# Help me!

• Jun 28th 2006, 10:15 AM
babygirl
Help me!
student has a car that gets, on average, 25.0 mi/gal of gasoline. She plans to spend the a year in Europe and take the car with her.
What should she expect the car's average gas mileage to be in kilometers per liter?
During the year there, she drove 6000 km. Assuming that gas costs $5.00/gal in Europe, how much did she spend on fuel? (Compute to the nearest dollar) • Jun 28th 2006, 01:54 PM syrivaci Km to Miles Quote: Originally Posted by babygirl student has a car that gets, on average, 25.0 mi/gal of gasoline. She plans to spend the a year in Europe and take the car with her. What should she expect the car's average gas mileage to be in kilometers per liter? During the year there, she drove 6000 km. Assuming that gas costs$5.00/gal in Europe, how much did she spend on fuel? (Compute to the nearest dollar)

Okay, I'm new here but let's see:
25miles/gallon
6000km
$5.00/gallon It seems the key to this puzzle is finding out how many miles in a km. Hope the following helps: http://www.metric-conversions.org/le...kilometers.htm Let us know how u get on • Jul 21st 2006, 03:08 AM thedarktemplar 25.0mi/gal of gasoline= 1.609km/gal of gasoline .: 6000km of travel= 6000/1.609= no. of gal of gasoline = ~3729.02 gallons ~3729.02 gallons=$18,645 (to nearest dollar) (3729.02 x 5)
• Jul 21st 2006, 07:57 PM
Quick
Quote:

Originally Posted by babygirl
student has a car that gets, on average, 25.0 mi/gal of gasoline. She plans to spend the a year in Europe and take the car with her.
What should she expect the car's average gas mileage to be in kilometers per liter?
During the year there, she drove 6000 km. Assuming that gas costs $5.00/gal in Europe, how much did she spend on fuel? (Compute to the nearest dollar) I would do this problem with ratios: When I convert measurements I write things out in ratios so that they equal 1, such as: $\frac{1y}{3ft}=\frac{1(3ft)}{3ft}=\frac{3ft}{3ft}= 1$ now let's try solving for how many feet are in 5 yards... $5y=5y$ $5y\times1=5y\times\frac{3ft}{1y}$ $5y=\frac{15fty}{1y}$ $5y=\frac{15ft\noty}{1\noty}$ $5y=15ft$ now let's do your problem: Student's car gets: $\frac{25m_i}{1g}$ Add conversions: $\frac{25m_i}{1g}\times1\times1=\frac{25m_i}{1g} \times \frac{1g}{3.785l}\times\frac{1km}{0.62m_i}$ $\frac{25m_i}{1g}=\frac{25m_i\times1g\times1km}{1g \times 3.785l\times0.62m_i}$ $\frac{25m_i}{1g}=\frac{25\not{m_i} \times \not{g} \times 1km}{\not{g}\times3.785l\times0.62\not{m_i}}$ $\frac{25m_i}{1g}=\frac{25\times1km}{3.785l\times0. 62}$ $\frac{25km}{2.347l}$ ~ $Q\!u\!i\!c\!k$ • Jul 23rd 2006, 01:58 AM earboth Quote: Originally Posted by babygirl student has a car that gets, on average, 25.0 mi/gal of gasoline. She plans to spend the a year in Europe and take the car with her. What should she expect the car's average gas mileage to be in kilometers per liter? During the year there, she drove 6000 km. Assuming that gas costs$5.00/gal in Europe, how much did she spend on fuel? (Compute to the nearest dollar)

Hello, babygirl,

you only have to change the imperial units into metric units, using the conversion factors:
$\frac{25 mi}{1 gal}=\frac{25 mi\cdot 1.609\frac{km}{mi}}{1 gal\cdot 3.785\frac{l}{gal}}\approx 10.6\frac{km}{l}$. That's the result Quick has already calculated.

She'll travel 6000 km. So she had to buy 6000 km / 10.6 (km/l) = 566 l.
566 l is the same as 566 / 3.785 (l/gal) = 149.54 gal.

So she has to pay: 149.54 * 5 $= 747.70$

By the way: Today in Germany 1 gal of gas costs 6.86 \$. So she will go a long distance by foot.

Greetings

EB
• Aug 19th 2006, 02:14 AM
CaptainBlack
Quote:

Originally Posted by earboth
Hello, babygirl,

you only have to change the imperial units into metric units, using the conversion factors:
$\frac{25 mi}{1 gal}=\frac{25 mi\cdot 1.609\frac{km}{mi}}{1 gal\cdot 3.785\frac{l}{gal}}\approx 10.6\frac{km}{l}$. That's the result Quick has already calculated.

Not imperial units, the imperial gallon is 20 fl-oz while the US gallon is 16 fl-oz.

You have used the correct conversion factor for US gallons to litres.

RonL