Find Total Distance

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mathdad

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John took a drive to town at an average rate of 40 mph. In the evening, he drove back at 30 mph. If he spent a total of 7 hours traveling, what is the distance traveled by John?

Solution:

Round trip distance = D

Going rate = 40 mph
Returning rate = 30 mph

Going time = x
Returning time = 7 - x

40x = 30(7 - x)

My problem here is the TIME set up. I do not understand why x is the going time and (7 - x) the returning time. Is the solution provided dividing 7 into two parts? If so, what WORDS in the problem reveal this information?
 

lookagain

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My problem here is the TIME set up. I do not understand why x is the going time and (7 - x) the returning time. Is the solution provided dividing 7 into two parts? If so, what WORDS in the problem reveal this information?
No, that is immaterial. It's fine if x is the return time and (7 - x) is the start time.
Yes, it divides seven hours into two parts. The words that stand out to me are
"he spent a total of 7 hours traveling."

That is similar to the discussion I had on another thread with you about two numbers that added to 90, where one part could be x, and the other part could
then be (90 - x).
 

mathdad

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No, that is immaterial. It's fine if x is the return time and (7 - x) is the start time.
Yes, it divides seven hours into two parts. The words that stand out to me are
"he spent a total of 7 hours traveling."

That is similar to the discussion I had on another thread with you about two numbers that added to 90, where one part could be x, and the other part could
then be (90 - x).
Nicely explained.
 

mathdad

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40x = 30(7 - x)

40x = 210 - 30x

40x + 30x = 210

70x = 210

x = 210/70

x = 3

John traveled 3 total miles.
 

MarkFL

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40x = 30(7 - x)

40x = 210 - 30x

40x + 30x = 210

70x = 210

x = 210/70

x = 3

John traveled 3 total miles.
You are using \(x\) to represent time, not distance. You have found that John took 3 hours to travel one way at 40 mph.
 

mathdad

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You are using \(x\) to represent time, not distance. You have found that John took 3 hours to travel one way at 40 mph.
Yes, I made a mistake.

Round trip distance = D

Going rate = 40 mph
Returning rate = 30 mph

Going time = x
Returning time = 7 - x

40x = 30(7 - x)

x = 3

40(3) = 120

Or

30(7 - 3)

30(4) = 120

D = 120 miles
 

MarkFL

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You found the one way distance.

\(\displaystyle D=40t+30(7-t)=10t+210\)

Now, given that:

\(\displaystyle 40t=30(7-t)\implies t=3\)

We find:

\(\displaystyle D=10(3)+210=240\)
 

mathdad

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You found the one way distance.

\(\displaystyle D=40t+30(7-t)=10t+210\)

Now, given that:

\(\displaystyle 40t=30(7-t)\implies t=3\)

We find:

\(\displaystyle D=10(3)+210=240\)
Wow! I got it wrong. Amazing. Oh boy!!
 

Otis

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... 40(3) = 120

Or

... 30(4) = 120 ...
Use AND, not OR

That is, John is going AND returning, not going OR returning.

D = 120 + 120

😎
 

MarkFL

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mathdad

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mathdad

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Use AND, not OR

That is, John is going AND returning, not going OR returning.

D = 120 + 120

😎
And not or, what? English lessons now???
 

Otis

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And not or, what? English lessons now???
No, it's a math lesson.

In math, AND and OR are logical operators. In this exercise, there are two trips (one going; one returning).

AND means both.

OR means just one.

You were thinking OR (one trip's distance or the other), when you ought to have thought AND (both trips together).

😎
 

mathdad

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No, it's a math lesson.

In math, AND and OR are logical operators. In this exercise, there are two trips (one going; one returning).

AND means both.

OR means just one.

You were thinking OR (one trip's distance or the other), when you ought to have thought AND (both trips together).

😎
Thank you for clearing that up for me.
 
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