Math problem that's confusing me. (+2 more)

xvillexvalox · May 23, 2007

[hide= Problem 1 (boat) solved by tzone92. (THANKS!)]

A boat travels upstream for 15 miles against a current of 3 mph, then it travels downstream 12 miles with the same current. If the 27-mile trip took 2.25 hours, how fast was the boat traveling in still water?

I set this up:

x = Speed in still water

With current: 12 Miles ... x+3 mph ... 12/(x+3) hours

 No current: 15 Miles ... x-3 mph ... 15/(x-3) hours

My equation work:


12/(x+3) + 15/(x-3) = 9/4

 48(x-3) + 60(x+3) = 9(x+3)(x-3)

    48x-144+60x+180= 9x^2 -81

           (108x+36= 9x^2 -81)/9

              12x+4= x^2 -9

        x^2 +12x+4 = -9

        x^2 +12x.. = -13  ***

        x^2 +12x+36=-13+36 ***

           (x+6)^2 = 23

               x+6 = (+-)sq root(23)

*** Completing the square technique

And then I get 2 negetive answers, which can't be applied to the problem.

What did I do wrong? Is my starting equation correct?

I believe you set it up the right way, but you forgot to make x^2 negative at this step

12x+4= x^2 -9

        x^2 +12x+4 = -9

So let me rework it here:


12/(x+3) + 15/(x-3) = 9/4

 48(x-3) + 60(x+3) = 9(x+3)(x-3)

    48x-144+60x+180= 9x^2 -81

           (108x+36= 9x^2 -81)/9

              12x+4= x^2 -9

       - x^2 +12x+4 = -9

      x^2-12x-4=9

      x^2-12x=13

      x^2-12x+36=49

       (x-6)^2=49

        x-6=(+-)sq.root 49

        x-6=(+-)7

         x=13

[/hide]

Problem2, I'm pretty sure I did this one right

[hide]

Mary left home at 12 noon adn drove to visit her mom who lives 20 miles away. After she arrived she stayed for 2 hours. Afterwards she drove to her freind Sue's house who lives 15 miles from her mom. She arrived at Sue's at 3 o'clock. If her speed heading for Sue's was 5 mph faster than her speed going to her mom's, find her average speed on her way to Sue's.


To mom: 20 Miles ..... x mph ... 20/x hours

To Sue: 15 Miles ... x+5 mph ... 15/x+5 hours

Equation work:


20/x + 15/(x+5) = 3

     20x+100+15x= 3x^2+15x

                0=3x^2-20-100



QUADRATIC FORMULA

x= [20 (-+)sq.root(400-4(-100)(3))]/6

x= [20(-+)40]/6

x= (-10/3) , 10

Average speed to Sue's = 15 mph[/hide]

[hide= Problem 3, easy-ish, pretty sure I did it right]

A square has a side of 5 feet. How much should each of its sides be increased in order to double its area?

Start off: The 5x5 square has an area of 25.

Does this look correct?


(5+x)^2=50

  5+x=(-+)5[sq.root(2)]

     x=(-+)5[sq.root(2)]-5

     x=2.07 , -12.07

My answer: You would need to add 2.07 to both sides to double its area.

[/hide]

Nazgul740 · May 23, 2007

I havn't done this type of math in ages... but you need to plug the time in somwhere...

And i think your first code thingy is wrong... i just don't see where... something about its 15-12/x(-+3) Im not sure its sopposed to be divided... not sure though

I would set it up with the same distances equaling the time then working out the average rate and that would give you part of the answer then you would have to subtract the 3 extra miles from the answer..

Meh i dunno :XD: I took technical math in college this year so all my applications went down the drain :uhh: :ohnoes:

xvillexvalox · May 23, 2007

I havn't done this type of math in ages... but you need to plug the time in somwhere...

And i think your first code thingy is wrong... i just don't see where... something about its 15-12/x(-+3) Im not sure its sopposed to be divided... not sure though

I would set it up with the same distances equaling the time then working out the average rate and that would give you part of the answer then you would have to subtract the 3 extra miles from the answer..

Meh i dunno :XD: I took technical math in college this year so all my applications went down the drain :uhh: :ohnoes:

I divided it because time = distance/rate

and I don't know where you're going with setting the distances equal to eachother...

tzone92 · May 23, 2007

I believe you set it up the right way, but you forgot to make x^2 negative at this step

12x+4= x^2 -9

        x^2 +12x+4 = -9

So let me rework it here:


12/(x+3) + 15/(x-3) = 9/4

 48(x-3) + 60(x+3) = 9(x+3)(x-3)

    48x-144+60x+180= 9x^2 -81

           (108x+36= 9x^2 -81)/9

              12x+4= x^2 -9

       - x^2 +12x+4 = -9

      x^2-12x-4=9

      x^2-12x=13

      x^2-12x+36=49

       (x-6)^2=49

        x-6=(+-)sq.root 49

        x-6=(+-)7

         x=13

xvillexvalox · May 23, 2007

I believe you set it up the right way, but you forgot to make x^2 negative at this step

12x+4= x^2 -9

        x^2 +12x+4 = -9

So let me rework it here:


12/(x+3) + 15/(x-3) = 9/4

 48(x-3) + 60(x+3) = 9(x+3)(x-3)

    48x-144+60x+180= 9x^2 -81

           (108x+36= 9x^2 -81)/9

              12x+4= x^2 -9

       - x^2 +12x+4 = -9

      x^2-12x-4=9

      x^2-12x=13

      x^2-12x+36=49

       (x-6)^2=49

        x-6=(+-)sq.root 49

        x-6=(+-)7

         x=13

You're right! <3:

I might be putting up a similiar problem soon to.

tzone92 · May 23, 2007

Hooray, I helped someone!

tzone92 · May 23, 2007

I'll check these ones, they look like they are most likely correct

A square has a side of 5 feet. How much should each of its sides be increased in order to double its area?

Start off: The 5x5 square has an area of 25.


x+5=sq.root 50

x+5=5(sq.root 2)

x=5(sq. root 2)-5

x=(+-)7.071-5

x=2.071

_________________

Mary left home at 12 noon adn drove to visit her mom who lives 20 miles away. After she arrived she stayed for 2 hours. Afterwards she drove to her freind Sue's house who lives 15 miles from her mom. She arrived at Sue's at 3 o'clock. If her speed heading for Sue's was 5 mph faster than her speed going to her mom's, find her average speed on her way to Sue's.


speed to Sue's=x

speed to mom's=x-5

total hours spent driving=1

total distance=35



x+(x-5)=35

2x=30

x=15

_____________________

Yes, it looks like those two are right.

assassin_696 · May 23, 2007

Giving it a quick glance over, they look spot on :)

darkforaster · May 23, 2007

problem 2 is wrong. if she was going to sue's 5 mph faster than she went to her mom's then her speed can't be 15 mph. If it was then she went to her mother's house at 10mph. bUt since her mother is 20 miles away it takes her 2 hours. She then stays at her Mom's house for another 2 hours.

So by the time she left for Sue's, it was 4:00. Sorry, but it's wrong. Re calculate. i'll help!

edit: let me think for a bit. hmm.... yes!

*phew!* 38 mph. She gets there at 3.00 with 0.04 minutes added on.

That was hard. I'll do the calculations in a seperate post.

~Summary~

Her average speed is around 38 mph. If you want it exact. It goes into decimals. But rounded it leads to that answer.

if all those calculations were for nothing. I'll go mad.

LordSoultar · May 23, 2007

For the second problem you messed up because you forgot that although her total trip was 3 hours, she only spent 1 hour of it traveling because she had been at her mom's for 2 hours.

So the new problem would look like this:

20/(x-5) + 15/x = 1

20(x-5)(x)/(x-5) + 15(x)(x-5)/x = 1(x)(x-5)

20x + 15(x-5) = x^2 - 5x

20x + 15x - 75 = x^2 - 5x

35x = x^2 - 5x + 75

0 = x^2 - 40x + 75

Quadratic Formula:

x = (40 +/- sqrt(1600 - 300))/2

x = (40 +/- sqrt(1300))/2

x = (40 +/- 10*sqrt(13))/2

x = 20 +/- 5*sqrt(13)

x ~= {1.97, 38.03}

Edit: Going to try and solve it a few other ways since this seems like an odd answer, will post if I find it to be wrong.

Ignore the edit, what made it seem odd was having two positive answers and one of them being obviously incorrect. The 38.03 is correct though, because if 1.97 was her speed on the way to Sue's then her speed on the way to her mom's house would be -3.03 m/h which is not possible.

darkforaster · May 24, 2007

20/(x-5) + 15/x = 1

20(x-5)(x)/(x-5) + 15(x)(x-5)/x = 1(x)(x-5)

20x + 15(x-5) = x^2 - 5x

20x + 15x - 75 = x^2 - 5x

35x = x^2 - 5x + 75

0 = x^2 - 40x + 75

Quadratic Formula:

x = (40 +/- sqrt(1600 - 300))/2

x = (40 +/- sqrt(1300))/2

x = (40 +/- 10*sqrt(13))/2

x = 20 +/- 5*sqrt(13)

x ~= {1.97, 38.03}

Ignore the edit, what made it seem odd was having two positive answers and one of them being obviously incorrect. The 38.03 is correct though, because if 1.97 was her speed on the way to Sue's then her speed on the way to her mom's house would be -3.03 m/h which is not possible.

thanks for doing the sum straight after I've posted the answer to the same thing. :wink: :evil:

fortunately. you've written the quadratic formula etc. I just pointed out what the answer truly was. :-w

captindeath · May 24, 2007

the third one is correct and the second one works out when I looked over the work.

Ginger_Warrior · May 25, 2007

Firstly, I don't know what a mathematic is. I can however help you with mathematics.

My blast on American renaming of the English Dictionary over... I can't understand how Q2 is right, as you haven't taken into consideration the two hours she spent at her Mum's, as mentioned in the question. I'll show you with a simple check:

MPH to Sue's = 15, therefore MPH to Mum's = 15-5 = 10

Hours = Miles/MPH

Hours to Sue's = 15/15 = 1, Hours to Mum's = 20/10 = 2

Hours Total = 1 + 2 + 2 (the two she spent at her mother's) = 5

She was only supposed to be out for 3, so 5 is wrong

Now, I've tried doing this for myself but I keep getting strange answers like 15+5(13)^1/2, which comes very close to three hours, but not quite. So I'm unsure. Either way, I strongly feel the answer 15 is wrong. If it helps, my answers are nearer 35mph than 15mph.

Q3 is right, but you've made it much more complex than it needs to be lol:

Area needed = 2(5x5) = 50

Side = 50^1/2

Difference between new side and old side = 50^1/2 - 5 = 2.07 feet

Then again, if you get the right answer, does it really matter?

As for Q1, looks too physics-y for my liking, so I won't even attempt that.

But like I said (and if someone wants to show me how Q2 is right, please do PM me) I think the answer of 15mph is wrong.

EDIT: Sorry, I didn't see that others had already seen this error. I apologise if it seemed like I ripped off your work.

zach312 · May 25, 2007

#3 is completely correct, just did it myself a different way and it's correct.

Math problem that's confusing me. (+2 more)

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