Difference between revisions of "1985 AJHSME Problem 13"
Coolmath34 (talk | contribs) (Created page with "== Problem == If you walk for <math>45</math> minutes at a rate of <math>4 \text{ mph}</math> and then run for <math>30</math> minutes at a rate of <math>10\text{ mph,}</math...") |
(→In-depth Solution by BoundlessBrain!) |
||
| (One intermediate revision by the same user not shown) | |||
| Line 5: | Line 5: | ||
<math>\text{(A)}\ 3.5\text{ miles} \qquad \text{(B)}\ 8\text{ miles} \qquad \text{(C)}\ 9\text{ miles} \qquad \text{(D)}\ 25\frac{1}{3}\text{ miles} \qquad \text{(E)}\ 480\text{ miles}</math> | <math>\text{(A)}\ 3.5\text{ miles} \qquad \text{(B)}\ 8\text{ miles} \qquad \text{(C)}\ 9\text{ miles} \qquad \text{(D)}\ 25\frac{1}{3}\text{ miles} \qquad \text{(E)}\ 480\text{ miles}</math> | ||
| − | == Solution == | + | ==In-depth Solution by BoundlessBrain!== |
| + | https://youtu.be/varobXGlcdU | ||
| + | |||
| + | ==Solution== | ||
Distance is equal to rate multiplied by time: | Distance is equal to rate multiplied by time: | ||
<cmath>d = \frac{45}{60} \cdot 4 + \frac{30}{60} \cdot 10 = \boxed{8}.</cmath> | <cmath>d = \frac{45}{60} \cdot 4 + \frac{30}{60} \cdot 10 = \boxed{8}.</cmath> | ||
The answer is <math>\text{(B)}.</math> | The answer is <math>\text{(B)}.</math> | ||
Latest revision as of 15:02, 4 July 2023
Problem
If you walk for 
minutes at a rate of 
and then run for 
minutes at a rate of 
how many miles will you have gone at the end of one hour and 
minutes?
In-depth Solution by BoundlessBrain!
Solution
Distance is equal to rate multiplied by time:
![\[d = \frac{45}{60} \cdot 4 + \frac{30}{60} \cdot 10 = \boxed{8}.\]](http://latex.artofproblemsolving.com/1/4/d/14dc9b0fee61d322a782528c39619e3acdd06fd1.png)
The answer is 
