DESPERATE URGENT AGLEBRA 3 WORD problems HELP NEEDED!?

Question

1.One weekend Chad rode his bike on a mountian trail that was 16 miles long. The next weekend he biked on a 36 mile road course. His average speed on the road course was 6 mph faster than his average speed on the mt. trail. If it took Chad 1/2 hour longer to complete the mt. trail than the road course, what was his average speed on the road course?
I GOT, 9.85MPH.

2.Two math students were working on their packets. One student could finish 5 more problems per hour than the other. The slower student completed 45 problems and the faster student completed 55. It took the slower students 15 minutes more than the faster to complete his problems. How many problems per hour could the faster student complete? (Think of this as a motion problem: D=RT)
I GOT 20.

3.In a certian factory, Team A can build a product in 4 hours less than Team B. When both teams work togther, they can biuld the product in 3 hrs. 45 mins. How long does it take Team B to build the product alone?
I GOT 10 HOURS

Answer 1

1) Let r be the average rate on the on the road course and t be the time on the road course (since we're looking for the road information, not the mountain trail information).

Then his rate on the mountain trail would be (r - 6) and his time on the mountain trail would be t + 1/2. This leads to the two equations:

16 = (r - 6)(t + 1/2)
36 = rt

I'd take the second equation and solve it for t:

36/t = r

And plug it in for t in the first:

16 = (r - 6)(36/r + 1/2)

Multiply through by 2r to make things nicer:

32r = (r - 6)(36/r + 1/2)2r = (r - 6)(72 + r)

FOIL on the right:

32r = 72r + r² - 432 - 6r

Get it all together on the right:

0 = r² + 34r - 432

Solve using the quadratic formula:

r = (-34 ± √(34² - 4(1)(-432)))/(2(1))
r = (-34 ± √2884)/2
r = (-34 ± 2√721)/2
r = (-17 ± √721)
r ≈ 9.85 mph

So I agree with your answer.

2) In this one, D is the number of problems, and R is the rate of completing problems.

We're looking for the rate of the faster student, so we'll let that be r, and the amount of time for the faster student will be t. Then the rate for the slower student will be (r - 5) and the time for the slower student is (t + 1/4), since 15 minutes is 1/4 hour. So the equations would be:

45 = (r - 5)(t + 1/4)
55 = rt

Again solving the second equation for t and subsituting:

55/r = t

45 = (r - 5)(55/r + 1/4)
180r = (r - 5)(55/r + 1/4)(4r) = (r - 5)(220 + r)
180r = 220r + r² - 1100 - 5r
0 = r² + 35r - 1100
0 = (r + 55)(r - 20)

So r = 20, and you're right again.

3) I always find these easier to reason out than to write an equation for. Since we're looking for Team B's time to build the product alone, we'll call that t.

If Team A can build a product in 4 hours less than Team B, that means Team A is working at a rate of 1 product per (t - 4) hours, or 1/(t - 4) products per hour.

Likewise, Team B works at a rate of 1 product per t hours, or 1/t products per hour.

We know that in 3.75 hours, working together, they make 1 product.

So that means that team A produces:

3.75 hours × 1 product/(t - 4 hours) = 3.75/(t - 4) products

And team B produces:

3.75 hours × 1 product/(t hours) = 3.75/t products

And the sum of the two is 1 whole product:

1 = 3.75/(t - 4) + 3.75/t

Multiply through by t(t - 4):

t(t - 4) = 3.75t + 3.75(t - 4)
t² - 4t = 3.75t + 3.75t - 15
t² - 11.5t + 15 = 0
(t - 10)(t - 1.5) = 0

The 2 solutions for this are t = 10 and t = 1.5. But since we know that Team A would finish 4 hours faster than Team B, we can throw out t = 1.5, as it would produce a negative time for Team A.

So again, you're right; t = 10.

Good job!

Answer 2

1. Let t = time taken in hours:

36/t - 16/(t+0.5) = 6

36/t - 6t/t = 16/(t+0.5)
(36-6t)/t = 16/(t+0.5)
cross multiplying
16t = -6t^2 + 33t + 18
6t^2 - 17t - 18 = 0
use quadratic formula to solve:
t = 3.654 hours

Road course speed:
36/3.654 = 9.85MPH

b) r = rate of slower student t = time of faster student in hours
r(t+0.25) = 45
(r+5)(t) = 55

rt + 0.25r = 45
rt + 5t = 55
substitute
t = 55/r+5
55r/(r+5) + 0.25r = 45
55r + 0.25r(r+5) = 45(r+5)
.25r^2 + 11.25r - 225 = 0
r^2 + 45r - 900 = 0
(r+60)(r-15) = 0 r = 15....rate of faster student r+5 = 20


3) products per hour for team A = x
products per hour for team B = y
t = time in hours
product rate x time = # of products

x(t-4) = 1
yt = 1
x(3.75) + y(3.75) = 1
t = 1/y

x/y - 4x =1
x - 4xy = y
x = y/(1-4y)

y/(1-4y)(3.75) + 3.75y = 1
3.75y + 3.75y(1-4y) = (1-4y)
15y^2-11.5y+1=0
y = 2/3 or 0.1
if 2/3 is used...a negative answer is found for x...so not valid
y=0.1

(0.1)t = 1

t = 10 hours for team B.

Why did you need help again? :)

Answer 3

it is the fourth desperate algebra question you have asked this nighttime. attempt doing all your guy or woman homework, rather of spending plenty time attempting to get somebody else to do it for you.