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805+ (Hard)|   Word Problems|      
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Bunuel
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Some people form a joint account for one year with the condition that 07 Nov 2019, 01:44
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95% (hard)

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55% (03:19) correct 45% (03:24) wrong based on 192 sessions

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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 At the end of the year there is an amount of Rs. 4860 in the account before paying of the withdrawn amounts.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


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B
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Some people form a joint account for one year with the condition that 07 Nov 2019, 20:59
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Bunuel
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 At the end of the year there is an amount of Rs. 4860 in the account before paying of the withdrawn amounts.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


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Let there be x members in the beginning..
So the duration can be divided in 3 parts..
(I) First 6 months -- number is x, so money per month = x*x, and for 6 months = \(6*x^2\)
(II) Next 3 months -- number reduces to \(\frac{3x}{4}\), so money per month = \((\frac{3x}{4})^2\), and for 3 months = \(3*(\frac{3x}{4})^2=\frac{27x^2}{16}\)
(III) Last 3 months -- number reduces to \(\frac{3x}{4}*\frac{2}{3}=\frac{x}{2}\), so money per month = \((\frac{x}{2})^2\), and for 3 months = \(3*(\frac{x}{2})^2=\frac{3x^2}{4}\)

Total = \(6x^2+\frac{27x^2}{16}+\frac{3x^2}{4}=4860........\frac{135x^2}{16}=4860........x^2=4860*\frac{16}{135}=36*16.....x=6*4=24\)

B
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Some people form a joint account for one year with the condition that 07 Nov 2019, 21:07
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Bunuel
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


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You can also make use of choices..

A. 12...
First 6 months = \(6*12*12\).... next 3 months=\(3*(12*\frac{3}{4})^2=3*9*9\)...last 3 months = \(3*(9*\frac{2}{3})^2=3*6*6\)..
Total = \(864+243+108=1215\neq{4860}\)

B. 24...
First 6 months = \(6*24*24\).... next 3 months=\(3*(24*\frac{3}{4})^2=3*18*18\)...last 3 months = \(3*(18*\frac{2}{3})^2=3*12*12\)..
Total = \(4860\)..
Also a quick look at choice A .. A is 12 and B is 2 times and we are dealing with x^2 only and in each term, so our answer here will be (2x)^2= 4* value for x = 4*1215=4860

No need to go furhter

B
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Some people form a joint account for one year with the condition that 27 Jul 2020, 04:28
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This question is excellent, however, you might want to rephrase this line:

"Also, the person who withdraws from the account before the end of the year get his/her amount at the end of the year."

The wording is causing some confusion.
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Some people form a joint account for one year with the condition that 27 Jul 2020, 05:14
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Solution


Given

    • Every month each member deposits an amount equal to the number of members in the account in that month
    • The person who withdraws from the account before the end of the year gets 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.
    • Of the monsters who can spit fireballs, 8 also blow freezing wind.
    • At the end of the year, there is an amount of Rs. 4860 in the account before paying off the withdrawn amounts.

To Find

• Number of members in the beginning.

Approach and Working Out

    • After 6 months \(\frac{1}{4}\)th leaves.
      o Remaining = \(\frac{3}{4}\)th.
    • After another 3 months \(\frac{1}{3}\)th of \(\frac{3}{4}\)th left.
      o This is \(\frac{1}{4}\)th of the total.
      o Remaining = 1 – \(\frac{1}{4}\) – \(\frac{1}{4}\)= \(\frac{1}{2}\)
    • From this I can see the answer is a multiple of 4.
    • Let me take the number to be 4x.
      o 6 months = 6 × \((4x)^2\)
        = \(96x^2\)
      o Next 3 months = 3× \((3x)^2\)
        = \(27x^2\)
      o End 3 months = 3 × (2x)^2.
        = \(12x^2\)
    • Total amount = 96\(x^2\) + 27\(x^2\) + 12\(x^2\) = 4860
      Or, 135\(x^2\) = 4860
      Or, \(x^2\) = 36
      Or, x = 6
    • Answer = 4x = 24

Correct Answer: Option B
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Some people form a joint account for one year with the condition that 12 Jan 2026, 10:38
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