Problem solving

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Edhelms1890

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Rene is going to the lake to visit some friends. If the lake is 60 miles away. And Rene is driving at 40 miles per hour the entire time. How long will it take her to get to the lake? A: 50 minutes B: 70 minutes. C: 90 minutes. D: 110 minutes
 

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My advice:
Forget it's a "maths" question from a textbook.
Pretend it is YOU going to the lake.
Draw a line representing 60km. Where will you be after 1 hour? How long will it take to travel the rest of the way to the lake? How long in total?
 
Cat, You are correct that students forget their everyday math when they walk into a math class or read a math book. It amazes me how students think that there are no numbers between say 2 and 3. When they go into a store the prices are not only $1 or $2 or $3 or... And even if they were then after sales tax the final price would not $1 or $2 or $3 or.... So they know as well as everyone else that there are numbers other than 1,2,3 etc.
I recently had a student tell me that the numbers between 2 and 3 are 2.1, 2.2, ..., 2.9. I could only imagine what store she shopped at!

Your advice telling the student to pretend that it is them going to the lake is quite good. I had a neighbor who was terrible at algebra but got the hardest problem on a test correct because that problem interested them. If I remember correctly it was a question about purchasing a CD player.
 
Edhelms1890 said:
Rene is going to the lake to visit some friends. If the lake is 60 miles away. And Rene is driving at 40 miles per hour the entire time. How long will it take her to get to the lake? A: 50 minutes B: 70 minutes. C: 90 minutes. D: 110 minutes
Click to expand...
What is the math-topic associated with this question!

It matters - the average diameter of lake Huron is 200 miles.
 
Jomo said:
Cat, You are correct that students forget their everyday math when they walk into a math class or read a math book. It amazes me how students think that there are no numbers between say 2 and 3. When they go into a store the prices are not only $1 or $2 or $3 or... And even if they were then after sales tax the final price would not $1 or $2 or $3 or.... So they know as well as everyone else that there are numbers other than 1,2,3 etc.
I recently had a student tell me that the numbers between 2 and 3 are 2.1, 2.2, ..., 2.9. I could only imagine what store she shopped at!

Your advice telling the student to pretend that it is them going to the lake is quite good. I had a neighbor who was terrible at algebra but got the hardest problem on a test correct because that problem interested them. If I remember correctly it was a question about purchasing a CD player.
Click to expand...
Jomo,
When teaching adding decimals (with 2 dec places), I find if you put a dollar sign in front, the students have no problems!
 
Harry_the_cat said:
Jomo,
When teaching adding decimals (with 2 dec places), I find if you put a dollar sign in front, the students have no problems!
Click to expand...
That is amazing that they understand when you use the $ sign! Until ....you ask them the value of .50 cents.
I can not count how many stores sell fruit for .50 cents and how many students tell me that .50 cents is 50 cents. When so many people make the same mistakes I feel that it must be the teacher's fault. I taught math for years at the community college level and regardless of which I course I was teaching at some point we talked about .50 cents vs 50 cents. I just did not want my students not knowing the difference.
 
Another way of thinking about it: The lake is 60 miles away. Rene is driving at 40 \(\displaystyle \frac{miles}{hour}\). You are given "miles" and want to find "hours". You need to cancel the "miles".

\(\displaystyle miles\left(\frac{hours}{miles}\right)= hours\). That is, you want to multiply by "hours/miles" which is the same as dividing by "miles/hour".
 
How I like to think about it is that you want to convert 60 miles to time. And in this problem you are given that 40miles = 1 hour so \(\displaystyle \frac{40 miles}{1hour} = \frac{1 hour}{40 miles} = 1\).

So 60 miles = 60 miles * 1 = 60 miles * \(\displaystyle \frac{1 hour}{40 miles}\) = ....
 
Jomo said:
How I like to think about it is that you want to convert 60 miles to time. And in this problem you are given that 40miles = 1 hour so \(\displaystyle \frac{40 miles}{1hour} = \frac{1 hour}{40 miles} = 1\).

So 60 miles = 60 miles * 1 = 60 miles * \(\displaystyle \frac{1 hour}{40 miles}\) = ....
Click to expand...


\(\displaystyle 40 \ miles \ne 1 \ hour. \ \ \) Rather, 40 miles corresponds to 1 hour.

\(\displaystyle distance \ = \ rate*time\)

\(\displaystyle \dfrac{distance}{rate} \ = \ time\)

\(\displaystyle \dfrac{60 \ miles}{(40 \ miles/hr)} \ = \ time\)

\(\displaystyle \dfrac{60 \ miles}{1} \ \div \ \dfrac{40 \ miles}{1 \ hour} \ = \ time\)

\(\displaystyle \dfrac{60 \ miles}{1} * \dfrac{1 \ hour}{40 \ miles} \ = \ ?\)
 
lookagain said:
\(\displaystyle 40 \ miles \ne 1 \ hour. \ \ \) Rather, 40 miles corresponds to 1 hour.

\(\displaystyle distance \ = \ rate*time\)

\(\displaystyle \dfrac{distance}{rate} \ = \ time\)

\(\displaystyle \dfrac{60 \ miles}{(40 \ miles/hr)} \ = \ time\)

\(\displaystyle \dfrac{60 \ miles}{1} \ \div \ \dfrac{40 \ miles}{1 \ hour} \ = \ time\)

\(\displaystyle \dfrac{60 \ miles}{1} * \dfrac{1 \ hour}{40 \ miles} \ = \ ?\)
Click to expand...
I disagree with you here for a couple of reasons. 1st I never said that 40 miles equal 1 hour, but rather I said was in this problem 40 miles equals 1 hour.
How much cleaner can a problem be explained than by saying that an expression equals itself times 1. I used a fact that almost everybody knows and you are going to zoom into whether or not (in this problem) 40 miles equals 1 hour or 40 miles corresponds to 1 hour.
Get a life.
 
Incorrect. Yes, you did. It's right here:

Jomo said:
40miles = 1 hour
Click to expand...

The problem never "gave that."

40 miles never equals 1 hour, but you stated that, and then you incorrectly set a fraction that is not equal to 1 equal to 1.


And, \(\displaystyle \dfrac{1 \ hour}{40 \ miles} \ne 1\)


And, your train of supposed equalities

60 miles = 60 miles * 1 = 60 miles * \(\displaystyle \frac{1 hour}{40 miles}\) = ....

has the conclusion that 60 miles equals \(\displaystyle \ \dfrac{3}{2}hours \ \)!
 
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