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Mathematics 7 Online
OpenStudy (flvskidd):

HELP Frank drove to the mountains last weekend. There was heavy traffic on the way there, and the trip took 8 hours. When Frank drove home, there was no traffic and the trip only took 5 hours. If his average rate was 21 miles per hour faster on the trip home, how far away does Frank live from the mountains? Do not do any rounding.

OpenStudy (evoker):

Probably easiest to solve for his speed first and then the distance.

OpenStudy (evoker):

d=r*t and same distance so first rate times time equals the second rate times time.

OpenStudy (flvskidd):

@EVOKER sorry i was gone

OpenStudy (flvskidd):

can you help @wolf1728

OpenStudy (wolf1728):

Yes I think I can

OpenStudy (wolf1728):

Average rate was 21 miles per hour Total time = 8 + 5 = 13 hours distance = rate * time d = 21 mph * 13 = 273 (which is twice the distance) so distance = 136.5 miles Trip to the mountains is 8 hours and distance = 136.5 miles so rate = 17.0625 mph Trip back - 5 hours; 136.5 miles = 27.3 mph Average speed = 17.0625 * 8 + 27.3 * 5 = 273/13 hours = 21 mph

OpenStudy (flvskidd):

so it's 136.5

OpenStudy (wolf1728):

yes 136.5 miles

OpenStudy (flvskidd):

it was wrong

OpenStudy (wolf1728):

it said it is wrong?

OpenStudy (flvskidd):

yeah

OpenStudy (wolf1728):

Oh, it says the average rate was 21 miles per hour FASTER on the way home. (I thought the average rate was 21 miles per hour - my mistake) Guess I'll recalculate

OpenStudy (wolf1728):

distance = rate * time let's say distance = x rate on the way up = x/8 rate on way back = x/5 x/5 = x/8 + 21 21 = x/5 -x/8 21 = 8x/40 -5x/40 21 = 3x/40 840 = 3x x = 280 miles

OpenStudy (flvskidd):

IT WAS RIGHT.

OpenStudy (flvskidd):

Thank you

OpenStudy (wolf1728):

you are welcome - and I'm glad I could redeem myself :-)

OpenStudy (flvskidd):

Can you do one more?

OpenStudy (wolf1728):

Okay - go ahead

OpenStudy (flvskidd):

Two trains leave the station at the same time, one heading east and the other west. The eastbound train travels 10 miles per hour slower than the westbound train. If the two trains are 450 miles apart after 3 hours, what is the rate of the eastbound train? Do not do any rounding.

OpenStudy (wolf1728):

distance = rate * time 450/3 hours = 150 miles per hour (COMBINED rate) one train is 10 mph faster so rate = 80 mph and slower = 70 mph Check: 80* 3 = 240 miles and 70*3 = 210 miles 240+210= 450 miles So that checks out 80 mph and 70 mph

OpenStudy (wolf1728):

70 mph eastbound train

OpenStudy (flvskidd):

Right again

OpenStudy (flvskidd):

Thank you for the help

OpenStudy (wolf1728):

Okay - I'm glad I was right on my first try!! LOL u r welcome

OpenStudy (flvskidd):

Lol

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