can someone help me with this word problem?

Question

i had an algebra II exam today, and my teacher put a distance problem on it. ok that's cool..but it was a really weird one! it goes like this:
a woman traveled part of a 185 distance at 50mi/h & the other part at 55 mi/h. Find the distance she traveled at 50 mi/h if the trip was 3 hours & 30 minutes long.

Answer 1

Simultaneous equations:
Time at 50mph = x Time at 55mph = y
we know that x + y = 3.5
and we know that 50x + 55y = 185

So express x in terms of y using first equation.
x + y = 3.5
x = 3.5 - y

Substitute into second equation:

50x + 55y = 185
(3.5 - y).50 + 55y = 185

Solve to find y

3.5.50 - 50y + 55y = 185
5y = 10
y = 2

Substitute solution for y into first equation:
x + 2 = 3.5
Solve
x = 1.5

Your answer is 1.5 hours at 50mph which equals 75miles.

Answer 2

distance = rate * time

total distance here = 185 miles

so the formula would be
185 = 50mi/hr*(time going 50) + 55mi/hr*(time going 55)
total time = 3.5 hours = time going 50 + time going 55

so time going 50 = 3.5 - time going 55

plug this back in
185 = 50 * (3.5 - time going 55) + 55*(time going 55)
solve for time going 55 = T
185 = 50*(3.5 - T) + 55*(T)
185 = 175 -50T +55T
10 = 5T
T = 2

so time going 55 is equal to 2 hours

that means the time going 50 was 3.5-2 = 1.5 hours

Answer 3

OK, lets see if this makes sense to you. Let

x = time driven at 50 mph
y = time driven at 55 mph and

(1) x + y = 3.5 hours (time in hours) so

50x = distance driven at 50 mph
55y = distance driven at 55 mph and

(2) 50x + 55y = 185

If you solve the system of equations (1) and (2) you should get
x = 1.5 and y = 2, which means she traveled 75 miles at 50 mph.

Answer 4

It only seems weird until you consider that you're solving for times, not disttances as is usually the case. She traveled for time t1 at 50 mph, and time t2 at 55 mph. Distance is rate x time, so
50 mph * t1 + 55 mph * t2 = 185 miles
Further, you know the total time:
t1 + t2 = 3.5 hours (keep all times in hours to make things simple)

Now it's fairly simple:
t2 = 3.5 hours - t1

50 * t1 + 55 * (3.5 - t1) = 185
50 t1 - 55 t1 + 192.5 = 185
7.5 = 5t1
1.5 = t1
therefore t2 = 2

So she traveled 1.5 hours at 50 mph for a distance of 75 miles, and 2 hours at 55 mph for a distance of 110 miles.

75 miles + 110 miles = 185 miles, the total distance.

Answer 5

No time to give complete answer....
Let x km = the first part of the 185 distance.
Therefore 185 - x = the second part of the distance traversed.
D=TS

D(total) = D1 + D2 = 3.5 hours

Answer 6

She traveled 1 and 1/2 hours on 50mph and 2 hours on 55mph. Its just a matter of guess and check. you see multiply 1.5*50=75mi
2*55=110mi
75mi+110mi=185mi
So she traveled at 50mph for an hour and a half

Answer 7

you have this eq. to solve:
50t+55(3.5-t)=185
-5t+192.5=185
5t=7.5
t=1.5

She traveled for 1.5 hrs at 50mi/hr

Answer 8

50a+55b=185
a+b=3.5
a=305-b
175-50b+55b=185
5b=10
b=2
a=1.5