MATH SOLVE

2 months ago

Q:
# PLEASE HELP:How do I solve this:If james goes 15 miles faster per hour, he will go 260 miles in 20 minutes less than normal. What is the original rate?

Accepted Solution

A:

15 mi/h= 15mi/60 min= 1/4 (mi/min)

x is original speed (mi/min)

t is original time

x*t =260

(x+1/4)=5x/4 is new speed

t-20 - new time

(x+1/4)(t-20)=260

xt = (x+1/4)(t-20)

xt=xt + t/4 - 20x - 5

t/4=20x+5

t=80x +20

We can substitute t=80x +20 into equation x*t =260

x*(80x+20)=260

80x²+20x-260=0

4x²+x-13=0

D=b² - 4ac = 1+4*4*13= 209

x=(-1+/-√209)/2*4

x=1.68 mi/min

1.68mi/min * 60 min/1h≈100 mi/h

Check

260/100=2.6h time with old speed

260/(115)=2.26 h time with new speed

2.6-2.26 = 0.34 h difference between old and new time

0.34h*60min/1h≈20 min difference between old and new time in minutes

x is original speed (mi/min)

t is original time

x*t =260

(x+1/4)=5x/4 is new speed

t-20 - new time

(x+1/4)(t-20)=260

xt = (x+1/4)(t-20)

xt=xt + t/4 - 20x - 5

t/4=20x+5

t=80x +20

We can substitute t=80x +20 into equation x*t =260

x*(80x+20)=260

80x²+20x-260=0

4x²+x-13=0

D=b² - 4ac = 1+4*4*13= 209

x=(-1+/-√209)/2*4

x=1.68 mi/min

1.68mi/min * 60 min/1h≈100 mi/h

Check

260/100=2.6h time with old speed

260/(115)=2.26 h time with new speed

2.6-2.26 = 0.34 h difference between old and new time

0.34h*60min/1h≈20 min difference between old and new time in minutes