METHOD 1: Strategy: Temporarily disregard the first 1/5 mile.
Dan pays $2.00 for the first fifth mile and $7.60 for the rest of the ride. Each fifth of a mile costs 40¢, and 760 ÷ 40 = 19. The $7.60 pays for a ride of 19 fifths of a mile. Adding 1/5 mile for the first $2.00, the longest ride Dan can afford is 20/5 or 4 miles.
METHOD 2: Strategy: Make a table.
The cost increases by $.40 for each fifth of a mile. When the desired amount is reached, the taxi will have traveled a length of 20/5 or 4 miles.
METHOD 3: Strategy: Use algebra.
Let N = the number of fifths of a mile after the first fifth mile that Dan can ride.
2.00 + .40N = 9.60
.40N = 7.60 (Subtract 2.00 from each side)
N = 19 (Divide each side by .40)
Add the first fifth of a mile, bringing the length of the ride to 20/5 or 4 miles.
Dan pays $2.00 for the first fifth mile and $7.60 for the rest of the ride. Each fifth of a mile costs 40¢, and 760 ÷ 40 = 19. The $7.60 pays for a ride of 19 fifths of a mile. Adding 1/5 mile for the first $2.00, the longest ride Dan can afford is 20/5 or 4 miles.
METHOD 2: Strategy: Make a table.
The cost increases by $.40 for each fifth of a mile. When the desired amount is reached, the taxi will have traveled a length of 20/5 or 4 miles.
METHOD 3: Strategy: Use algebra.
Let N = the number of fifths of a mile after the first fifth mile that Dan can ride.
2.00 + .40N = 9.60
.40N = 7.60 (Subtract 2.00 from each side)
N = 19 (Divide each side by .40)
Add the first fifth of a mile, bringing the length of the ride to 20/5 or 4 miles.