A Jogger ran 7 miles, decreased her speed by 1 mile per hour, and than ran another 8 miles. if her total hows were 2 23/30 , find her speed in each run

Question

Please help I seem to always get stuck, on these word problems, thank you!!! A Jogger ran 7 miles, decreased her speed by 1 mile per hour, and than ran another 8 miles. if her total hows were 2 23/30 , find her speed in each run

Answer 1

distance=rate*time

7=r*t

8=(r-1)(2 23/30-t)

from 7=rt we get r=7/t

substitute into the second equation

8=([7/t]-1)(83/30-t)  (I changed the mixed number into an improper fraction)

multiply the binomials

8=(7/t)(83/30)-(1)(83/30)-(t)(7/t)+(1)(t)

8=(581/30t)-(83/30)-7+t

multiply both sides by 30t

240t=581-83t-210t+30t²

240t=581-293t+30t²

0=-240t+581-293t+30t²

0=581-533t+30t²

30t²-533t+581=0

factor the polynomial

30*581=2*3*5*7*83=(2*3*83)*(5*7)=498*35 and 498+35=533, the coefficient of the middle term

(6t     )(5t     )=0 so far...

the 6 must multiply the 83 as shown above and the 5 multiplies the 7

(6t   7)(5t   83)=0 so far...

(6t-7)(5t-83)=0  

you can't use 5t-83 because the time would be too great

6t-7=0

6t=7

t=7/6 hours

now find the speeds...

7=rt

7=r*(7/6)

multiply both sides by 6/7

(6/7)(7)=r(7/6)(6/7)

6=r

r=6 mph for the 7 mile run

she reduced her speed by 1 mile per hour for the 8 mile run so her speed is 6-1=5 mph for the 8 mile run

check:

8=5([83/30]-[7/6])

8=5([83/30]-[35/30)

8=5(48/30)

8=5(8/5)

8=8

 

 

Answer 2

Distance = (Rate)(Time), so Time = Distance/Rate

 

let x = original speed

 

Time to go the first 7 miles + Time to go the last 8 miles=2 23/30 hr

 

7/x + 8/(x-1) = 83/30

 

Multiply by the LCD, 30x(x-1):

 

210(x-1) + 240x = 83x(x-1)

 

210x - 210 + 240x = 83x² - 83x

 

83x² -533x + 210 = 0

 

x =   533 ± √214369  = 6 mph, 0.42 mph

               166

 

x can't be 0.42, otherwise x - 1 would be negative

 

So, x = 6 mph

 

The jogger ran  6 mph for the first 7 miles and ran 5 mph for the last 8 miles.