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Does the product jkmn equals 1?

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Does the product jkmn equals 1?  [#permalink]

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New post Updated on: 31 Mar 2014, 00:29
1
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A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

65% (01:01) correct 35% (01:01) wrong based on 234 sessions

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Does the product jkmn equals 1?

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

(2) \(mn>7\)

Why statement 1 is not sufficient please? In any situation ( +ve on -ve) I think statement 1 is sufficient to answer the question but that's not the answer. Can someone please help?

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Originally posted by enigma123 on 30 Mar 2014, 14:44.
Last edited by Bunuel on 31 Mar 2014, 00:29, edited 1 time in total.
Edited the question.
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Re: Does the product jkmn equals 1  [#permalink]

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New post 30 Mar 2014, 19:37
1
First statement suggests that jk=mn
but value of jk or mn is not known Not sufficient

second statement is mn>7 but value of jk is not known
if jk is 1/7.. jkmn will equal 1

together since jk = mn, so if mn=8 then jk=8
jkmn cannot be 1
so C
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Re: Does the product jkmn equals 1?  [#permalink]

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New post 31 Mar 2014, 00:43
enigma123 wrote:
Does the product jkmn equals 1?

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

(2) \(mn>7\)

Why statement 1 is not sufficient please? In any situation ( +ve on -ve) I think statement 1 is sufficient to answer the question but that's not the answer. Can someone please help?


Does the product jkmn equals 1?

(1) \(\frac{jk}{mn} = 1\) --> \(jk=mn\). The question becomes: does \((mn)^2=1\). We don't know that. Not sufficient.

(2) \(mn>7\). If \(mn=8\) and \(jk=\frac{1}{8}\), then \(jkmn=1\) but if \(mn=8\) and \(jk
\neq{\frac{1}{8}}\), then \(jkmn\neq{1}\). Not sufficient.

(1)+(2) Since from (1) \(jk=mn\) and from (2) \(mn>7\), then \(jk>7\) too. Now, the product of two multiples (jk and mn) each of which is greater than 7 cannot equal to 1. Sufficient.

Answer: C.

Similar question to practice: does-the-product-jkmn-24606.html

Hope it helps.
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Re: Does the product jkmn equals 1?  [#permalink]

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New post 30 Sep 2018, 03:18
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.
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Re: Does the product jkmn equals 1?  [#permalink]

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New post 30 Sep 2018, 07:59
enigma123 wrote:
Does the product jkmn equals 1?

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

(2) \(mn>7\)

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

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

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

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


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

Regards,
Fabio.
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Re: Does the product jkmn equals 1? &nbs [#permalink] 30 Sep 2018, 07:59
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