Difference between revisions of "2010 AMC 10A Problems/Problem 13"

m (Problem)
m
 
(5 intermediate revisions by 3 users not shown)
Line 1: Line 1:
==Problem==
+
== Problem ==
 +
 
 
Angelina drove at an average rate of <math>80</math> kmh and then stopped <math>20</math> minutes for gas. After the stop, she drove at an average rate of <math>100</math> kmh. Altogether she drove <math>250</math> km in a total trip time of <math>3</math> hours including the stop. Which equation could be used to solve for the time <math>t</math> in hours that she drove before her stop?
 
Angelina drove at an average rate of <math>80</math> kmh and then stopped <math>20</math> minutes for gas. After the stop, she drove at an average rate of <math>100</math> kmh. Altogether she drove <math>250</math> km in a total trip time of <math>3</math> hours including the stop. Which equation could be used to solve for the time <math>t</math> in hours that she drove before her stop?
  
<math>\textbf{(A)}\ 80t+100(\frac{8}{3}-t)=250 \qquad \textbf{(B)}\ 80t=250 \qquad \textbf{(C)}\ 100t=250 \qquad \textbf{(D)}\ 90t=250 \qquad \textbf{(E)}\ 80(\frac{8}{3}-t)+100t=250</math>
+
<math>\textbf{(A)}\ 80t+100\left(\frac{8}{3}-t\right)=250 \qquad \textbf{(B)}\ 80t=250 \qquad \textbf{(C)}\ 100t=250 \qquad \textbf{(D)}\ 90t=250 \qquad \textbf{(E)}\ 80\left(\frac{8}{3}-t\right)+100t=250</math>
 
 
==Solution==
 
  
The answer is <math>A</math> because she drove at <math>80</math> kmh for <math>t</math> hours (the amount of time before the stop), and 100 kmh for <math>\frac{8}{3}-t</math> because she wasn't driving for <math>20</math> minutes, or <math>\frac{1}{3}</math> hours. Multiplying by <math>t</math> gives the total distance, which is <math>250</math> kms. Therefore, the answer is <math>80t+100(\frac{8}{3}-t)=250</math> <math>\Rightarrow</math> <math>\boxed{(A)}</math>
+
== Solution ==
  
 +
The answer is <math>A</math> because she drove at <math>80</math> kmh for <math>t</math> hours (the amount of time before the stop), and <math>100</math> kmh for <math>\frac{8}{3}-t</math> because she wasn't driving for <math>20</math> minutes, or <math>\frac{1}{3}</math> hours. Multiplying by <math>t</math> gives the total distance, which is <math>250</math> km. Therefore, the answer is <math>80t+100\left(\frac{8}{3}-t\right)=250</math> <math>\Rightarrow</math> <math>\boxed{(A)}</math>
  
 
== See Also ==
 
== See Also ==

Latest revision as of 13:01, 29 October 2024

Problem

Angelina drove at an average rate of $80$ kmh and then stopped $20$ minutes for gas. After the stop, she drove at an average rate of $100$ kmh. Altogether she drove $250$ km in a total trip time of $3$ hours including the stop. Which equation could be used to solve for the time $t$ in hours that she drove before her stop?

$\textbf{(A)}\ 80t+100\left(\frac{8}{3}-t\right)=250 \qquad \textbf{(B)}\ 80t=250 \qquad \textbf{(C)}\ 100t=250 \qquad \textbf{(D)}\ 90t=250 \qquad \textbf{(E)}\ 80\left(\frac{8}{3}-t\right)+100t=250$

Solution

The answer is $A$ because she drove at $80$ kmh for $t$ hours (the amount of time before the stop), and $100$ kmh for $\frac{8}{3}-t$ because she wasn't driving for $20$ minutes, or $\frac{1}{3}$ hours. Multiplying by $t$ gives the total distance, which is $250$ km. Therefore, the answer is $80t+100\left(\frac{8}{3}-t\right)=250$ $\Rightarrow$ $\boxed{(A)}$

See Also

2010 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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 AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png