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

Revision as of 12:26, 25 January 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?

$[asy] // Diagram by TheMathGuyd size(8cm); // this is an important stickman to the left of the origin pair C=midpoint((-0.5,0.5)--(-0.6,0.05)); draw((-0.5,0.5)--(-0.6,0.05)); // Head to butt draw((-0.64,0.16)--(-0.7,0.2)--C--(-0.47,0.2)--(-0.4,0.22)); // LH-C-RH draw((-0.6,0.05)--(-0.55,-0.1)--(-0.57,-0.25)); draw((-0.6,0.05)--(-0.68,-0.12)--(-0.8,-0.20)); filldraw(circle((-0.5,0.5),0.1),white,black); int i; real d,s; // gap and side d=0.2; s=1-2*d; for(i=0; i<10; i=i+1) { //dot((i,0), red); //marks to start filldraw((i+d,d)--(i+1-d,d)--(i+1-d,1-d)--(i+d,1-d)--cycle, lightgrey, black); filldraw(conj((i+d,d))--conj((i+1-d,d))--conj((i+1-d,1-d))--conj((i+d,1-d))--cycle,lightgrey,black); } fill((5+d,-d/2)--(6-d,-d/2)--(6-d,d/2)--(5+d,d/2)--cycle,lightred); draw((0,0)--(5,0)--(5,1)--(6,1)--(6,0)--(10.1,0),black); label("School", (10,0),E, Draw()); [/asy]$

$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 4.2 \qquad \textbf{(C)}\ 4.5 \qquad \textbf{(D)}\ 4.8 \qquad \textbf{(E)}\ 5$

Solution 1

Note that Viswam walks at a constant speed $60$ 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 $7 * 60/5 = 84$ blocks per hour to get to school on time, from the time he starts his detour. Since he normally walks $1/2$ mile, which is equal to $10$ blocks, $1$ mile is equal to $20$ blocks. Therefore, he must walk at $84/20 = 4.2$ mph from the time he starts his detour, to get to school on time, so the answer is \boxed{(B)}.

pianoboy (Edits by ILoveMath31415926535

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

Animated Video Solution

https://youtu.be/iSEwbNKrvWw

~Star League (https://starleague.us)

Video Solution by Magic Square

https://youtu.be/-N46BeEKaCQ?t=4153

@@ Line 39: / Line 39: @@
 ==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>7/5 x 3 = 21/5 = \boxed{\textbf{(B)}\ 4.2}</math>
+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>7/5 \times 3 = 21/5 = \boxed{\textbf{(B)}\ 4.2}</math>
 Solution by [[User:ILoveMath31415926535|ILoveMath31415926535]]

2023 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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