SOLUTION: Betty ann drove from city A TO B a distance of 250 miles she increased her speed by 10 mi/hr for the 360 mile trip from B TO C. if the total trip took 11 hours what was her sp

Question 1135050: Betty ann drove from city A TO B a distance of 250 miles she increased her speed by 10 mi/hr for the 360 mile trip from B TO C.

if the total trip took 11 hours what was her speed from city A to city B.
Found 4 solutions by greenestamps, swincher4391, Theo, MathTherapy:
Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!

Let x be her speed from A to B; then x+10 is her speed from B to C.

The time from A to B is distance divided by rate, 250/x; the time from B to C is 360/(x+10).

Write and solve an equation that says the total time for the trip is 11 hours:

Finishing solving the problem using the formal algebra is good practice; I leave that to you.

However, if you just need to get to the solution as quickly as possible, trial and error is best. Since the total time for the trip is a whole number of hours, look for speeds of x and x+10 that make both 250/x and 360/(x+10) whole numbers....
Answer by swincher4391(1107)   (Show Source): You can put this solution on YOUR website!
250 = r*t
360 = (r+10)*(11-t)
360 = 11r -rt + 110 - 10t
360 = 11r - 250 + 110 - 10t
360 = 11r -140 - 10t
We want to find r. Solve t in terms of r.
250/r = t
360 = 11r - 140 - 10(250/r)
360 = 11r - 140 - 2500/r
360 = (11r^2 - 140r - 2500)/r
360r = 11r^2 - 140r - 2500
11r^2 - 500r - 2500 = 0
(500 +- sqrt((-500)^2 - 4(11)(-2500))) / 22
(500 +- sqrt(360000))/22
(500 + 600) / 22 = 1100 / 22 = 50
Hence, 50 mph.

Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
total trip took 11 hours.

assume it took x hours to do the 250 mile leg.

the 350 mile leg would then take 11 - x hours.

her rate for the first leg can be assumed to be y.

her rate for the second leg can then be assumed to be y + 10.

rate * time = distance.

for the first leg, you get x * y = 250

for the second leg, you get (11 - x) * (y + 10) = 350

multiply the factors in the second leg equation to get:

11y + 110 - xy - 10x = 350

in the first leg equation, solve for y to get y = 250 / x

in the second leg equation, replace y with 250 / x to get:

11 * 250 / x + 110 - x * 250 / x - 10 * x = 350

simplify to get 2750 / x + 110 - 250 - 10 * x = 350

combine like terms to get 2750 / x - 140 - 10 * x = 350

multiply both sides of this equation by x to get 2750 - 140x - 10x^2 = 350x

subtract 350x from both sides of this equation to get 2750 - 490x - 10x^2 = 0

divide both sides of this equation by 10 to get 275 - 49x - x^2 = 0

multiply both sides of this equation by -1 to get -275 + 49x + x^2 = 0

rearrange the terms in descending order of degree to get x^2 + 49x - 275 = 0

factor this quadratic equation to get x = -54.084624384974 or x = 5.084624384974.

x can't be negative so x has to be 5.084624384974.

for the first leg x * y = 250.

solve for y to get y = 250 / x = 49.16784035

x * y in the first leg becomes 5.084624384974 * 49.1678403 = 250.

for the second leg, 11 - x equals 5.915375615 and y + 10 equals 59.16784035

(11 - x) * (y + 10) becomes 5.915375615 * 59.16784035 = 350 miles.

a lot of arithmetic was required to get the answer and you had to use the quadratic formula to factor the quadratic equation but the answer looks good.

your solution is that the speed from city A to B was 49.167840 miles per hour.

round your answer as required.

Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!

Betty ann drove from city A TO B a distance of 250 miles she increased her speed by 10 mi/hr for the 360 mile trip from B TO C.

if the total trip took 11 hours what was her speed from city A to city B.

Let speed from A to B. be S
Then speed from B to C is, S + 10
We then get the following TIME equation:
250(S + 10) + 360S = 11S(S + 10) ------- Multiplying by LCD, S(S + 10)

11S(S - 50) + 50(S - 50) = 0
(S - 50)((11S + 50) = 0
Speed from A to B, or OR
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