MATHCOUNTS
https://www.mathcounts.org
enCentral IN Chapter Competition
https://www.mathcounts.org/central-chapter-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>Wed, 21 Aug 2019 13:59:49 +0000KCottrell243668 at https://www.mathcounts.orghttps://www.mathcounts.org/central-chapter-competition#commentsSummer Plans
https://www.mathcounts.org/resources/problem-of-the-week/summer-plans
<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="2019-08-19T11:45:00-07:00">August 19, 2019</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/general-math">General Math</a></div></div></div></div><div class="l-constrain l-constrain--prose">
<p>After polling each student in her homeroom about their summer plans, Mrs. Baker discovered that each student planned to participate in one or more of the following activities:</p>
<ol><li>Attend summer school</li>
<li>Work part-time</li>
<li>Attend sports camp</li>
</ol><p>Mrs. Baker learned that 17 of the students in her homeroom plan to participate in at least two of these activities, and exactly two students plan to participate in all three activities this summer. If the number of students participating in exactly two of these activities is equally divided among the three pairs of activities, how many students plan only to attend sports camp and work part-time this summer?</p>
<p> </p>
<p>If 1/4 of Mrs. Baker’s homeroom students who plan to attend summer school do not plan to participate in either of the two other activities, how many students plan only to attend summer school?</p>
<p> </p>
<p>According to this poll, 16 of Mrs. Baker’s homeroom students plan to participate in only one of these activities. If the number of students who plan only to attend sports camp is three times the number of students who plan only to work part-time, what is the total number of students in Mrs. Baker’s homeroom?</p>
<p> </p>
<div style="margin-left:-13.5pt;">
<p align="center"><strong>CHECK THE </strong><a href="https://www.mathcounts.org/resources/problem-archive"><strong>PROBLEM OF THE WEEK ARCHIVE</strong></a><br /><br /><strong>FOR SOLUTIONS TO PREVIOUS PROBLEMS</strong></p>
</div>
<p> </p>
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Mon, 19 Aug 2019 15:24:46 +0000michellelefrancois243505 at https://www.mathcounts.orghttps://www.mathcounts.org/resources/problem-of-the-week/summer-plans#commentsConstruction Projects
https://www.mathcounts.org/resources/problem-of-the-week/construction-projects-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="2019-08-12T09:45:00-07:00">August 12, 2019</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/logic">Logic</a></div></div></div></div><div class="l-constrain l-constrain--prose">
<p>A particular construction crew places orange barrels on both sides of a road that is under construction such that the centers of adjacent barrels on the same side of the road are 15 feet apart. If the crew does this for a 1.5 mile stretch of roadway, how many barrels will be placed on the two sides of the road in total?</p>
<div>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>There are 1.5 </em>×<em> 5280 = 7920 feet on each side of the road. We start with one barrel at the "zero" mark and then add 920 </em><em>÷</em><em> 15 = 528 barrels on each side of the road, for a total of 528 + 1 = 529 barrels per side. So for both sides there are a total of 2 </em>× <em>529 = </em></span><span style="color:#272561;"><em><strong>1058</strong> barrels</em></span><span style="color:#19a2bf;"><em>.</em></span></p>
</div>
<p>When planning how long this project will take, the construction company considers that it took 5 workers 7 days, while working as quickly as possible, to complete the same job on a 2640-foot stretch of road way. If they want this job (on a 1.5 mile stretch of road) completed in 7 days, what is the minimum number of workers they will need?</p>
<div>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>Notice that 2640 ft is equivalent to 0.5 miles. Thus, the 1.5 mile job is 3 times as long and will require 3 times as many workers to complete it in the same amount of time. Therefore, it will take 5 </em>×<em>3 = </em></span><span style="color:#272561;"><em><strong>15</strong> workers</em></span><span style="color:#19a2bf;"><em>.</em></span></p>
</div>
<p>If only 5 workers were available for the first 2 days, how many additional workers will be needed during the last 5 days so that the job can be completed on time?</p>
<div>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>Each worker works at a rate of 2640 </em><em>÷</em><em> (5 </em><em>×</em><em> 7) = 528/7 ft completed per day. Thus during the first 2 days, (528/7)(5 workers)(2 days) = 5280/7 ft will be completed if only 5 workers are working. The 1.5 mile stretch is equivalent to 1.5 </em><em>×</em><em> 5280 = 7920 ft, thus after the first 2 days there are 7920 – (5280/7) = 50,160/7 ft left to complete in 5 days. That means that (50,160/7)/5 = 10,032/7 ft must be completed each day. Since each worker works at a rate of 528/7 feet per day, they will need a total of (10,032/7)/(528/7) = 19 people working for the last 5 days to complete the job on time. They already have 5 people on site, to that is an additional 19 – 5 = </em></span><span style="color:#272561;"><em><strong>14</strong> people</em></span><span style="color:#19a2bf;"><em>.</em></span></p>
</div>
<p> </p>
<div style="margin-left:-13.5pt;">
<p align="center"> <strong>♦</strong> <span style="color:#ff8c00;"><strong>Page 1</strong></span><strong> of the linked PDF contains PROBLEMS & SOLUTIONS.</strong></p>
<p align="center"><span style="color:#ff8c00;"><strong>Page 2</strong></span> <strong>contains ONLY PROBLEMS.</strong> <strong>♦</strong></p>
</div>
<p style="text-align: center;"> </p>
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Mon, 19 Aug 2019 14:52:35 +0000michellelefrancois243501 at https://www.mathcounts.orghttps://www.mathcounts.org/resources/problem-of-the-week/construction-projects-0#commentsAlgenite (2019 Winner)
https://www.mathcounts.org/resources/video-library/math-video-challenge/algenite-2019-winner
<div class="field field-name-field-video-embed-code field-type-video-embed-field field-label-hidden field--video-embed-code"><div class="field-items"><div class="field-item even">
<div class="embedded-video">
<div class="player">
<iframe class="" width="455px" height="260px" src="//www.youtube.com/embed/jojcSDpItcg?width%3D455px%26amp%3Bheight%3D260px%26amp%3Btheme%3Ddark%26amp%3Bautoplay%3D0%26amp%3Brel%3D0%26amp%3Bshowinfo%3D1%26amp%3Bmodestbranding%3D1%26amp%3Biv_load_policy%3D1%26amp%3Bautohide%3D2%26amp%3Bvq%3Dhd720%26amp%3Bwmode%3Dopaque" frameborder="0" allowfullscreen></iframe> </div>
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<p>When four students accidentally switch minds with each other and get trapped inside the video game <em>Algenite</em>, they must work together to create a potion that will switch everything back to normal. Will their problem-solving skills succeed or will their “switcharoo” become permanent?</p>
<p>"Algenite," the First Place Video in the 2018-2019 Math Video Challenge, was created by Team Switcharoo Crew of Georgia.</p>
<p><a href="https://www.mathcounts.org/math-video-challenge-0">Learn more about the Math Video Challenge here</a>.</p>
</div>
<div class="field field-name-field-video-category field-type-taxonomy-term-reference field-label-inline clearfix field--video-category"><div class="field-label">Video Category: </div><div class="field-items"><div class="field-item even"><a href="/resources/video-library/math-video-challenge">Math Video Challenge</a></div></div></div>Tue, 06 Aug 2019 20:27:35 +0000amanda.naar242864 at https://www.mathcounts.orghttps://www.mathcounts.org/resources/video-library/math-video-challenge/algenite-2019-winner#commentsGrowing Pains
https://www.mathcounts.org/resources/problem-of-the-week/growing-pains
<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="2019-07-29T08:00:00-07:00">July 29, 2019</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/measurement">Measurement</a>, <a href="/problem-topic-content-area/statistics-data">Statistics & Data</a></div></div></div></div><div class="l-constrain l-constrain--prose">
<div style="float:right;"> </div>
<p>On September 28th, at Ciara’s appointment for her annual physical exam, she measured 3 feet 11.5 inches tall. At her last appointment on May 31st, she measured 4 feet 3 inches. Based on this, what was the average number of inches Ciara grew each day from September 28th to May 31st? Express your answer as a common fraction.</p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>First let’s determine how many inches Ciara grew from September 28th to May 31st. We know that 3 feet 11½ inches is ½ inch less than 4 feet. So, Ciara grew ½ + 3 = 3½ inches. We are asked to determine the average daily rate of growth. Thus, we need to count the number of days from September 28th to May 31st. There are 2 more days in September. There are 31 days in each of October, December, January, March and May. November and April each have 30 days and February has 28 days, assuming it’s not a leap year. That adds up to 2 + 5(31) + 2(30) + 28 = 245 days. Ciara grew, on average, 3½ <span style="color:#19a2bf;"><em>÷</em></span> 245 = 7/2 × 1/245 = </em></span><span style="color:#272561;"><em><strong>1/70 </strong></em></span><span style="color:#19a2bf;"><em>inch per day. (Note: The difference in the average daily inches of growth in a non-leap year and a leap year is neglegible since 3.5/245 = 1/70 ≈ 0.143, and 3.5/246 = 7/492 ≈ 0.142. The exact difference is 1/70 - 7/492 = 246/17,220 - 245/17,220 = 1/17,220 inch.)</em></span></p>
<p>School ends on June 20th this year, and Ciara wants to ride the Shocker roller coaster at least once before returning to school in the fall. Unfortunately, that roller coaster has a minimum height restriction of 4.5 feet. Assuming that Ciara maintained the same average rate of growth as in the previous problem from June 1st to June 20th, how many more inches does she need to grow during her summer break in order to meet the minimum height restriction and ride the Shocker? Express your answer as a mixed number in simplest form.</p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>On May 31st Ciara’s height was 4 feet 3 inches. From the date of her last doctor’s visit to the last day of school there are 20 full days (1 full day = 1 twenty-four hour period) so, using the result from the previous problem, we can determine that she will have grown 1/70 (20) = 20/70 = 2/7 inch during that time. Thus, when school ends her heigh will be 4 feet 3 2/7 inches. Since the minimum height required to ride the shocker is 4 feet 6 inches and since 3 2/7 inches = 23/7 inches, Ciara will need to grow an additional 6 – 23/7 = 42/7 – 23/7 = 19/7 = </em></span><span style="color:#272561;"><em><strong>2 5/7</strong></em></span><span style="color:#19a2bf;"><em> inches.</em></span></p>
<p>Ciara really wants to ride the Shocker before school begins on September 6th. Maggie told Ciara that her brother’s girlfriend’s best friend’s sister said that practicing Yoga makes you taller. Assume that Ciara grows the expected number of inches from June 20th to September 5th. How many more inches will Ciara need to grow from practicing Yoga to meet the minimum height restrictions and ride the Shocker before September 6th? Express your answer as a decimal to the nearest tenth.</p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>From the last day of school to the last day of summer break there are 10 + 2(31) + 5 = 77 full days. Again, using the result of the first problem we see that during that time, Ciara can expect to grow 1/70 × 77 = 77/70 = 1.1 inches. We also know from the previous problem that she needs to increase her height by 2 5/7≈ 2.714286 inches to meet the minimum height required to ride the Shocker. So, she needs practicing yoga to help her grow an additional 2.714286 – 1.1 = 1.614286 ≈ </em></span><span style="color:#272561;"><em><strong>1.6</strong></em></span><span style="color:#19a2bf;"><em> inches.</em></span></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, 29 Jul 2019 16:03:52 +0000kera.johnson242738 at https://www.mathcounts.orghttps://www.mathcounts.org/resources/problem-of-the-week/growing-pains#commentsVacation Plans
https://www.mathcounts.org/resources/problem-of-the-week/vacation-plans-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="2019-07-15T12:15:00-07:00">July 15, 2019</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/probability-counting-combinatorics">Probability, Counting & Combinatorics</a></div></div></div></div><div class="l-constrain l-constrain--prose">
<p>The Walton and the Brady families plan to vacation together in an 8-bedroom vacation home. John and Olivia will bring six children — boys, Jason, Ben and Jim-Bob, and girls, Mary Ellen, Erin and Elizabeth. Mike and Carol also will bring six children — boys, Greg, Peter and Bobby, and girls, Marcia, Jan and Cindy. John and Olivia will sleep in one of the bedrooms, and Mike and Carol will stay in another bedroom. Each of the remaining six bedrooms each will be occupied by a pair of children, with six boys using three bedrooms and six girls using three bedrooms.</p>
<p>To assign roommates, Mr. Brady has written the numbers 1, 2, 3 and the letters A, B, C on six slips of paper. After the slips are folded and placed in a bowl, each of the boys will randomly select a folded paper from the bowl, without replacement, and the roommate pairings will be 1 with A, 2 with B and 3 with C. All six slips, then, will be refolded and returned to the bowl so that this process can be repeated for the girls.</p>
<p>Using this method, how many distinct sets of roommate pairings are possible for the boys?</p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>Each distinct set of roommate pairings will consist of 3 pairs of boys. There are <sub>6</sub>C<sub>2</sub> = 15 ways of selecting the first roommate pairing. There are <sub>4</sub>C<sub>2</sub> = 6 ways of selecting the second pairing. That leaves just <sub>2</sub>C<sub>2</sub> = 1 option for the last pairing of the set. That means there are 15 × 6 × 1 = 90 sets of roommate pairings. But they are not all distinct because each distinct set of pairings can occur in 3! = 6 different ways. So there are 90 ÷ 6 = </em></span><span style="color:#272561;"><em><strong>15</strong> </em></span><span style="color:#19a2bf;"><em>distinct pairings.</em></span></p>
<p>What is the percent probability that Jim-Bob will <em>not</em> room with one of his brothers?</p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>Jim-Bob is equally likely to be paired with any of the other five boys. Three of those boys are not his brothers. So, the probability that Jim-Bob will <strong>not </strong>room with one of his brothers is 3/5 = </em></span><span style="color:#272561;"><em><strong>60</strong></em></span><span style="color:#19a2bf;"><em>%.</em></span></p>
<p>Marcia and Mary Ellen both are 13 years old, and Erin, Jan, Cindy and Elizabeth are 12, 10, 8 and 6 years old, respectively. In what fraction of the distinct roommate pairings for the girls is the sum of their ages is 23? Express your answer as a common fraction.</p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>There are <sub>6</sub>C<sub>2</sub> = 15 distinct roommate pairings for the girls. The pairing of Marcia with Jan and the pairing of Mary Ellen and Jan are the two roommate pairings for which the sum of their ages is 23. Those two pairings represent<strong> </strong></em></span><span style="color:#272561;"><em><strong>2/15</strong></em></span><span style="color:#19a2bf;"><em> of the distinct roommate pairings for the girls.</em></span></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, 15 Jul 2019 13:57:15 +0000kera.johnson242547 at https://www.mathcounts.orghttps://www.mathcounts.org/resources/problem-of-the-week/vacation-plans-0#commentsSummer Jobs
https://www.mathcounts.org/resources/problem-of-the-week/summer-jobs-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="2019-07-08T06:45:00-07:00">July 8, 2019</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/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>This summer Owen and Jack will both have part time jobs. Owen will be working as a lifeguard and Jack will be working as a server at a local restaurant. Owen plans to work 15 hours each week and will be making $7 per hour. Jack will make an hourly rate of $2.10 plus 18% of his sales in tips. If Jack averages $30 in sales per hour, how many hours will he need to work in order to make the same amount per week as Owen?</p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>Owen will make $7/hour </em><em>×</em><em> 15 hours/week = $105/week. Jack will make $2.10/hour + 0.18 </em><em>×</em><em> $30/hour = $7.50/hour. In order to make the same amount per week as Owen, Jack will need to work $105/week ÷ $7.50/hour = </em></span><span style="display: none;"> </span><span style="color:#272561;"><strong><em><span id="cke_bm_610S" style="display: none;"> </span>14 </em></strong></span><span style="display: none;"> </span><span style="color:#19a2bf;"><em><span id="cke_bm_610E" style="display: none;"> </span>hours/week.</em></span></p>
<p>Owen and Jack will both work 15-hour weeks, but Jack’s paychecks will be reduced by 6% for state income tax. What will be the absolute difference in Owen and Jack’s weekly pay?</p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>Owen will be making $105 per week, as calculated in the previous solution. Jack will be making $7.50/hour </em><em>×</em><em> 15 hours = $112.50 but will then pay a 6% state income tax. Jack will actually take home (1 − 0.06) </em><em>×</em><em> $112.50 = 0.94 </em><em>×</em><em> $112.50 = $105.75 per week. Jack will make $<span id="cke_bm_611S" style="display: none;"> </span><span id="cke_bm_614S" style="display: none;"> </span></em></span><span style="color:#272561;"><strong><em><span id="cke_bm_616S" style="display: none;"> </span>0.75<span id="cke_bm_616E" style="display: none;"> </span><span id="cke_bm_614E" style="display: none;"> </span><span id="cke_bm_611E" style="display: none;"> </span></em></strong></span><span style="color:#19a2bf;"> <em>more per week than Owen.</em></span></p>
<p>At the end of the summer, Jack and Owen each will have worked for 12 weeks. They plan to combine their earnings in a joint account to save to buy a used car when they get their licenses. If the account earns 4% interest annually, what is the total amount Jack and Owen will have in their savings account in two years when they get their licenses?</p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>Jack will earn $105.75 per week, and Owen will earn $105 per week. Collectively, at the end of the summer, they will have earned 12 </em><em>×</em><em> ($105.75 + $105) = 12 </em><em>×</em><em> $210.75 = $2529. After one year they will have 1.04 </em><em>× </em><em>$2529 = $2630.16 in their account. After two years they will have 1.04 </em><em>× </em><em>$2630.16 = $</em></span><span style="display: none;"> </span><strong><em><span id="cke_bm_612S" style="display: none;"> </span><span id="cke_bm_613S" style="display: none;"> </span><span id="cke_bm_617S" style="display: none;"> </span>2<span style="color:#272561;">735.37</span><span id="cke_bm_613E" style="display: none;"> </span><span id="cke_bm_612E" style="display: none;"> </span></em></strong><span style="display: none;"> </span><span style="color:#19a2bf;"><em><span id="cke_bm_617E" style="display: none;"> </span>.</em></span></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, 08 Jul 2019 15:39:13 +0000kera.johnson242482 at https://www.mathcounts.orghttps://www.mathcounts.org/resources/problem-of-the-week/summer-jobs-0#commentsIndependence Day 2019
https://www.mathcounts.org/resources/problem-of-the-week/independence-day-2019-0
<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="2019-07-01T07:45:00-07:00">July 1, 2019</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/problem-solving-misc">Problem Solving (Misc.)</a></div></div></div></div><div class="l-constrain l-constrain--prose">
<p>Thursday is the 4th of July... here are a few problems relating to Independence Day!! While walking around at their town’s Independence Day Festival, Latrease and Hannah decide to buy lunch. Latrease bought two hot dogs and a soda for a total of $7.50. Hannah bought a hot dog, a soda, and a bag of chips for $5.50. If a bag of chips costs $1.00, how much does one hot dog cost?</p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>Let h, s and c represent the costs of a hot dog, a soda and a bag of chips, respectively. From the problem we can set up the following equations: 2h + s = 7.50 and h + s + c = 5.50. We are told that a bag of chips costs one dollar, so c = 1. Substituting, we get h + s + 1 = 5.50, and h + s = 4.50. Now we can subtract the equations 2h + s = 7.50 and h + s = 4.50 to get 2h + s - (h + s) = 7.50 - 4.50 <span style="color:#19a2bf;"><em>→</em></span> h = 3.00. Thus, one hot dog costs $</em></span><strong><span style="color:#272561;"><em>3.00</em></span></strong><span style="color:#19a2bf;"><em>.</em></span></p>
<p>Frank is going to put on a fireworks show for his neighborhood. He bought 15 different fireworks to set off. If he plans to end with a specific sequence of five fireworks and he only wants to set off one firework at a time, how many distinct orders are there in which he could set off the fireworks?</p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>Since he has already determined a specific sequence for the last 5 fireworks, there are 10 slots within the order to be filled. Thus, there are a total of 10! = 10(9)(8)(7)(6)(5)(4)(3)(2)(1) = </em></span><strong><span style="color:#272561;"><em>3,628,800</em></span></strong><span style="color:#19a2bf;"><em> orders in which the fireworks could be set off.</em></span></p>
<p>Victoria and Hugo bought 42 cups of lemonade for the 10 guests they expected to have at their 4th of July party plus themselves. If 13 guests end up coming to the party, what is the minimum number of additional cups of lemonade that Victoria and Hugo must buy in order to have the planned amount of lemonade available to each guest and themselves? Express your answer as a decimal to the nearest tenth.</p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>The original amount of lemonade Victoria and Hugo wanted to have available for each person was 42/(10 + 2) = 3.5 cups. If three additional people showed up they would need an additional (3.5)(3) = </em></span><strong><span style="color:#272561;"><em>10.5</em></span></strong><span style="color:#19a2bf;"><em> cups of lemonade.</em></span></p>
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Mon, 01 Jul 2019 15:46:46 +0000kera.johnson242444 at https://www.mathcounts.orghttps://www.mathcounts.org/resources/problem-of-the-week/independence-day-2019-0#commentsThirst Quencher
https://www.mathcounts.org/resources/problem-of-the-week/thirst-quencher-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="2019-06-17T06:15:00-07:00">June 17, 2019</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/measurement">Measurement</a>, <a href="/problem-topic-content-area/proportional-reasoning">Proportional Reasoning</a></div></div></div></div><div class="l-constrain l-constrain--prose">
<p>In the summer months, a common problem people suffer from is dehydration – not enough water in their body. Waiting until you feel thirsty is not a good indicator that it’s time to drink water. In fact, once thirst has settled in, you are already dehydrated! A general guideline is that people should drink eight 8-ounce glasses of water per day. According to this guideline, how many gallons of water should a person drink in a day, given that 1 gallon = 4 quarts and 1 quart = 32 ounces? Express your answer as a decimal to the nearest tenth.</p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>Eight 8-ounce glasses of water 8 <em><em>×</em></em> 8 = 64 ounces of water. Since it takes exactly double that amount, or 128 ounces, to make a gallon, the guideline for water consumption is </em></span><span style="color:#272561;"><em><strong>0.5</strong></em></span><span style="color:#19a2bf;"><em> gallons of water per day.</em></span></p>
<p>When dehydrated, your body needs water. Juices and sodas are not the optimal beverages for replenishing your body with the water it needs. And if you take cost into account, you’ll definitely want to grab a glass of water! It is estimated that 4000 glasses of tap water cost the same as a six-pack of soda. If a six-pack of soda costs $2.99, how many glasses of water would have a cost of 10¢? Express your answer to the nearest whole number.</p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>If a six-pack of soda costs $2.99, then a glass of water would cost 2.99 <em>÷</em> 4000 = 0.0007475 cents. To determine how many glasses have a cost of 10 cents, we multiply the ratio of glasses to cents by 0.10 to get (1/0.0007475)(0.10) ≈ 133.779. So, the number of glasses of water that have a cost of 10 cents is about </em></span><span style="color:#272561;"><em><strong>134</strong> </em></span><span style="color:#19a2bf;"><em>glasses of water.</em></span></p>
<p>Lake Tahoe is the second deepest lake in the U.S. and it holds 40 trillion gallons of water – enough to cover the state of California to a depth of 14 inches! Given that 1 ft<sup>3</sup> = 7.48 gallons and 1 mile = 5280 ft, how many square miles are in the area of California? Express your answer to the nearest thousand.</p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>First, 40 trillion gallons of water is equivalent to [(40 <em><em>×</em></em> 10<sup>12</sup>)/7.48] = [(10 <em><em>×</em></em> 10<sup>12</sup>)/1.87] ft<sup>3</sup> of water. Next, we are told that this water would have depth 14 inches <em><em>×</em></em> 1/12 = 7/6 feet. At that depth, [(10 <em><em>×</em></em> 10<sup>12</sup>)/1.87] ft<sup>3</sup> of water would cover [((10 <em><em>×</em></em> 10<sup>12</sup>)/1.87)/(7/6)] = [((10 <em><em>×</em></em> 10<sup>12</sup>)/1.87)(6/7)] ft<sup>2</sup>. Finally, since (1 mi)<sup>2</sup> = (5280 ft)<sup>2</sup>, it follows that the number of square miles covered would be [((10 <em>×</em> 10<sup>12</sup>)/1.87)(6/7)]/(5280)<sup>2</sup> ≈ 164,415,8790 mi<sup>2</sup>, which means that the area of California is about </em></span><span style="color:#272561;"><em><strong>164,000</strong> </em></span><span style="color:#19a2bf;"><em>mi<sup>2</sup>.</em></span></p>
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Mon, 24 Jun 2019 16:02:00 +0000kera.johnson242404 at https://www.mathcounts.orghttps://www.mathcounts.org/resources/problem-of-the-week/thirst-quencher-0#commentsCamsie's New Job
https://www.mathcounts.org/resources/problem-of-the-week/camsies-new-job-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="2019-06-24T07:45:00-07:00">June 24, 2019</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/number-theory">Number Theory</a></div></div></div></div><div class="l-constrain l-constrain--prose">
<p>Camsie is starting a new job where she earns <em>n</em> dollars on her<em> n</em>th day of work. What is the total amount that Camsie will earn for the first ten days of work at her new job?</p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>On each of the first 10 days of work at her new job, Camsie will earn 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10 dollars, respectively. That’s a total of 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 10(1 + 10)/2 = 5(11) = </em></span><span style="color:#272561;"><em><strong><em>55 </em></strong></em></span><span style="color:#19a2bf;"><em>dollars.</em></span></p>
<p>How many days will Camsie have worked at her new job when her cumulative earnings total 595 dollars?</p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>Camsie's cumulative earnings for working the first n days at her new job equal the sum of the integers from 1 to n. Since 1 + 2 + 3 + <span style="color:#19a2bf;"><em>…</em></span> + n = (n/2)(n + 1), we can solve the equation (n/2)(n + 1) = 595 </em>→<em> n(n + 1) = 1190 →</em><em> n<sup>2</sup> + n – 1190 = 0. Factoring the quadratic yields (n – 34)(n + 35) = 0. So, n – 34 = 0 </em>→<em> n = 34 and n + 35 = 0 </em>→<em> n = –35. Therefore, Camsie’s cumulative earnings will total 595 dollars when she has worked </em></span><span style="color:#272561;"><em><strong><em>34 </em></strong></em></span><span style="color:#19a2bf;"><em>days.</em></span></p>
<p>What is the minimum number of days Camsie must work at her new job for her cumulative earnings to exceed 1000 dollars?</p>
<p style="margin-left: 40px;"><span style="color:#19a2bf;"><em>Using the same expression for the sum of the integers from 1 to n, we have the following inequality: (n/2)(n + 1) > 1000 </em>→ <em>n(n + 1) > 2000. Simplifying yields the quadratic expression n<sup>2</sup> + n – 2000, which cannot be factored. However, we are essentially</em><em> looking for two consecutive integers, n and n + 1, that have a product greater than 2000. Since 40<sup>2</sup> = 1600 and 50<sup>2</sup> = 2500, we know the integers are between 40 and 50. The square of 45 is 2025, so we check 44 × 45 and 45 × 46 and get 1980 and 2070, respectively. Therefore, we conclude that in order for Camsie’s cumulative earnings to exceed 1000 dollars, she must work at least </em></span><span style="color:#272561;"><em><strong><em>45 </em></strong></em></span><span style="color:#19a2bf;"><em>days.</em></span></p>
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Mon, 24 Jun 2019 15:25:41 +0000kera.johnson242402 at https://www.mathcounts.orghttps://www.mathcounts.org/resources/problem-of-the-week/camsies-new-job-0#comments