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20 Jul 2022, 08:00
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B
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Mbappe withdraws \(\frac{100}{x}\%\) of the amount in his account each time he visits the bank, where \(x\) is an integer greater than 1. If, after 5 visits, he has less than \(\frac{1}{x}^{th}\) of the initial amount left in the bank, what is the range of possible values for \(x\)?
A. 1
B. 2
C. 3
D. 4
E. 5
A. 1
B. 2
C. 3
D. 4
E. 5
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28 Aug 2024, 03:04
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Official Solution:
Mbappe withdraws \(\frac{100}{x}\%\) of the amount in his account each time he visits the bank, where \(x\) is an integer greater than 1. If, after 5 visits, he has less than \(\frac{1}{x}^{th}\) of the initial amount left in the bank, what is the range of possible values for \(x\)?
A. 1
B. 2
C. 3
D. 4
E. 5
Since \(\frac{100}{x}\%\) of the money is withdrawn each time, after each visit, \((1 - \frac{100}{x}*\frac{1}{100})= (1 - \frac{1}{x})\) of the previous amount will remain in the bank. For example, if 10% is withdrawn each time, after each visit, \((1 - 10*\frac{1}{100})= (1 - \frac{1}{10})=\frac{9}{10}\) of the previous amount will remain in the bank.
Assuming the initial amount in the bank is A, then after 5 visits, the amount would be \(A(1 - \frac{1}{x})^5\). We are told that this is less than \(\frac{1}{x}\) of the initial amount, so we have:
\(A(1 - \frac{1}{x})^5 < A*\frac{1}{x}\)
\((1 - \frac{1}{x})^5 < \frac{1}{x}\)
\((\frac{x-1}{x})^5 < \frac{1}{x}\)
\(\frac{(x-1)^5}{x^5} < \frac{1}{x}\)
\((x-1)^5 < x^4\)
As \(x\) increases, the left-hand side of the inequality grows faster than the right-hand side, so the inequality will only hold for smaller values of \(x\). Let's test some values. For \(x = 5\), the inequality does not hold: \(4^5 > 5^4\). However, if \(x\) is 4 or less, it does hold. Since \(x\) is an integer greater than 1, it can be 2, 3, or 4, making the range of possible values for \(x\) equal to \(4 - 2 = 2\).
Answer: B
Mbappe withdraws \(\frac{100}{x}\%\) of the amount in his account each time he visits the bank, where \(x\) is an integer greater than 1. If, after 5 visits, he has less than \(\frac{1}{x}^{th}\) of the initial amount left in the bank, what is the range of possible values for \(x\)?
A. 1
B. 2
C. 3
D. 4
E. 5
Since \(\frac{100}{x}\%\) of the money is withdrawn each time, after each visit, \((1 - \frac{100}{x}*\frac{1}{100})= (1 - \frac{1}{x})\) of the previous amount will remain in the bank. For example, if 10% is withdrawn each time, after each visit, \((1 - 10*\frac{1}{100})= (1 - \frac{1}{10})=\frac{9}{10}\) of the previous amount will remain in the bank.
Assuming the initial amount in the bank is A, then after 5 visits, the amount would be \(A(1 - \frac{1}{x})^5\). We are told that this is less than \(\frac{1}{x}\) of the initial amount, so we have:
\(A(1 - \frac{1}{x})^5 < A*\frac{1}{x}\)
\((1 - \frac{1}{x})^5 < \frac{1}{x}\)
\((\frac{x-1}{x})^5 < \frac{1}{x}\)
\(\frac{(x-1)^5}{x^5} < \frac{1}{x}\)
\((x-1)^5 < x^4\)
Answer: B
General Discussion
20 Jul 2022, 09:04
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Given: Mbappe withdraws \(\frac{100}{x}\%\) of his money each time he visits the bank, where \(x > 1\).
Asked: If after 5 visits, he has less than \(\frac{1}{x}^{th}\) of the initial amount in the bank, what is the range of all possible value of \(x\) ?
Let the initial amount in the bank = M
Mbappe withdraws \(\frac{100}{x}\%\) of his money each time he visits the bank = M/x
Money left after 1st withdrawal = M - M/x = M(x-1)/x
Money left after 2nd withdrawal = M(x-1)^2/x^2
Similarily money left after 5th withdrawal = M(x-1)^5/x^5 < M/x
(x-1)^5/x^5 - 1/x < 0
(x^4 - (x-1)^5)/x^5 > 0
x^4 - (x-1)^5 > 0
x^4 > (x-1)^5
If x = 2 ; x^4 = 2^4 = 16; (x-1)^5 = 1; x^4 > (x-1)^5
If x =3; x^4 = 3^4 = 81 ; (x-1)^5 = 2^5 = 32; x^4 > (x-1)^5
If x =4; x^4 = 4^4 = 256; (x-1)^5 = 3^5 = 243; x^4 > (x-1)^5
If x = 5 ; 5^4 = 625 ; 4^5= 1024; x^4 < (x-1)^5
Range of values of x = 4 - 2 = 2
I am assuming x> 1 ; where x is an integer.
IMO B
Asked: If after 5 visits, he has less than \(\frac{1}{x}^{th}\) of the initial amount in the bank, what is the range of all possible value of \(x\) ?
Let the initial amount in the bank = M
Mbappe withdraws \(\frac{100}{x}\%\) of his money each time he visits the bank = M/x
Money left after 1st withdrawal = M - M/x = M(x-1)/x
Money left after 2nd withdrawal = M(x-1)^2/x^2
Similarily money left after 5th withdrawal = M(x-1)^5/x^5 < M/x
(x-1)^5/x^5 - 1/x < 0
(x^4 - (x-1)^5)/x^5 > 0
x^4 - (x-1)^5 > 0
x^4 > (x-1)^5
If x = 2 ; x^4 = 2^4 = 16; (x-1)^5 = 1; x^4 > (x-1)^5
If x =3; x^4 = 3^4 = 81 ; (x-1)^5 = 2^5 = 32; x^4 > (x-1)^5
If x =4; x^4 = 4^4 = 256; (x-1)^5 = 3^5 = 243; x^4 > (x-1)^5
If x = 5 ; 5^4 = 625 ; 4^5= 1024; x^4 < (x-1)^5
Range of values of x = 4 - 2 = 2
I am assuming x> 1 ; where x is an integer.
IMO B
