warm-up 1...warm-up 1 31. the table shown is an organized list of all 7 combinations 0of quarters,...

MATHCOUNTS 2018-2019 15

31. ��

32. ��

33. ��

34. ��

35. ��

36. ��

37. ��

38. ��

39. ��

40. ��

Warm-Up 1Bob has 40 cents in his pocket. If Bob has no pennies, how many different combinations of quarters, dimes and/or nickels could he have?

On the number line shown, what is the value of x − y ? Express your answer as a mixed number.

Ted flips a coin that is equally likely to land heads up or tails up. Ted flips the coin 10 times, and each time it lands heads up. What is the probability that the next flip

will also land heads up? Express your answer as a common fraction.

What is the value of 9 + 5 × 3 – 8 ÷ 2?

If two more than three times x is equal to five less than ten times x, what is the value of x ?

What is the volume of a rectangular prism of height 5 cm, width 7 cm and depth 3 cm?

What is the average of the prime numbers between 20 and 30?

How many lines of symmetry does an isosceles right triangle have?

What is the quotient when 1,000,000,000 is divided by 28 × 57?

Of 1000 people surveyed, one-third of the 630 people who reported owning a cat also own a dog. If each person surveyed owns a cat, a dog or both, how many own a dog?

combi-nations

8

y

10 12 14

x

3 cm

5 cm

7 cm

cm3

lines

people

MATHCOUNTS 2018-201916

41. ��

42. ��

43. ��

44. ��

45. ��

46. ��

47. ��

48. ��

49. ��

50. ��

Warm-Up 2If the value of x is 10, what is the value of 3x + 4x + 5x ?

In the figure, two regular pentagons have been attached to a regular hexagon to create a 12-gon. If each regular polygon has side length 3 feet, what is the perimeter of the resulting 12-gon?

A speed of 60 miles per hour is equal to 88 feet per second. If the speed limit in a school zone is 15 miles per hour, what is the speed limit in this zone in feet per second?

How many divisors of 64 are perfect squares?

If ab = 2c, bd = c, b ≠ 0 and d = 16, what is the value of a?

All of the streets in Tom’s city are on a regular rectangular grid and run east-west or north-south. Tom starts at the intersection of Poyntz Avenue

and Eleventh Street and walks 2 blocks east, then 3 blocks north, then 4 blocks east, then 5 blocks north. Tom then returns to his starting location

by walking 6 blocks west, then 8 blocks south. What is the area, in square blocks, enclosed by Tom’s path?

What is the value of +−

31

21

41

51 ? Express your answer as a common fraction.

What is the arithmetic mean of 23 ,

79 ,

14 and

516? Express your answer as a common

fraction.

If a and b are real numbers such that a + b = a − b and a ≠ b, what is the value of a b a b ab

a b

2 2+ + –– ?

Given that 75% of a certain number is 88, what is 38 of the number?

LK

JF

E D

IH

GC

BA

feet

ft/s

divisors

squareblocks


171. ��

172. ��

173. ��

174. ��

175. ��

176. ��

177. ��

178. ��

179. ��

180. ��

A rectangular sheet of paper is cut in half perpendicular to the longer side. One half is discarded, and the other half is cut into thirds as shown. Two of the thirds are discarded, and the remaining third is cut into fourths by vertical lines. Three of the fourths are discarded, and the remaining fourth is cut into fifths by horizontal lines. Four of the fifths are discarded, and the remaining fifth is cut into sixths by vertical lines. Five of the sixths are discarded, leaving a square piece with side length 2 cm. What is the perimeter of the original sheet of paper?

In 1969, the Apollo 10 mission set the record for the fastest crewed space travel, at 39,897 km/h. At that speed, how many days would it take to travel 54.6 million km, the minimum distance from Earth to Mars? Express your answer to the nearest

whole number.

Yeong multiplies two-digit positive integers AB and CD. If the digits A, B, C and D are all distinct, what is the greatest possible value of the product?

Karli paid $3.00 for lunch every day she attended school. During a six-week period, Karli attended school every Monday through Friday, with the exception of one school holiday. What is the total amount that Karli spent on school lunches during this six-week period?

What is the sum of the mean, median, mode and range of the numbers 15, 33, 24, 10, 20 and 24?

A certain race car consumes 1.3 gallons of fuel during each lap of a race. If each lap is 2.5 miles, and the entire race is 500 miles, how much fuel does the race car consume from start to finish?

One television screen measures 56 inches long and 33 inches wide. A smaller, geometrically similar television screen measures 48 inches long. What is the width of the smaller screen? Express your answer as a decimal to the nearest tenth.

Lindsay starts at the peak of a mountain, and it takes her 50 minutes to hike 15,000 feet. What was her average walking speed, in miles per hour, given 1 mile = 5280 feet? Express your answer as a decimal to the nearest tenth.

Let a be the arithmetic mean of 3.27 and 17.95. Let b be the product of 32.7 and 0.4382. Let c be the quotient of 2.637 and 0.316. Let d be the absolute difference between 793.241 and 804.3692. What is the numerical value of the median of a, b, c and d ? Express your answer as a decimal to the nearest hundredth.

Jessie makes butterfly wings by shading four sections of a regular hexagon as shown. If the hexagon has side length 4 3 units, what is the area of the shaded region? Express your answer in simplest radical form.

cm

days

$

gallons

inches

mi/h

units2

Workout 1


25. When a point is rotated 90 degrees clockwise about the origin, in general, the x-coordinate of the image is the y-coordinate of the pre-image, and the y-coordinate of the image is the opposite of the x-coordinate of the pre-image. The image of D(−5, −3) rotated 90 degrees clockwise about (0, 0) is D′(−3, 5).

26. When rotating a figure about a point that is not the origin, it helps to think of the figure’s location as if the rotation point were the origin. Once the location of the image (after rotation) is determined, we can rewrite its coordinates in terms of the actual origin. If we think of F(5, 4) as (5 − 5, 4 − 4) = (0, 0), then we need to think of the coordinates of E(3, −1) as (3 − 5, −1 − 4) = (−2, −5). When a point is rotated 90 degrees counterclockwise about the origin, in general, the x-coordinate of the image is the opposite of the y-coordinate of the pre-image, and the y-coordinate of the image is the x-coordinate of the pre-image. So, the image of (−2, −5) rotated 90 degrees counterclockwise about (0, 0) is (5, −2). In terms of the actual origin, the image has coordinates (5 + 5, −2 + 4) = (10, 2).

27. When a point is dilated by a scale factor of k about the origin, we multiply both the x- and y-coordinate by k to determine the coordinates of the image. So, the image of G(−2, 3) dilated by a factor of 2/3 about (0, 0) is G′((−2)(2/3), (3)(2/3)) = (−4/3, 2). The image of H(4, 7) dilated by a factor of 2/3 about (0, 0) is H′((4)(2/3), (7)(2/3)) = (8/3, 14/3). The sum of the coordinates of G′ and H′ is −4/3 + 2 + 8/3 + 14/3 = (−4 + 6 + 8 + 14)/3 = 24/3 = 8. Alternatively, we could multiply the sum of the coordinates of G and H by 2/3 to get (2/3)(−2 + 3 + 4 + 7) = (2/3)(12) = 8.

28. Similar to rotating a figure about a point that is not the origin, when dilating a figure about a point that is not the origin, let’s think of the figure’s location as if the center of dilation were the origin. Once the location of the image (after dilation) is determined, we’ll rewrite its coordinates in terms of the actual origin. If we think of K(2, 2) as ( 2 − 2, 2 − 2) = (0, 0), then we need to think of the coordinates of J(4, 8) as ( 4 − 2, 8 − 2) = (2, 6). The image of J(4, 8) dilated by 3/2 about (0, 0) is ((2)(3/2), (6)(3/2)) = (3, 9). In terms of the actual origin, the image has coordinates J′(3 + 2, 9 + 2) = (5, 11). The product of the coordinates of J′ is 5 × 11 = 55.

29. If we think of M(3, 2) as (3 − 3, 2 − 2) = (0, 0), then we need to think of the coordinates of L(−2, 4) as (−2 − 3, 4 − 2) = (−5, 2). The image of (−5, 2) rotated 90 degrees clockwise about (0, 0) is (2, 5). The image of (2, 5) dilated 3/2 with center of dilation (0, 0) is ((2)(3/2), (5)(3/2)) = (3, 15/2). The area of the triangle doesn’t change if we translate it back to the right 3 and up 2, so let’s use the current vertices L(−5, 2), M(0, 0) and N(3, 15/2) to determine its area. Using the shoelace method (as shown), we have (1/2)[(−5 × 0 + 0 × 15/2 + 3 × 2) − (2 × 0 + 0 × 3 + 15/2 × (−5)] = (1/2)(6 + 75/2) = (1/2)(12 + 75)/2 = (1/2)(87/2) = 87/4. So the area of DLMN is 87/4 units2.

30. The line of reflection y = x − 2 has slope 1. The point R(−5, 3) and its reflection across this line must both be points on a line that is perpendicular to the line of reflection. The perpendicular line containing R and R′ has slope −1 and is given by the equation y − 3 = −1(x − (−5)) y − 3 = −x − 5 y = −x − 2. The intersection of these two lines is the midpoint of segment RR′. Setting the two expressions for y equal to each other and solving for x, we get x − 2 = −x − 2 2x = 0 x = 0. Substituting back into the equation y = x − 2, we see that y = 0 − 2 = −2. So, the midpoint of segment RR′ has coordinates (0, −2). Using the midpoint formula, we can determine the coordinates of R′ as follows: 0 = (x − 5)/2 0 = x − 5 x = 5 and −2 = (y + 3)/2 −4 = y + 3 y = −7. The image R′(5, −7) rotated 90 degrees clockwise about (0, 0) is R″(−7, −5). The distance from R to R″ is √[(−7 + 5)2 + (−5 − 3)2] = √(4 + 64) = √68 = 2√17 units.

Warm-Up 1

31. The table shown is an organized list of all 7 combinations of quarters, dimes and/or nickels that have a total value of 40 cents.

32. The scale of the number line is 1/6, so x = 13 3 ⁄ 6 and y = 9 ² ⁄ 6. The difference is x − y = 13 3 ⁄ 6 − 9 ² ⁄ 6 = 4 ¹⁄ 6.

33. The coin that Ted is flipping is equally likely to land heads up or tails up. Since each flip is an independent event, which means that the previous flips have no influence over the next flip, the probability that the next flip will land heads up is 1/2.

34. The scientific order of operations requires that we do the multiplication and division in order from left to right before we do the addition and subtraction in order from left to right. The result is 9 + (5 × 3) – (8 ÷ 2) = 9 + 15 – 4 = 24 – 4 = 20.

35. Translating the English to algebra, we get 2 + 3x = 10x − 5. Now, solve for x by first adding 5 to each side of the equation to get 7 + 3x = 10x. Then subtract 3x from each side to get 7 = 7x. Finally, divide each side by 7 to get 1 = x, or x = 1.

36. The formula for the volume of a rectangular prism is given by a formula usually expressed as V = lwh, which is the product of the length, the width and the height of the prism. In this case, we are given the height, the width and the depth, a slightly different way to name the three dimensions of the prism. The volume is thus V = (3)(5)(7) = 105 cm3.

37. There are only two prime numbers between 20 and 30, namely 23 and 29. Their average is (23 + 29)/2 = 52/2 = 26.

38. There is only 1 line of symmetry in an isosceles right triangle. The line of symmetry is shown as a dotted line in the figure.

QDN

008

016

024

032

040

103

111

Shoelace Method: The area of a closed figure with vertices A(x1, y1), B(x2, y2) and C(x3, y3) is

It’s easy to remember these products if you notice that the arrowsconnecting factors resemble shoelaces.

(x1y2 + x2y3 + x3y1) − (x2y1 + x3y2 + x1y3)

2

Ax1

y1

Ax1

y1

Bx2

y2

Cx3

y3


39. The number 1,000,000,000 can be written as 109 = [(2)(5)]9 = (29)(59). If we divide this by (28)(57), we get [(29)(59)]/[(28)(57)] = (29 − 8)(59 – 7) = (21)(52) = (2)(25) = 50.

40. One-third of the 630 people who own a cat also own a dog, which accounts for 630/3 = 210 people. Since each person surveyed owns a cat, a dog or both, there are 1000 – 630 = 370 people who own a dog but not a cat. So, a total of 210 + 370 = 580 people own a dog.

Warm-Up 2

41. Substituting 10 for x, we get (3)(10) + (4)(10) + (5)(10) = (3 + 4 + 5)(10) = (12)(10) = 120.

42. The 12-gon has 12 sides, each of length 3 feet. Thus, the perimeter is (12)(3) = 36 feet.

43. The school zone speed limit of 15 mi/h is one-quarter of 60 mi/h = 88 ft/s. So, 15 mi/h is equivalent to 88/4 = 22 ft/s.

44. The number 64 is a power of the prime number 2, and the divisors of 64 are all powers of 2: 1, 2, 4, 8, 16, 32 and 64. Of these factors, 1, 4, 16 and 64 are the 4 divisors that are perfect squares.

45. Since d = 16, we can rewrite bd = c as c = 16b. Since c = 16b, we can rewrite ab = 2c as ab = (2)(16b) = 32b. The value of a must be 32.

46. We can think of the figure that Tom’s path enclosed as a 6-by-8 rectangle with a 3-by-4 rectangle missing, as shown. The area is (6)(8) − (3)(4) = 48 – 12 = 36 square blocks. You might also calculate the area of this region by decomposing it into a 5-by-6 rectangle with area (5)(6) = 30 and a 3-by-2 rectangle with area (3)(2) = 6 to get a total area of 30 + 6 = 36 square blocks.

47. (1/2 + 1/3) ÷ (1/4 − 1/5) = (3/6 + 2/6) ÷ (5/20 − 4/20) = (5/6) ÷ (1/20) = (5/6)(20/1) = (5/3)(10/1) = 50/3.

48. The sum of the four fractions is 2/3 + 7/9 + 1/4 + 5/16 = (6 + 7)/9 + (4 + 5)/16 = 13/9 + 9/16 = (208 + 81)/144 = 289/144. To find the arithmetic mean of the four numbers, we need to divide this sum by 4 or, equivalently, multiply by 1/4. The result is (289/144)(1/4) = 289/576.

49. If we subtract a from both sides of the equation a + b = a − b, we get b = –b. The only number equal to its own opposite is zero. So, in order for the sum and difference of a and b to be equal, the value of b must be zero. We can simplify the expression as follows: (a2b + a + b − ab2)/(a − b) = a2(0) + a + 0 − a(02)/(a − 0) = a/a = 1.

50. Seventy-five percent of a number is the same as 3/4 = 6/8 of the number. Since 88 is 6/8 of the number and since 3/8 is half of 6/8, 3/8 of the number must be half of 88, which is 44.

Warm-Up 3

51. If two-thirds of Sammi’s jelly beans are yellow, then the 5 red and 7 orange jelly beans must account for the remaining one-third of the jelly beans. There must be (5 + 7)(3) = (12)(3) = 36 jelly beans in Sammi’s cup.

52. In the sum 0.49 + 0.53 + 0.55 + 0.47 + 0.48, each addend is about 0.50. So, the sum is about (5)(0.50) = 2.50. To find the exact total, let's examine by how much each addend exceeds or falls short of 0.50. We have, in hundredths, −1, +3, +5, −3 and −2, giving us a net of +2 hundredths. Therefore, the actual total is 2.50 + 0.02 = 2.52.

53. Each outfit will include Allie’s pair of jeans. For each of the four shirts, there are three sweaters Allie can wear, resulting in (4)(3) = 12 outfits so far. Each of these 12 combinations of shirt and sweater can be worn with either of the two scarves, so there are (12)(2) = 24 outfits. The Fundamental Counting Principle states that if independent events M and N can occur in m ways and n ways, respectively, then the event M followed by N can occur in m × n ways. So, we can get the same answer directly by multiplying (1)(4)(3)(2) = 24 outfits.

54. Since it takes Destiny 3/4 minute to run 1/8 mile, it will take her (8)(3/4) = 6 minutes to run a mile. Since it takes her 4 minutes to walk 1/4 mile, it will take her (4)(4) = 16 minutes to walk a mile. So, to run a mile and then walk a mile, it will take her 6 + 16 = 22 minutes.

55. The difference of two perfect squares x2 − y2 can be rewritten as the product (x + y)(x − y). That means the expression 272 − 232 can be rewritten as the product (27 + 23)(27 − 23) = (50)(4) = 200.

56. When two chords intersect in a circle, the point of intersection cuts each chord into two segment, and the product of the lengths of the two segments of one chord is equal to the product of the lengths of the two segments of the other chord. In this case chords BD and CE intersect at A, with AB = 3 units, AD = 6 units and AC = 8 units. So, (3)(6) = (8)(AE) AE = 18/8 = 2 ¼ units.

5

432


165. Whatever the ratio of the side lengths, if the diagonal of Joe’s screen is twice as long as Daniel’s, then Joe’s screen has 22 = (2)(2) = 4 times the area, as shown in the figure. This means the ratio of the area of Daniel’s screen to Joe’s is 1/4.

166. There are 25 of the smallest size rhombi in the figure. Four of the smallest size rhombi combined give us the next size, of which there are 16 rhombi in this figure. Nine of the smallest size rhombi combined give us the next size, of which there are 9 rhombi in this figure. Sixteen of the smallest size rhombi combined give us the next size, of which there are 4 rhombi in this figure. Finally, 25 of the smallest size rhombi combined give us the largest, of which there is 1 rhombus in this figure. That’s a total of 25 + 16 + 9 + 4 + 1 = 55 rhombi.

167. The decimal expansion of 3/7, which repeats after 6 digits, is written in “bar notation” as 0.428571. Since the remainder is 3 when 15 is divided by 6, or 15 is “3 mod 6”, it follows that the 15th digit to the right of the decimal point will be the 3rd digit of the repeating pattern, which is 8.

168. Since we want the least possible value of c and since the integers must be different, we will rewrite the list as 6, 7, 8, c, c + 1, c + 2, 34. The sum of these numbers is 6 + 7 + 8 + c + c + 1 + c + 2 + 34 = 3c + 58 = 7c, since the mean of the numbers is c. Solving the equation 3c + 58 = 7c, we get 4c = 58, and c = 14.5. We cannot lower c to the integer 14 while maintaining a mean of 14, but we can raise it to 15 if we raise one or two of the other unknown values to compensate. Some possible sets are {6, 7, 8, 15, 17, 18, 34}, {6, 7, 9, 15, 16, 18, 34}, {6, 7, 10, 15, 16, 17, 34} and {6, 8, 9, 15, 16, 17, 34}. After some trial and error, we see that the smallest possible value of c is 15.

169. The figure shows an area model for the probability of two continuous variables. For ordered pairs (x, y) with 0 ≤ x ≤ 10 and 0 ≤ y ≤ 10, if the point is located within the shaded region, then the values x and y have an absolute difference less than 3; otherwise their absolute difference is greater than or equal to 3. The 10-by-10 square has area 102 = 100 units2, and the unshaded region has area 2(1/2)(7)(7) = 49 units2, meaning the shaded region has area 100 − 49 = 51 units2. The probability that (x, y) is located within the shaded region, then, is 51/100.

170. Among the digits 1, 2, 3 and 4, the only digits that differ by more than two are 1 and 4. So, the permutations we must exclude are of the form 14_ _, _ _14, _14_, 41_ _, _ _ 41 and _41_. There are two ways to order the digits 2 and 3 for each of these. So, of the 4! = 24 permutations of the digits 1, 2, 3 and 4, we must exclude the (6)(2) = 12 permutations in which 1 and 4 are adjacent. That leaves 24 − 12 = 12 permutations as having no adjacent digits that differ by more than two.

Workout 1

171. The cuts to the rectangular sheet of paper are made both vertically and horizontally. The vertical cuts reduce the length to one-half of one-fourth of one-sixth of the original length, so we have 1/2 × 1/4 × 1/6 × original length = 2 cm. So, the length of the original sheet is (2)(4)(6)(2) = 96 cm. Similarly, the horizontal cuts reduce the width to one-third of one-fifth of the original width, so we have 1/3 × 1/5 × original width = 2 cm. So, the width of the original sheet is (3)(5)(2) = 30 cm. The perimeter of the original sheet is 2(96 + 30) = (2)(126) = 252 cm.

172. Based on the information provided, it would take about 54,600,000 km ×

1 hour

39, 897 km ×

1 day

24 hours

≈ 57 days.

173. To get the greatest possible product AB × CD, we need A, B, C and D to be as large as possible, namely chosen from the set {6, 7, 8, 9}. Since A and C are tens digits, we’ll chose those digits from the subset {8, 9}. So, the greatest product will be 9B × 8C (or 8B × 9C). Since there are only two options for choosing B and C from the subset {6, 7}, we’ll calculate both possible products to see which is greater. We get (96)(87) = 8352 and (97)(86) = 8342. The greatest product, then, is 8352.

174. Six full weeks of school lunches would be 30 lunches at $3.00 each for a total of (30)(3) = $90.00. Excluding one lunch to account for the holiday, we see that Karli spent 90 − 3 = $87 or 87.00.

175. From least to greatest, the numbers are 10, 15, 20, 24, 24 and 33. Their sum is 126, so the mean is 126/6 = 21. The median is the average of 20 and 24, which is 22. The mode is the value that occurs the most, which is 24, and the range is 33 – 10 = 23. The sum of the mean, median, mode and range is 21 + 22 + 24 + 23 = 90.

176. Since each lap is 2.5 miles, the 500-mile race is a total of 500/2.5 = 200 laps. Since the car consumes 1.3 gallons of fuel per lap, during the entire race it consumes (200)(1.3) = 260 gallons.

177. The two television screens are geometrically similar, so their length-to-width ratios must be equal. We can set up the proportion 56/33 = 48/w, where w represents the width of the smaller screen. Solving for w yields 56w = (33)(48) w = (33)(48)/56 ≈ 28.3 inches.

178. Lindsay’s average walking speed was 15,000 feet

50 minutes

×

60 minutes

1 hour

×

1 mile

5280 feet

≈ 3.4 mi/h.

179. Evaluating each variable, we get the average a = (3.27 + 17.95)/2 = 10.6100, the product b = (32.7)(0.4382) ≈ 14.3291, the quotient c = 2.637/0.316 ≈ 8.34494 and the absolute difference d = 804.3692 − 793.241 = 11.1282. From least to greatest, the values are 8.34494, 10.6100, 11.1282 and 14.3291. The median of these numbers is the average of the two middle values, which is (10.6100 + 11.1282)/2 ≈ 10.87.

20 4 6 8 10 x

y

2

46

810


180. Recall that the area, in square units, of an equilateral triangle of side length s is s 2 × √3/4. Since the hexagon is divided into six congruent equilateral triangles of side length 4√3 units, four of which are shaded, it follows that the shaded region has area (4)(4√3)2 (√3/4) = 48√3 units2.

Workout 2

181. According to Heron’s formula, the area of a triangle with side lengths a, b and c equals s s a s b s c( )( )( )− − − , where s = (a + b + c)/2, which is the semiperimeter of the triangle. For this triangle, s = (16 + 30 + 34)/2 = 80/2 = 40, so its area is √[(40)(40 − 16)(40 − 30)(40 − 34)] = √[(40)(24)(10)(6)] = √57,600 = 240 units2. Alternatively, you may recognize that the side lengths 16, 30 and 34 are a multiple of the 8-15-17 Pythagorean Triple, meaning the triangle in question is a right triangle with legs of length 16 and 30. Its area is (1/2)(16)(30) = 240 units2.

182. The radius of the cylinder is 14/2 = 7 inches, so its volume is V = πr 2h = π(72)(12) = 588π in3, which equals (588π)/231 ≈ 8 gallons.

183. We will evaluate each greatest common divisor (GCD) and least common multiple (LCM) expression starting with the innermost parentheses and then working outward. First, we evaluate 2 @ 7 = LCM(2, 7). Since 2 and 7 are relatively prime, it follows that the LCM is 2 × 7, so (2 @ 7)2 = 22 × 72. Next, let’s evaluate (22 × 72) # 42 = GCD(22 × 72, 42). Since 42 = 2 × 3 × 7, it follows that the GCD is 2 × 7. Finally, we evaluate (2 × 7) # 105 = GCD(2 × 7, 105). Since 105 = 3 × 5 × 7, it follows that the GCD is 7.

184. During the 4 o’clock hour, the sum of the digits displayed is 10 for these 6 1-minute intervals: 4:06, 4:15, 4:24, 4:33, 4:42 and 4:51. During the 5 o’clock hour, the sum of the digits displayed is 10 for these 6 1-minute intervals: 5:05, 5:14, 5:23, 5:32, 5:41 and 5:50. The total is 12 minutes.

185. The density of water is 1 gram per cubic centimeter, and 1 kilogram = 1000 grams, so 1 kilogram of water would occupy 1000 cm3 of space. When the same mass of water freezes, it will take up more space. Since the density of ice is 91.67% that of water, or 0.9167 gram/cm3, the space occupied by the frozen 1 kilogram = 1000 grams of water is 1000/0.9167 ≈ 1091 cm3.

186. Evaluating aek + bfg + cdh − ceg − afh − bdk for a = 2, b = −5, c = 3, d = 0, e = 4, f = −6, g = −1, h = 8 and k = 7, we get (2)(4)(7) + (−5)(−6)(−1) + (3)(0)(8) − (3)(4)(−1) − (2)(−6)(8) − (−5)(0)(7) = 56 + (−30) + 0 − (−12) − (−96) − 0 = 56 − 30 + 12 + 96 = 134.

187. The figure shows trapezoid ABCD with diagonals AC and BD that are perpendicular to each other and with AC = BD = CB. So, triangle AED is an isosceles right triangle, with m EAD = mEDA = 45 degrees. In addition, triangles ACB and DBC are congruent isosceles triangles, and mCAB = mCBA = mBDC = mBCD = x degrees. We are asked to find mCBA + mBCD = 2x. Since the sum of the interior angles of a quadrilateral is 360 degrees, it follows that 4x + 45 + 45 = 360 4x = 270 2x = 135 degrees.

188. The 7 subsets with a sum of 15 are {15}, {10, 4, 1}, {10, 3, 2}, {8, 6, 1}, {8, 4, 3}, {8, 4, 2, 1} and {6, 4, 3, 2}.

189. If the shaded octagon has side length s, then the four unshaded isosceles right triangles each have legs of length s/√2 = (s√2)/2 units, and the square has side length (s√2)/2 + s + (s√2)/2 = s(√2 + 1) units. Together, two of the unshaded triangles form a square of side length (s√2)/2 units. So, the four unshaded triangles have a total area of 2 × ((s√2)/2)2 = s 2 units2. Since the square has area (s(√2 + 1))2 = s 2(3 + 2√2) units2, then s 2/(s 2(3 + 2√2)) = 1/(3 + 2√2) ≈ 0.17 = 17% of the figure is unshaded, meaning the shaded octagon accounts for 100 − 17 = 83% of the figure.

190. Spike digs 8 holes in 3 hours, or 8/3 holes per hour. Similarly, Butch digs 7/4 holes per hour, and Lucky digs 6/5 holes per hour. Working together, they would dig 8/3 + 7/4 + 6/5 = 337/60 holes in an hour. At this rate, digging 3 holes would take Spike, Butch and Lucky 3/(337/60) = 3(60/337) = 180/337 hour, which is equivalent to (180/337)(60) ≈ 32 minutes.

Workout 3

191. Let s be the side length of the cube. Since q represents the sum of the edge lengths, q = 12s. Since A represents its total surface area, A = 6s 2. The volume of a cube is V = s 3, and we are told that qA = kV, so (12s)(6s 2) = k(s 3) 72s 3 = ks 3. Thus, k = 72.

192. Let R represent the radius of the sphere, and let r represent the radius of the cylinder. The surface area of the sphere is 4πR 2, and the surface area of the cylinder is 2πr 2 + 2πrh. Since the cylinder’s height is equal to its diameter, or twice its radius, we can rewrite the cylinder’s surface area as 2πr 2 + (2πr)(2r) = 2πr 2 + 4πr 2 = 6πr 2. We know that the sphere has six times the surface area of the cylinder, so 4πR 2 = (6)(6πr 2) R 2 = 9r 2 R = 3r. The cylinder has volume πr 2h = πr 2(2r) = 2πr 3, and the sphere has volume (4/3)πR 3 = (4/3)π(3r)3 = 36πr 3. The ratio of the volume of the cylinder to the volume of the sphere is 2πr 3/(36πr 3) = 1/18.

193. All the triangles in the figure are isosceles right triangles. Let n be the leg length of the smallest triangle. We will categorize and count the triangles based on their leg lengths. There are 32 triangles with legs of length n. There are 18 triangles with legs of length 2n. There are 8 triangles with legs of length 3n. There are 2 triangles with legs of length 4n. That’s a total of 32 + 18 + 8 + 2 = 60 triangles. Alternatively, another counting method is to examine the lattice points of the five rows, starting at the leftmost lattice point of the top row and counting the number of triangles for which that point is the right-angle vertex. The totals, in order, for all five rows are: 4 + 3 + 2 + 1 + 0 = 10, 3 + 4 + 3 + 2 + 1 = 13, 2 + 3 + 4 + 3 + 2 = 14, 1 + 2 + 3 + 4 + 3 = 13 and 0 + 1 + 2 + 3 + 4 = 10. That’s a total of 10 + 13 + 14 + 13 + 10 = 60 different triangles.

E

D

C B

A

warm-up 1...warm-up 1 31. the table shown is an organized list of all 7 combinations 0of quarters,...

Documents