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## Distance Time | AIME I, 2012 | Question 4

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2012 based on Distance Time.

## Distance Time – AIME 2012

When they meet at the milepost, Sparky has been ridden for n miles total. Assume Butch rides Sparky for a miles, and Sundance rides for n-a miles. Thus, we can set up an equation, given that Sparky takes $\frac{1}{6}$ hours per mile, Butch takes $\frac{1}{4}$ hours per mile, and Sundance takes $\frac{2}{5}$ hours per mile.

• is 107
• is 279
• is 840
• cannot be determined from the given information

### Key Concepts

Time

Distance

Speed

AIME, 2012, Question 4

Problem Solving Strategies by Arther Engel

## Try with Hints

After meeting at milepost, Sparky for n miles. Let Butch with Sparky for a miles Sundance with Sparky for n-a miles.

Then
$\frac{a}{6} + \frac{n-a}{4}$ = $\frac{n-a}{6} + \frac{2a}{5}$ implies that $a = \frac{5}{19}n$

Then integral value of n is 19 and a = 5 and $t = \frac{13}{3}$ hours that is 260 minutes. Then $19 + 260 = {279}$.

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## Logic and speed | AIME I, 2008 | Question 3

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2008 based on Logic and Speed.

## Logic and Speed – AIME I, 2008

Ed and Sue bike at equal and constant rates and they swim at equal and constant rates. The same way they jog at equal and constant rates. ed covers 74 kms after biking for 2 hrs, jogging for 3 hrs and swimming for 4 hrs while sue covers 91 kms after jogging for 2 hrs swimming for 3 hrs and biking for 4 hrs. Their biking jogging and swimming rates are whole numbers of km/hr, find the sum of the squares of Ed’s biking jogging and swimming rates.

• is 107
• is 314
• is 840
• cannot be determined from the given information

### Key Concepts

Logic

Speed

Integers

AIME I, 2008, Question 3

Elementary Number Theory by David Burton

## Try with Hints

Let a,b, c be biking jogging and swimming rates then 2a+3b+4c=74 first eqn and 4a+2b+3c=91 second eqn subtracting second from first eqn gives 2a-b-c=17 third eqn

third eqn multiplied by 3 + first eqn gives 8a+c=125 gives $a \leq 15$ third eqn multiplied by 4 +first eqn gives 10a-b=142 gives $a \gt 14$

then a=15 and b=8, c=5 and $a^{2} +b^{2} + c^{2}$=225+64+25=314.

Categories

## Competency in Focus: Time and Distance calculation

This problem from American Mathematics Contest 8 (AMC 8, 2018) is based on calculation of time and distance. It is Question no. 6 of the AMC 8 2018 Problem series.[/et_pb_text][et_pb_text _builder_version=”4.2.2″ text_font=”Raleway|300|||||||” text_text_color=”#ffffff” header_font=”Raleway|300|||||||” header_text_color=”#e2e2e2″ background_color=”#0c71c3″ custom_margin=”10px||10px||false|false” custom_padding=”10px|10px|10px|10px|false|false” border_radii=”on|5px|5px|5px|5px” box_shadow_style=”preset3″ inline_fonts=”Aclonica”]

## Next understand the problem

[/et_pb_text][et_pb_text _builder_version=”4.2.2″ 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″]On a trip to the beach, Anh traveled 50 miles on the highway and 10 miles on a coastal access road. He drove three times as fast on the highway as on the coastal road. If Anh spent 30 minutes driving on the coastal road, how many minutes did his entire trip take? $\textbf{(A) }50\qquad\textbf{(B) }70\qquad\textbf{(C) }80\qquad\textbf{(D) }90\qquad \textbf{(E) }100$[/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=”4.3.1″ toggle_font=”||||||||” body_font=”Raleway||||||||” text_orientation=”center” custom_margin=”10px||10px” hover_enabled=”0″][et_pb_accordion_item title=”Source of the problem” open=”on” _builder_version=”4.3.1″ hover_enabled=”0″]

### American Mathematical Contest 2018, AMC 8 Problem 6

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### Basic Time and Distance problem with an easy interpretation from AMC 8 – 2018 – Problem 6

[/et_pb_accordion_item][et_pb_accordion_item title=”Difficulty Level” _builder_version=”4.2.2″ open=”off”]5/10[/et_pb_accordion_item][et_pb_accordion_item title=”Suggested Book” _builder_version=”4.2.2″ open=”off”]Challenges and Thrills in Pre College Mathematics Excursion Of Mathematics

[/et_pb_text][et_pb_tabs _builder_version=”4.2.2″][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=”4.2.2″]Speed = $\frac {distance}{time}$  This can be the first hint for this sum. It is one of the important formula in science. Try to use it in this sum……..[/et_pb_tab][et_pb_tab title=”HINT 2″ _builder_version=”4.2.2″]So if we use the previous hint the speed would be  r = $\frac {d}{t}$ so , r = $\frac {10}{0.5}$    r = 20 mph.[/et_pb_tab][et_pb_tab title=”HINT 3″ _builder_version=”4.2.2″]His speed on the highway then is $60$ mph. He drives $50$ miles, so he drives for $\frac{5}{6}$ hours, which is equal to $50$ minutes. Note : 60 miles\hour is equal to 1 mile\minute[/et_pb_tab][et_pb_tab title=”HINT 4″ _builder_version=”4.2.2″]I think you already got the answer but if not here is the last hint. The total amount of minutes spent on his trip is  =80 minutes[/et_pb_tab][/et_pb_tabs][/et_pb_column][/et_pb_row][/et_pb_section][et_pb_section fb_built=”1″ fullwidth=”on” _builder_version=”4.2.2″ global_module=”50833″][et_pb_fullwidth_header title=”AMC – AIME Program” button_one_text=”Learn More” button_one_url=”https://www.cheenta.com/amc-aime-usamo-math-olympiad-program/” header_image_url=”https://www.cheenta.com/wp-content/uploads/2018/03/matholympiad.png” _builder_version=”4.2.2″ title_level=”h2″ background_color=”#00457a” custom_button_one=”on” button_one_text_color=”#44580e” button_one_bg_color=”#ffffff” button_one_border_color=”#ffffff” button_one_border_radius=”5px”]