Difference between revisions of "2023 AMC 8 Problems/Problem 15"

Revision as of 21:45, 16 February 2023

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 $7/5 \times 3 = 21/5 = \boxed{\textbf{(B)}\ 4.2}$

Solution by ILoveMath31415926535

Solution 3 (Cheap)

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

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/1bA7fD7Lg54?t=1221

@@ Line 52: / Line 52: @@
 ==Video Solution by Magic Square==
 https://youtu.be/-N46BeEKaCQ?t=4153
+==Video Solution by Interstigation==
+https://youtu.be/1bA7fD7Lg54?t=1221
 ==See Also==
 {{AMC8 box|year=2023|num-b=14|num-a=16}}
 {{MAA Notice}}

2023 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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All AJHSME/AMC 8 Problems and Solutions

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