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 July 17th, 2014, 02:58 AM #1 Senior Member   Joined: Jan 2012 Posts: 725 Thanks: 7 Find the normal average rate at which Isatu walks to school. Isatu walks a distance of 1.5km to school everyday; her rate of walking being always constant. On a certain day, owing to ill health, she had to reduce her walking rate by 0.5km/ h and as a result, she took extra 6 minutes to reach the school than usual. Find the normal average rate at which Isatu walks to school. Time t = 6/60=1/10h Let her normal walking rate be x. If she reduce it by 0.5km/h, the walking rate becomes (x-0.5)km/h when she was sick. I can't go further here, I need help please. July 17th, 2014, 05:27 AM #2 Senior Member   Joined: Nov 2013 From: Baku Posts: 502 Thanks: 56 Math Focus: Geometry Velocity is v, time is t. vt=1.5 and (v-0.5)(t+0.1)=1.5 6 min. is 0.1 hours. We get t=(v-0.5)/5 and we know t=1.5/v. Rest is easy. Change your profile picture. July 17th, 2014, 08:21 AM #3 Senior Member   Joined: Apr 2014 From: Europa Posts: 575 Thanks: 176 You forgot change your profile picture. $\displaystyle \color{blue}{Read \ "Decalogue"... \\\;\\ and\ let \ you \ read .}$ July 17th, 2014, 11:20 AM   #4
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 Originally Posted by aurel5 You forgot change your profile picture. $\displaystyle \color{blue}{Read \ "Decalogue"... \\\;\\ and\ let \ you \ read .}$
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 Originally Posted by tahirimanov Velocity is v, time is t. vt=1.5 and (v-0.5)(t+0.1)=1.5 6 min. is 0.1 hours. We get t=(v-0.5)/5 and we know t=1.5/v. Rest is easy. Change your profile picture.
Why do you want me to change my profile picture? July 17th, 2014, 09:27 PM   #5
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 Originally Posted by tahirimanov We get t=(v-0.5)/5 and we know t=1.5/v.
How did you get the two equations above? July 18th, 2014, 12:41 AM   #6
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 Originally Posted by Chikis Why do you want me to change my profile picture?

There is no place .

So, it's better let us, profane humanity and unconfessed,

to look calm and safe the sanctity of mathematics. July 18th, 2014, 06:23 AM #7 Senior Member   Joined: Nov 2013 From: Baku Posts: 502 Thanks: 56 Math Focus: Geometry (v-0.5)(t+0.1)=1.5 therefor vt+0.1v-0.5t-0.05=1.5 and we know vt=1.5 therefore 0.1v-o.5t-0.05=0 therefore v-5t-0.5=0 and t=1.5/v therefore v-7.5/v-0.5=0 v^2 - 0.5v - 7.5 = 0 July 19th, 2014, 09:00 PM   #8
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 Originally Posted by tahirimanov $(v-0.5)(t+0.1)=1.5$ therefor $vt+0.1v-0.5t-0.05=1.5$ and we know$\displaystyle { vt=1.5\ therefore\ 0.1v-0.5t-0.05=0\ therefore\ v-5t-0.5=0 and t=\frac{1.5}{v} \ therefore \ v-\frac{7.5}{v}-0.5=0}$ $v^2-0.5v-7.5= 0$
Quote:
 Originally Posted by tahirimanov $v^2-0.5v-7.5= 0$
I can see clearly that the equation above cannot be solved using factorization method. Or could it be solved using factorization method?
Using quadratic formula, we will come to a point where we have $\displaystyle \frac{0.5\pm\sqrt{30.25}}{2}$
= $\displaystyle \frac{0.5pm5.5}{2}\rightarrow \\ v=3 \;or -2.5$
So normal rate = $3km/h$

Last edited by Chikis; July 19th, 2014 at 09:19 PM. July 19th, 2014, 09:15 PM #9 Senior Member   Joined: Nov 2013 From: Baku Posts: 502 Thanks: 56 Math Focus: Geometry -2.5 is also valid solution, if Isatu walks backwards. July 19th, 2014, 10:22 PM   #10
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 Originally Posted by Chikis I can see clearly that the equation above cannot be solved using factorization method. Or could it be solved using factorization method?
Yes; multiply by 2:
2v^2 - v - 15 = 0
Now factor:
(2v + 5)(v - 3) = 0 Tags average, find, isatu, normal, rate, school, walks Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Shayna Calculus 1 June 4th, 2014 05:51 PM headsupman Algebra 1 February 6th, 2013 02:59 PM ChristinaScience Calculus 2 October 19th, 2011 11:45 PM ChristinaScience Calculus 6 October 19th, 2011 11:21 PM symmetry Algebra 6 June 12th, 2007 01:53 PM

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