Hard Algebra Word Problem

Q3. A boy can row a boat at a constant rate of 5 mi/hr in still water, as indicated in the figure. He rows upstream for 21 minutes and then rows downstream, returning to his starting point in another 11 minutes.

Q4. A salesperson purchased an automobile that was advertised as averaging 25 mi/gal in the city and 40 mi/gal on the highway. A recent sales trip that covered 2700 miles required 78 gallons of gasoline. Assuming that the advertised mileage estimates were correct, how many miles were driven in the city?

upstream
d = (21/60) (5) - (21/60) w
60d = 105 - 21 w
downstream
d = (11/60) (5) + (11/60) w
60d = 55 + 11 w
now equate them
105-21w=55+11w
32 w = 50
w = 50/32 mi/hr = 1 18/32 mph
check - does ?
21(160/32 - 50/32) = 11(160/32+50/32)
21(110/32) = 11(210/32)
yes - problem is solved correctly

x in the city at 25 mpg, 2700-x at 40 mpg
x/25 + (2700-x)/40 = 78 gallons use 1000 as LCD
40x + 25(2700-x) = 78000
x = 700 miles in the city leaving 2000 on the highway
40x +

What's the question for Q3

q4

let g = gallons

you have


25g + 40(78-g) = 2700 (78 is the total amount of gas minus what you used at 25mpg

expand

25g + 3120 - 40g = 2700

add like terms

-15g +3120 = 2700

subtract 3120 from both sides

-15g +3120 - 3120 = 2700 -3120

that give you

-15g = -420

divide by -15

-15g/-15 = -420/-15

g = 28


so you used 28 gallon in the city

25 * 28 = 700 miles