SOLUTION: Dorothy drives 10 kilometers, then increases her speed by 10 kilometers per hour and drives another 25 kilometers. Find her original speed if she drove for 45 minutes.
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Question 102852: Dorothy drives 10 kilometers, then increases her speed by 10 kilometers per hour and drives another 25 kilometers. Find her original speed if she drove for 45 minutes.
Please help me with this!! Thanks!! I can't understand how to answer this!!
Found 2 solutions by ankor@dixie-net.com, scott8148:
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Dorothy drives 10 kilometers, then increases her speed by 10 kilometers per hour and drives another 25 kilometers. Find her original speed if she drove for 45 minutes.
:
Let s = original speed
then
Increased speed = (s+10)
:
Write a time equation: Time = distance/speed; Change 45 min to hrs: 45/60 = .75 hr
:
Time for 10 km + time for 25 km = .75 hr
+ = .75
:
Multiply equation by s(s+10) and you have:
10(s+10) + 25s = .75(s(s+10)
:
10s + 100 + 25s = .75s^2 + 7.5s
:
35s + 100 = .75s^2 + 7.5s
:
Arrange as a quadratic equation:
.75s^2 + 7.5s - 35s - 100 = 0
:
.75s^2 - 27.5s - 100 = 0
:
Use the quadratic formula to solve for s: a=.75; b=-27.5; c=-100
:
:
:
:
:
; we only want the positive solution here
:
:
s = 40 mph is her speed for the 1st 10 km (original speed
:
:
Check our solution using the time equation:
+ = .75 as given
:
:
Did this help you understand this?
Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
time=distance/rate ... let x=speed ... 45/60=.75
(10/x)+(25/(x+10))=.75 ... 10(x+10)+25(x)=.75(x)(x+10) ... 40x+400+100x=3x^2+30x
3x^2-110x-400=0 ... (3x+10)(x-40)=0 ... x=-10/3 and x=40 ... negative value not realistic
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