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Old 11-01-2004, 04:57 PM
skate_mate Australia skate_mate is offline
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Average Speeds [Maths]

This is an extension Maths assignment that I recieved a few days ago, it seems to be really simple. Thats why I'm posting here; it seems to be too simple. Maybe there's a trick question somewhere? Help would be much appreciated.

Quote:
Drink Driving

madame Vite, the wife of the Delirian ambassador to Pandemonium, wanted to visit her sister Madame Joyeux, so she set out from Decorum to drive to Atrium where Madame Joyeux lived. Madame Vite is usually a catious driver and her n-powered limousine maintained an exact 40 kilometres an hour on the outward journey. While she was at Atrium however Henry Weinbauer came to visit Madame Joyeux. Madame Vite inbibed rather freely of her choice lillipilli red, and in consequence drove home at 60km/h. What was her average speed for the journey to and from Atrium?
This has an obvious and all-too easy answer. The average velocity from the journey to Atrium (40km/h) additioned to the average velocity on the journey back (60km/h) and then divided by 2 (averaged) to find the average speed for the total journey (50km/h).
Answer = 50km/h.

Is this right? It just seems way too easy. :-S
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Old 11-01-2004, 08:14 PM
Rem Rem is a male Canada Rem is offline
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Re: Average Speeds [Maths]

yea your right... Well thats how you do averages... How about this one:

Nicole has 12 000$ to invest, she placed a part of it in a obligation giving her 8% interest and another in an account giving her 9% interest ... After a year the interest was 1043$ how much did she invest in both accounts
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Old 11-01-2004, 08:38 PM
Tohopekaliga Tohopekaliga is online now
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Re: Average Speeds [Maths]

I don't see a trick question there at all.
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Old 11-01-2004, 11:55 PM
Tsukie Tsukie is offline
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Re: Average Speeds [Maths]

Sounds right.

I almost tried to make a correction, but this doesn't relate to physics. I assume that velocity was involved just as an example, not as a physics term.
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Old 11-02-2004, 05:59 AM
skate_mate Australia skate_mate is offline
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Re: Average Speeds [Maths]

Yeah, sorry, Tsu. You're right about the velocity term, I wrote that early this morning. >_>

Anyway, I can't see a trick question either O_o;; This really is the easiest Maths assignment I've done in my life.
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Old 11-04-2004, 07:44 PM
ice_23 ice_23 is offline
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Re: Average Speeds [Maths]

Quote:
Originally Posted by SchattenReïter
yea your right... Well thats how you do averages... How about this one:

Nicole has 12 000$ to invest, she placed a part of it in a obligation giving her 8% interest and another in an account giving her 9% interest ... After a year the interest was 1043$ how much did she invest in both accounts
How is the interest compounded for each account? Monthly? Yearly? Immediately? Your question is solvable if you know that.

If it's yearly, as I suspect, then the solution is:

Amount in 8% interest bearing account: $3704
Amount in 9% interest bearing account: $8296

Hope this helps.

-Ice
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Old 11-04-2004, 09:02 PM
Dekrem Dekrem is a male United States Dekrem is offline
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Re: Average Speeds [Maths]

Average speed is the change in distance over the change in time, so it's not necessarily the two speeds over two. You might want to recheck that.
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Old 11-04-2004, 09:21 PM
Rem Rem is a male Canada Rem is offline
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Re: Average Speeds [Maths]

Quote:
Originally Posted by ice_23
How is the interest compounded for each account? Monthly? Yearly? Immediately? Your question is solvable if you know that.

If it's yearly, as I suspect, then the solution is:

Amount in 8% interest bearing account: $3704
Amount in 9% interest bearing account: $8296

Hope this helps.

-Ice

thanx for your help but thats not how we learned to do it... the total interest was **looks** 1043$ so what you do is this

X represents 9% Y reps 8%

x+y=12000
0,09x + 0,08 y = 1043
(x+y=12000) x-8
9x+8y = 104300
-8x-8y= -96000
_____________
x=8300
y=3700 So she placed 8300 at 9% Int and 3700 at 8% int... I prolly made a mistake somewhere in there but thats it... welcome to math 11-2 intro..
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Old 11-04-2004, 09:30 PM
Tsukie Tsukie is offline
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Re: Average Speeds [Maths]

Quote:
Originally Posted by DekuKid
Average speed is the change in distance over the change in time, so it's not necessarily the two speeds over two. You might want to recheck that.
That's exactly what I was thinking.

But, this is in a math class, not physics. Velocity is simply used as an example in this question. If you notice, no other variable except velocity was given. The amount of time it took is neither given nor can it be determined, thus the average velocity formula cannot be used.
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Old 11-05-2004, 12:47 AM
Dekrem Dekrem is a male United States Dekrem is offline
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Re: Average Speeds [Maths]

Let's say then that the distance between the two towns is 100 km:

Time = Distance / Speed
(100 km)/(60 km/hr) = 5/3 hr
(100 km)/(40 km/hr) = 5/2 hr
Total Time = 5/3 hrs + 5/2 hrs = 25/6 hrs = 4 1/6 hrs

From what you said before, it can be inferred that the total time should be 4 hrs. But it isn't.

Because the speeds aren't the same both ways, the time isn't the same both ways.
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Old 11-08-2004, 05:34 PM
ice_23 ice_23 is offline
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Re: Average Speeds [Maths]

Quote:
Originally Posted by SchattenReïter
thanx for your help but thats not how we learned to do it... the total interest was **looks** 1043$ so what you do is this

X represents 9% Y reps 8%

x+y=12000
0,09x + 0,08 y = 1043
(x+y=12000) x-8
9x+8y = 104300
-8x-8y= -96000
_____________
x=8300
y=3700 So she placed 8300 at 9% Int and 3700 at 8% int... I prolly made a mistake somewhere in there but thats it... welcome to math 11-2 intro..
Your first two equations are correct and your answer is correct. That is how I solved the problem; but I did the answers in my head initially, hence the slight error.

-Ice
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Old 11-18-2004, 12:17 AM
skate_mate Australia skate_mate is offline
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Re: Average Speeds [Maths]

Quote:
Let's say then that the distance between the two towns is 100 km:

Time = Distance / Speed
(100 km)/(60 km/hr) = 5/3 hr
(100 km)/(40 km/hr) = 5/2 hr
Total Time = 5/3 hrs + 5/2 hrs = 25/6 hrs = 4 1/6 hrs

From what you said before, it can be inferred that the total time should be 4 hrs. But it isn't.

Because the speeds aren't the same both ways, the time isn't the same both ways.
DekuKid, you're 100% correct. The correct answer was 48km/h. I was wrong because of the reasons DekuKid stated. You can check this easily by substituting 120km as the distance and working it out from there.
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