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

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

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New post 25 Feb 2015, 03:21
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A
B
C
D
E

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Question Stats:

81% (01:15) correct 19% (01:44) wrong based on 203 sessions

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Re: The lengths of the sides of triangle ABC are such that AB = x + 2, BC  [#permalink]

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New post 25 Feb 2015, 05:41
1
(1) x - y = 1 | move y to the right side
x = y+1 | add +2
x+2 = y+3
AB = y-1 +4
AB = AC + 4
hence AB is largest side and angle ACB is largest
SUFFICIENT

(2) y = 8
AC = 7
AB+BC>AC
x+2+x-3>7
2x>8
x>4

so, AB>6 and BC>1
and also, AB>BC because x+2>x-3
AB can be 7, so AB=AC and there will be 2 largest angles is isosceles triangle
or AB can be 8 or larger and angle ACB will be largest
INSUFFICIENT

answer is A
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Re: The lengths of the sides of triangle ABC are such that AB = x + 2, BC  [#permalink]

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New post 25 Feb 2015, 07:17
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
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Re: The lengths of the sides of triangle ABC are such that AB = x + 2, BC  [#permalink]

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New post 25 Feb 2015, 23:15
1
A side of a triangle can't be negative.
So atleast we need x to be greater than 3.
Using A we can easy say that side AB is the greatest.
But using B we don't know the values of x,thus not knowing the values of other two sides of the triangle.So B can't give us the answer.

Hence A is sufficient.
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Re: The lengths of the sides of triangle ABC are such that AB = x + 2, BC  [#permalink]

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New post 02 Mar 2015, 07:24
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Re: The lengths of the sides of triangle ABC are such that AB = x + 2, BC  [#permalink]

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New post 07 Jan 2018, 14:22
Bunuel wrote:
The lengths of the sides of triangle ABC are such that AB = x + 2, BC = x - 3, and AC = y - 1. Which angle in triangle ABC is the largest?

(1) x - y = 1

(2) y = 8

Kudos for a correct 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
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Re: The lengths of the sides of triangle ABC are such that AB = x + 2, BC  [#permalink]

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New post 30 Sep 2018, 14:34
Bunuel wrote:
The lengths of the sides of triangle ABC are such that AB = x + 2, BC = x - 3, and AC = y - 1. Which angle in triangle ABC is the largest?

(1) x - y = 1

(2) y = 8

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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Re: The lengths of the sides of triangle ABC are such that AB = x + 2, BC   [#permalink] 30 Sep 2018, 14:34
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