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AMC 8 USA Math Olympiad

Calculating the median of observations AMC 8 2014 Problem 24

Try this beautiful problem from AMC 8. It is based on calculating the median of even number of observations. We provide sequential hints so that you can try the problem.

[et_pb_section fb_built=”1″ _builder_version=”4.0″][et_pb_row _builder_version=”3.25″][et_pb_column type=”4_4″ _builder_version=”3.25″ custom_padding=”|||” custom_padding__hover=”|||”][et_pb_text _builder_version=”3.27.4″ text_font=”Raleway|300|||||||” text_text_color=”#ffffff” header_font=”Raleway|300|||||||” header_text_color=”#e2e2e2″ background_color=”#0c71c3″ custom_padding=”20px|20px|20px|20px” border_radii=”on|5px|5px|5px|5px” box_shadow_style=”preset3″]

What are we learning ?

[/et_pb_text][et_pb_text _builder_version=”4.1″ text_font=”Raleway||||||||” text_font_size=”20px” text_letter_spacing=”1px” text_line_height=”1.5em” background_color=”#f4f4f4″ custom_margin=”10px||10px” custom_padding=”10px|20px|10px|20px” hover_enabled=”0″ box_shadow_style=”preset2″]Competency in Focus: Calculating the median of even number of observations  This problem from American Mathematics contest (AMC 8, 2014) is based on maximising median of even number of observations. [/et_pb_text][et_pb_text _builder_version=”3.27.4″ text_font=”Raleway|300|||||||” text_text_color=”#ffffff” header_font=”Raleway|300|||||||” header_text_color=”#e2e2e2″ background_color=”#0c71c3″ custom_padding=”20px|20px|20px|20px” border_radii=”on|5px|5px|5px|5px” box_shadow_style=”preset3″]

First look at the knowledge graph.

[/et_pb_text][et_pb_image src=”https://www.cheenta.com/wp-content/uploads/2020/01/amc-8-2014-problem-number-24-1.png” align=”center” force_fullwidth=”on” _builder_version=”4.0.9″ min_height=”252px” height=”316px” max_height=”411px”][/et_pb_image][et_pb_text _builder_version=”3.27.4″ text_font=”Raleway|300|||||||” text_text_color=”#ffffff” header_font=”Raleway|300|||||||” header_text_color=”#e2e2e2″ background_color=”#0c71c3″ custom_padding=”20px|20px|20px|20px” border_radii=”on|5px|5px|5px|5px” box_shadow_style=”preset3″]

Next understand the problem

[/et_pb_text][et_pb_text _builder_version=”4.0.9″ text_font=”Raleway||||||||” text_font_size=”20px” text_letter_spacing=”1px” text_line_height=”1.5em” background_color=”#f4f4f4″ custom_margin=”10px||10px” custom_padding=”10px|20px|10px|20px” box_shadow_style=”preset2″]One day the Beverage Barn sold $252$ cans of soda to $100$ customers, and every customer bought at least one can of soda. What is the maximum possible median number of cans of soda bought per customer on that day?[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version=”4.0″][et_pb_column type=”4_4″ _builder_version=”3.25″ custom_padding=”|||” custom_padding__hover=”|||”][et_pb_accordion open_toggle_text_color=”#0c71c3″ _builder_version=”4.0.9″ toggle_font=”||||||||” body_font=”Raleway||||||||” text_orientation=”center” custom_margin=”10px||10px”][et_pb_accordion_item title=”Source of the problem” open=”on” _builder_version=”4.0.9″]American Mathematical Contest 2014, AMC 8 Problem 24[/et_pb_accordion_item][et_pb_accordion_item title=”Key Competency” _builder_version=”4.0.9″ open=”off”]Calculation of median for even number of observations .[/et_pb_accordion_item][et_pb_accordion_item title=”Difficulty Level” _builder_version=”4.0.9″ open=”off”]

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Start with hints

[/et_pb_text][et_pb_tabs _builder_version=”4.0.9″][et_pb_tab title=”HINT 0″ _builder_version=”4.0.9″]Do you really need a hint? Try it first![/et_pb_tab][et_pb_tab title=”HINT 1″ _builder_version=”4.0.9″]Suppose the numbers of cans purchased by the 100 customers are listed in increasing order.Now median is the average of the 50th and 51th numbers in the ordered list.  [/et_pb_tab][et_pb_tab title=”HINT 2″ _builder_version=”4.0.9″]Now look -How can you maximise the median ?  [/et_pb_tab][et_pb_tab title=”HINT 3″ _builder_version=”4.0.9″]In order to maximize the median, we need to make the first half of the numbers as small as possible. To minimise the median ,minimise the first 49 numbers by taking them all to be 1.As given  every customer bought at least one can of soda.[/et_pb_tab][et_pb_tab title=”HINT 4″ _builder_version=”4.0.9″] To minimize the first 49, they would each have one can. Subtracting these $49$ cans from the $252$ cans gives us $203$ cans left to divide among $51$ people.If the 50th number is 4,then the sum of all 100 numbers would  at least 49+51.4=253, which is too large .If instead the 50th number is 3 and the following numbers all equal 4 ,then the sum of the 100 numbers is 49+3+50.4=252.Now it’s fine right![/et_pb_tab][et_pb_tab title=”HINT 5″ _builder_version=”4.0.9″]Thus the median is  the average of $3$ and $4$ is $3.5$. Thus our answer is 3.5.[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version=”3.27.4″ text_font=”Raleway|300|||||||” text_text_color=”#ffffff” header_font=”Raleway|300|||||||” header_text_color=”#e2e2e2″ background_color=”#0c71c3″ min_height=”12px” custom_margin=”50px||50px” custom_padding=”20px|20px|20px|20px” border_radii=”on|5px|5px|5px|5px” box_shadow_style=”preset3″]

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Similar Problems

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