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World problem - another one!!

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World problem - another one!! [#permalink]

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New post 26 Aug 2011, 02:19
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A store selld newspaper A at price of $1 and B at price of $1.25. In a certain day, if r percent of the total income is from A, and p percent of the total copies sold are A, in terms of r, p=?
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Re: World problem - another one!! [#permalink]

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New post 26 Aug 2011, 02:30
A store selld newspaper A at price of $1 and B at price of $1.25. In a certain day, if r percent of the total income is from A, and p percent of the total copies sold are A, in terms of r, p=?


x-number to total copies sold

income from A = Qty of A* 1,

r/100* x (1+1.25) = p/100*x*1

income from A in terms of r = Income from A in terms of p

2.25 r = p
r= p/2.25
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Re: World problem - another one!! [#permalink]

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New post 26 Aug 2011, 05:59
DeeptiM wrote:
A store selld newspaper A at price of $1 and B at price of $1.25. In a certain day, if r percent of the total income is from A, and p percent of the total copies sold are A, in terms of r, p=?


DeeptiM:
Please take care while posting question. This question looks as if you have formatted it in your own words; the options are also missing. Please mention the source of the question.

A's revenue: p*1=p
B's revenue: (100-p)*1.25=125-1.25p
Total revenue=p+125-1.25p=125-0.25p

We know, A's revenue is r% of the total revenue

\(\frac{p}{125-0.25p}=\frac{r}{100}\)

OR

\(r=\frac{100p}{125-0.25p}\)

Likewise, please solve to get p in terms of r. This looks time consuming to me.
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Re: World problem - another one!! [#permalink]

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New post 26 Aug 2011, 09:31
fluke wrote:
DeeptiM wrote:
A store selld newspaper A at price of $1 and B at price of $1.25. In a certain day, if r percent of the total income is from A, and p percent of the total copies sold are A, in terms of r, p=?


DeeptiM:
Please take care while posting question. This question looks as if you have formatted it in your own words; the options are also missing. Please mention the source of the question.

A's revenue: p*1=p
B's revenue: (100-p)*1.25=125-1.25p
Total revenue=p+125-1.25p=125-0.25p

We know, A's revenue is r% of the total revenue

\(\frac{p}{125-0.25p}=\frac{r}{100}\)

OR

\(r=\frac{100p}{125-0.25p}\)

Likewise, please solve to get p in terms of r. This looks time consuming to me.


Hi Fluke, Thanks a lot for the help.

I have posted the question in exactly similar terms in which it was presented to me..with no options however, the answer you derived is exactly what i have..

The question is from one of my practice set.
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Re: World problem - another one!! [#permalink]

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New post 26 Aug 2011, 13:08
DeeptiM wrote:
Hi Fluke, Thanks a lot for the help.

I have posted the question in exactly similar terms in which it was presented to me..with no options however, the answer you derived is exactly what i have..

The question is from one of my practice set.


Really!!! My bad then. This really sounded strange!! Good that the answer matched luckily and it helped. thanks for posting.
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Re: World problem - another one!! [#permalink]

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New post 28 Oct 2011, 15:54
I didn't get it.
P is the percent copies ....
To find the A' and B' income, why u multiple "p"? shouldn't we multiple the real numbers, not percent, of copies sold?
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Re: World problem - another one!! [#permalink]

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New post 29 Oct 2011, 02:12
fluke wrote:
DeeptiM wrote:
A store selld newspaper A at price of $1 and B at price of $1.25. In a certain day, if r percent of the total income is from A, and p percent of the total copies sold are A, in terms of r, p=?


DeeptiM:
Please take care while posting question. This question looks as if you have formatted it in your own words; the options are also missing. Please mention the source of the question.

A's revenue: p*1=p
B's revenue: (100-p)*1.25=125-1.25p
Total revenue=p+125-1.25p=125-0.25p

We know, A's revenue is r% of the total revenue

\(\frac{p}{125-0.25p}=\frac{r}{100}\)

OR

\(r=\frac{100p}{125-0.25p}\)

Likewise, please solve to get p in terms of r. This looks time consuming to me.



Hi Fluke
Could you please explain it in detail.. Please elaborate your answer..
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Re: World problem - another one!! [#permalink]

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New post 29 Oct 2011, 06:27
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ashimasood wrote:
fluke wrote:
DeeptiM wrote:
A store selld newspaper A at price of $1 and B at price of $1.25. In a certain day, if r percent of the total income is from A, and p percent of the total copies sold are A, in terms of r, p=?


DeeptiM:
Please take care while posting question. This question looks as if you have formatted it in your own words; the options are also missing. Please mention the source of the question.

A's revenue: p*1=p
B's revenue: (100-p)*1.25=125-1.25p
Total revenue=p+125-1.25p=125-0.25p

We know, A's revenue is r% of the total revenue

\(\frac{p}{125-0.25p}=\frac{r}{100}\)

OR

\(r=\frac{100p}{125-0.25p}\)

Likewise, please solve to get p in terms of r. This looks time consuming to me.



Hi Fluke
Could you please explain it in detail.. Please elaborate your answer..


Which part you didn't understand?

p percent of total copies sold were "A". If p=20%, then B must have sold 80% i.e. (100-p)%

If A sold p percent; B must have sold (100-p)%.

Now, let's suppose the number of copies sold be 100.

A sold = p
B sold = 100-p

Money collected from A's sale=p*($1)=p
Money collected from B's sale=(100-p)*($1.25)=125-1.25p
Total income=p+125-1.25p=125-0.25p

Now,
In a certain day, if r percent of the total income is from A;

r% of total income= (r/100)*(125-0.25p)

A's income= p

\(p=(\frac{r}{100})*(125-0.25p)\)

\(100p=125r-0.25pr\)

\(100p+0.25pr=125r\)

\(10000p+25pr=12500r\)

\(400p+pr=500r\)

\(p(400+r)=500r\)

\(p=\frac{500r}{(400+r)}\)
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Re: World problem - another one!! [#permalink]

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New post 29 Oct 2011, 11:37
Got it! Thanks a lot fluke. :)
Re: World problem - another one!!   [#permalink] 29 Oct 2011, 11:37
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World problem - another one!!

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