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Some people form a joint account for one year with the condi

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Some people form a joint account for one year with the condi [#permalink] New post 03 Sep 2013, 07:03
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Some people form a joint account for one year with the condition that every month each member deposits an amount equal to the number of members in the account in that month. Also, the person who withdraws from the account before the end of the year get his/her amount at the end of the year. After 6 months, 1/4th of the people withdraw and 1/3rd of the remaining withdraw after 3 months. At the end of the year there is an amount of Rs. 4860 in the account before paying of the withdrawn amounts. Find out the number of people in the beginning of the year.

A. 12
B. 24
C. 36
D. 48
E. 60
[Reveal] Spoiler: OA

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Last edited by Bunuel on 03 Sep 2013, 07:08, edited 1 time in total.
Renamed the topic and edited the tags.
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Re: Some people form a joint account for one year with the condi [#permalink] New post 03 Sep 2013, 17:41
Qoofi wrote:
Some people form a joint account for one year with the condition that every month each member deposits an amount equal to the number of members in the account in that month. Also, the person who withdraws from the account before the end of the year get his/her amount at the end of the year. After 6 months, 1/4th of the people withdraw and 1/3rd of the remaining withdraw after 3 months. At the end of the year there is an amount of Rs. 4860 in the account before paying of the withdrawn amounts. Find out the number of people in the beginning of the year.

A. 12
B. 24
C. 36
D. 48
E. 60


Need the solution, cant reach to B as the solution
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Re: Some people form a joint account for one year with the condi [#permalink] New post 03 Sep 2013, 19:42
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N people in the start.

first 6 months there would 6.N.N money deposited.
1/4 members are gone so 3/4 left
Then next 3 month 3.3N/4.3N/4

then 1/3 of remaining gone = 3N/4*2/3 = N/2
so money deposited 3.N/2.N/2

total = 6N^2+27/16N^2+3/4N^2 = 4860
solve for N = 24
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Re: Some people form a joint account for one year with the condi [#permalink] New post 03 Sep 2013, 23:30
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Let x be the number of people in the beginning.
Amount deposited for 6 months = 6*x*x = 6x^2

Number of people for the next 3 months = x -x/4 = 3x/4

Amount deposited for the next 3 months = 3 * (3x/4)^2

Number of people for the last 3 months = 3x/4 - (1/3 * 3x/4) = x/2

Amount deposited for the last 3 months = 3* (x/2)^2

Total amount = 6x^2 + 27x^2/16 + 3x^2/4 = 4860

x= 24 (option B)
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Re: Some people form a joint account for one year with the condi [#permalink] New post 08 Nov 2014, 08:15
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Re: Some people form a joint account for one year with the condi [#permalink] New post 14 Apr 2015, 04:37
Qoofi wrote:
Let x be the number of people in the beginning.
Amount deposited for 6 months = 6*x*x = 6x^2

Number of people for the next 3 months = x -x/4 = 3x/4

Amount deposited for the next 3 months = 3 * (3x/4)^2

Number of people for the last 3 months = 3x/4 - (1/3 * 3x/4) = x/2

Amount deposited for the last 3 months = 3* (x/2)^2

Total amount = 6x^2 + 27x^2/16 + 3x^2/4 = 4860

x= 24 (option B)


Why are we multiplying things two times(^2)?
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Re: Some people form a joint account for one year with the condi [#permalink] New post 15 Apr 2015, 00:00
ssriva2 wrote:
Why are we multiplying things two times(^2)?


Because at the end of each month, each person deposits an amount equal to the number of people. For example, if there are 24 people, each would deposit $24 (or whatever currency the question is asking), meaning a total of 24*24 dollars is deposited. So if you start out with \(n\) people:

> After month 1: (\(n\) x \(n\)) deposited = \(n^2\)
> After month 2: +\(n^2\) deposited
> After month 3: +\(n^2\) deposited
> After month 4: +\(n^2\) deposited
> After month 5: +\(n^2\) deposited
> After month 6: +\(n^2\) deposited

** Here, 1/4 of people leave and you are left with \(\frac{3}{4}n\) people, each depositing \(\frac{3}{4}n\) dollars **

> After month 7: +(\(\frac{3}{4}n\)) x (\(\frac{3}{4}n\)) deposited = \(\frac{9}{16}n^2\)
> After month 8: +\(\frac{9}{16}n^2\) deposited
> After month 9: +\(\frac{9}{16}n^2\) deposited

** Here, 1/3 of the remaining people leave and you are left with \((\frac{2}{3})(\frac{3}{4}n)\) people, each depositing \((\frac{2}{3})(\frac{3}{4}n)\) dollars **
> After month 10: + (\(\frac{2}{3})(\frac{3}{4}n\)) x \((\frac{2}{3})(\frac{3}{4}n)\) = \(\frac{1}{4}n^2\)
> After month 11: +\(\frac{1}{4}n^2\)
> After month 12: +\(\frac{1}{4}n^2\)

Put it all together and for the 12-month period, you get:

\(6(n^2) + 3(\frac{9}{16}n^2) + 3 (\frac{1}{4})n^2 = 4860\)

Finally, solve to get:

\(n=24\)

Answer: B
Re: Some people form a joint account for one year with the condi   [#permalink] 15 Apr 2015, 00:00
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