Solve using algebra.

Question

Yvette walks at 4 mph to the
grocery store, shops for an hour,
then takes a bus averaging 24 mph
back home. If the entire trip takes
2.75 hours, how far away is the
store?

Answer 1

Distance = rate(time)

Let t = time to walk to the store

 

d = 4t   to walk to the store

 

Riding the bus back we have the d=rt given by 24 mph multiplied by (2.75 - 1 - t)

because we have to discount the hour spent shopping and the time spent walking to the store to relate

the time riding back from the store in terms of the amount of time spent walking to the store

 

d = 24(1.75 -t)  riding bus back

 

These two distances must be equal so:

 

4t = 24(1.75-t)

4t = 42 - 24t

28t = 42

t = 42/28

t = 1.5 hours to walk to the store

 

distance to the store = 4t = 4(1.5) = 6 miles

Answer 2

distance=rate*time

d=4*t     (to)

d=24*(2.75-t)     (from)

set the equations equal to each other

4t=24(2.75-t)

4t=66-24t

4t+24t=66

28t=66

t=66/28

from the first equation...

d=4*(66/28)

d=66/7

d=9 3/7 miles

checking with the bus...

d=24(2.75-66/28)

d=24(11/4-66/28)

d=24(77/28-66/28)

d=24(11/28)

d=264/28

d=9 12/28

d=9 3/7 miles again

 

Answer 3

distance = x      time = distance/ speed

total time :   (x/4) + (x/24)  = 2.75 - 1

                                7x/24 =  1.75

                                      x  = 6 miles