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30 Oct 2018, 18:35
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93% (01:03) correct
7%
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[Math Revolution GMAT math practice question]
\(m\) and \(n\) are two different numbers selected from the integers between \(11\) and \(20\), inclusive. What is the maximum value of \(\frac{mn}{(m-n)}\)?
\(A. 320\)
\(B. 340\)
\(C. 360\)
\(D. 380\)
\(E. 400\)
\(m\) and \(n\) are two different numbers selected from the integers between \(11\) and \(20\), inclusive. What is the maximum value of \(\frac{mn}{(m-n)}\)?
\(A. 320\)
\(B. 340\)
\(C. 360\)
\(D. 380\)
\(E. 400\)
30 Oct 2018, 19:00
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MathRevolution
Beautiful problem, Max. Congrats!
\(m,n\,\,{\rm{distinct}}\,\, \in \left\{ {11,12, \ldots ,20} \right\}\)
\(? = \max \left( {{{mn} \over {m - n}}} \right)\)
\(\left( {\rm{I}} \right)\,\,m > n\,\,\,\,\left( {{\rm{otherwise}}\,\,{{mn} \over {m - n}} < 0\,\,,\,\,{\rm{not}}\,\,\max } \right)\)
\(\left( {{\rm{II}}} \right)\,\,0 < {{mn} \over {m - n}} \le \max \left( {{{mn} \over {m - n}}} \right)\,\,\,\,\, \Leftrightarrow \,\,\,\,\,{1 \over n} - {1 \over m} = {{m - n} \over {mn}} \ge {\left[ {\max \left( {{{mn} \over {m - n}}} \right)} \right]^{ - 1}}\)
\(\left( {{\rm{III}}} \right)\,\,\,?\,\,\,\, \Leftrightarrow \,\,\,\,\min \left( {{1 \over n} - {1 \over m}} \right)\,\,\,\,\, \Leftrightarrow \,\,\,\left( {m,n} \right) = \left( {20,19} \right)\)
\(\left( {{\rm{IV}}} \right)\,\,\,? = {\left( {{{20 - 19} \over {20 \cdot 19}}} \right)^{ - 1}} = 20 \cdot 19 = 380\)
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
01 Nov 2018, 00:36
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=>
To obtain the maximum value for \(\frac{mn}{(m-n)}\), \(m-n\) must be the smallest possible positive number and \(mn\) must be the largest possible product.
\(mn\) is largest when both \(m\) and \(n\) are largest, that is, when \(m = 20\) and \(n – 19\). In this case, \(m – n = 1\) is as small as possible and \(mn = 20*19 = 380\) is as large as possible.
Thus, the maximum value of \(\frac{mn}{(m-n)}\) is \(\frac{20*19}{(20-19)} = \frac{380}{1} = 380.\)
Therefore, D is the answer.
Answer: D
To obtain the maximum value for \(\frac{mn}{(m-n)}\), \(m-n\) must be the smallest possible positive number and \(mn\) must be the largest possible product.
\(mn\) is largest when both \(m\) and \(n\) are largest, that is, when \(m = 20\) and \(n – 19\). In this case, \(m – n = 1\) is as small as possible and \(mn = 20*19 = 380\) is as large as possible.
Thus, the maximum value of \(\frac{mn}{(m-n)}\) is \(\frac{20*19}{(20-19)} = \frac{380}{1} = 380.\)
Therefore, D is the answer.
Answer: D
