Difference between revisions of "2023 AMC 8 Problems/Problem 15"
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+ | ==Problem== | ||
+ | Viswam walks half a mile to get to school each day. His route consists of <math>10</math> city blocks of equal length and he takes <math>1</math> minute to walk each block. Today, after walking <math>5</math> blocks, Viswam discovers he has to make a detour, walking <math>3</math> blocks of equal length instead of <math>1</math> block to reach the next corner. From the time he starts his detour, at what speed, in mph, must he walk, in order to get to school at his usual time? | ||
+ | <asy> | ||
+ | // Diagram by TheMathGuyd | ||
+ | size(13cm); | ||
+ | // this is an important stickman to the left of the origin | ||
+ | pair C=midpoint((-0.5,0.5)--(-0.6,0.05)); | ||
+ | draw((-0.5,0.5)--(-0.6,0.05)); // Head to butt | ||
+ | draw((-0.64,0.16)--(-0.7,0.2)--C--(-0.47,0.2)--(-0.4,0.22)); // LH-C-RH | ||
+ | draw((-0.6,0.05)--(-0.55,-0.1)--(-0.57,-0.25)); | ||
+ | draw((-0.6,0.05)--(-0.68,-0.12)--(-0.8,-0.20)); | ||
+ | |||
+ | filldraw(circle((-0.5,0.5),0.1),white,black); | ||
+ | |||
+ | int i; | ||
+ | real d,s; // gap and side | ||
+ | d=0.2; s=1-2*d; | ||
+ | for(i=0; i<10; i=i+1) | ||
+ | { | ||
+ | //dot((i,0), red); //marks to start | ||
+ | filldraw((i+d,d)--(i+1-d,d)--(i+1-d,1-d)--(i+d,1-d)--cycle, lightgrey, black); | ||
+ | filldraw(conj((i+d,d))--conj((i+1-d,d))--conj((i+1-d,1-d))--conj((i+d,1-d))--cycle,lightgrey,black); | ||
+ | } | ||
+ | |||
+ | fill((5+d,-d/2)--(6-d,-d/2)--(6-d,d/2)--(5+d,d/2)--cycle,lightred); | ||
+ | |||
+ | draw((0,0)--(5,0)--(5,1)--(6,1)--(6,0)--(10.1,0),deepblue+linewidth(1.25)); //Who even noticed | ||
+ | label("School", (10,0),E, Draw()); | ||
+ | </asy> | ||
+ | <math>\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 4.2 \qquad \textbf{(C)}\ 4.5 \qquad \textbf{(D)}\ 4.8 \qquad \textbf{(E)}\ 5</math> | ||
+ | |||
+ | (NOTE: THE FOLLOWING DIAGRAM WAS NOT SHOWN DURING THE ACTUAL EXAM, BUT IS NOW HERE TO GUIDE STUDENTS IN PICTURING THE PROBLEM) | ||
+ | |||
+ | ==Solution 1== | ||
+ | Note that Viswam walks at a constant speed of <math>60</math> blocks per hour as he takes <math>1</math> minute to walk each block. After walking <math>5</math> blocks, he has taken <math>5</math> minutes, and he has <math>5</math> minutes remaining, to walk <math>7</math> blocks. Therefore, he must walk at a speed of <math>7 \cdot 60 \div 5 = 84</math> blocks per hour to get to school on time, from the time he starts his detour. Since he normally walks <math>\frac{1}{2}</math> mile, which is equal to <math>10</math> blocks, <math>1</math> mile is equal to <math>20</math> blocks. Therefore, he must walk at <math>84 \div 20 = 4.2</math> mph from the time he starts his detour to get to school on time, so the answer is <math>\boxed{\textbf{(B)}\ 4.2}</math>. | ||
+ | |||
+ | ~pianoboy (Edits by [[User:ILoveMath31415926535|ILoveMath31415926535]], apex304 and MrThinker) | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Viswam walks <math>10</math> blocks, or half a mile, in <math>10</math> minutes. Therefore, he walks at a rate of <math>3</math> mph. From the time he takes his detour, he must travel <math>7</math> blocks instead of <math>5</math>. Our final equation is <math>\frac{7}{5} \times 3 = \frac{21}{5} = \boxed{\textbf{(B)}\ 4.2}</math> | ||
+ | |||
+ | -[[User:ILoveMath31415926535|ILoveMath31415926535]] | ||
+ | |||
+ | Note: in the second half of his journey, he is travelling 7/5 as far as he originally expected to, so he needs to travel at 7/5 the speed in order to get there on time. | ||
+ | |||
+ | —[[User:wescarroll|wescarroll]] | ||
+ | |||
+ | ==Solution 3 (Cheap)== | ||
+ | |||
+ | We can cheese this problem. | ||
+ | |||
+ | Notice that Viswam will need to walk <math>7</math> blocks during the second half as opposed to his normal <math>5</math> blocks. Since rate is equal to distance over time, this implies that the final answer will likely be a multiple of <math>7</math>, since you will need to convert <math>7</math> blocks to miles. The only answer choice that is a multiple of <math>7</math> is <math>\boxed{\textbf{(B)}\ 4.2}</math>. | ||
+ | |||
+ | ==Solution 4== | ||
+ | |||
+ | To travel <math>\frac{1}{2}</math> of a mile in total, each block must be <math>\frac{1}{20}</math> of a mile long. Since Viswam takes <math>1</math> minute to walk along each block, it would take him <math>10</math> minutes normally. | ||
+ | |||
+ | Viswam has already travelled for <math>5</math> mins by the time he encounters the detour, so he must travel <math>7</math> block lengths in the remaining <math>5</math> minutes. The distance he has to travel is <math>7\cdot \frac{1}{20} = \frac{7}{20}</math> of a mile. Therefore, | ||
+ | |||
+ | <cmath>\frac{7}{20} \mathrm{\ miles} = r\cdot 5\mathrm{\ mins}.</cmath> | ||
+ | <cmath>\frac{7\mathrm{\ miles}}{100\mathrm{\ mins}} = r.</cmath> | ||
+ | |||
+ | As <math>60</math> mins equals <math>1</math> hour, we set up the following proportion: | ||
+ | |||
+ | <cmath>r = \frac{7\mathrm{\ miles}}{100\mathrm{\ mins}} = \frac{m\mathrm{\ miles}}{60\mathrm{\ mins}}.</cmath> | ||
+ | |||
+ | Cross multiplying and cancelling units yields | ||
+ | |||
+ | <cmath>m = \frac{7\cdot 60}{100},</cmath> | ||
+ | |||
+ | or <math>\boxed{\textbf{(B)}\ 4.2}</math>. | ||
+ | |||
+ | -Benedict T (countmath1) | ||
+ | |||
+ | ==Solution 5== | ||
+ | When calculating the speed of his normal route, we get: | ||
+ | |||
+ | <math>\frac{0.5 \mathrm{miles}}{10 \mathrm{minutes}}= \frac{3 \mathrm{miles}}{60 \mathrm{minutes}}</math> | ||
+ | |||
+ | Keep in mind that his route is <math>10</math> blocks long. Since we count <math>7</math> blocks when the detour starts, than this would mean that he has <math>\frac{7}{10}</math> miles left to walk, considering the normal amount of blocks that he usually walks. Multiplying by <math>6</math> to get the rate, we get, <math>\frac{42}{10}=4.2</math> | ||
+ | |||
+ | Therefore, the answer is <math>\boxed{\textbf{(B)}\ 4.2}</math>. | ||
+ | |||
+ | ==Solution 6== | ||
+ | |||
+ | We can start by splitting the walk into two sections: The 5-block section before the detour and the 7-block section during and after the detour. | ||
+ | |||
+ | Since Viswam is walking at his normal rate for the first 5, it took him 5 minutes to walk that much. Since it must take him a total of 10 min in order for him to reach on time, he has 5 min for the second section of his walk. | ||
+ | |||
+ | We can figure out that 1 block equals 0.05 mi by dividing 10 by 0.5. Now, we can convert 7 blocks per 5 min into 4.2 miles per 1 hour. So, we arrive at <math>\boxed{\textbf{(B)}\ 4.2}</math>. | ||
+ | |||
+ | ~CoOlPoTaToEs | ||
+ | |||
+ | ==Video Solution by Math-X (Let's first Understand the question)== | ||
+ | https://youtu.be/Ku_c1YHnLt0?si=sWlX8leEM3rALWLK&t=2686 ~Math-X | ||
+ | |||
+ | |||
+ | ==Video Solution (CREATIVE THINKING!!!)== | ||
+ | https://youtu.be/3UrNUd-s59o | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
==Animated Video Solution== | ==Animated Video Solution== | ||
https://youtu.be/iSEwbNKrvWw | https://youtu.be/iSEwbNKrvWw | ||
~Star League (https://starleague.us) | ~Star League (https://starleague.us) | ||
+ | |||
+ | ==Video Solution by Magic Square== | ||
+ | https://youtu.be/-N46BeEKaCQ?t=4153 | ||
+ | ==Video Solution by Interstigation== | ||
+ | https://youtu.be/DBqko2xATxs&t=1626 | ||
+ | |||
+ | ==Video Solution by harungurcan== | ||
+ | https://www.youtube.com/watch?v=otNV1MviJsA&t=20s | ||
+ | |||
+ | ~harungurcan | ||
+ | |||
+ | |||
+ | ==See Also== | ||
+ | {{AMC8 box|year=2023|num-b=14|num-a=16}} | ||
+ | {{MAA Notice}} |
Revision as of 17:51, 11 May 2024
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3 (Cheap)
- 5 Solution 4
- 6 Solution 5
- 7 Solution 6
- 8 Video Solution by Math-X (Let's first Understand the question)
- 9 Video Solution (CREATIVE THINKING!!!)
- 10 Animated Video Solution
- 11 Video Solution by Magic Square
- 12 Video Solution by Interstigation
- 13 Video Solution by harungurcan
- 14 See Also
Problem
Viswam walks half a mile to get to school each day. His route consists of city blocks of equal length and he takes
minute to walk each block. Today, after walking
blocks, Viswam discovers he has to make a detour, walking
blocks of equal length instead of
block to reach the next corner. From the time he starts his detour, at what speed, in mph, must he walk, in order to get to school at his usual time?
(NOTE: THE FOLLOWING DIAGRAM WAS NOT SHOWN DURING THE ACTUAL EXAM, BUT IS NOW HERE TO GUIDE STUDENTS IN PICTURING THE PROBLEM)
Solution 1
Note that Viswam walks at a constant speed of blocks per hour as he takes
minute to walk each block. After walking
blocks, he has taken
minutes, and he has
minutes remaining, to walk
blocks. Therefore, he must walk at a speed of
blocks per hour to get to school on time, from the time he starts his detour. Since he normally walks
mile, which is equal to
blocks,
mile is equal to
blocks. Therefore, he must walk at
mph from the time he starts his detour to get to school on time, so the answer is
.
~pianoboy (Edits by ILoveMath31415926535, apex304 and MrThinker)
Solution 2
Viswam walks blocks, or half a mile, in
minutes. Therefore, he walks at a rate of
mph. From the time he takes his detour, he must travel
blocks instead of
. Our final equation is
Note: in the second half of his journey, he is travelling 7/5 as far as he originally expected to, so he needs to travel at 7/5 the speed in order to get there on time.
Solution 3 (Cheap)
We can cheese this problem.
Notice that Viswam will need to walk blocks during the second half as opposed to his normal
blocks. Since rate is equal to distance over time, this implies that the final answer will likely be a multiple of
, since you will need to convert
blocks to miles. The only answer choice that is a multiple of
is
.
Solution 4
To travel of a mile in total, each block must be
of a mile long. Since Viswam takes
minute to walk along each block, it would take him
minutes normally.
Viswam has already travelled for mins by the time he encounters the detour, so he must travel
block lengths in the remaining
minutes. The distance he has to travel is
of a mile. Therefore,
As mins equals
hour, we set up the following proportion:
Cross multiplying and cancelling units yields
or .
-Benedict T (countmath1)
Solution 5
When calculating the speed of his normal route, we get:
Keep in mind that his route is blocks long. Since we count
blocks when the detour starts, than this would mean that he has
miles left to walk, considering the normal amount of blocks that he usually walks. Multiplying by
to get the rate, we get,
Therefore, the answer is .
Solution 6
We can start by splitting the walk into two sections: The 5-block section before the detour and the 7-block section during and after the detour.
Since Viswam is walking at his normal rate for the first 5, it took him 5 minutes to walk that much. Since it must take him a total of 10 min in order for him to reach on time, he has 5 min for the second section of his walk.
We can figure out that 1 block equals 0.05 mi by dividing 10 by 0.5. Now, we can convert 7 blocks per 5 min into 4.2 miles per 1 hour. So, we arrive at .
~CoOlPoTaToEs
Video Solution by Math-X (Let's first Understand the question)
https://youtu.be/Ku_c1YHnLt0?si=sWlX8leEM3rALWLK&t=2686 ~Math-X
Video Solution (CREATIVE THINKING!!!)
~Education, the Study of Everything
Animated Video Solution
~Star League (https://starleague.us)
Video Solution by Magic Square
https://youtu.be/-N46BeEKaCQ?t=4153
Video Solution by Interstigation
https://youtu.be/DBqko2xATxs&t=1626
Video Solution by harungurcan
https://www.youtube.com/watch?v=otNV1MviJsA&t=20s
~harungurcan
See Also
2023 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.