Does the product jkmn equals 1?

1- Jk=mn
which states that they both are equal. here comes two cases: 1st case- Both JK and mn = 1 then there multiplication will be 1 but if they are =2 then there multiplication will be 4. hence Not sufficient.

2- mn>7 not sufficient does not tells anything about J.k which can be 1/7 or an integer.

1+2- 2nd statement makes sure M.N is greater than 7 and 1st statement states that J.K is equal to M.N. so there multiplication cant be 1.

Please comment if I am wrong at something.

\(jkmn\,\,\mathop = \limits^? \,\,1\)

\(\left( 1 \right)\,\,\,{{jk} \over {mn}} = 1\,\,\,\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,\left( {j,k,m,n} \right) = \left( {1,1,1,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr \\
\,{\rm{Take}}\,\,\left( {j,k,m,n} \right) = \left( {2,2,2,2} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr} \right.\)

\(\left( 2 \right)\,\,\,mn > 7\,\,\,\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,\left( {j,k,m,n} \right) = \left( {1,8,1,8} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr \\
\,{\rm{Take}}\,\,\left( {j,k,m,n} \right) = \left( {{1 \over 8},1,1,8} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.\)

\(\left( {1 + 2} \right)\,\,\,\,\,?\,\,\,:\,\,\,jkmn\,\, = \,\,\left( {jk} \right)\left( {mn} \right)\,\,\,\mathop = \limits^{\left( 1 \right)} \,\,\,{\left( {mn} \right)^2}\,\,\,\mathop > \limits^{\left( 2 \right)} \,\,\,{7^{\,2}}\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,\,\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

Solution:
Pre Analysis:

• We are asked if \(jkmn=1\) or not

Statement 1: \(\frac{jk}{mn} = 1\)

• According to this statement, \(jk=mn\)
• \(jkmn=1\) is possible if \(j=k=m=n=1\) and not possible otherwise
• Thus, statement 1 alone is not sufficient and we can eliminate options A and D

Statement 2: \(mn>7\)

• Let us assume \(mn=8\) and \(jk=\frac{1}{8}\), then \(jkmn=1\) is possible
• However, we are not sure of this
• Thus, statement 2 alone is also not sufficient

Combining:

• Upon combining we get that jk = mn and mn > 7
• This means jk is also greater than 7
• Since both \(jk>7\) and \(mn>7\), we can be sure that \(jkmn≠1\)

Hence the right answer is Option C