The lengths of the sides of triangle ABC are such that AB = x + 2, BC

1) x-y=1 gives us relation of difference between two that means y is more closer to the the zero than x
since triangle sides are positive non zero values, x>3 and y>1 which rules out the -ve values and we can have only positive value above 3 and 1 for x and y respectively.
so y= x-1 which gives sides as x+2,x-2,x-3. Angle opposite largest side is largest. Sufficient.

2) does not give any relation between x and y insufficient.

answer: A

MAGOOSH OFFICIAL SOLUTION:

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question, and then recheck the question.

An angle from which the longest side is accross is the largest.
Since x + 2 > x - 3 and AB > BC, we need to figure which one between x + 2 and y - 1 is largest.

By the condition 1), x - y = 1 or x = y + 1 and AB = x + 2 = y + 3 > y - 1 = AC.
The angle C accross from AB is the largest angle.

Since the condition 2) doesn't have any clue about x, it is not sufficient.

Therefore, the answer is A

In any triangle, a greater (length of) side is always "facing" a greater (measure of) angle, and vice-versa. Hence:

\(?\,\,\,\,:\,\,\,{\text{greater}}\,\,{\text{angle}}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,?\,\,\,\,:\,\,\,{\text{greater}}\,\,{\text{side}}\)

\(x + 2\,\,\mathop > \limits^{{\text{always}}} \,\,\,x - 3\,\,\,\,\,\, \Rightarrow \,\,\,\,\,AB > BC\,\,\,\,\left( * \right)\)

\(\left( * \right)\,\,\,\, \Rightarrow \,\,\,\,?\,\,\,:\,\,\,\max \left\{ {AB,AC} \right\} = \,\,\boxed{\,\,\max \left\{ {x + 2\,\,,\,\,y - 1} \right\}\,\,}\)


\(\left( 1 \right)\,\,x - y = 1\,\,\,\,\mathop \Rightarrow \limits^{{\text{FOCUS}}\,\,!} \,\,\,\,\,x + 2 = 1 + y + 2\,\,\,\,\mathop > \limits^{{\text{always}}} \,\,\,\,y - 1\,\,\,\,\, \Rightarrow \,\,\,\,{\text{SUFF}}.\)


\(\left( 2 \right)\,\,\,?\,\,:\,\,\,\max \left\{ {x + 2\,\,,\,\,8 - 1} \right\}\,\,\,\,\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,x = 6\,\,\,\, \Rightarrow \,\,\,\,\Delta = \left( {8,3,7} \right)\,\,{\rm{exists}}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,? = AB\,\,\,\, \hfill \cr \\
\,{\rm{Take}}\,\,x = 4.5\,\,\,\, \Rightarrow \,\,\,\,\Delta = \left( {6.5,1.5,7} \right)\,\,{\rm{exists}}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,? = AC\,\, \hfill \cr} \right.\)


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

Regards,
Fabio.

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