Algebra math word problems

Question

Joe and Nancy decide to take a road trip. The first 96 miles of the drive is pretty easy, while the last 100 miles of the drive is filled with curves. They drove at an average of 9 miles per hour faster for the first 96 miles of the trip. The entire trip took 4 hours. How fast were Joe and Nancy driving for the first 96 miles of the trip?

Answer 1

Let x = speed for the first 96 miles

     x - 9 = speed for the last 100 miles

 

Distance = (rate)(time), so time = (distance)/(rate)

 

(time for first part of trip) + (time for last part of trip) = 4 hours

 

So, (96)/(x) + (100)/(x - 9) = 4

 

Multiply both sides by the LCD (x)(x - 9) to get 

 

96(x - 9) + 100x = 4x(x - 9)

 

96x - 864 +100x = 4x² - 36x

 

4x² - 232x + 864 = 0

 

Divide both sides by 4 to obtain x² - 58x +216 = 0

 

(x - 4)(x - 54) = 0

 

x = 4 or x = 54

 

Note that x can't equal 4 because, in that case, x - 9 would be negative, so x = 54 miles per hour.

 

 

 

Answer 2

distance=rate*time

d=rt

96=(r+9)(t)

100=(r)(4-t)

96/(r+9)=t from the first equation

substitute this into the second equation

100=(r)(4-[96/(r+9)])

100=4r-96r/(r+9)

multiply both sides by (r+9)

100(r+9)=4r(r+9)-(r+9)(96r)/(r+9)

100r+900=4r²+36r-96r

100r+900=4r²-60r

get all terms on one side of the equation

4r²-60r-100r-900=0

4r²-160r-900=0

divide both sides by 4

r²-40r-225=0

factor  (225=3*3*5*5=(3*3*5)(5)=45*5 and 45-5=40)

(r-45)(r+5)=0

r-45=0

r=45 mph

r+9=45+9=54 mph for the first 96 miles

 

 

Answer 3

Sydnee - I've got an answer for you, but it is not simple, so you'll have to read carefully and I'll try my hardest to make it readable. 

 

This problem depends on the relationship Distance = (Rate)(Time) 

 

That relationship is true during the first (faster) part of their trip and the second (slower) part of their trip. This can be summarized like this: 

 

D_{1 =} R_{1 *} T_{1    and} D₂ = R₂ * T₂

 

Since that expression has lots of different variables in it we want to find some relationships so that we can express some of the variables in terms of others. 

 

The problem tells us that D1 = 96 and D2 = 100 so we can substitute those numbers. 

They also tell us that "they drove an average of 9 mph faster during D1". So we can say 

R1 = R2 + 9 .... or R2 = R1 - 9 

We also know that since the total trip took 4 hours, T1 + T2 = 4 

 

Here's where we do some fancy substitution. Ready? 

 

Take D = R * T and solve for T ..... T = D/R

 

Since we know that T1 = D1/R1   AND   we know that T2 = D2/R2 we can add these two different times together to get the total time (which we know is 4 hours). 

Let's list the variables again: 

D1 = 96

R1 = R 

D2 = 100 

R2 =  R - 9 

 

(D1/R1) + (D2/R2) = Total time

(96/R) + (100/R-9) = 4 

 

To simplify this a little, we can factor out a 4 from both sides: 

4 (24/r   +   25/r-9) = 4 

 

>>>>  divide by 4 on both sides to get

 

24/r   +   25/r-9   = 1

 

now make a common denominator of r(r-9) 

 

24(r-9)    +    25r     =    r(r-9)

  r(r-9)         r(r-9)          r(r-9)

 

24r - 216  +25r    =  r² - 9r

 

Collect like terms on one side of the equals sign and factor: 

 

r²  -  58r + 216 = 0

(r - 54)(r -4) = 0 

Use zero product property.... to solve for r. 

r - 54 = 0 so r = 54     and r -4 = 0, so r = 4

 

Only one of these factors makes sense.... the 54. 

Thats R1! So R2 = 54-9 = 45 mph

 

you can check whether the answer makes sense by using D = R*T 

 

D1 = 96 = 54(T1)   so   T1 =  96/54  = 1.78 hours

D2 = 100 = 45(T2)  so   T2 = 100/45 = 2.22 hours. 

 

and T1 + T2 = 1.78 + 2.22 = 4

 

Yeah! everything checks!