7th grade math problem
 

For you mathematicians:

You drive 135 miles in 3 hours. You drive at a speed of 55 mph part of the time, and 40 mph the other part of the time. How much time do you spend at each speed?

66 min @55 for 60 miles
114 min @40. For 76 miles

If you round to full min. You will actually cover 136 miles in 3 hrs at these time/speeds.

Show your work, or no credit! SmileWavy

How about 55 for 1 hour and 40 for 2

Yeah Doug methinks thou dost try waaaaay too hard.

distance = rate * time
t1 = time at 55 mph t2 = time at 40

t1 * 55 + t2 * 40 = 135

t1 + t2 = 3 or t2 = 3 - t1

substitute 2 into 1
t1 * 55 + (3 - t1) * 40 = 135
t1 * 55 - 40 * t1 = 15
t1 = 1

Now calculate the redshift for those speeds and how much you have travelled into the future after the three hours have elapsed in your frame of reference.

I am going to show Ricks work because I like his answer

55mph x 1hr = 55 miles
40mph x 2hrs = 80miles

1 + 2 = 3 hrs
55+80 = 135

The marketing side is just trying to go the extra mile.
The math side of me says, don't do math with an ipad calculator and no paper while it's late at night.

Of course, my daughter is in algebra, so I was thinking in terms of two algebra equations, substituting, etc. But maybe Rick's Occum's Razor method is the way to go.

Is this a real 7th grade problem of your kid? Where did it come from?

Rick-l has the analytical solution.

That said, a keen eye will see that 135 in 3h is an average of 45 and that you can get there by trial and error. More 40 mph time is needed, so let's try 2*40 + 55 and voila ...

G

Simple, once you figure it out correctly....

d = rt

Part of the 135 mile trip was taken at 55 MPH and the other part was taken at 40 MPH. So we have two related distance equations to solve.

Let's solve for the 55 MPH part of the trip.

The part of the trip taken at 55 MPH equates to this:

d = 55t

The part of the trip taken at 40 MPH, relative to the other part, looks like this:

135 - d = 40(3 - t)

Add them together: 135 = 55t + 40(3 - t)

Now simplify and solve for t:

135 = 55t + 120 - 40t

135 = 15t + 120

15 = 15t

t = 1

So, the 55 MPH part of the trip took 1 hour so the 40 MPH part took 2 hours. 1 hour at 55 MPH and 2 hours at 40 MPH equal 135 miles.

We can prove it by solving it the other way around.

The part of the trip taken at 40 MPH equates to this:

d = 40t

The part of the trip taken at 55 MPH, relative to the other part, looks like this:

135 - d = 55(3 - t)

Add them together: 135 = 40t + 55(3 - t)

Now simplify and solve for t:

135 = 40t + 165 - 55t

135 = -15t + 165

-30 = -15t

t = 2

So, the 40 MPH part of the trip took 2 hours which matches up perfectly with what we did above.

Most 7th graders would say f-this and go over to the Chest Thread. I recommend you all remove your pocket protectors and do the same.

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What, can't we do both??

By the way, I never got a word problem like this in 7th grade....I am not sure I got one like this in high school either!

Actially, kids these days would probably just google the question as a first step in figuring out the problem.

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This statement makes no sense:

How did I get from zero to 55? Or from 55 to 40? Clearly, there must have been a nearly infinite amount of additional speeds that were driven. Besides, when have I ever driven 135 miles an not exceeded 55 mph? The answer is: "never."

I would submit the question back to the "teacher" for revision. There are far too many people that blindly accept this type of indoctrinated interrogation from so-called "authority" figures, without questioning that "authority." Leads to very bad things.

Johnny's mother had three children.
The eldest is named April.
The second eldest is named May.
What is the name of the third child?

Johnny?

ding! ding! ding!

If you took all the Pickles eaten in the world each year and laid them end to end:

People Would Think You Were Crazy........

I was thinking how I've never run into this, or any of the other stupid word problems we were graded on since...

...I left high school.

^ +1