pat drove from home to school in 3 hours. On the way back, because of the traffic, it took him 1 hour longer and the average speed was 2 mph slower. What was his speed of driving to school ?
a. 4 mph
b. 6 mph
c. 8 mph
d. 10 mph
please explain too
thank you !
xD
2007-07-05 13:01:31 · 10 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
pat took 3 hrs to drive to school and an hour longer to drive home, therefore took him 4 hours. Iets say his original mph is x then:
3x = 4(x-2) since he was 2mph slower on the way back and the distance is the same.
Simplifying gives:
3x = 4x-8
3x+8 = 4x
8 = x
Therefore answer is c) 8 mph
Hope this has helped! :)
2007-07-05 13:10:34 · answer #1 · answered by wycombew2001 2 · 1⤊ 0⤋
The distance from home to school is the same, regardless of the traffic. Let "d" be this distance. Let "s" be the speed we're looking for. Since speed is distance over time, we have s = d/3 for the time without traffic, and (s-2) = d/4 for the time with traffic. You have two equations with two unknowns, so solve this for s. (Hint: solve both for d and set the two equations equal to each other)
2007-07-05 20:07:54 · answer #2 · answered by Anonymous · 1⤊ 0⤋
let x be teh speed of Pat on the way to school. It took him 3 hrs to get there.
distance = speed * time
d = 3x
On the way back home, it took him 1 hr longer, that means it took him 4hrs total to get home. The speed was x - 2
d = 4(x - 2)
you have two equations
d = 3x
d = 4(x - 2)
3x = 4(x - 2)
3x = 4x - 8
-x = -8
x = 8
his speed was 8 mi/hr
2007-07-05 20:07:16 · answer #3 · answered by 7 · 2⤊ 0⤋
OK let
X=speed traveling to school
y=distance to school
So 3x=y since speed x time = distance
the distance home is the same as to school so
y=distance
But speed is 2mph slower and travel time is 1 hour longer.
we know the speed is x-2
since speed x time = distance
(x-2)4=y
now since we know y=3x and (x-2)4 we can solve for x
3x=4(x-2)
3x=4x-8
x=8
2007-07-05 20:13:39 · answer #4 · answered by scotts1870 3 · 0⤊ 0⤋
4(v-2)=3v
v=8 mph
2007-07-05 20:11:04 · answer #5 · answered by Alberd 4 · 1⤊ 0⤋
Let the speed while going to school be x mph.Hence his speed while coming back was x-2 mph
by the problem
3x=4(x-2)
or 3x=4x-8
or,3x-4x=-8
or x=8
Therefore speed while going to school ws 8 mph
Hence answer c is correct
2007-07-05 20:09:30 · answer #6 · answered by alpha 7 · 1⤊ 0⤋
3s = 4(s - 2), since the distances to and from are equal.
3s = 4s - 8
8 = s
speed = 8 mph
2007-07-05 20:07:08 · answer #7 · answered by John V 6 · 2⤊ 0⤋
Let the distance from home to school be x
Speed=Distance/Time
Let the speed to school be s
s=x/3
s-2=x/4
Therefore using substitution
(x/3)-2=(x/4)
(x/3)-(x/4)=2
x/12=2
x=24
Now s=x/3 so
s=24/3
s=8mph..answer (c)
2007-07-05 20:09:18 · answer #8 · answered by Anonymous · 1⤊ 0⤋
r =rate going to school
So 3 r = distance to school
On the way home 4(r-2) = distance to school
So 3r = 4(r-2)=4r-8
-r = -8
r =8 = rate going to school.
2007-07-05 20:11:12 · answer #9 · answered by ironduke8159 7 · 1⤊ 0⤋
this problem cant be solved with out the distance from his house to school
2007-07-05 20:05:57 · answer #10 · answered by joshua macmiller 3 · 0⤊ 5⤋