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09 Dec 2021, 04:30
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
29% (02:20) correct
71%
(02:11)
wrong
based on 41
sessions
History
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Not Attempted Yet
A person who has Rs. 729 decides to play a game of betting. He wins three times and loses three times, the wins and losses occurring in random order. The chance of a win is equal to the chance of a loss and he wins and loses by the same percentage of the amount by which he plays a particular round. The change in the final amount which he has with him with respect to the initial investment is Rs. 217. What is the wager (stake) as a fraction of the money remaining with him at the time of each betting?
A. \(\frac{1}{9}\)
B. \(\frac{1}{3}\)
C. \(\frac{2}{3}\)
D. \(\frac{3}{4}\)
E. Cannot be determined
Are You Up For the Challenge: 700 Level Questions
A. \(\frac{1}{9}\)
B. \(\frac{1}{3}\)
C. \(\frac{2}{3}\)
D. \(\frac{3}{4}\)
E. Cannot be determined
Are You Up For the Challenge: 700 Level Questions
25 Dec 2021, 06:53
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We know that the person had lost trice and won trice.
Let's say the person either won x% of the amount or the person lost x %.
If the person wins, he / she the amount the person has =
(Remaining Amount) * [1 + \(\frac{x}{100}\)]
If the person wins, he / she the amount the person has =
(Remaining Amount) * [1 - \(\frac{x}{100}\)]
Because the person wins thrice and looses thrice the factors [1 + \(\frac{x}{100}\)] & [1 - \(\frac{x}{100}\)] will be multiplied thrice.
For ex:
[1 - \(\frac{x}{100}\)] * [1 - \(\frac{x}{100}\)] * [1 - \(\frac{x}{100}\)] * [1 + \(\frac{x}{100}\)] * [1 + \(\frac{x}{100}\)] * [1 + \(\frac{x}{100}\)]
Note: The order of win or lose is not given, however it does not matter. Because the factors will remain the same.
The multiplication is in the form of (a+b)*(a-b) and we see that there are three pairs of the same.
Hence we can reduce the multiplication as
[(a+b)(a-b)] * [(a+b)(a-b)] * [(a+b)(a-b)]
\((a^{2} - b^{2}) * (a^{2} - b^{2}) * (a^{2} - b^{2}) \)
\((a^{2} - b^{2})^{3} \)
Hence the problem now becomes
729 * [1 - (\(\frac{x^{2} }{ 10^{4}})^3\) ] = 217
1 - (\(\frac{x^{2} }{ 10^{4}})^3\) = 217 / 729
Taking cube root on both sides
1 - (\(\frac{x^{2} }{ 10^{4}})\) = 2 / 3
Note: \(6^{3}\) = 216, hence the same has been taken for approximation purpose.
1 - 2 / 3 = \(\frac{x^{2} }{ 10^{4}}\)
= 1 / 3
Let's say the person either won x% of the amount or the person lost x %.
If the person wins, he / she the amount the person has =
(Remaining Amount) * [1 + \(\frac{x}{100}\)]
If the person wins, he / she the amount the person has =
(Remaining Amount) * [1 - \(\frac{x}{100}\)]
Because the person wins thrice and looses thrice the factors [1 + \(\frac{x}{100}\)] & [1 - \(\frac{x}{100}\)] will be multiplied thrice.
For ex:
[1 - \(\frac{x}{100}\)] * [1 - \(\frac{x}{100}\)] * [1 - \(\frac{x}{100}\)] * [1 + \(\frac{x}{100}\)] * [1 + \(\frac{x}{100}\)] * [1 + \(\frac{x}{100}\)]
Note: The order of win or lose is not given, however it does not matter. Because the factors will remain the same.
The multiplication is in the form of (a+b)*(a-b) and we see that there are three pairs of the same.
Hence we can reduce the multiplication as
[(a+b)(a-b)] * [(a+b)(a-b)] * [(a+b)(a-b)]
\((a^{2} - b^{2}) * (a^{2} - b^{2}) * (a^{2} - b^{2}) \)
\((a^{2} - b^{2})^{3} \)
Hence the problem now becomes
729 * [1 - (\(\frac{x^{2} }{ 10^{4}})^3\) ] = 217
1 - (\(\frac{x^{2} }{ 10^{4}})^3\) = 217 / 729
Taking cube root on both sides
1 - (\(\frac{x^{2} }{ 10^{4}})\) = 2 / 3
Note: \(6^{3}\) = 216, hence the same has been taken for approximation purpose.
1 - 2 / 3 = \(\frac{x^{2} }{ 10^{4}}\)
= 1 / 3
25 Aug 2023, 06:28
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gmatophobia
All good up to here, but the left side yields the cumulative while the right is the change, which are not the same things.
The correct formulation is:
729(1+X)^3(1-X)^3 = some cumulative number
Now, we know the change is 217 and it seems likely we'll be taking cube roots, so is the right side
729+217=946 or 729-217=512 ?
A familiarity with cubes indicates 9^3=729 and 8^3=512, but no cube root for 946, so the equation seems to be:
729(1+X)^3(1-X)^3 = 512 or
[(1+X)(1-X)]^3 = 512/729 and
1-X^2 = 8/9 or
X^2 = 1-(8/9) = 1/9 and
X=1/3
Edit:
Actually this answer doesn't seem responsive to the question, which seems to be asking for the percentage wagered each time.
This answer assumes the percentage wagered is 100% and actually provides the win/loss percentage.
This question should be reviewed for accuracy.
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