My Math Forum Find the normal average rate at which Isatu walks to school.

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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. $\displaystyle \color{blue}{Read \ "Decalogue"... \\\;\\ and\ let \ you \ read .}$
Quote:
 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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Quote:
 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

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