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Forum index » Science and Technology » Math » Undergraduate
NEED HELP ONCE AGAIN
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Missy
science forum beginner


Joined: 02 Jul 2006
Posts: 5

PostPosted: Sun Jul 16, 2006 2:16 am    Post subject: NEED HELP ONCE AGAIN Reply with quote

Please help I need to know how to solve each of the following
step-by-step:

1. Daria can wash and detail 3 cars in 2 hours. Larry can wash and
detail the same 3 cars in
1.5 hours. About how long will it take to wash and detail the 3
cars if Daria and Larry worked
together?



2. 100 people will attend the theatre if tickets cost $40 each. For
each $5 increase in price, 10
fewer people will attend. what price will deliver the maximum
dollar sales?



3. How many 3-person groups can be formed in a club with 8 people?
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William Elliot
science forum Guru


Joined: 24 Mar 2005
Posts: 1906

PostPosted: Sun Jul 16, 2006 4:15 am    Post subject: Re: NEED HELP ONCE AGAIN Reply with quote

TO DEMAND ATTENTION MISSY RUDDY SHOUTED:

Quote:
Please help I need to know how to solve each of the following
step-by-step:

Are you wanting us to do your homework?

That we do not. We offer advise, hints and help.

Quote:
1. Daria can wash and detail 3 cars in 2 hours. Larry can wash and
detail the same 3 cars in
1.5 hours. About how long will it take to wash and detail the 3
cars if Daria and Larry worked
together?

How many cars per hour can D do?

How many cars per hour can L do?
How many cars per hour can D and L do together?
Can you take it from there?

Quote:
2. 100 people will attend the theatre if tickets cost $40 each. For
each $5 increase in price, 10
fewer people will attend. what price will deliver the maximum
dollar sales?

Let p be number of people. c cost per ticket. t take. g gouge
t = (p - 10g/5)(c + g) = (100 - 2g)(40 + g)

There, I've done the hard part. Let's see what you can do with the
straight forward easy stuff. Beware I charge for financial advise.

Quote:
3. How many 3-person groups can be formed in a club with 8 people?

Let see what you accomplished and we'll help you from there.
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jasen
science forum beginner


Joined: 28 Jun 2006
Posts: 16

PostPosted: Sun Jul 16, 2006 8:41 am    Post subject: Re: NEED HELP ONCE AGAIN Reply with quote

On 2006-07-16, Missy <tawanda.green@yahoo.com> wrote:
Quote:
Please help I need to know how to solve each of the following
step-by-step:

1. Daria can wash and detail 3 cars in 2 hours. Larry can wash and
detail the same 3 cars in
1.5 hours. About how long will it take to wash and detail the 3
cars if Daria and Larry worked
together?

how many cars can each, and both, do in 6 hours.

Quote:
2. 100 people will attend the theatre if tickets cost $40 each. For
each $5 increase in price, 10
fewer people will attend. what price will deliver the maximum
dollar sales?

takings = price * ( 180 - ( price * 2 ) )

solve for d.takings / d.price = 0

Quote:
3. How many 3-person groups can be formed in a club with 8 people?

8!/(3!*(8-3)!)

--

Bye.
Jasen
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G.E. Ivey
science forum Guru


Joined: 29 Apr 2005
Posts: 308

PostPosted: Sun Jul 16, 2006 11:52 am    Post subject: Re: NEED HELP ONCE AGAIN Reply with quote

Quote:
On 2006-07-16, Missy <tawanda.green@yahoo.com> wrote:
Please help I need to know how to solve each of the
following
step-by-step:

1. Daria can wash and detail 3 cars in 2 hours.
Larry can wash and
detail the same 3 cars in
1.5 hours. About how long will it take to wash
and detail the 3
cars if Daria and Larry worked
together?

how many cars can each, and both, do in 6 hours.

Oh, that's nice! I confess I would have calculated "number of cars per hour" but that's harder.


Missy, do you see where he got "6 hours"? We are told how many cars Daria can wash in two hours: multiples of 2 hours are 2, 4, 6, 8, etc. We are told how many cars Larry can wash in 1.5 hours. Multiples of 1.5 hours are 1.5, 3, 4.5, 6, etc. They match up, and we can calculate how many cars each can wash in 6 hours. Once you know that, divide both cars and 6 hours by a number that will give 3 cars.

Quote:
2. 100 people will attend the theatre if tickets
cost $40 each. For
each $5 increase in price, 10
fewer people will attend. what price will
deliver the maximum
dollar sales?

takings = price * ( 180 - ( price * 2 ) )

solve for d.takings / d.price = 0

From the first problem, I suspect that this is not a calculus problem and that method is not appropriate (especially if you don't know what "d.takings/d.price" means!). I'm going to write T for "takings" and P for "price" to simplify the writing.
T= P(180- 2P)= 180P- 2P^2= 2(90P- P^2). Graphically, that is a parabola opening downward (because of the "-" in "-P^2"). The highest point (maximum dollar sales or "takings") will be at the vertex of the parabola. Complete the square to find the vertex.

Quote:

3. How many 3-person groups can be formed in a
club with 8 people?

8!/(3!*(8-3)!)

Missy, you probably have that formula in you book. Here's how you could work it out yourself: Choose one person to be in a group- there are 8 possible choices (any one of the 8 people in the club), 8 different ways you could do that. Now there are 7 people left "unchosen" so there are 7 ways to choose the next person. Now there are 6 people left unchosen, so there are 6 ways to choose the 3rd person. There are 8*7*6 ways to make those 3 choices. BUT!! Those are chosen in a particular order. That counts, for example, {Bill, Mary, Fred}(chosen in that order) as different from {Mary, Bill, Fred} (chosen in THAT order. Since it doesn't matter in which order they were chosen, we need to divide by the number of different orders, 3!= 3(2)(1).

The number of different 3 person groups is (8*7*6)/(3*2*1)= (8*7*6)/6= 8*7= 42.

There is no standard short notation for something like 8*7*6 but there is for 8*7*6*5*4*3*2*1= 8! so we can write 8*7*6= (8*7*6*5*4*3*2*1)/(5*4*3*2*1)= 8!/5!= 8!/(8-3)! . That's where 8!/(3!(8-5)!) comes from.
Quote:
--

Bye.
Jasen
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Badger
science forum beginner


Joined: 07 May 2006
Posts: 38

PostPosted: Mon Jul 17, 2006 3:31 am    Post subject: Re: NEED HELP ONCE AGAIN Reply with quote

On Sun, 16 Jul 2006 07:52:01 EDT, "G.E. Ivey"
<george.ivey@gallaudet.edu> wrote:

[snip]

Quote:
There is no standard short notation for something like 8*7*6 but
there is for 8*7*6*5*4*3*2*1= 8! so ...

Although perhaps not appropriate for this thread, for various forms of
factorials and notation see

<http://mathworld.wolfram.com/topics/Factorials.html>,

e.g. the falling and rising factorial and the Pochhammer symbol.
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Dave L. Renfro
science forum Guru


Joined: 29 Apr 2005
Posts: 570

PostPosted: Mon Jul 17, 2006 7:28 pm    Post subject: Re: NEED HELP ONCE AGAIN Reply with quote

Missy wrote (in part):

Quote:
Please help I need to know how to solve each of the
following step-by-step:

1. Daria can wash and detail 3 cars in 2 hours.
Larry can wash and detail the same 3 cars in 1.5 hours.
About how long will it take to wash and detail the
3 cars if Daria and Larry worked together?

I've always found it easy to do this type of problem by
considering them to be a generalized distance-rate-time
problem.

distance -- number of cars washed
rate -- cars washed per hour
time -- hours

The fundamental equation is: distance = (rate)(time),
or D = RT for short. The fundamental principle is
that you add the rates when they're working together.
For example, if you throw a rock out of a car window,
the speed of the rock is the throwing speed plus the
car speed.

Let R_d be Daria's rate and R_l be Larry's rate.

D = RT for Daria: (3 cars) = (R_d)(2 hours)

D = RT for Larry: (3 cars) = (R_l)(1.5 hours)

D = RT when working together: (3 cars) = (R_d + R_l)(t)

Quote:
From the first two equations we get R_d = 3/2 cars/hour
and R_l = 2 cars/hour. Plugging these values into the

third equation gives

(3 cars) = (3/2 cars/hour + 2 cars/hour)(t),

and now just solve for t.

The same method can be used when the combined rate
is given along with one of the individual rates
(but not the other rate), when more than two rates
are being considered, when one of the times is an
unknown (with enough other information given), etc.

Dave L. Renfro
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