MATHCOUNTS
https://www.mathcounts.org
enIL State Competition Competition
https://www.mathcounts.org/il-state-competition-competition
<div class="field field-name-field-competition-type-of-data field-type-list-text field-label-above field--competition-type-of-data"><div class="field-label">What best describes this competition data set?: </div><div class="field-items"><div class="field-item even">Official data</div></div></div>Mon, 06 Jul 2020 20:16:07 +0000CStaten305069 at https://www.mathcounts.orghttps://www.mathcounts.org/il-state-competition-competition#commentsThe Fourth of July
https://www.mathcounts.org/resources/problem-of-the-week/fourth-july
<div class="field field-name-field-date-of-problem field-type-datetime field-label-hidden field--date-of-problem"><div class="field-items"><div class="field-item even"><span class="date-display-single" property="dc:date" datatype="xsd:dateTime" content="2020-07-06T10:00:00-07:00">July 6, 2020</span></div></div></div><div class="field field-name-field-problem-category field-type-taxonomy-term-reference field-label-inline clearfix field--problem-category"><div class="field-label">Topics/Content Areas: </div><div class="field-items"><div class="field-item even"><div class="textformatter-list"><a href="/problem-topic-content-area/percents-fractions">Percents & Fractions</a>, <a href="/problem-topic-content-area/plane-geometry">Plane Geometry</a>, <a href="/problem-topic-content-area/probability-counting-combinatorics">Probability, Counting & Combinatorics</a>, <a href="/problem-topic-content-area/proportional-reasoning">Proportional Reasoning</a></div></div></div></div><div class="l-constrain l-constrain--prose">
<p><span style="font-size:12pt"><span style="font-family:"Times New Roman",serif"><span style="font-family:"Calibri",sans-serif">When most of us think of the Fourth of July, we think of fireworks! Over the years, fireworks have become more and more elaborate. However, this is also the time of year when safety agencies are concerned about the improper use of fireworks and the possible injuries that can occur. Fireworks are safer now than they used to be but can still cause very serious injuries. In 1990, there were 17.8 injuries per 100,000 pounds of fireworks. In 1999, that number had fallen approximately 30.5%. What was the number of injuries per 100,000 pounds of fireworks in 1999? </span></span></span></p>
<p> </p>
<p><span style="font-size:12pt"><span style="font-family:"Times New Roman",serif"><span style="font-family:"Calibri",sans-serif">Fifty-six men signed the Declaration of Independence, though the signing was not actually on the Fourth of July. Being the President of the Congress, John Hancock, one of the five signers from Massachusetts, was the first to sign the Declaration. If the other men were to sign the document in random order, what is the probability that the next two signers would also be from Massachusetts? Express your answer as a common fraction.</span></span></span></p>
<p> </p>
<p><span style="font-size:12pt"><span style="font-family:"Times New Roman",serif"><span style="font-family:"Calibri",sans-serif">Though most people are familiar with John Hancock’s “slightly large” signature on the Declaration of Independence, it is not so well-known that Thomas Jefferson was the primary author of the document. Jefferson is one of the four men whose faces were permanently carved in stone at Mount Rushmore. In the carving, each of Jefferson’s eyes is 11 feet across. How many square yards, to the nearest whole number, are covered by his two eyes?</span></span></span></p>
<p> </p>
<p style="text-align: center;"><strong>CHECK THE <a href="https://www.mathcounts.org/resources/problem-archive"> PROBLEM OF THE WEEK ARCHIVE </a><br /><br />
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Mon, 06 Jul 2020 14:41:29 +0000michellelefrancois305062 at https://www.mathcounts.orghttps://www.mathcounts.org/resources/problem-of-the-week/fourth-july#commentsYard Work
https://www.mathcounts.org/resources/problem-of-the-week/yard-work-0
<div class="field field-name-field-date-of-problem field-type-datetime field-label-hidden field--date-of-problem"><div class="field-items"><div class="field-item even"><span class="date-display-single" property="dc:date" datatype="xsd:dateTime" content="2020-06-29T10:00:00-07:00">June 29, 2020</span></div></div></div><div class="field field-name-field-problem-category field-type-taxonomy-term-reference field-label-inline clearfix field--problem-category"><div class="field-label">Topics/Content Areas: </div><div class="field-items"><div class="field-item even"><div class="textformatter-list"><a href="/problem-topic-content-area/plane-geometry">Plane Geometry</a>, <a href="/problem-topic-content-area/probability-counting-combinatorics">Probability, Counting & Combinatorics</a>, <a href="/problem-topic-content-area/solid-geometry">Solid Geometry</a></div></div></div></div><div class="l-constrain l-constrain--prose">
<p style="margin-bottom:11px"><span style="font-size:11pt"><span style="line-height:normal"><span style="font-family:Calibri,sans-serif">Stephen spent this past weekend helping his mother get some yard work done. His first project was to build a rectangular planter with the exterior dimensions of 6 inches by 12 inches by 18 inches. He originally planned to use wood that was 1 inch thick. However, he mistakenly bought wood that was 2 inches thick. How many fewer cubic inches of soil are required to fill the planter made with 2-inch-thick wood than would have been required to fill the planter had it been made with the 1-inch-thick wood?</span></span></span></p>
<p style="margin-bottom: 11px; margin-left: 40px;"><span style="color:#19a2bf;"><em>Using 2-inch-thick wood, the 6 inch by 12 inch by 18 inch planter's interior is 2 inches smaller in height and 4 inches smaller in both length and width. Thus, the planter made with 2-inch-thick wood has an internal volume of (6 - 2)(12 - 4)(18 - 4) = 448 cubic inches. Had the planter been made with 1-inch-thick wood, the height would have only been decreased by 1 inch and the length and width would have each decreased by 2 inches. Thus, the internal volume of the planter made with 1-inch-thick wood would have been (6 - 1)(12 - 2)(18 - 2) = 800 cubic inches. This is a difference of 800 - 448 = </em></span><span style="color:#272561;"><em><strong>352 </strong></em></span><span style="color:#19a2bf;"><em>cubic inches. </em></span></p>
<p style="margin-bottom:11px"><span style="font-size:11pt"><span style="line-height:normal"><span style="font-family:Calibri,sans-serif">Next on Stephen’s task list was spreading mulch in his mother’s new garden. The garden is in the shape of an isosceles trapezoid. The large base is 10 feet long and the short base is 6 feet long. In the center of the garden is a tree with a diameter of 2 feet. If Stephen spreads mulch 2 inches thick, he would spread 8 – π/6 cubic feet. What is the altitude, in feet, of the trapezoid?</span></span></span></p>
<p style="margin-bottom: 11px; margin-left: 40px;"><span style="color:#19a2bf;"><em>The area of a trapezoid is equal to the average of the two bases times the height. Thus, the area of the garden is ((10 + 6)/2)(h) and to get the volume, we just multiply this area by the height (i.e. the depth of the mulch), which is 2 inches or 1/6 ft. So, the volume of the garden, when disregarding the tree, is (4/3)h. Now, to take the tree into consideration, we must subtract the volume of the tree trunk that will be buried in mulch, which is 1<sup>2</sup><span style="font-size:11.0pt"><span style="line-height:107%"><span style="font-family:"Calibri",sans-serif">π</span></span></span>(1/6) = <span style="font-size:11.0pt"><span style="line-height:107%"><span style="font-family:"Calibri",sans-serif">π</span></span></span>/6 cubic feet. Thus, (4/3)h - <span style="font-size:11.0pt"><span style="line-height:107%"><span style="font-family:"Calibri",sans-serif">π</span></span></span>/6 = 8 - <span style="font-size:11.0pt"><span style="line-height:107%"><span style="font-family:"Calibri",sans-serif">π</span></span></span>/6. Solving for h, we find that the height of the trapezoid must be </em></span><span style="color:#272561;"><em><strong>6</strong></em></span><span style="color:#19a2bf;"><em> feet. </em></span></p>
<p style="margin-bottom:11px"><span style="font-size:11pt"><span style="line-height:normal"><span style="font-family:Calibri,sans-serif">Finally, Stephen’s mother sent him to the store to buy flowers. She told him to by 6 pansies but did not specify which color(s). When he got to the store, he saw that they had four different colors in stock and that there were 20 plants of each color. How many distinct combinations of colors of pansies could he have bought?</span></span></span></p>
<p style="margin-left: 40px;"><em><span style="color:#19a2bf;">If Stephen bought all one color of any pansy, there would be 4 ways he could do this. If Stephen bought 2 colors of pansies, he could buy 5 of one color and 1 of another color (4 × 3 = 12 combinations), 4 of one color and 2 of another (12 color combinations) or 3 of one color and 3 of another (6 combinations), which is a total of 12 + 12 + 6 = 30 combinations. If Stephen bought all four colors, we can envision the first four flowers selected being one of each color (let's say A, B, C and D), leaving two flowers that can be any combination of the 4 colors (AA, AB, AC, AD, BB, BC, BD, CC, CD, DD), thus making 10 possible color combinations. Finally, if he buys only 3 of the colors, let's assume the first three are one of each of three colors (let's say A, B and C). This means that there are 4 ways he can choose the first three flowers and the last three flowers can be chosen 10 ways (AAA, AAB, AAC, ABB, ABC, ACC, BBB, BBC, BCC, CCC). Thus, there are 4(10) = 40 ways three colors can be chosen. This gives a total of 4 + 30 + 10 + 40 = </span><span style="color:#272561;"><strong>84 </strong></span><span style="color:#19a2bf;">possible combinations. </span></em></p>
<p style="margin-left: 40px; text-align: center;"><span style="font-size:12pt"><span style="font-family:"Times New Roman",serif"><strong><span style="font-size:11.0pt"><span style="font-family:"Segoe UI Symbol",sans-serif"><span style="color:#3b3838">♦</span></span></span></strong><strong> </strong><strong><span style="font-size:9.0pt"><span style="font-family:"Akzidenz Grotesk BE Regular""><span style="color:#ff8a3b">Page 1</span></span></span></strong><strong><span style="font-size:9.0pt"><span style="font-family:"Akzidenz Grotesk BE Regular""> <span style="color:#3b3838">of the linked PDF contains PROBLEMS & SOLUTIONS.</span></span></span></strong></span></span></p>
<p style="margin-left: 40px; text-align: center;"><span style="font-size:12pt"><span style="font-family:"Times New Roman",serif"><strong><span style="font-size:9.0pt"><span style="font-family:"Akzidenz Grotesk BE Regular""><span style="color:#ff8a3b">Page 2</span><strong> </strong><span style="color:#3b3838">contains ONLY PROBLEMS. </span></span></span><span style="font-size:11.0pt"><span style="color:#3b3838">♦</span></span></strong></span></span></p>
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Mon, 06 Jul 2020 14:17:38 +0000michellelefrancois305060 at https://www.mathcounts.orghttps://www.mathcounts.org/resources/problem-of-the-week/yard-work-0#commentsPerfect Days in June
https://www.mathcounts.org/resources/problem-of-the-week/perfect-days-june-0
<div class="field field-name-field-date-of-problem field-type-datetime field-label-hidden field--date-of-problem"><div class="field-items"><div class="field-item even"><span class="date-display-single" property="dc:date" datatype="xsd:dateTime" content="2020-06-22T10:00:00-07:00">June 22, 2020</span></div></div></div><div class="field field-name-field-problem-category field-type-taxonomy-term-reference field-label-inline clearfix field--problem-category"><div class="field-label">Topics/Content Areas: </div><div class="field-items"><div class="field-item even"><div class="textformatter-list"><a href="/problem-topic-content-area/number-theory">Number Theory</a></div></div></div></div><div class="l-constrain l-constrain--prose">
<p style="margin-bottom:11px"><span style="font-size:11pt"><span style="line-height:107%"><span style="font-family:Calibri,sans-serif"><i>The following problem of the week was submitted by Tim Ramsey of the Singapore American School in 1999.</i></span></span></span></p>
<p style="margin-bottom:11px"><span style="font-size:11pt"><span style="line-height:107%"><span style="font-family:Calibri,sans-serif">Some would say that a day during late June would definitely be perfect. Halfway between the drenching showers of April and the brutal heat of August. To mathematicians, one particular day in June is perfect: June 28. When written in the form mm/dd, June 28 is written as 6/28, and both 6 and 28 are called <i>perfect numbers</i>. A perfect number is a number whose proper factors (that is, the factors other than the number itself) add up to the number. What are the proper factors of 6, and what is the sum of these proper factors?</span></span></span></p>
<p style="margin-bottom: 11px; margin-left: 40px;"><span style="color:#19a2bf;"><em>The proper factors of 6 are </em></span><span style="color:#272561;"><em><strong>1</strong></em></span><span style="color:#19a2bf;"><em>, </em></span><span style="color:#272561;"><em><strong>2</strong></em></span><span style="color:#19a2bf;"><em> and </em></span><span style="color:#272561;"><em><strong>3</strong></em></span><span style="color:#19a2bf;"><em>. (Note that 6 is a factor of 6, but it is not a proper factor of 6.) The sum of the factors is 1 + 2 + 3 = </em></span><span style="color:#272561;"><em><strong>6</strong></em></span><span style="color:#19a2bf;"><em>, which is why 6 is considered a perfect number.</em></span></p>
<p style="margin-bottom:11px"><span style="font-size:11pt"><span style="line-height:107%"><span style="font-family:Calibri,sans-serif">When was the last time that the day, month and year were all perfect numbers? When is the next time? (Hint: it may be easier to answer this question after you see the next question about Mersenne primes and perfect numbers.)</span></span></span></p>
<p style="margin-bottom: 11px; margin-left: 40px;"><span style="color:#19a2bf;"><em>After 6 and 28, perfect numbers get fairly scarce. The next two smallest perfect numbers are 496 and 8128. So, the last perfect date happened on </em></span><span style="color:#272561;"><em><strong>6/28/496</strong></em></span><span style="color:#19a2bf;"><em>, and the next one will occur on </em></span><span style="color:#272561;"><em><strong>6/28/8128</strong></em></span><span style="color:#19a2bf;"><em>.</em></span></p>
<p style="margin-bottom:11px"><span style="font-size:11pt"><span style="line-height:107%"><span style="font-family:Calibri,sans-serif">The formula for deriving perfect numbers is (2<i><sup>p</sup></i><sup></sup><sup>– 1</sup>)(2<i><sup>p</sup></i> – 1) where <i>p</i> and (2<i><sup>p</sup></i> – 1) are prime. If <i>p</i> and (2<i>p</i> – 1) are prime, then (2<i><sup>p</sup></i> – 1) is called a <i>Mersenne Prime</i>. Find the first prime number that is not a Mersenne Prime.</span></span></span></p>
<p style="margin-bottom: 11px; margin-left: 40px;"><span style="color:#19a2bf;"><em>The sequence of prime numbers is 2, 3, 5, 7, 11, 13, ... The first prime number, </em></span><span style="color:#272561;"><em><strong>2</strong></em></span><span style="color:#19a2bf;"><em>, is not a Mersenne prime, because it cannot be written in the form (2<sup>p</sup> – 1). Note, however, that 3 is a Mersenne prime, because 2<sup>2</sup> – 1 = 3. </em></span></p>
<p style="margin-bottom:11px"><span style="font-size:11pt"><span style="line-height:107%"><span style="font-family:Calibri,sans-serif">The largest known Mersenne Prime is 2<sup>3021377</sup> – 1, and its associated perfect number has 909,526 digits. What is the units digit of this Mersenne Prime, and what is the units digit of the associated perfect number?</span></span></span></p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>The units digit of 2<sup>1</sup> is 2; of 2<sup>2</sup> is 4; of 2<sup>3</sup> is 8; of 2<sup>4</sup> is 6; of 2<sup>5</sup> is 2; of 2<sup>6</sup> is 4; and so on. Notice that the pattern of units digits is 2, 4, 8, 6, 2, 4, 8, 6 and repeats every fourth term. Hence, the units digit of 2<sup>3021377</sup> can be found by dividing the exponent by 4, noting the remainder r, and finding the units digit of 2<sup>r</sup>. When 3,021,377 is divided by 4, the quotient is 755,344, remainder 1. Because the units digit of 2<sup>1</sup> is 2, the units digit of 2<sup>3021377</sup> is also 2, and the units digit of 2<sup>3021377</sup> – 1 is therefore </em></span><span style="color:#272561;"><em><strong>1</strong></em></span><span style="color:#19a2bf;"><em>. The associated perfect number is (2<sup>3021377 – 1</sup>)(2<sup>3021377</sup> – 1). The units digit of the first term, 2<sup>3021377 – 1</sup>, is 6. This follows from the fact that the units digit of 2<sup>3021377</sup> was 2, and we need the units digit of one fewer power. And, we saw above that the units digit of 2<sup>3021377</sup> – 1 was 1. Hence, the units digit of this perfect number will be 6 × 1 = </em></span><span style="color:#272561;"><em><strong>6</strong></em></span><span style="color:#19a2bf;"><em>. </em></span></p>
<p style="margin-left: 40px; text-align: center;"><span style="font-size:12pt"><span style="font-family:"Times New Roman",serif"><strong><span style="font-size:11.0pt"><span style="font-family:"Segoe UI Symbol",sans-serif"><span style="color:#3b3838">♦</span></span></span></strong><strong> </strong><strong><span style="font-size:9.0pt"><span style="font-family:"Akzidenz Grotesk BE Regular""><span style="color:#ff8a3b">Page 1</span></span></span></strong><strong><span style="font-size:9.0pt"><span style="font-family:"Akzidenz Grotesk BE Regular""> <span style="color:#3b3838">of the linked PDF contains PROBLEMS & SOLUTIONS.</span></span></span></strong></span></span></p>
<p style="margin-left: 40px; text-align: center;"><span style="font-size:12pt"><span style="font-family:"Times New Roman",serif"><strong><span style="font-size:9.0pt"><span style="font-family:"Akzidenz Grotesk BE Regular""><span style="color:#ff8a3b">Page 2</span><strong> </strong><span style="color:#3b3838">contains ONLY PROBLEMS. </span></span></span><span style="font-size:11.0pt"><span style="color:#3b3838">♦</span></span><span style="font-size:9.0pt"><span style="font-family:"Akzidenz Grotesk BE Regular""><span style="color:#3b3838"> </span></span></span></strong></span></span></p>
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Mon, 29 Jun 2020 13:45:27 +0000michellelefrancois304976 at https://www.mathcounts.orghttps://www.mathcounts.org/resources/problem-of-the-week/perfect-days-june-0#commentsCentral Iowa Chapter Competition
https://www.mathcounts.org/central-iowa-chapter-competition-0
<div class="field field-name-field-competition-type-of-data field-type-list-text field-label-above field--competition-type-of-data"><div class="field-label">What best describes this competition data set?: </div><div class="field-items"><div class="field-item even">Official data</div></div></div>Wed, 24 Jun 2020 20:26:43 +0000isuleuch304941 at https://www.mathcounts.orghttps://www.mathcounts.org/central-iowa-chapter-competition-0#commentsA Tribute to Flag Day
https://www.mathcounts.org/resources/problem-of-the-week/tribute-flag-day-0
<div class="field field-name-field-date-of-problem field-type-datetime field-label-hidden field--date-of-problem"><div class="field-items"><div class="field-item even"><span class="date-display-single" property="dc:date" datatype="xsd:dateTime" content="2020-06-15T10:00:00-07:00">June 15, 2020</span></div></div></div><div class="field field-name-field-problem-category field-type-taxonomy-term-reference field-label-inline clearfix field--problem-category"><div class="field-label">Topics/Content Areas: </div><div class="field-items"><div class="field-item even"><div class="textformatter-list"><a href="/problem-topic-content-area/percents-fractions">Percents & Fractions</a>, <a href="/problem-topic-content-area/plane-geometry">Plane Geometry</a></div></div></div></div><div class="l-constrain l-constrain--prose">
<p style="margin-top: 3px;"><span style="font-size:13pt"><span style="line-height:107%"><span style="font-family:"Calibri Light",sans-serif"><span style="font-weight:normal"><span style="font-size:11.0pt"><span style="line-height:107%"><span style="font-family:"Calibri",sans-serif">Yesterday, Sunday, June 14<sup>th</sup>, the United States celebrated Flag Day. If you check out the many websites devoted to Flag Day, you are sure to learn a lot about the history of the holiday and the history of the flag, the Pledge of Allegiance and the Star Spangled Banner. Francis Scott Key wrote the Star Spangled Banner in 1814 when he saw the enormous American flag (made by Mary Pickersgill) blowing in the wind approximately 8 miles away. Assuming the flag was visible to anyone within a 15,000-yard radius of the flag, how many square miles was the total area from which the flag could be viewed? Give your answer to the nearest whole square mile.</span></span></span></span></span></span></span></p>
<p style="margin-right: -24px; margin-bottom: 16px; margin-left: 40px;"><span style="color:#19a2bf;"><em>If anyone within a 15,000-yard radius can see the flag, then we are trying to find the area of a circle with a radius of 15,000 yards. Since we are going to have to change this into miles eventually, let's go ahead and do it now so that we don't have to work with such large numbers. There are 1760 yards in 1 mile, so dividing by 1760 gives us a circle with a radius of 8.522 miles. Therefore, the flag is visible to anyone within <span style="font-size:11.0pt"><span style="line-height:107%"><span style="font-family:"Calibri",sans-serif">π</span></span></span>(8.522)<sup>2</sup> = </em></span><span style="color:#272561;"><em><strong>228 </strong></em></span><span style="color:#19a2bf;"><em>square miles, to the nearest whole square mile. </em></span></p>
<p><span style="font-size:12pt"><span style="font-family:"Times New Roman",serif"><span style="font-size:11.0pt"><span style="font-family:"Calibri",sans-serif">You’ve probably seen US flags in a variety of different sizes, but they are all geometrically similar and look the same because there are strict guidelines concerning the proportions of the different parts of the flag. A flag has a total of 13 stripes of equal width, starting with red at the top and alternating with white all the way down. There is also a rectangular region in the upper left-hand corner that is blue with white stars. Assuming that a flag is being displayed horizontally, all of the dimensions of the flag are compared to its width (top to bottom) measurement. Assume that the width is 1 unit, the length of the flag (right to left) is 1.9 units, the width of the blue rectangle is 7/13 of a unit (7 of the 13 stripes) and the length of the blue rectangle is .76 units long. Using these proportions, what percent, to the nearest whole number, of the surface area of the flag is red? </span></span></span></span></p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>Looking at a flag while doing the problem will probably be helpful. Since we are working with the red parts of the flag, we are only concerned with the 7 red stripes. Remember that every stripe has a width of 1/13 units. However, notice 4 of the red stripes are shorter than the other three. The long three are rectangles that are 1.9 units by 1/13 units, so the total area of them is 3 × 1.9 × (1/13) = .438462 square units. The shorter 4 rectangles are (1.9 - .76) units by 1/13 units, so the total area of them is 4 × (1.9 - .76) × (1/13) = .350769 square units. Together, the 7 red stripes are .789231 square units. The area of the entire flag is 1 by 1.9, which gives an area of 1.9 square units. So, dividing .789231 ÷ 1.9 gives that red makes up </em></span><span style="color:#272561;"><em><strong>42%</strong></em></span><span style="color:#19a2bf;"><em> of the U.S. flag, to the nearest whole percent. </em></span></p>
<p><span style="font-size:12pt"><span style="font-family:"Times New Roman",serif"><span style="font-size:11.0pt"><span style="font-family:"Calibri",sans-serif">On many of the Flag Day websites, there are specific guidelines concerning the displaying of and care of a U.S. flag. Many people put flags out for certain holidays, Flag Day being one of the most popular. Suppose you have 48 little flags each attached to 3-foot stakes and want to place them around the perimeter of your garden in the front yard. Your garden is a perfect square measuring 8 yards by 8 yards and you want to equally space the flags around its perimeter. If you place the first flag in the front, right corner and continue placing them around the entire garden, how many feet apart do you have to place them and how many flags will be along the back side of the garden?</span></span></span></span></p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>First, the perimeter of the garden is 4 × 8 = 32 yards. If there are 48 flags, then we need to take 32 and divide by 48, giving us 2/3 yards or 2 feet. So, we need to put a flag every two feet starting at the front right corner. To find the number of flags per side of the garden, it may appear that you just have to divide 48 by 4 and see that there are 12 flags per side, but the flags in the corners make a difference. Think of it as we are starting at the front right corner and putting in the corner flag along with 11 other flags on the first side. Moving to the second side, we are also starting with the corner and placing 11 more flags along that side. Notice that the corner flag we started the second side with is also on the first side. So, each side of the garden actually has </em></span><span style="color:#272561;"><em><strong>13 </strong></em></span><span style="color:#19a2bf;"><em>flags.</em></span></p>
<p style="margin-left: 40px;"> </p>
<p style="margin-left: 40px; text-align: center;"><span style="font-size:12pt"><span style="font-family:"Times New Roman",serif"><strong><span style="font-size:11.0pt"><span style="font-family:"Segoe UI Symbol",sans-serif"><span style="color:#3b3838">♦</span></span></span></strong><strong> </strong><strong><span style="font-size:9.0pt"><span style="font-family:"Akzidenz Grotesk BE Regular""><span style="color:#ff8a3b">Page 1</span></span></span></strong><strong><span style="font-size:9.0pt"><span style="font-family:"Akzidenz Grotesk BE Regular""> <span style="color:#3b3838">of the linked PDF contains PROBLEMS & SOLUTIONS.</span></span></span></strong></span></span></p>
<p style="margin-left: 40px; text-align: center;"><span style="font-size:12pt"><span style="font-family:"Times New Roman",serif"><strong><span style="font-size:9.0pt"><span style="font-family:"Akzidenz Grotesk BE Regular""><span style="color:#ff8a3b">Page 2</span><strong> </strong><span style="color:#3b3838">contains ONLY PROBLEMS. </span></span></span><span style="font-size:11.0pt"><span style="color:#3b3838">♦</span></span></strong></span></span></p>
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Mon, 22 Jun 2020 12:54:27 +0000michellelefrancois304897 at https://www.mathcounts.orghttps://www.mathcounts.org/resources/problem-of-the-week/tribute-flag-day-0#commentsRoller Coaster Madness
https://www.mathcounts.org/resources/problem-of-the-week/roller-coaster-madness-1
<div class="field field-name-field-date-of-problem field-type-datetime field-label-hidden field--date-of-problem"><div class="field-items"><div class="field-item even"><span class="date-display-single" property="dc:date" datatype="xsd:dateTime" content="2020-06-08T10:00:00-07:00">June 8, 2020</span></div></div></div><div class="field field-name-field-problem-category field-type-taxonomy-term-reference field-label-inline clearfix field--problem-category"><div class="field-label">Topics/Content Areas: </div><div class="field-items"><div class="field-item even"><div class="textformatter-list"><a href="/problem-topic-content-area/general-math">General Math</a></div></div></div></div><div class="l-constrain l-constrain--prose">
<p>The duration of a ride on the Shocker roller coaster is 1 minute 30 seconds. The Shocker has only one train consisting of a series of linked cars, each of which carries a pair of passengers. It takes each group of passengers exactly 30 seconds to board the train and exactly 30 seconds to exit the train. In one hour 480 people board the train, complete the ride and exit the train. Assuming there is never a seat left vacant, how many linked cars are in the Shocker’s train?</p>
<p style="margin-left: 40px;"><em><span style="color:#19a2bf;">Each group of passengers boards the train, completes the ride and exits the train, which takes a combined total of 0.5 + 1.5 + 0.5 = 2.5 minutes. In 60 minutes this occurs 60 ÷ 2.5 = 24 times for 24 groups of passengers. If 480 people board, ride and exit the train in 60 minutes, it follows that each group consists of 480 ÷ 24 = 20 passengers. Since passengers are seated two to a car, the train must consist of 20 ÷ 2 = </span><span style="color:#272561;"><strong>10</strong></span></em><span style="color:#19a2bf;"><em> cars.</em> </span></p>
<p>Marlo is waiting in the long line to ride the Shocker. He reaches a point where a sign is posted which reads, “Wait time from this point: 20 minutes.” Assuming, once again, that there is never a seat left vacant and there are no gaps in the line, what is the furthest position (i.e. 1st, 2nd, etc.) Marlo can occupy in the line at this point?</p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>If 20 minutes will pass before Marlo boards the train, that means 20 ÷ 2.5 = 8 groups of passengers will board, ride and exit the Shocker. From his current position, Marlo will be in the 9th group to board the ride. Since no seat is ever left vacant, there are 8 × 20 = 160 passengers who will ride the Shocker in the time it takes Marlo to move from his current position in the queue to board the train. The 9th group will have passengers who were 161st through 180th in line from the point where Marlo read the sign. Therefore, the furthest position Marlo can occupy in the line at the point where he encounters the sign is </em></span><span style="color:#272561;"><em><strong>180th</strong></em></span><span style="color:#19a2bf;"><em>. </em></span></p>
<p>The maximum speed of the Shocker is 65 mph. Its average speed, however, is a mere 8 mph due to the extremely slow rate at which the roller coaster ascends to the top of the first hill from which there is a 175 ft drop. If the ride on the Shocker takes 1 minute 30 seconds to be completed, how many feet long is the entire track? (Recall 1 mile = 5280 feet)</p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>We can use the distance formula d = rt to solve this problem. We are told that r = 8 mph. The duration of the ride is 1.5 min, which equals 1.5/60 = 0.025 h. Substituting, we get d = 8(.025) = 0.2 mi. Thus, the length of the Shocker's track is 0.2(5280) = </em></span><span style="color:#272671;"><em><strong>1056</strong></em></span><span style="color:#19a2bf;"><em> ft.</em></span></p>
<p style="text-align: center;"><span style="font-size:12pt"><span style="font-family:"Times New Roman",serif"><strong><span style="font-size:11.0pt"><span style="font-family:"Segoe UI Symbol",sans-serif"><span style="color:#3b3838">♦</span></span></span></strong><strong> </strong><strong><span style="font-size:9.0pt"><span style="font-family:"Akzidenz Grotesk BE Regular""><span style="color:#ff8a3b">Page 1</span></span></span></strong><strong><span style="font-size:9.0pt"><span style="font-family:"Akzidenz Grotesk BE Regular""> <span style="color:#3b3838">of the linked PDF contains PROBLEMS & SOLUTIONS.</span></span></span></strong></span></span></p>
<p style="text-align: center;"><span style="font-size:12pt"><span style="font-family:"Times New Roman",serif"><strong><span style="font-size:9.0pt"><span style="font-family:"Akzidenz Grotesk BE Regular""><span style="color:#ff8a3b">Page 2</span><strong> </strong><span style="color:#3b3838">contains ONLY PROBLEMS. </span></span></span><span style="font-size:11.0pt"><span style="color:#3b3838">♦</span></span></strong></span></span></p>
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Mon, 15 Jun 2020 13:13:23 +0000michellelefrancois304784 at https://www.mathcounts.orghttps://www.mathcounts.org/resources/problem-of-the-week/roller-coaster-madness-1#commentsOn the Road Again
https://www.mathcounts.org/resources/problem-of-the-week/road-again-1
<div class="field field-name-field-date-of-problem field-type-datetime field-label-hidden field--date-of-problem"><div class="field-items"><div class="field-item even"><span class="date-display-single" property="dc:date" datatype="xsd:dateTime" content="2020-06-01T10:00:00-07:00">June 1, 2020</span></div></div></div><div class="field field-name-field-problem-category field-type-taxonomy-term-reference field-label-inline clearfix field--problem-category"><div class="field-label">Topics/Content Areas: </div><div class="field-items"><div class="field-item even"><div class="textformatter-list"><a href="/problem-topic-content-area/general-math">General Math</a>, <a href="/problem-topic-content-area/percents-fractions">Percents & Fractions</a>, <a href="/problem-topic-content-area/proportional-reasoning">Proportional Reasoning</a></div></div></div></div><div class="l-constrain l-constrain--prose">
<p>Each summer Radolpho, Lucia and Mya, along with their families, spend two weeks on vacation at their shared lake house.</p>
<p>Mya travels the furthest distance to get to the lake house from her home. Mya’s truck has a fuel tank that holds 12 gallons of gas, and she can drive an average of 22 miles per gallon of gas. If Mya leaves home with a full tank of gas, drives directly to the lake house, and arrives with exactly 1/8 tank of gas remaining, how far does Mya travel from her home to the lake house?</p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><i><span style="font-size:11.0pt"><span style="line-height:107%"><span style="font-family:"Calibri",sans-serif">Since Mya can drive an average of 22 miles per gallon of gas, with 12 gallons of gas, she can travel a total of 12 </span></span></span><span style="font-size:11.0pt"><span style="line-height:107%"><span style="font-family:"Calibri",sans-serif">×</span></span></span><span style="font-size:11.0pt"><span style="line-height:107%"><span style="font-family:"Calibri",sans-serif"> 22 = 264 miles. Since she uses only 1 – 1/8 = 7/8 of the full tank of gas, she must travel only 7/8 of this distance. Therefore, the distance Mya travels from her home to the lake house is (7/8) </span></span></span><span style="font-size:11.0pt"><span style="line-height:107%"><span style="font-family:"Calibri",sans-serif">×</span></span></span><span style="font-size:11.0pt"><span style="line-height:107%"><span style="font-family:"Calibri",sans-serif"> 264 = </span></span></span></i></span><span style="color:#272561;"><i><span style="font-size:11.0pt"><span style="line-height:107%"><span style="font-family:"Calibri",sans-serif"><strong>231 miles</strong></span></span></span></i></span><span style="color:#19a2bf;"><i><span style="font-size:11.0pt"><span style="line-height:107%"><span style="font-family:"Calibri",sans-serif">. </span></span></span></i></span></p>
<p>Typically, it takes Lucia 2 hours to drive 132 miles from her home to the lake house. But this year Lucia is towing a trailer with two jet skis, and the trip takes 20% longer than usual. What is the ratio of Lucia’s typical average speed to Lucia’s average speed when towing the jet skis? Express your answer as a common fraction.</p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><i><span style="font-size:11.0pt"><span style="line-height:107%"><span style="font-family:"Calibri",sans-serif">Towing the jet skis takes 20% longer than the typical 2 hours, or 1.2 × 2 = 2.4 hours. Traveling 132 miles in 2.4 hours means Lucia is traveling at an average speed of 132/2.4 = 55 mi/h. Typically, Lucia can travel 132 miles in 2 hours; that is an average speed of 132/2 = 66 mi/h. The ratio of Lucia’s typical average speed to her average speed when towing the jet skis then is 66/55 = </span></span></span></i></span><span style="color:#272561;"><i><span style="font-size:11.0pt"><span style="line-height:107%"><span style="font-family:"Calibri",sans-serif"><strong>6/5</strong></span></span></span></i></span><span style="color:#19a2bf;"><i><span style="font-size:11.0pt"><span style="line-height:107%"><span style="font-family:"Calibri",sans-serif">. </span></span></span></i></span></p>
<p>At 9:35 a.m. Radolpho passes a highway sign that indicates the next rest area is 36 miles from his current location. The speed limit on the highway is 55 mi/h, but after Radolpho has driven 3/4 of the distance to the rest area, the speed limit is reduced to 40 mi/h due to ongoing construction. If Radolpho does not exceed the speed limit at any time by more than 5 mi/h, what is the earliest time he will reach the rest area?</p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><i><span style="font-size:11.0pt"><span style="line-height:107%"><span style="font-family:"Calibri",sans-serif">If the total distance traveled is 36 miles, we know that 3/4 of that distance is 3/4 × 36 = 27 miles. Since we are told that Radolpho at no time exceeds the speed limit by more than 5 mi/h, he will go a maximum of 55 + 5 = 60 mi/h for the first 27 miles. We also know that he will go a maximum of 40 + 5 = 45 mi/h for the remaining 36 – 27 = 9 miles. If Radolpho travels at a maximum speed of 60 mi/h, that means he is traveling at a maximum speed of 1 mile per minute. Therefore, at that rate, it will take him a minimum of 27 minutes to travel the first 27 miles. Using the formula d = rt, we can determine that the time it takes Radolpho to travel 9 miles at a maximum speed of 45 mi/h is 9 = 45t </span><span style="font-family:Wingdings">à</span></span></span><span style="font-size:11.0pt"><span style="line-height:107%"><span style="font-family:"Calibri",sans-serif"> t = 0.2 hours. That means it will take Radolpho a minimum of 0.2 </span></span></span><span style="font-size:11.0pt"><span style="line-height:107%"><span style="font-family:"Calibri",sans-serif">×</span></span></span><span style="font-size:11.0pt"><span style="line-height:107%"><span style="font-family:"Calibri",sans-serif"> 60 = 12 minutes to travel the final 9 miles to the rest area. It follows that it will take Radolpho a minimum of 27 + 12 = 39 minutes to get from the location of the highway sign to the rest area. If he passes the sign at 9:35 a.m., the earliest he can get to the rest area is 39 minutes later at </span></span></span></i></span><span style="color:#272561;"><i><span style="font-size:11.0pt"><span style="line-height:107%"><span style="font-family:"Calibri",sans-serif"><strong>10:14 a.m.</strong></span></span></span></i></span></p>
<p style="margin-left: 40px; text-align: center;"><span style="font-size:12pt"><span style="font-family:"Times New Roman",serif"><strong><span style="font-size:11.0pt"><span style="font-family:"Segoe UI Symbol",sans-serif"><span style="color:#3b3838">♦</span></span></span></strong><strong> </strong><strong><span style="font-size:9.0pt"><span style="font-family:"Akzidenz Grotesk BE Regular""><span style="color:#ff8a3b">Page 1</span></span></span></strong><strong><span style="font-size:9.0pt"><span style="font-family:"Akzidenz Grotesk BE Regular""> <span style="color:#3b3838">of the linked PDF contains PROBLEMS & SOLUTIONS.</span></span></span></strong></span></span></p>
<p style="margin-left: 40px; text-align: center;"><strong><span style="font-size:9.0pt"><span style="line-height:107%"><span style="font-family:"Akzidenz Grotesk BE Regular""><span style="color:#ff8a3b">Page 2</span></span></span></span></strong><strong> </strong><strong><span style="font-size:9.0pt"><span style="line-height:107%"><span style="font-family:"Akzidenz Grotesk BE Regular""><span style="color:#3b3838">contains ONLY PROBLEMS.</span></span></span></span></strong><strong> </strong><strong><span style="font-size:11.0pt"><span style="line-height:107%"><span style="font-family:"Calibri",sans-serif"><span style="color:#3b3838">♦</span></span></span></span></strong></p>
<p style="text-align: center;"> </p>
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Mon, 08 Jun 2020 13:30:25 +0000michellelefrancois304741 at https://www.mathcounts.orghttps://www.mathcounts.org/resources/problem-of-the-week/road-again-1#commentsThe Unofficial Start of Summer
https://www.mathcounts.org/resources/problem-of-the-week/unofficial-start-summer-0
<div class="field field-name-field-date-of-problem field-type-datetime field-label-hidden field--date-of-problem"><div class="field-items"><div class="field-item even"><span class="date-display-single" property="dc:date" datatype="xsd:dateTime" content="2020-05-25T10:00:00-07:00">May 25, 2020</span></div></div></div><div class="field field-name-field-problem-category field-type-taxonomy-term-reference field-label-inline clearfix field--problem-category"><div class="field-label">Topics/Content Areas: </div><div class="field-items"><div class="field-item even"><div class="textformatter-list"><a href="/problem-topic-content-area/algebraic-expressions-equations">Algebraic Expressions & Equations</a>, <a href="/problem-topic-content-area/probability-counting-combinatorics">Probability, Counting & Combinatorics</a>, <a href="/problem-topic-content-area/proportional-reasoning">Proportional Reasoning</a></div></div></div></div><div class="l-constrain l-constrain--prose">
<p>Memorial Day has always been the unofficial start of summer. Here are some problems about some things we love in hot weather!</p>
<p>Two local supermarkets sell watermelons grown at McDowell’s farm. But Biggy Market sells their watermelons for $3.99 each, regardless of size, whereas Harmless Trail Market sells their watermelons for $0.19 per pound. For what weight will a watermelon at Harmless Trail cost the same as any watermelon at Biggy? If the largest watermelon grown at McDowell’s farm weighs 18 pounds, at which place should you buy your watermelons?</p>
<p style="margin-left: 40px;"><i><span style="color:#19a2bf;">We can rewrite $3.99 as 399 cents. So, we want to know the number of pounds, w, for which 19w = 399. Dividing by 19 on both sides of the equation gives w = </span><span style="color:#272561;"><b>21 pounds</b></span><span style="color:#19a2bf;"> as the weight at which a watermelon will cost the same at both supermarkets. If the largest watermelon grown at McDowell’s farm weighs 18 pounds, then you should buy your watermelons at </span><span style="color:#272561;"><strong>Harmless Trail Market</strong></span><span style="color:#19a2bf;">. </span></i></p>
<p>The SPF (sun protection factor) listed on sunscreen is an indication of how long you will be protected from sunburn when wearing the sunscreen. The amount of time you’re protected is proportional to the SPF. If wearing SPF 8 sunscreen will protect you for 40 minutes, how long will SPF 30 sunscreen protect you?</p>
<p style="margin-left: 40px;"><i><span style="color:#19a2bf;">Because the time protected is proportional to the SPF, the equation 8/40 = 30/t represents the scenario. Cross multiplying gives 8t = 1200, so t = 150 minutes. So, SPF 30 sunscreen will protect you for </span><span style="color:#272561;"><b>2 hours, 30 minutes</b></span><span style="color:#19a2bf;">. </span></i></p>
<p>A soda manufacturer knows that sales of soda increase during the summer. To make sure that they get a large portion of the sales, they decide to offer a contest throughout the spring. The contest is pretty simple: when you open your bottle of soda, if it says, “You win!” under the cap, you get another bottle of their soda for free. The chance of winning a free soda is 1 in 12. Lucky Lucy bought five sodas and won a free one with four of the caps. What’s the probability of that happening? Theoretically, how many sodas would you have to buy before having a greater than 50% chance of winning a free soda?</p>
<p style="margin-left: 40px;"><i><span style="color:#19a2bf;">The probability of winning is 1/12, and the probability of not winning is 11/12. So, the probability of winning exactly four times in five tries is (1/12)<sup>4</sup>(11/12) = </span><span style="color:#272561;"><b>11/248,832</b></span><span style="color:#19a2bf;">, or speaking technically, highly unlikely.</span></i></p>
<p style="margin-left: 40px;"><i><span style="color:#19a2bf;">There is an 11/12 chance of not winning on each bottle. So, the probability of not winning when purchasing n bottles is (11/12)<sup>n</sup>. We need to find the number of n for which (11/12)<sup>n</sup> is less than 50%, because then the probability of winning (1 minus that value) will be greater than 50%. Using a calculator and some guess and check work, it can be found that (11/12)<sup>n</sup> is less than 0.50 when n ></span><span style="color:#19a2bf;"> 7, so a minimum of </span><span style="color:#272561;"><strong>8 bottles</strong></span><span style="color:#19a2bf;"> need to be purchased to have a greater than 50% chance of winning.</span></i></p>
<p> </p>
<p style="text-align: center;"><strong>♦ <span style="color:#f47321;"> Page 1 </span> of the linked PDF contains PROBLEMS & SOLUTIONS.<br>
<span style="color:#f47321;">Page 2 </span> contains ONLY PROBLEMS. ♦ </strong></p>
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Mon, 01 Jun 2020 15:31:25 +0000michellelefrancois304680 at https://www.mathcounts.orghttps://www.mathcounts.org/resources/problem-of-the-week/unofficial-start-summer-0#commentsPlanning for Next Year
https://www.mathcounts.org/resources/problem-of-the-week/planning-next-year-0
<div class="field field-name-field-date-of-problem field-type-datetime field-label-hidden field--date-of-problem"><div class="field-items"><div class="field-item even"><span class="date-display-single" property="dc:date" datatype="xsd:dateTime" content="2020-05-18T12:00:00-07:00">May 18, 2020</span></div></div></div><div class="field field-name-field-problem-category field-type-taxonomy-term-reference field-label-inline clearfix field--problem-category"><div class="field-label">Topics/Content Areas: </div><div class="field-items"><div class="field-item even"><div class="textformatter-list"><a href="/problem-topic-content-area/general-math">General Math</a>, <a href="/problem-topic-content-area/percents-fractions">Percents & Fractions</a></div></div></div></div><div class="l-constrain l-constrain--prose">
<p style="margin-bottom:11px"><span style="font-size:11pt"><span style="line-height:107%"><span style="font-family:Calibri,sans-serif">Before Kyra’s school moved to distance learning because of the COVID-19 outbreak, her school building was overcrowded due to a lot of new registered students. Classroom space was a real issue. In looking ahead to next year, the administration figures that if teachers did not stay in their classrooms during their planning/off periods and allow other teachers to come in to use the room, it might free up space and the administration could hire more teachers. This year each teacher teaches 5 class periods per day, but the school day is 7 full class periods long. If there are 43 teachers in the school, what is the fewest number of classrooms that are needed, assuming any teacher can teach in any classroom during any period of the day and there are no other scheduling restrictions?</span></span></span></p>
<p style="margin-bottom: 11px; margin-left: 40px;"><span style="color:#19a2bf;"><em>If there are 43 teachers and they are each teaching for 5 class periods, that is 43 × 5 = 215 classes being taught. The school day lasts for 7 class periods, so each room can be used for 7 classes. That means that with perfect scheduling, the 215 classes being taught would need 215 ÷ 7 = 30.7 or </em></span><span style="color:#272561;"><em><strong>31 classrooms</strong></em></span><span style="color:#19a2bf;"><em>. </em></span></p>
<p style="margin-bottom:11px"><span style="font-size:11pt"><span style="line-height:107%"><span style="font-family:Calibri,sans-serif">As the administration looks ahead to next year, Kyra is concentrating on finishing up this year well. Students at her school receive 4 quarter grades, a mid-term exam grade and a final exam grade for each class. The quarter grades are all weighted equally, the mid-term exam grade is double the weight of a quarter grade, and the final exam grade is four times the weight of a quarter grade. What fraction of the final course grade is the final exam grade? Express your answer as a common fraction.</span></span></span></p>
<p style="margin-bottom: 11px; margin-left: 40px;"><span style="color:#19a2bf;"><em>The grades are Q, Q, Q, Q, M, F, but each one is not worth 1/6 of the final grade. Since the mid-term exam is worth double a quarter grade, let's replace M with two Qs. We'll underline the Qs so that we remember where they came from. Now, we have Q, Q, Q, Q, <u>Q</u>, <u>Q</u>, F. The final exam grade is worth four times a quarter grade, so let's replace the F with four Qs, but this time, we'll make them un-italicized: Q, Q, Q, Q, <u>Q</u>, <u>Q</u>, </em>Q, Q, Q, Q. <em>Rather than dealing with six scores that had different weights, we now have 10 scores that are all weighted the same. Four of them are representing the final exam grade, so that grade is worth 4/10 or </em></span><span style="color:#272561;"><em><strong>2/5</strong></em></span><span style="color:#19a2bf;"><em> of the final course grade. </em></span></p>
<p style="margin-bottom:11px"><span style="font-size:11pt"><span style="line-height:107%"><span style="font-family:Calibri,sans-serif">Knowing that the final course grade is determined in the manner described above, Kyra can figure out what grade she needs to earn on the final exam to earn at least a 90% for her final course grade. These are her other grades she has earned so far: Q1 = 92%, Q2 = 85%, Q3 = 88%, Q4 = 84%, and mid-term exam = 89%. What is the least whole percent she must earn on the final exam? </span></span></span></p>
<p style="margin-bottom: 11px; margin-left: 40px;"><span style="color:#19a2bf;"><em>Using our solution from the previous problem, we can fill in some numbers. Rather than working with Q, Q, Q, Q, <u>Q</u>, <u>Q</u>, </em>Q, Q, Q, Q, <em>we have 92, 85, 88, 84, <u>89</u>, <u>89</u>, Q, Q, Q, Q. We want the average of these 10 scores to be at least 90, so we have the equation (92 + 85 + 88 + 84 + 89 + 89 + Q + Q + Q + Q) ÷ 10 <span style="font-size:11.0pt"><span style="line-height:107%"><span style="font-family:"Calibri",sans-serif">≥</span></span></span> 90, which simplifies to 527 + 4Q <span style="font-size:11.0pt"><span style="line-height:107%"><span style="font-family:"Calibri",sans-serif">≥</span></span></span> 900, which further simplifies to 4Q <span style="font-size:11.0pt"><span style="line-height:107%"><span style="font-family:"Calibri",sans-serif">≥</span></span></span> 373. Dividing both sides by 4, we see that Q <span style="font-size:11.0pt"><span style="line-height:107%"><span style="font-family:"Calibri",sans-serif">≥</span></span></span> 93.25 or 94, to the nearest whole number that would put the average above 90. Kyra must therefore earn a </em></span><span style="color:#272561;"><em><strong>94% </strong></em></span><span style="color:#19a2bf;"><em>on her final exam to earn a 90% for the final course grade. </em></span></p>
<p style="margin-bottom: 11px; margin-left: 40px; text-align: center;"><span style="font-size:12pt"><span style="font-family:"Times New Roman",serif"><strong><span style="font-size:11.0pt"><span style="font-family:"Segoe UI Symbol",sans-serif"><span style="color:#3b3838">♦</span></span></span></strong><strong> </strong><strong><span style="font-size:9.0pt"><span style="font-family:"Akzidenz Grotesk BE Regular""><span style="color:#ff8a3b">Page 1</span></span></span></strong><strong><span style="font-size:9.0pt"><span style="font-family:"Akzidenz Grotesk BE Regular""> <span style="color:#3b3838">of the linked PDF contains PROBLEMS & SOLUTIONS.</span></span></span></strong></span></span></p>
<p style="margin-bottom: 11px; margin-left: 40px; text-align: center;"><span style="font-size:12pt"><span style="font-family:"Times New Roman",serif"><strong><span style="font-size:9.0pt"><span style="font-family:"Akzidenz Grotesk BE Regular""><span style="color:#ff8a3b">Page 2</span><strong> </strong><span style="color:#3b3838">contains ONLY PROBLEMS. </span></span></span><span style="font-size:11.0pt"><span style="color:#3b3838">♦</span></span></strong></span></span></p>
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Mon, 25 May 2020 13:08:03 +0000michellelefrancois304671 at https://www.mathcounts.orghttps://www.mathcounts.org/resources/problem-of-the-week/planning-next-year-0#comments