Does the product jkmn equals 1?

Question

(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?

athem · Answer

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

Bunuel · Answer

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 eq{\frac{1}{8}} , then jkmn eq{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.

garvitbh11 · Answer

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.

fskilnik · Answer

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.

SaquibHGMATWhiz · Answer

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

Does the product jkmn equals 1?

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Does the product jkmn equals 1?

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