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2023 AMC 8 Problems/Problem 15

Revision as of 15:59, 27 December 2023 by London system (talk | contribs) (→‎Solution 2)

Problem

Viswam walks half a mile to get to school each day. His route consists of $10$ city blocks of equal length and he takes $1$ minute to walk each block. Today, after walking $5$ blocks, Viswam discovers he has to make a detour, walking $3$ blocks of equal length instead of $1$ 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] $\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 4.2 \qquad \textbf{(C)}\ 4.5 \qquad \textbf{(D)}\ 4.8 \qquad \textbf{(E)}\ 5$

Solution 1

Note that Viswam walks at a constant speed of $60$ blocks per hour as he takes $1$ minute to walk each block. After walking $5$ blocks, he has taken $5$ minutes, and he has $5$ minutes remaining, to walk $7$ blocks. Therefore, he must walk at a speed of $7 \cdot 60 \div 5 = 84$ blocks per hour to get to school on time, from the time he starts his detour. Since he normally walks $\frac{1}{2}$ mile, which is equal to $10$ blocks, $1$ mile is equal to $20$ blocks. Therefore, he must walk at $84 \div 20 = 4.2$ mph from the time he starts his detour to get to school on time, so the answer is $\boxed{\textbf{(B)}\ 4.2}$.

~pianoboy (Edits by ILoveMath31415926535, apex304 and MrThinker)

Solution 2

Viswam walks $10$ blocks, or half a mile, in $10$ minutes. Therefore, he walks at a rate of $3$ mph. From the time he takes his detour, he must travel $7$ blocks instead of $5$. Our final equation is $\frac{7}{5} \times 3 = \frac{21}{5} = \boxed{\textbf{(B)} \ 4.2 }}$ (Error compiling LaTeX. Unknown error_msg)

-ILoveMath31415926535

Solution 3 (Cheap)

We can cheese this problem.

Notice that Viswam will need to walk $7$ blocks during the second half as opposed to his normal $5$ blocks. Since rate is equal to distance over time, this implies that the final answer will likely be a multiple of $7$, since you will need to convert $7$ blocks to miles. The only answer choice that is a multiple of $7$ is $\boxed{\textbf{(B)}\ 4.2}$.

Solution 4

To travel $\frac{1}{2}$ of a mile in total, each block must be $\frac{1}{20}$ of a mile long. Since Viswam takes $1$ minute to walk along each block, it would take him $10$ minutes normally.

Viswam has already travelled for $5$ mins by the time he encounters the detour, so he must travel $7$ block lengths in the remaining $5$ minutes. The distance he has to travel is $7\cdot \frac{1}{20} = \frac{7}{20}$ of a mile. Therefore,

\[\frac{7}{20} \mathrm{\ miles} = r\cdot 5\mathrm{\ mins}.\] \[\frac{7\mathrm{\ miles}}{100\mathrm{\ mins}} = r.\]

As $60$ mins equals $1$ hour, we set up the following proportion:

\[r = \frac{7\mathrm{\ miles}}{100\mathrm{\ mins}} = \frac{m\mathrm{\ miles}}{60\mathrm{\ mins}}.\]

Cross multiplying and cancelling units yields

\[m = \frac{7\cdot 60}{100},\]

or $\boxed{\textbf{(B)}\ 4.2}$.

-Benedict T (countmath1)

Solution 5

When calculating the speed of his normal route, we get: $\frac{0.5 miles}{10 minutes}= \frac{3 miles}{60 minutes}$

Keep in mind that his route is $10$ blocks long. Since we count $7$ blocks when the detour starts, than this would mean that he has $\frac{7}{10}$ miles left to walk, considering the normal amount of blocks that he usually walks. Multiplying by $6$ to get the rate, we get, $\frac{42}{10}=4.2$

Therefore, the answer is $\boxed{\textbf{(B)}\ 4.2}$.

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

https://youtu.be/iSEwbNKrvWw

~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

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