Question = is x/140 an integer?
$$\frac{x}{140}=\ \frac{x}{4\cdot5\cdot7}=\ \frac{x}{2^2\cdot5\cdot7}$$
For x to be an integer it must have 2 power of 2 and 1 power each of 5 and 7
Statement 1 = LCM of x and y = 360
= 5* 8 * 9
$$=5\ \cdot\ 2^3\ \cdot\ 3^2$$
$$=2^3\ \cdot\ 3^2\ \cdot\ 5$$
This means there is no powers of 7 in x and y and we are able to answer with all certainty that,
$$\frac{x}{140}\ is\ not\ an\ integer,\ hence\ statement\ 1\ is\ SUFFICIENT$$

Statement 2 = GCF of x and y = 40
GCF is same thing as HCF
Therefore, HCF of x and y = 40
=5 * 8
$$=5\ \cdot\ 2^3$$
$$=2^3\ \cdot\ 5$$

The HCF just provides us with the common powers and it doesn't provide enough information about the powers of other prime numbers.
Hence, Statement 2 is INSUFFICIENT.

Option A is CORRECT.

AbhishekRyu wrote:is x/140 an integer?
1) LCM of x & y is 360
2) GCF of x & y is 40

$${x \over {{2^2} \cdot 5 \cdot 7}}\,\,\,\mathop = \limits^? \,\,\,{\mathop{\rm int}} $$
$$\left( 1 \right)\,\,\,LCM\left( {x,y} \right) = 360\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\{ \matrix{
\,x\,\,{\mathop{\rm int}} \,\,\,\,({\rm{implicitly}}) \hfill \cr
\,x\,\,{\rm{is}}\,{\rm{a}}\,{\rm{factor}}\,{\rm{of}}\,\,360\,\,\,\,\, \Rightarrow \,\,\,\,\,{{{2^3} \cdot {3^2} \cdot 5} \over x} = {\mathop{\rm int}} \,\,\,\left( * \right) \hfill \cr} \right.$$
$$\left( * \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,7\,\,{\rm{is}}\,\,{\rm{not}}\,\,{\rm{a}}\,\,{\rm{factor}}\,\,{\rm{of}}\,\,x\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,{x \over {{2^2} \cdot 5 \cdot 7}}\,\,\, \ne \,\,\,{\mathop{\rm int}} \,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,\,\,$$

$$\left( 2 \right)\,\,\,GCF\left( {x,y} \right) = 40 = {2^3} \cdot 5\,\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {{2^3} \cdot 5 \cdot 7,{2^3} \cdot 5} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {{2^3} \cdot 5,{2^3} \cdot 5 \cdot 7} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr} \right.$$

The correct answer is therefore (A).


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

Regards,
Fabio.