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Warm-Up 5 1. ________ The number 169 can be written as the sum of three positive perfect squares. What is the greatest of these perfect squares? 2. ________ The world‛s tallest mountain, vehicles Greatest Depth Attained by Submersibles measured from its base, is Mauna 36000 35790 Loa. Only 13,680 feet of this Number of Feet 30000 33,356-foot mountain are visible above the water; the rest of the 24000 19700 19700 mountain is under the Paciﬁc Ocean. 18000 Of the seven submersible vehicles in 13000 the graph, how many could get to the 12000 9775 bottom of Mauna Loa on the ocean 6000 3000 3300 ﬂoor? re V-1 ana lvin Tow son ste he a ie y sp DSR Cy A ep - J Tr th De rgo Ba A 3. ________ If x = −2, what is the value of 2x − 7? 4. ________ The ratio of boys to girls in a class of 35 students is 3:4. How many more girls than girls boys are in the class? (x, y) 5. ________ On a coordinate grid, the triangle is translated from position A to position B. What is the value of x + y ? B (2, 3) (9, 4) A (0, 0) (4, 0) 6. ________ The diagonals of a rhombus measure 10 meters and 24 meters and are perpendicular meters bisectors of each other. What is the perimeter of the rhombus? 7. ________ The Fibonacci sequence is 1, 1, 2, 3, 5, 8, … . Each number after the ﬁrst two numbers is the sum of the preceding two numbers. What is the ﬁrst perfect square greater than 1 to occur in this sequence? 8. ________ How many toothpicks would you need in order to tpks build the sixth ﬁgure in the pattern? The ﬁrst three ﬁgures of the pattern are shown here. The ﬁrst ﬁgure is made from four toothpicks. 9. ________ When four standard six-sided dice are rolled, what is the most likely sum of the four numbers? 10. _______ A rectangular frame measuring 10” by 15” has ﬁve square holes for photos. Four of the holes are each 2” by 2”, and one hole is 3” by 3”. What portion of the frame‛s area was not cut away for the photo holes? Express your answer as a common fraction. MATHCOUNTS 2005-2006 29 Warm-Up 5 Answers 1. 144 (E, G) 5. 14 (M, P, T) 8. 64 (P, T) 2. 3 (E) 6. 52 (F, M) 9. 14 (S, T) 5 3. –11 (F) 7. 144 (T) 10. (F) 6 4. 5 (F, G) Solution - Problem #8 We have been given the ﬁrst three ﬁgures of a pattern, and we now must determine how many toothpicks would be needed to create the sixth ﬁgure. We could certainly continue to draw the ﬁgures until we get to the sixth ﬁgure in the pattern, but perhaps there is a quicker, more efﬁcient way to proceed. We can see that for each successive ﬁgure, a square region is added to the top, bottom, right and left “arms” of the ﬁgure. However, notice that though a square region with four sides is formed, only three toothpicks are necessary to create each new square region. Since four square regions are added each time, and each one requires three toothpicks, we can see that we are adding 4 × 3 = 12 toothpicks at each successive stage of the pattern. The pattern starts with four toothpicks. The second ﬁgure requires us to add 12 toothpicks. The third ﬁgure requires us to add 12 more toothpicks. To get to the sixth ﬁgure of the pattern, we will have to add 12 toothpicks a total of ﬁve times, which is a grand total of 4 + 12(5) = 64 toothpicks. How many toothpicks would be required to create the 100th ﬁgure of the pattern? Figure 1 Representation - Problem #4 We are told that the ratio of boys to girls in a class of 35 students is 3:4. This means that for every three boys in the class, there are four girls. We could draw boys and girls, but to make this a little Figure 2 easier, we will use common symbols to represent a boy (arrow) and a girl (plus). Figure 1 shows our ratio of three boys to four girls. The glitch, though, is that we have only seven students. We need 35 students. Let‛s add on three more boys and four more girls. We still have only 14 students, as shown in Figure 2. Notice that to keep our ratio correct, we must add on three boys and four girls each time, Figure 3 which is seven students. It will take ﬁve groups of these seven students to arrive at our class total of 35 students (Figure 3). So there are ﬁve groups of these three boys, which is a total of 5 × 3 = 15 boys. There are also ﬁve groups of these four girls, which is a total of 5 × 4 = 20 girls. There are then 20 − 15 = 5 more girls than boys in the class. A shortcut would have been to see that each group of seven students has one more girl than boy. Since there are ﬁve groups of seven, there are ﬁve more girls than boys. Many times when we are working with ratios we can skip right to an algebraic representation of the situation. A ratio of three boys to four girls in a class of 35 students translates to 3x + 4x = 35. (We don‛t know by which factor to increase the number of boys and girls, but we know that we need to increase them both by the same factor.) Our equation simpliﬁes to 7x = 35, which is x = 5. Notice this relates to the previous solution in that x is the number of groups of seven students needed to make up the class of 35 students. Continuing to use 3x for the number of boys and 4x for the number of girls, there are 4x − 3x = x more girls than boys, and we have already determined x = 5. 30 MATHCOUNTS 2005-2006 Warm-Up 6 1. ________ The set {2, 3, 4, 5, 9} contains ﬁve of the factors of which one of the following four integers? 480 1260 1890 3888 $ 2. ________ The table below shows the average amount paid to winners of the Kentucky Derby races from 1976 through 2004. For example, there were four years that 15 horses started the race, and the ﬁrst-place winners for those years averaged $527,175. For the 29 races shown, what is the average winnings amount associated with the median number of starters? No. of Frequency First Prize No. of Frequency First Prize Starters Starters 9 1 $165,200 16 3 $688,467 10 1 $228,650 17 3 $680,600 11 1 $186,900 18 2 $1,299,900 13 3 $452,450 19 6 $777,750 14 1 $628,800 20 3 $605,967 15 4 $527,175 21 1 $317,200 −3 3 −3 3. ________ What is the 11th term in the geometric sequence 3, , , , 2 4 8 … ? Express your answer as a common fraction. 4. ________ Goran has a standard deck of 52 cards. He considers an ace to have a value of 1, a jack has a value of 11, a queen has a value of 12, a king has a value of 13, and all other cards are their face value. What is the probability that a randomly selected card will have an even value? Express your answer as a common fraction. 12’ 5. sq ft ________ Jen and Jerry need to carpet their L-shaped room, having dimensions as shown. The carpet they chose is a solid color and is sold in rolls 10’ that are 12 feet wide. How many square feet of wasted carpet will 4’ they have if the minimum length from the roll is used and only one 10’ seam is permitted? 6. ________ What is the 100th term of the sequence 2, 3, 5, 6, 7, 10, 11, … which consists of all of the positive integers that are neither perfect squares nor perfect cubes? 7. ________ What is the integer value of x that minimizes the absolute value of (6x 2 – 2x )? 8. ________ Cheryl, Don, Oleg, Shandra and Hans each have exactly one pet. Their ﬁve pets are ways a ferret, parrot, dog, cat and ﬁsh, but not necessarily in that order. In how many different ways can the ﬁve friends be paired with the ﬁve animals? 9. ________ The mean of a collection of ﬁve positive integers is 7. If the collection has a unique mode, what is the greatest possible member of the collection? _______ Maria sees these ﬁrst four ﬁgures of a sequence in her textbook. The ﬁrst ﬁgure 10. Figure is made from 6 toothpicks. Maria has 41 toothpicks. What is the largest ﬁgure of this sequence Maria can create if the sequence were to continue forever? ... Figure 1 Figure 2 Figure 3 Figure 4 MATHCOUNTS 2005-2006 31 Warm-Up 6 Answers 1. 1260 (E, F, G, P) 5. 8 (F, M) 8. 120 (F, M, T) 2. 680,600 (T) 6. 112 (P, T) 9. 31 (G, M, T) 3 3. (F, P, T) 7. 0 (F, G) 10. 8 (F, M, P, T) 1024 6 4. (M, T) 13 Solution - Problem #1 We know 2, 3, 4, 5 and 9 are factors of the number we need to ﬁnd. We also know that the number we need to ﬁnd is either 480, 1260, 1890 or 3888. Immediately, you may notice that the ﬁrst three options each end with a zero for a units digit. This is possible only if the number has a factor of 10, which is really a factor of 2 and a factor of 5. We see that the number we are searching for must have factors of 2 and 5, so we know it must end in a zero. Therefore, we know 3888 is not our answer. We also know 9 must be a factor of our answer. A number has a factor of 9 only if its digit-sum is divisible by 9. Looking at 480, which has a digit-sum of 4 + 8 + 0 = 12, we see that 12 is not divisible by 9, so neither is 480. Let‛s perform the same test on our last two possibilities: 1260 and 1890. The integer 1260 has a digit-sum of 1 + 2 + 6 + 0 = 9, and that is divisible by 9. The integer 1890 has a digit-sum of 1 + 8 + 9 + 0 = 18, which also is divisible by 9. So far we have not been able to rule out 1260 or 1890 as our answer. Notice that by passing the tests for divisibility by 10 and 9, we are ensured that these two integers are divisible by 2, 3 and 5, also. That leaves the factor of 4. An integer is divisible by 4 if the two-digit number formed by the tens and units digit is divisible by 4. Looking at 1260, we just need to see if 60 is divisible by 4, which it is. For 1890, we need to check 90 for a factor of 4, and we see that it is not divisible by 4. Now we know that only 1260 is divisible by 2, 3, 4, 5 and 9. Representation - Problem #3 For a geometric sequence, a common ratio is multiplied with each term to determine the next term. If this common ratio is positive, then the terms will stay positive or stay negative. If the common ratio is negative, the terms will alternate between positive and negative values. If we are multiplying each term by a common ratio with an absolute value greater than one, the absolute value of the terms will get larger. However, if we are multiplying each term by a common ratio with an absolute value less than one, the absolute value of the terms will get smaller. For the geometric sequence in problem #3, we go from 3 to −3 . What ratio was multiplied with 3 to get −3 ? We can 2 2 solve 3x = −3 . Dividing both sides by 3 yields x = −1 . From here we can see that each term has 2 2 been multiplied by −1 to get the next term. Because this common ratio is negative, our terms will 2 alternate between positive and negative values. Because this common ratio has an absolute value less than 1, the absolute values of our terms will get smaller. 0 Here is a visual representation of the terms of the sequence. The vertical line represents 0, and the terms of the sequence 3 -3 are shown from top to bottom. Notice how the segments 2 3 representing the terms are each half the size of the previous 4 segment, and the values are alternating from one side of zero -3 8 to the other. If we continue this pattern, we can determine the 3 16 11 term of the sequence. Notice, also, the numerator is always th 3 while the denominators are successive powers of 2. The 11 th term will be on the right side of 0 (positive) and will have a 3 numerator of 3 and a denominator of 210. This is 1024 . A shortcut for determining our 11th term is to see that we will have to multiply our ﬁrst term by 10 this common ratio of −1 a total of 10 times, which is 3 × ( −1 )10 = 3 × ( −1) = 3 × 1024 = 1024 . 2 2 210 1 3 32 MATHCOUNTS 2005-2006 Workout 3 miles 1. ________ The surface of Lake Michigan is 22,300 square miles. If the surface of Lake Michigan were a perfect circular region, what would its circumference be? Express your answer to the nearest ten. $ 2. ________ The six-person Droz family spends $14.27 per person weekly at The Corner Grocers. If The Corner Grocers makes a net proﬁt of 5% of the amounts customers pay, what is the net proﬁt they will make on the Drozes‛ purchases over a six-week period? 3. ________ Lo has 300 yuan, Maria has 175 euros and Jorge has 500 pesos. If the three of them line up in order of wealth, such that the person Currency In USD Per USD holding the greatest value Chinese Yuan 0.120 8.333 in currency is at the front, Euro 1.320 0.758 in what order will they be standing, front to back? Mexican Peso 0.090 11.111 $ 4. ________ At the Surf‛s Up! Restaurant, a 15% tip is automatically added to Jarett‛s bill because there are more than six people in his party. On his check, the included tip is $27. What is the total cost of the check including the tip? pos 5. ________ In how many different positions can a 2 by 1 rectangle be placed on the 8 by 8 square board so that the rectangle covers exactly two squares of the square board? 6. ________ If x < −4.38, what is the greatest possible integer value of x ? pounds 7. ________ A 75-pound bag of feed for livestock contains 30 pounds of bone meal, 30 pounds of limestone and 15 pounds of salt. If Mandelbrot Feed Co. wishes to sell this same mixture in 100-pound bags, how many pounds of salt should be in each bag? 8. ________ The graph shows the daily lbs 6 Up to amount of food needed to feed Number of Cups of Food 5 2 Months a puppy based on the puppy‛s maximum adult weight. For 4 which maximum adult weight 2-4 Months does a puppy have the greatest 3 percent increase of food intake from “Up to 2 Months” to 2 4-6 Months “4-6 Months?” 1 0 30 50 80 Maximum Adult Weight (lbs) 9. ________ Water weighs approximately 62.4 lbs/ft3. How much does the water in a full 5‛ by 3‛ lbs by 2‛ tank in the shape of a rectangular prism weigh? 10. _______ To get from one corner of a 90‛ by 90‛ square plaza to the diagonally opposite % corner, what percent shorter is it to walk diagonally across than to walk along two of its sides? Express your answer to the nearest whole percent. MATHCOUNTS 2005-2006 33 Workout 3 Answers 1. 530 (F) 5. 112 (M, P, S) 8. 50 (E) 2. 25.69 (M) 6. −5 (E, G, M) 9. 1872 (F) 3. Maria, Jorge, Lo (C) 7. 20 (F, M, P) 10. 29 (F, S) 4. 207 (F, M) Solution - Problem #5 We need to be sure we do not leave any positions for the 2 by 1 rectangle out of our count, while also ensuring that we do not count any positions more than once. (For our picture, arrows will be easier to see than rectangles.) Our arrows may only be placed in a horizontal position or a vertical position if they must cover two unit squares. Starting with the upper left square of our grid, we can place only two arrows that cover it while also remaining fully within the grid. One is vertical and one is horizontal. If we move to the second square of the top row, again, we have only two arrows that cover it, remain fully within the grid and do not go back to cover the ﬁrst square. Remember, we do not want to recount an arrow that we have already counted. If we continue this process, we will see that each of the ﬁrst seven squares in the ﬁrst row yields two positions for a 2 by 1 arrow. However, the last square of the row will give us only one new arrow position. This is a total of 15 positions for arrows that cover a square in the ﬁrst row. These are shown in the ﬁgure to the right. If we move to the second row, we will see that again we get 15 new positions for the 2 by 1 arrow. Remember that none of these positions may cover a square in the ﬁrst row because those positions have already been counted. Each of our ﬁrst seven rows will yield 15 new positions with this counting technique. However, the eighth row is different. There is only one position for the arrow that will cover the ﬁrst square of the bottom row and no squares from the row above. The same is true for the second square in this row. There is only one position for a 2 by 1 arrow that does not cover any squares above it or to the left of it. Each of the ﬁrst seven squares of this ﬁnal row yields one new position. However, there are no new positions for the ﬁnal square of the grid. This is a grand total of 7(15) + 7 = 105 + 7 = 112 positions for a 2 by 1 arrow (rectangle) in the 8 by 8 square board. Solution - Problem #10 To walk from one corner of the 90‛ by 90‛ square to the opposite corner, along the sides of the square, we would walk a total of 180‛. If we walk across the square, this distance will be shorter. The diagonal path across the square plaza splits our square into two isosceles right triangles with the diagonal path as the hypotenuse of both triangles. We can use the Pythagorean Theorem to determine the distance of this path. Let d be the length of our diagonal path. We have d2 = 902 + 902, which is d2 = 16,200 or d ≈ 127.28 feet. This is a difference of 180 − 127.28 = 52.72 feet, which is 52.72 ÷ 180 = 29% off the initial length. (Since we were working with an isosceles right triangle, we also could see that the diagonal path was equal to 90 2 ≈ 127.28 feet since the hypotenuse of a 45-45-90 triangle is the product of the length of the leg and 2 .) Because we have a formula for the diagonal path, we could check if our answer would remain the same for a square plaza of any size. Letting s be the side length of a square plaza, the long way from corner to opposite corner is 2s feet, and the diagonal route is s 2 feet. This is shorter by 2s − s 2 feet, which is a decrease of (2s − s 2 ) ÷ 2s = (2 − 2 )s ÷ 2s = (2 − 2 ) ÷ 2 = 29% for any value of s. 34 MATHCOUNTS 2005-2006 Warm-Up 7 1. ________ Annie estimated that her household recycles ﬁve cubic feet of materials per cu ft collection day. There are 26 collection days per year in her town of 16,000 households. If every household is similar to Annie‛s, how much recycled material is collected in one year in Annie‛s town? 2. ________ The distance from the earth to the sun is 93,000,000 miles, and light travels at sec 186,000 miles per second. How many seconds does light from the sun take to reach the earth? 3. ________ Jorge bought a TV at a 15% discount. Later he found that there should have been $ a 20% discount, so the store gave him the $17 he was owed. What was the price of the TV before any discount? 4. ________ The whole numbers are arranged in an array as shown. If the pattern continues going down one column and up the next, what is the number at the bottom of the 19th column? ... 1 8 9 16 ... 2 7 10 15 3 6 11 14 4 5 12 13 5. ________ A bracelet is made by stringing together patterns R R four beads. Each bead is either red or green. How many different color patterns R G G R are possible for the bracelet, where patterns are considered the same if turning one will G G produce the other, as shown here? 4 4 4 6. ________ A rectangular 7-inch by 12-inch picture is framed with a 7 mat so that four inches of the mat are seen on the top and on each of the sides. Five inches of the mat are seen on the bottom. What fraction of the ﬁnished product‛s area 12 (picture and mat) is the picture? Express your answer as a common fraction. 5 7. ________ The numbers 1, 3, 6, 10, … are called triangular numbers, as shown geometrically here. What is the 20th triangular number? 1 3 6 10 8. ________ If a # b = a 2 + b and a @ b = b − a , what is the value of ((1 # 3) @ 2)? 9. ________ If y = 2x + 1, which of the following equations is true? A) x = 2y + 1 B) x = (½)y + 1 C) x = (½)y − ½ D) x = (½)y + ½ 10. _______ A special deck of cards consists of green cards and yellow cards. The odds of yel cds randomly selecting a green card from the deck are 3:5. If there are 40 cards in the deck, how many yellow cards are in the deck? MATHCOUNTS 2005-2006 35 Warm-Up 7 Answers 1. 2,080,000 (F, M) 5. 6 (E, M, P, T) 8. −2 (F) 4 2. 500 (F) 6. (F, M) 9. C (E, F, G) 15 3. 340 (F, M) 7. 210 (F, P, S, T) 10. 25 (G) 4. 76 (F, P, S, T) Solution - Problem #4 We could continue writing out the numbers and ﬁlling in the entries through the 19th column, but perhaps there is a more efﬁcient way of solving this problem. If we follow the progression of the consecutive whole numbers with our pencil, we can see that the whole numbers are written consecutively and seem to go down one column, then up the next column, then 1 8 9 16 down the following column, and this pattern of alternating 2 7 10 15 up and down columns is continued. Since we want to know 3 6 11 14 what number is at the bottom of the 19th column, let‛s look 4 5 12 13 at the numbers at the bottom of the columns we already know. They are 4, 5, 12 and 13. These four numbers don‛t seem to make a very obvious pattern, but we can see that there is a difference of 8 between the ﬁrst and third numbers and between the second and fourth numbers. Notice this also is true with the numbers at the top of the columns (1, 8, 9 and 16). We can see that this would continue since every grouping of two columns has eight consecutive whole numbers, with the greatest one appearing at the top of the second column. The number at the top of the sixth column, which will be the end of the third grouping of eight consecutive integers, will be 3 × 8 = 24. The number at the top of the 18th column, which will be the end of the ninth grouping of eight consecutive integers, will be 9 × 8 = 72. Starting down the 19th column, we‛ll then have the numbers 73, 74, 75 and ﬁnally, 76. Representation - Problem #7 To generate the triangular numbers, we continue to add on consecutive integers to our growing total. So the triangular numbers are 1, (add 2) 3, (add 3) 6, (add 4) 10, (add 5) 15, (add 6) 21, (add 7) 28, etc. It could take a long time to get to the 20th one, and there are many opportunities to make a mistake in our calculations. Noticing the visual representation of the triangular numbers on the preceding page, we see that the third triangular number is the number of dots needed for a right triangle made from three rows of dots, with three dots in the bottom row. The fourth triangular number would be the number of dots needed for a right triangle made from four rows of dots, with four dots in the bottom row. The 20th triangular number would then be the number of dots needed for a right triangle made from 20 rows of dots, with 20 dots in the bottom row. Notice there will be 20 dots in the bottom row and 1 dot in the top row, for a total of 21 dots for these two rows. The total number of dots in the second row (2 dots) and in the second-to-last row (19 dots) is also 21 dots. This pairing continues with the third row and third-to-last row, the fourth row and fourth-to-last row, etc. Since there are 20 rows, there will be 10 of these pairings of rows that each have a total of 21 dots. This is a grand total of 10(21) = 210 dots. Another way to look at the representation of the triangular numbers is to take the triangular representation and turn it into a rectangle. Notice the identical triangle has been ﬂipped and moved up to complete a rectangle. The second triangular number produces a rectangle with two columns of three rows, or 6 dots. Half of this is 3. The third triangular number produces a rectangle with three columns and four rows, or 12 dots. Half of this is 6. The 20th triangular number would produce a rectangle with 20 columns and 21 rows, or 420 dots. Half of this is 210. 36 MATHCOUNTS 2005-2006 Warm-Up 8 $ 1. ________ A resort hotel charges a regular rate of $100 per night, but Wednesday nights are $150, and Saturday nights and Sunday nights are each $200. What is the price difference between the most expensive and cheapest possible stays of four consecutive nights? % 2. ________ There are three times the number of orange ﬁsh as blue ﬁsh in a tank at the pet store, and there are no other ﬁsh. If Kaya randomly pulls out one ﬁsh from the tank, what is the probability that it is orange? Express your answer as a percent. sq cm 3. ________ A large square region is divided into four congruent, non- overlapping square regions. If the perimeter of one of these small square regions is 60 cm, what is the area of the large square? 4. ________ Mary will pick a positive integer less than 80 that is a multiple of 7. Susan will pick a positive integer less than 80 that is a multiple of 9. What is the probability that they both will pick the same number? Express your answer as a common fraction. hours 5. ________ A certain amount of work can be done by seven identical machines in 30 hours. How long will it take 10 of these same machines to do the same amount of work? 6. ________ How many 1” cubes will pack into a rectangular box with dimensions 8” by 6” by 6.5”? cubes : 7. ________ Changee is scheduling a.m. Avge Time of Event Event use of the pool for the (min:sec) 2008 Summer Olympic Women‛s Indiv Freestyle 2:23 Games. He has decided to allow 15 minutes between Women‛s Indiv Butterﬂy 3:20 events, and he will use the Women‛s Indiv Backstroke 3:10 historic average time for Women‛s Indiv Breaststroke 2:56 each event rounded up to the next whole minute for Women‛s Relay 9:27 the length of each event. Men‛s Indiv Freestyle 2:10 He needs to schedule these 10 events, in order, Men‛s Indiv Butterﬂy 3:01 beginning at 8 a.m. At Men‛s Indiv Backstroke 2:54 what time will the men‛s Men‛s Indiv Breaststroke 2:45 relay begin, according to his schedule? Men‛s Relay 8:15 °F 9 8. ________ Each degree in the Celsius temperature system is 5 of a degree in the Fahrenheit system, and 32°F = 0°C. What is the Fahrenheit temperature when the Celsius temperature is 45°? 1 9. ________ What number should be added to both the numerator and the denominator of 5 to 4 get a fraction equivalent to 5 ? 10. _______ A rectangle has a perimeter of 38 cm. Three-fourths of its width is equal to one- cm ﬁfth of its length. What is the measure of the length? MATHCOUNTS 2005-2006 37 Warm-Up 8 Answers 1. 150 (E, G, T) 5. 21 (F, S) 8. 113 (F) 2. 75 (F, M) 6. 288 (M, S) 9. 15 (G, M, T) 3. 900 (F, M) 7. 10:52 (T) 10. 15 (F, G, M) 1 4. (C, F, G, M, T) 88 Solution - Problem #5 We can see that seven machines each working 30 hours is equivalent to 7 × 30 = 210 hours of machine work. (It‛s the same as one machine working for 210 hours.) We need to keep 210 hours of machine work, but now we are going to have 10 machines to do the work. The 210 hours would need to be spread evenly over the 10 machines, which results in 210 ÷ 10 = 21 hours of work for each of the 10 machines. Representation - Problem #9 We are asked to ﬁnd the number that should be added to both the numerator and denominator 1 4 of 5 to get a fraction equivalent to 5 . We can apply algebra to the situation and see where that 1+ x gets us. Let x be the number we are looking for. Then 5 + x = 5 ; 5(1 + x) = 4(5 + x); 5 + 5x = 20 + 4x; 4 and x = 15. To check our work, we would see that our numerator would become 16 while our 16 denominator would become 20, and 20 = 5 .4 Is there a way to do this without all of the messy algebra? Is there a way to see what is happening throughout the process? Let‛s take a look at a representation of 5 (row 1), and what 1 happens as we add numbers to the numerator and denominator. Row one is showing 5 of the squares as shaded. If we add 1 1 to both the numerator and denominator of 5 , we get 1 2 6 , which is represented in the second row. If we add a 2 to 3 both the numerator and denominator of 5 , we get 7 . The 1 representation of this is in row 3. Notice what never changes in the representations: There are always four squares not shaded. This means that when we have added on enough squares to get 4 to a representation that is equivalent to 5 , there will still be four squares that are not shaded. These four squares will then represent 5 of that ﬁnal row. If 4 squares are 5 of the ﬁnal 1 1 row, how many squares are in the ﬁnal row? There must be 4 × 5 = 20. If there are 20 squares in the row, we must have added on 15 squares to the original ﬁve squares in row 1. Representation - Problem #10 This one sounds very messy, and we have lots of fractions to take into consideration. However, let‛s not go right to an equation. Instead, let‛s take a look at the following representation. We are told that one-ﬁfth of the length, or x, is equal to three-fourths of the width. That leaves one- fourth of the width, and it must be equal to a third of x. Since the perimeter is 38 cm, we know length + width = 19 cm, and we can perform a fairly simple calculation for x: 5x + 1 3 x = 19; 6 3 x = 19 or 19 x = 19; ( 19 ) 19 x = 19( 19 ); x = 3. 1 1 3 3 Length 3 3 Remember that we are looking for the length, which is 5x or 15 cm. x x x x x Width x 1 3 x 38 MATHCOUNTS 2005-2006 Workout 4 1. ________ Thirty-two square 8” tiles are laid next to a 12‛ by 18‛ % rectangular pool in the design shown. What percent of the perimeter of the pool does not have tile next to it? Express your answer to the nearest whole number. 2. ________ John‛s piggy bank had $1.20 when he went to bed on Monday. On Tuesday morning cents he put three coins in the bank. He put in three more coins in the afternoon and then three more in the evening. That night there was a total of $2.20 in his bank. If no coin is worth more than 25 cents, what is the greatest amount John could have put in the bank on Tuesday evening? New York Survey 3. ________ Ryan made the doughnut graph shown here using the results sq cm Soda vs. Pop of a survey. If the radius of the inner circle is 3 cm and SODA 67% the radius of the outer circle is 6 cm, what is the area of the region representing Pop? Express your answer as a Terms for decimal to the nearest tenth. Soft Drinks OTHER COKE 2% POP 2% 29% $ 4. ________ A new motorcycle costs $6000. It loses 20% of its value by the end of the ﬁrst year. Each year thereafter, it loses 10% of its remaining value from the end of the previous year. How much is the motorcycle worth after ﬁve years? 5. ________ Some farmers use circular irrigation systems, creating green circles in their ﬁelds. % If a circular sprinkler at the center of a square ﬁeld sprays as much of the ﬁeld as possible, without spraying past the boundaries of the ﬁeld, what percent of the ﬁeld is not watered? Express your answer to the nearest tenth. 6. ________ A player has a batting average of exactly .250. What is the greatest amount his average could increase with one more hit? Express your answer as a decimal to the nearest thousandth. $ 7. ________ The amount of $72 is to be divided among Mr. Abbondanzio, Ms. Barta and Ms. Conders in the ratio 6:2:1, respectively. How much should Ms. Barta receive? 8. ________ What is the least positive fraction whose numerator is two less than a perfect square and whose denominator is one more than the same perfect square? 6 9 8 6 9. ________ Jared earned grades of 10 , 15 , 11 , 10 and 8 on his quizzes in history. What was the % 15 average of his ﬁve scores on these quizzes? Express your answer as a percent to the nearest whole percent. 10. _______ A ½-mile long circus train is traveling at the rate of 10 mph when it reaches a tunnel min two miles long. How many minutes was it from the time when the front of the engine entered the tunnel until the rear of the caboose left the tunnel? MATHCOUNTS 2005-2006 39 Workout 4 Answers 2 1. 84 (F, P) 5. 21.5 (F, M) 8. (G, M, P, T) 5 2. 60 (G, T) 6. 0.150 (G, M) 9. 69 (C) 3. 24.6 (F) 7. 16 (M, T) 10. 15 (F, M) 4. 3149.28 (M, P, T) Solution - Problem #10 This problem is a little tricky because we have to ﬁgure out how far the train has traveled from the time the front engine entered the tunnel to the time the rear of the caboose left the tunnel. The tunnel is two miles long, but the train has actually traveled farther than this. The front of the train entered the tunnel, went two miles, exited the tunnel, and then went another half-mile before the end of the caboose was completely out of the tunnel. We need to ﬁnd the number of minutes it took for this train to travel 2.5 miles. Using the equation Rate × Time = Distance, and knowing that the rate was 10 mph and the distance was 2.5 miles, we can set up an equation for the time (T) needed to complete the distance: 10 × T = 2.5 or 10T = 2.5. Dividing both sides of the equation by 10, we have T = 0.25. Remember that our rate was in mph, our distance ½ mile 2 miles was in miles, and so the time is in hours. The problem asked for the number of minutes, and 0.25 hours is a quarter of an hour, which is 15 minutes. Representation - Problem #4 We can approach this problem by ﬁguring out the value of the motorcycle for each year until we reach the end of the ﬁfth year. We know the motorcycle loses 20% of its value after the ﬁrst year, so it keeps 80% of its value. This is (0.80)($6000) = $4800. After this ﬁrst year, the motorcycle loses 10% of its value each year, so it will keep 90% of its value. After the second year, the motorcycle will be worth (0.90)($4800) = $4320. After the third year, its value will be (0.90)($4320) = $3888. After the fourth year, the value is (0.90)($3888) = $3499.20. And ﬁnally, after the ﬁfth year, the value of the motorcycle is (0.90)($3499.20) = $3149.28. In this case we can see that we are performing the same multiplication process four times over the ﬁve-year period. (Remember that the motorcycle keeps a different portion of its value the ﬁrst year.) Because of this repeated multiplication, we could save a couple of steps by using exponents. The long procedure in the solution above is equivalent to performing the following calculation: ($6000)(0.80)(0.90)4. With the help of a calculator, we see this is again $3149.28. Again remembering that we start with a value of $6000, go to $4800 after the ﬁrst year, and then keep decreasing by 10%, we could come up with an equation that we could use for any year after the ﬁrst year. Using the same logic that resulted in the expression ($6000)(0.80)(0.90)4 for the value after the ﬁfth year, we can see that if we use the equation y = 4800(0.90)x−1, we can plug in any number of years (x) and determine the value of the car (y). This is helpful for answering questions like, “At the end of which year would the value of the car ﬁrst be below $2500?” We could enter the equation Y1 = 4800(0.90)x−1 into a graphing calculator, see the ordered pairs on the TABLE screen, and see that y is ﬁrst less than $2500 when x is 8. After eight years, the value of the car is about $2296, which is the ﬁrst time it is less than $2500. 40 MATHCOUNTS 2005-2006

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