearing_glasses.”

_ We_certainly_do_not_always_want_to_rely_on_Guess_&_Check,_but_this_is_a_great_problem_for_that_strategy.Weknowwehaveaclasswithhalfboysandhalfgirls.Rightthereweknowweneedtohaveanevennumberofkidsintheclass.We’regoingtodividethenumberofboysinhalfagaintofindthenumberofboyswearingglasses,andweknowathirdofthetotalclassiswearingglasses.Sinceweneedtodivideby2twiceandtherearethirdsinvolved,itmakessensetochooseaclass-sizethatisamultipleof12.Otherwise,wewon’tbeabletotakethefractionalpartsthatarerequired.Let’sstartwithaclassof24.Therearethen12boysand12girls.Halfoftheboyswearglasses,so6boyswearglasses.Athirdoftheclass,or8studentswearglasses.Thismeans8_-_6_=_2_of_those_students_are_girls.__Therefore,_2/12_=_1/6_of_the_girls_wear_glasses.

boys girls

boys girls

boys girls


Answers

1. 8 (C, M, P, T)

2. 28 (C, F, M)

3. 1 (C, F, T)

4. 324 (C, E, F, G, P, T)

Warm-Up 9

5. 4.50 (C, F, M, T)

6. 21 (C, G, M, P, T)

7._ 800 35 (C, F, M)

8._3

3 (C, F, M)

9. 2 (C, F, P, S, T)

10. 40* (C, F, G, P, T)

Solution/Representation - Problem #4_ We are asked to find two consecutive squares with a difference of 35. Let x 2 and (x + 1)2_be_those squares. The positive difference between these two squares is (x + 1)2_-_x 2 = 35. If we square the binomial on the left side of the equation, we have x 2_+_2x_+_1_-_x 2 = 35 which simplifies to_2x + 1 = 35 and finally x = 17. This means our squares were 172_and_182,_the_greatest_of_which_is__182 = 324.

Rather than setting up an equation, we can look at the pattern that forms when evaluating the differences of consecutive squares. Here we start with 0 and list the perfect squares in ascending order. The differences of the consecutive squares are written_to_the_right_side_of_our_list.__Notice_the_differences_are_the_odd_integers.__The_difference_between_32_and_22_is_the_3rd odd integer, 5. Likewise, the difference between 52_and_42 is the 5th_odd_integer,_9.__The_difference_between_n_2 and (n_- 1)2_is_the_nth_odd_integer. (It’s helpful to know the nth_even_integer_is_2n,_and_the_nth_odd_integer_is_2n_-1.) For Problem #4, we have x 2_- (x_- 1)2 = 35 and 35 is the 18th_odd_integer,_so_x = 18 and x 2 = 324.

*A solution is provided in the back of the book. However, it may be possible for students to find a more complex combination of_addition_and_multiplication_that_also_results_in_a_different,_correct_answer.


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0

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16

25

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9


AnswersWarm-Up 10

1. 10 (C)

2. 6 (C)

3. 350 (C, F, G, S)

4. 25 (C, F, M)

5. 5 (C, G, P, S)

6. 1.1 (C)

7. 90 (C, F, M)

8._ 1225

(C, F, M, T)

9. 60 (C, F, M, S)

10._10231024 (C, F, P, S)

Solution/Representation - Problem #4_ A_triangle_that_is_inscribed_in_this_semicircle_has_one_side_that_is_also_the_diameter_of_the_semicircle. The vertex of the angle opposite this side will be on the semicircle. This angle is inscribed_in_the_semicircle,_and_any_angle_inscribed_in_a_semicircle_has_a_measure_of_90_degrees.__Now_we_know_our_triangle_is_a_right_triangle_and_its_hypotenuse_is_10_units.__The_area_of_a_right_triangle_can_be_found_by_taking_half_the_product_of_its_leg_measurements.__However,_using_the_hypotenuse_as_the_base,_we_see_that_this_base_will_remain_the_same_length_regardless_of_its_height.__The_largest_area_will_come_from_the_triangle_with_the_highest_altitude,_which_we_see_is the triangle with its right angle in the middle of the semicircle’s arc. This then makes the legs of the triangle congruent, and we have a 45-45-90 triangle. The leg of a 45-45-90 triangle is the quotient of the hypotenuse and 2 ,_

which_is_ 10 10 2 5 222

= = _units.__We_still_do_not_know_the_length_of_the_altitude_to_the_hypotenuse,_

but_now_that_we_know_the_length_of_each_of_the_legs,_we_can_calculate_the_area_of_the_triangle_using_the_legs_as_the_base_and_height:_ 1

2( )(5 2)(5 2) = 25 square units.


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Answers

1._16 (C, E, M, T)

2. 13.31 (C, F, T)

3. 22 (C, F, M)

4. 20 (C, F)

Warm-Up 11

5. 5037 (C, F, P, S)

6. 2 (C, P, S)

7. 20 (C, F, G, M, T)

8. 132 (C, F, G, M)

9. 02/02/2004 (G, P, T)

10. A, B, D (E, G, S, T)

Solution/Representation - Problem #3 Getting from A to B requires us to go up 4 units and right 2 units. The same_is_true_getting_from_D_to_C,_so_these_two_sides_are_parallel_and_congruent.__Similarly, getting from B to C requires us to go up 1 unit and right 6 units. This is_the_same_route_used_to_get_from_A_to_D,_so_these_two_sides_are_parallel_and_congruent, and quadrilateral ABCD is a parallelogram. The area formula for a_parallelogram_is_A = bh, but in this case, there isn’t a horizontal or vertical side, so it is difficult to determine the base length and height. Notice in our second figure that we have cut off a triangle at the top and re-attached it at the bottom. Quadrilateral ABFE is also a parallelogram and has the same area as ABCD, but there is a horizontal side we can use as our base. What is the length of this base? Point E appears to be at (5.5, 0). Remember the slope from D to C was up 4 and right 2. This means we can find other points on the line by going up 2 and right 1; up 1 and right 0.5; etc. Each of these points keeps the ratio of rise/run intact. Additionally, we can go down 1 and left 0.5. We know from D to E is down 1 unit, and to remain on line CD, it must be 0.5 units to the left of point D, which gives us (0, 5.5). With a base of 5.5 units and a height of 4 units, we can calculate the area to be (5.5)(4) = 22 square units.

A different approach involves drawing rectangle AXCY with X(0, 5) and Y(8, 0). Then the areas of the regions that are inside AXCY but not within quadrilateral ABCD can be subtracted from the area of AXCY, leaving us with the area of ABCD.

A

BC

D

A

BC

DE

F


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AnswersWarm-Up 12

1. 17 (C, F, G, M)

2. 3 (C, F, M)

3. R (E, G, M)

4. 30 (E, F, G, M)

5. 223 (C, S)

6. 156.25 (C, F, M)

7._ 280 (C, S)

8. 19 (C, F, M)

9._136 (C, T)

10. 44,000 (C, F, T)

Solution/Representation - Problem #4 We know CA = CD because each of these segments is a radius of circle C. Likewise, BA = BD because each of these segments is a radius of circle B. All three sides of triangle BCD are congruent to the three sides of triangle ABC, so the triangles are congruent and triangle BCD is an equilateral triangle. Together, the triangles form rhombus ABDC. In a rhombus, the diagonals are perpendicular to_each_other_and_bisect_the_angles_of_the_rhombus.__Therefore,_diagonal_AD_bisects angle BAC. Because triangle ABC is equilateral, the measure of angle BAC is 60 degrees, and now we know the measure of angle BAD is 30_degrees.

There is a lot we can do with the figure for Problem #4. Here are some additional questions that_can_be_investigated:_ _ _ _ _ _ _ _ _ _

_

A B

C D

●If the radius of circle C is 5 units, what is the area of triangle ABC? ●What is the area of rhombus ABDC? ●What_is_the_length_of_arc_AC?_ _ _ _ _ _ _●What is the area of the total region covered by the figure?


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Answers


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1._ 10_ (C,_P,_S,_T)

2. 486 (C,F,G)

3. 5000 (C,F,M)

4. 145 (C,E,G,P,T)

Warm-Up 13

5._15 (C,F,P,T)

6. 30 (C,F,G,S)

7. 8 (C,F,G,T)

8. 280 (C,F,M)

9._94 (C,F,M)

10. 7.5 (C,F)

π

Solution/Representation - Problem #1 Someonemayhaveacalculatorthatwouldshowtheexactdecimalvalueof1÷210,butwe’renotsupposedtobeusingcalculatorstosolvethisproblem.Let’stakealookatsomeoftheexpressionsthathavesmallerexponentsinthedenominator.Mostpeopleareprettycomfortablewith1/21_and_1/22.Thesehavedecimalvaluesof0.5and0.25.Whenweprogressto1/23,_we_start_to_get_into_unfamiliar_territory_for_most_people.__Notice_that_rather_than_“starting_over”_and_dividing_1_by_8,_wecanuse1/23=1/22÷2.Dividing0.25by2,weseethatweget0.125.Whenwegoto1/24,_again,_thisisjust1/23÷2or0.125÷2=0.0625.Wecanseethattothispointeverydecimal_representation_is_gaining_a_digit_to_the_right_of_the_decimal_point,_andeachvaluehasa5initsright-mostlocation.After1/2,eachdecimalrepresentation_actually_has_“25”_as_its_right-most_digits.__Will_this_pattern_continue?__It_will.__When_we_do_the_division_by_2,_we_will_be_“bringing_down”_the_second-to-last_digit_2,_which_will_mean_we_are_dividing_our_original_2_into_an_even_number_with_a_ones_digit_of_2.__This_will_divide_evenly.__When_we_bring_down_the_last-digit_5_at_the_end_(it_will_be_by_itself)_and_divide_it_by_2,_the_divisionwon’thappenevenly,andwewillhavetobringdownanextrazero,resultinginanadditionaldigitinthequotient.Nowthatwe’reconvincedofthepattern,noticethat1/22_has_two_digits_to_the_right_of_the_decimal_point,_1/23hasthreedigitstotherightofthedecimalpoint,etc.Therefore,1/210_will_have_10_digits_to_the_right_of_the_decimal_point_(and_the_last_two_will_be_“25”).

2.0625.0_3__1__2_5

622

45

1__01__0


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Warm-Up 14

1._ 100_ (C,_M,_T)

2._1

24 _ (C,_M,_T)

3. 72 (C,F,M)

4._ 40_ (C)

5._ 3_ (P,_S,_T)

6. 10 (C,G,M,S,T)

7. 8 (M)

8._ 52_ (T)

9._ 25_ (C,_S)

10. 636 (C,F,P,T)

Solution/Representation - Problem #1_ We_will_let_x_be_the_number_of_crates_on_the_truck_before_any_deliveries_were_made.__At_the_firststop,1/4ofthecrateswereunloaded,whichlefthim(3/4)x_crates_still_on_the_truck.__At_thesecondstop,2/3oftheremainingcrateswereunloaded,leavinghimwith(1/3)((3/4)x)_crates_stillonthetruck.Athislaststopheunloaded80%,or4/5,ofthecrates,leavinghimwith(1/5)((1/3)((3/4)x))).__Because_we_are_only_multiplying,_the_associative_property_allows_us_to_multiply_thevaluesinanyorder.Considerthat(1/5)((1/3)(3/4))x=(1/5)(1/4)x=(1/20)x.__This_represents_thenumberofcratesstillonthetruck.Weweretoldthisamountis20crates,so(1/20)x_=_20_and_multiplying_both_sides_by_20_yields_x_=_400.__If_he_had_400_crates_at_the_beginning_and_unloaded__1/4ofthematthefirststop,heunloaded(1/4)(400)=100cratesatthefirststop.

Let’sseeifthisiseasier.Allowthefullrectangletorepresentthetotalnumberofcratesonthetruck.Dividingitintofourths,wecanshade1/4oftheareatorepresentthecratesthatwereunloadedatthefirststop.Atthesecondstopweneedtounload2/3oftheremainingcrates.Noticethatwealreadyhavethreeequalregionsremaininginthefirstfigure,soshadingintwoofthosetakescareofoursecondstop.Wenowhave1/4ofthecrateswestartedwith.Ifweunload80%ofthoseatthethirdstop,thatmeanswekeep20%=1/5,asshowninthethirdfigure.Wearetoldwehave20cratesleft.These20cratesareafifthofthewhiterectangleinthesecondfigure,sothatwhiterectangleinthesecondfigurerepresented20×5=100crates.Therefore,thefirst“fourth”thatweshadedalsorepresented100_crates.

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1. 81 (C,G,S)

2._ (0,_3)_ (M)

3. 72 (C,P,S)

4. 36 (C,F,M)

Warm-Up 15

5. 19 (C,F,M)

6. 31 (C,G,P,S,T)

7. 24 (C,F,M)

8. 25 (C,E,F,G,T)

9._1928 _ (C,_M,_T)

10. (1,7) (C,F,M)

Solution/Representation - Problem #2 Totherightwecangetasenseofwhatwe’retryingtodowiththis_problem.__We_are_rotating_the_point_(4,_1)_a_total_of_90_degrees_counterclockwise_about_the_center_point_(1,_0).__We_can_see_the_complete_circle_of_rotation_(dotted)_with_center_(1,_0)_that_goes_through_(4,_1),_and_we_know_that90degreesisaquarterofthecircle.Nowweneedtofigureouttheexactpoint_on_which_(4,_1)_will_land.__Since_we_are_rotating_90_degrees,_we_know_that_the_angle_formed_with_the_start_point_(4,_1),_center_point_(1,_0)_and_end_point_willbearightangle.Thisisshowninthesecondfigure.Iftheangleis

90_degrees,_then_the_two_sides_of_the_angle_are_perpendicular_and_have_slopes_that_are_opposite_reciprocals_of_each_other.__Notice_we_travel_up__1_unit_and_right_3_units_to_go_from_the_center_(1,_0)_to_the_point_(4,_1).__This_isaslopeof1/3.Tocreateasegmentperpendiculartothisoneitmusthaveaslopeof-3/1.Goingdown3unitsandrightoneunitdoesnotgetusgoinginthecorrectdirection,solet’sconsidertheslope3/(-1)wherewegoup_3_units_and_left_1_unit_from_the_center_(1,_0).__This_puts_us_at_the_point__(0, 3)_which_is_in_the_correct_direction.

90


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Warm-Up 16

1._ 12_ (C,_P,_S)

2. 180 (C,F,S)

3. 9 (C,F,G)

4. 30 (C,F,M,P,S)

5._23 (C,F,M)

6. 15 (C,G,P,T)

7. 8-2πor-2π+8 (C,F,M)

8. 25 (C,F,P,T)

9. 85,723 (C,E,G,P,T)

10._ 15_ (C,_S)

Solution/Representation - Problem #1_ This_problem_involves_an_arithmetic_sequence.__An_arithmetic_sequence_is_a_sequence_of_numbers_with_a_common_difference_between_consecutive_pairs_of_numbers;_the_numbers_increase_or_decrease_byaconstantamount.Forthissequencewehavebeengiventhefirstfourterms.Weseethattheyaredecreasing,sowemustbesubtractingaparticularamounteachtime.From1000to987,we_see_1000_-987=13wassubtracted.Togetfrom987to974weagainsubtract13;andonceagaintogofrom974to961.Thissequencecancontinueforever,butsoontheintegerswillbenegativeandvery,verysmall.Wehavebeenaskedtofindthesmallestpositiveintegeronthelist.How_many_times_can_we_subtract_13_before_the_terms_become_negative_integers?__This_question_is_represented_by_the_inequality_1000_-_13x_>_0,_where_x_is_an_integer__Adding_13x_to_both_sides_and_thendividingby13getsusto76.9>x,_which_tells_us_xmuststaysmallerthan76.9Thismeanswecanperformthesubtraction76times,whichresultsin1000-13(76)=1000-_988_=_12.__This_is_the_smallest_positive_integer_in_our_sequence.

_ If_we_have_access_to_a_graphing_calculator,_we_can_use_our_equation_above_(1000_-_13x_>_0)_in_the“Y=”field.EnteringY=1000-_13xand_then_selecting_the_TABLE_function_shows_us_each_of_the_terms_of_our_sequence_starting_at_x_=_0.__Again,_the_x_value_is_the_number_of_times_we_subtracted_13,_but_we_also_will_get_a_y_value_that_indicates_the_value_of_the_term.__Notice_that_when_x_=_2,_the_corresponding_y_value_is_the_third_term_of_the_sequence.__Scrolling_to_x=76,weseey_=_12.__The_77th_term_is_12,_and_this_is_the_least_positive_integer_in_the_sequence.


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1._ 35 (C,F,M)

2._ 32_ (C)

3._ (2,_-1)_ (C,_P,_S,_T)

4._ 247

_ (P,_T)

Warm-Up 17

5._ 6 1313

_ (C,_M)

6. 72 (F,M)

7. 6.15 (S,T)

8. 3.5×10-10_ (C,F,T)

9. 38 (G,P,T)

10._ 2 73 (F,M,P)

Solution/Representation - Problem #4 Let’sfirstfigureouthowmanyarrangementsofboys(Bs)andgirls(Gs)wecanhave.Wereallydon’tcarewhichboysaresittingincertainseats;it’sjustimportanttoknowwhichseatstheboys_are_sitting_in.__So_there_are_7C4=35arrangementsofBsandGsinthesevenseats.WewillrepresenttheplaceswhereBsandGssitwithgrayandwhiteregions.GrayisforBs,andwhiteisforGs.WhereagrayrectangleandawhiterectangletouchisaplacewhereaBissittingnexttoaG.Row1showstheonlytwowaysthereis_just_one_“meeting_point”_for_the_Bs_and_Gs(alloftheGsontheleftoralloftheBsontheleft).Row2showshowwecanhavetwomeetingpoints.NoticetherearefivedifferentwaystofillintheBsandGsshownbelowrow2.Therewouldbe12waystofillin_the_options_in_row_3;_there_are_9_ways_for_row4;thereare6waysforrow5;andthereisonlythe1wayforrow6.Thisgivesusour35_arrangements_with_a_total_of__ _2(1)+5(2)+12(3)+9(4)+6(5)+1(6)=120_meeting_points.__This_gives_us_an_average_of120÷35=24/7_meeting_points_per_arrangement.

B B B GGG B G G GB BB

B G G G BBB

B B B G BG

G

G G GBB B

G B GB BB

BBG

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6.


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Warm-Up 18

1. 31 (C,F,M)

2._1

182 (C,F,M,P)

3. 5 (E,G,M,T)

4. 3888 (F,P,T)

5._ 15_ (C,_P,_T)

6. 360 (C,F,M)

7. 288 (C,F)

8._6037 (C,F)

9. 12.1 (E,G)

10. 11 (C,F,M)

Solution/Representation - Problem #8 Alexruns26milesin6hours,soshecanrun26/6=13/3milesin1hour.Similarly,Bethanycanrun26/5milesin1hour,andCarolcanrun26/4=13/2milesin1hour.Ifthethreeofthemeachrunforanhourandthenadduptheirdistances,theywillrun13/3+26/5+13/2miles.Thecommondenominatoris30,sowecanfindtheirdistanceis130/30+156/30+195/30= 481/30milesin1hour.Let’sseehowwecanusethistodeterminetheamountoftimeitwouldtaketorun26miles...

481 30miles 1 mile 1 mile 26 miles 26 miles2630 481 ;30 30 30 60(26)(30)1 hour 26hours hours hourshours481 481 481 37481

× = × = = ._

_ We_also_could_consider_that_their_rate_of_speed_when_running_at_the_same_time_is_the_sum_oftheirindividualrates.Theirtotalrate,inmilesperhour,isthen(26/6)+(26/5)+(26/4)=481/30milesperhour.Usingtheformularate×time_=_distance,wecansetup(481/30)(time)_=_26.Multiplyingbothsidesoftheequationby30/481,wehavetime=(26)(30/481).Afactorof13dividesoutofthenumeratoranddenominatorforouranswerof(2)(30/37)=60/37_hours.

warm-up 1 - lexingtonjfrost/mc/2006 - 2007... · 54 mathcounts 2006-2007 answers standard...

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