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# World problem - another one!!

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Senior Manager
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World problem - another one!! [#permalink]

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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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Joined: 14 Mar 2011
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Re: World problem - another one!! [#permalink]

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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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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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Joined: 16 Feb 2011
Posts: 262
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Kudos [?]: 127 [0], given: 9

Re: World problem - another one!! [#permalink]

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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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Posts: 2021
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Kudos [?]: 1559 [0], given: 376

Re: World problem - another one!! [#permalink]

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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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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?
Intern
Joined: 03 Oct 2011
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Re: World problem - another one!! [#permalink]

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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
Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2021
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Kudos [?]: 1559 [1] , given: 376

Re: World problem - another one!! [#permalink]

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29 Oct 2011, 06:27
1
KUDOS
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

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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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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