A salesperson purchased an automobile that was advertised as averaging 25 mi/gal in the city and 40 mi/gal on?

Let the no of miles driven in the city be x miles

No of miles driven on the highway = ( 2300 - x) miles

x / 25 + ( 2300 -x ) /40 = 62

MULTIPLY by 200

8x + 11500 - 5x = 12400

3x = 900

x = 300

ANSWER 300 miles ( city) and 2000 miles ( highway)

CHECK

300 / 25 = 12 gallons

2000 /40 = 50 gallons

Total = 62 gallons

enable's say the # of city miles = C The # of highway miles = H you will could locate the two a sort of Given: the comprehensive # of miles is 2270, so: H + C = 2270 miles Get between the unknowns (H or C) on my own on one fringe of the equation: H = 2270 miles - C city mile fee is 25 mi/gal - this is the comparable as asserting that for each a million gallon you employ, you went 25 miles, or a million gal/25 mi Hwy mile fee is 40-one mi/gal - this is the comparable as a million gal/40-one mi [(city Miles) x (city Gallon/Mile)] + [(highway Miles) x (highway Gallon/Mil)] = comprehensive Gallons [(C) x (a million gal/25 mi)] + [(H) x (a million gal/40-one mi)] = sixty two gal (a million gal/25 mi)C + (a million gal/40-one mi)H = sixty two gal all of us understand that H = 2270 miles - C, so substitute that for the period of: (a million gal/25 mi)C + (a million gal/40-one mi) (2270 mi - C) = sixty two gal (a million gal/25 mi)C + (2270/40-one) gal - (a million gal/40-one mi)C = sixty two gal sparkling up for C by having a client-friendly denominator.