Gas Costs! A Moron Needs Help!

My fellow B&Bers,

I was trying to figure this out, and have been pulling out my hair. I figure the quickest way to get an answer is to post it on B&B!!

My question:

I have a 1998 Dodge Stratus (I/100 KM) = 6.4. If I am traveling 890 KM, and if gas is $0.80/L, how much will it cost in CDN?

I am having a hard time figuring this out. I need to calculate how much this will cost so I can get funding BEFORE I go. I am driving 890 KM round trip. I own a 1998 Dodge Stratus. If, for example, gas is 80 cents a litre and my car is l/100 km = 6.4, how much will it cost to get to and from my destination (445 KM one way = 890 KM round trip)? I am horrible at math and would appreciate any help on this one!

Thanks!

(890/100)*6.4*$0.80
8.9*6.4*$0.80
$45.57

Ha! Great, thanks!

Oh, actually someone just PM'd me and said $69.

Hmmmm..... this is so sad I can't figure it out!!

(6.4 liter/100 kilometers)(890 kilometers)($0.80/liter)
Liter cancels out
Kilometers cancels out
So you know you have the equation, well, not terribly wrong
Preservation of units...the one thing I still use from college.

(5696/100)($0.80) which is $45.568 or $45.57.

EDIT: Oops, left out the 6.4

ah.... now someone just told me it's $111. Yikes.

At least you showed your work!!

SA_bmatth said:

(1 liter/100 kilometers)(890 kilometers)($0.80/liter)
Liter cancels out
Kilometers cancels out
So you know you have the equation, well, not terribly wrong
Preservation of units...the one thing I still use from college.

(890/100)($0.80) which is $45.568 or $45.57.

Click to expand...

Hey, your way makes sense. And I appreciate it greatly!

Using the definition of the W,

n = W(n) eW(n)

The original problem:
xx = 2

Change to exponential form:
ex ln(x) = 2

Take ln both sides:
x ln(x) = ln(2)

Let y = 1/x:
-ln(y)/y = ln(2)

Let z = y/ln(2):
-ln(z/ln(2)) ln(2)/z = ln(2)

Divide both sides by -ln(2):
ln(z/ln(2))/z = -1

Move z over to the rhs:
-ln(z/ln(2)) = z

Exponentiate:
ln(2)/z = ez

Move the z to the rhs:
ln(2) = z ez

which, by definition of the W function, implies that z = W(ln(2)). And thus x = ln(2)/W(ln(2

Never mind.

carry a credit card.....

Paying is not a problem, I am just filling out all these forms for funding, which has to be done before I go. And since I've never had to figure it out before, I am a bit confused.

So far I've received totals of $45, $69 and $111.

Oh well.....

Maybe I'll just average them all out!!

slcsteve said:

Using the definition of the W,

n = W(n) eW(n)

The original problem:
xx = 2

Change to exponential form:
ex ln(x) = 2

Take ln both sides:
x ln(x) = ln(2)

Let y = 1/x:
-ln(y)/y = ln(2)

Let z = y/ln(2):
-ln(z/ln(2)) ln(2)/z = ln(2)

Divide both sides by -ln(2):
ln(z/ln(2))/z = -1

Move z over to the rhs:
-ln(z/ln(2)) = z

Exponentiate:
ln(2)/z = ez

Move the z to the rhs:
ln(2) = z ez

which, by definition of the W function, implies that z = W(ln(2)). And thus x = ln(2)/W(ln(2

Never mind.

Click to expand...

And all this time I thought ln(2) was 0.693

If your trip is being funded then definitely go with $111. You should have $65 and change left over for snacks.

SA_bmatth said:

If your trip is being funded then definitely go with $111. You should have $65 and change left over for snacks.

Click to expand...

and shave gear......don't forget the shave gear

mark tssb

SA_bmatth said:

If your trip is being funded then definitely go with $111. You should have $65 and change left over for snacks.

Click to expand...

So true.... but actually they'll be able to check easily, so I am trying NOT to look like the idiot that I am! And it's small scale, so asking for funding, for me anyway, is awkward.

It's not really that complicated if you break it down into the simplest terms.

You use 6.4 Liters of fuel for every 100 Kilometers driven?

If you drive 890 Kilometers you will have driven 100 Kilometers 8.9 times
(890÷100=8.9)

Meaning you will use 6.4Liters x 8.9 times = or 56.96 Liters used
(6.4x8.9=56.96)

56.96 liters @ $0.80/L = $45.568



Plus $65 for candy bars and soda pops brings you to $111.00

rhett121 said:

It's not really that complicated if you break it down into the simplest terms.

You use 6.4 Liters of fuel for every 100 Kilometers driven?

If you drive 890 Kilometers you will have driven 100 Kilometers 8.9 times
(890÷100=8.9)

Meaning you will use 6.4Liters x 8.9 times = or 56.96 Liters used
(6.4x8.9=56.96)

56.96 liters @ $0.80/L = $45.568



Plus $65 for candy bars and soda pops brings you to $111.00

Click to expand...

Thank you for breaking this down for me. I really appreciate it. You and SA_bmatth should be teachers!!

Here's a great response I was just sent:

If in fact you are getting 6.4L/100KM, then that is very good - exceptionally good, actually. The lower the number, the better the mileage. Anything under 9 is good, and under 8 is super. So that also makes me question whether it's really accurate. But anyways, I will assume it is.

If you burn 6.4L every 100 KM, you would burn 56.96L in 890 KM ( 6.4 * 890 / 100).

56.96L of gas @ 80¢/L = $45.57. Which I believe is the answer you came up with. So you don't suck at math after all.

~However~ - I would check that mileage to make sure it's really accurate. If that is what the manufacturer is quoting, then you're surely not going to get mileage that good, as they state the absolute optimal number. Also, you will no doubt have to do some city driving, idling, etc. So I think I would be on the safe side and budget considerably more than $46 for gas.

Hope that helps!



And, indeed, it did.

This would be much easier in miles per gallon ... but that's another thread!!