# What speed can he drive?

Gary had to buy a watch because the clock in his van was so unreliable. Unfortunately, he bought a used watch and it didn't work very well. He compared it to his good clocks in his house and found that his watch was fast: When his watch showed 1 minute had passed, it had really only been 57 seconds. One day, Gary was taking a trip to Monterey. He suspected that the speedometer in hi van was off, but he wasn't sure by how much. He found one of those mileage checks on the freeway. He drove 5 miles (according to the mileage signs, which you may assume are accurate) in 5 minutes 30 seconds according to his watch. He kept his speedometer on 53 miles per hour the whole time. If the speed limit is 65 miles per hour, what is the fastest he can drive according to his speedometer and avoid breaking the law? (Speedometer reads 0 mph when stopped).

Need a step by step explanation and answer to this problem.

• Any thoughts of your own? Commented Mar 25, 2014 at 16:37
• How fast was Gary going based off the mile markers and his watch? Commented Mar 25, 2014 at 16:37
• I know that 1 min= 57 seconds on his watch. If he times 5 miles = 5 min. 30 sec on his watch, it means he is really only going 5 min. 13.5 seconds? and if he is going 5 miles than for every mile it takes him 1 min. 3.37 sec for every mile. I don't really know where to go from here or if the changing the time is correct. Commented Mar 25, 2014 at 16:47
• You've changed the time correctly. Now you either have to convert 1 mile in 1 min and 3.37 sec to miles per hour or convert 53 miles per hour to miles per minute so you can compare the two. I recommend going to miles per hour to make the final part of the question easier. Commented Mar 25, 2014 at 16:51
• I am not sure on how to convert the two. I tried to convert it by changing the 53 mph to 60min/53miles * 1 mile/ 1min. 3.37sec but all the units would cancel out. Commented Mar 25, 2014 at 17:00

First, we start with classifying the data we know about. We know that $1$ minute on his incorrect watch is equal to $57$ seconds on his correct clock, that he traveled from where he was to Monterey at a with his speedometer on $53$ miles per hour, and it took him, according to his incorrect watch, $5.5$ minutes ($5$ minutes $30$ seconds) to move $5$ miles. To do some calculations, we can start by finding the time it took for him to go $5$ miles on the correct time: $$m=minutes, \ mi=miles, \ s=seconds, \ mph=miles \ per \ hour$$ $$\frac{60 \ s}{57 \ s}=\frac{330 \ s}{x}$$ $$60x=18810$$ $$x=313.5 \ s$$ $$=5 \ minutes \ 13.5 \ seconds(correct \ time)$$

Above, I decided to provide a proportion between the $1$-minute-to-$57$-seconds-ratio, from incorrect to correct, and the time of the correct time for him to move $5$ miles.

Next you find the speed of the correct time converted into miles per hour (mph):

$$\frac{5 \ mi}{5.5 \ m}$$ $$=0.909090\dots$$ $$0.909090\dots\times60$$ $$=54.545454\dots$$ $$=54.5 \ mph$$

So, according to the incorrect watch, the speed is approximately $54.5$ miles per hour. Now, on the finding the speed with an emphasis on the correct clock:

$$\frac{5 \ mi}{5.225 \ m}$$ $$=0.956937799$$ $$0.956937799\times60$$ $$=57.41626794$$ $$=57.4 \ mph$$

And now, according to the correct clock, the speed he is going at is approximately $57.4$ miles per hour.

In your question above, I don't think it specifies which data it wants, to solve this problem, I gave the solutions to the incorrect and correct times present by the problem.

Hope this helps.