If L 0, is 18K/L an integer?

according to statement A K^2/L^2 is integer but that doesn't mean that K & L are integer; square of 1.732 is 3 which is an integer but 1.732 itself is not a integer. hence statement A - i nsufficient

statment B - k=2l sufficeint we get a integer from it
so B must be the answer.

Solution:
Pre Analysis:

• Given \(L ≠ 0\)
• We are asked if \(\frac{18K}{L}\) is an integer or not

Statement 1: K^2/L^2 is an integer

According to this statement,
• \(\frac{k^2}{L^2}=integer\)
  or \(L^2=\frac{K^2}{integer}\)
  or \(L=\frac{K}{\sqrt{integer}}\)
• Plugging \(L=\frac{K}{\sqrt{integer}}\) in \(\frac{18K}{L}\), we get \(\frac{18\times K\times \sqrt{integer}}{K}\)
  or \(18\times \sqrt{integer}\) which we cannot be sure if it is integer or not
• Thus, statement 1 alone is not sufficient and we can eliminate options A and D

Statement 2: K - L = L

• According to this statement, \(2L=K\) or \(L=\frac{K}{2}\)
• Plugging \(L=\frac{K}{2}\) in \(\frac{18K}{L}\), we get \(\frac{18\times K\times 2}{K}=36\)
• It is definitely an integer

Hence the right answer is Option B

Official Solution:


If \(L \not= 0\), is \(\frac{18K}{L}\) an integer?

It is crucial to remember that we are not given any information about whether K and L are integers. Keep this consideration in mind when evaluating the statements.

(1) \(\frac{K^2}{L^2}\) is an integer.

If \(K=L=1\), then \(\frac{18K}{L}=18\) and the answer to the question is YES. However, if \(K=\sqrt{2}\) and \(L=1\), then \(\frac{18K}{L}=18\sqrt{2}\), which is not an integer, and the answer to the question is NO. This statement is not sufficient.

Note that if we were told that K and L are integers, the first statement would have been sufficient since \(\frac{K^2}{L^2}\) being an integer would imply that \(\frac{K}{L}\) is an integer. This is because the ratio of two integers (\(\frac{K}{L}\)) can only be an integer or a non-integer rational number, but it cannot be an irrational number. However, \(\frac{K}{L}\) cannot be a non-integer rational number, because its square would not be an integer. Therefore, \(\frac{K}{L}\) must be an integer, making \(\frac{18K}{L}\) an integer.

(2) \(K - L = L\). \(K=2L\).

In this case, \(\frac{18K}{L}=\frac{18(2L)}{L}=36\), which is an integer. Thus, the answer to the question is YES. This statement is sufficient.


Answer: B