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23 May 2019, 06:01
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
21% (03:00) correct
79%
(02:54)
wrong
based on 117
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Is x an integer?
(1) A triangle whose sides are of length x - 1, 2x – 3 and 6 - x units is an isosceles triangle
(2) A triangle whose sides are of length x, x + 1 and 2x – 1 units is a right-angled triangle
(1) A triangle whose sides are of length x - 1, 2x – 3 and 6 - x units is an isosceles triangle
(2) A triangle whose sides are of length x, x + 1 and 2x – 1 units is a right-angled triangle
23 May 2019, 13:00
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Gmatprep998
For a triangle to be an isosceles triangle with sides x-1 , 2x-3 , 6-x . Any two sides must be equal in length.
So, that means either of 3 conditions must be true.
(1) x-1 = 2x-3 - I
(2) 6-x = 2x-3 - II
(3) x-1 = 6-x - III
Solving I ,II, III
(1) x = 2 , sides of triangle 1,1,4
This case can be discarded as it violates the property of triangles that sum of any 2 sides of a triangle must be greater than the third side.
(2) x = 3 , sides of triangle 2,3,3
(3) x = 7/2 sides of triangle 5/2,4,5/2
Since 2 possible values of x , one integer(3) and one fraction (7/2).
Not Suff.
Second Statement
We can use the above 2 values and substitute in second statement.
When X = 3
Sides of triangle --> 3,4,5 - which is a right angle triangle
When x = 7/2
Sides of triangle --> 3.5 , 4.5 , 6 - NOT sides of right triangle as 3.5^2 + 4.5^2 <> 6^2
Hence using statement 1 and 2 , we can say that X is an integer.
Option C.
General Discussion
01 Dec 2020, 16:43
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Gmatprep998
Statement 1:
We need to check if in any case, the lengths give integer x results (and the lengths make sense).
x - 1 = 2x - 3 case: x = 2. The lengths would be 1, 1, 4. That cannot build a triangle.
x - 1 = 6 - x case: x = 7/2 = 3.5. The lengths would be 2.5, 4, 2.5. This is a viable triangle.
2x - 3 = 6 - x case: 3x = 9, x = 3. The lengths would be 3, 3, 2. This is a viable triangle.
Thus both the 2nd and 3rd triangles can be possible answers, x = 3.5 or x = 3 both viable. Insufficient.
Statement 2:
Note x and x + 1 have a strict relation, x must be less than x + 1. So we can only have either x + 1 or 2x - 1 as the hypotenuse.
Case 1: x + 1 is the hypotenuse. Then \(x^2 + (2x - 1)^2 = (x + 1)^2\) and we get \(x^2 + 4x^2 - 4x + 1 = x^2 + 2x + 1\)
After simplifying we get \(4x^2 -6x = 0\) and the meaningful solution is \(x = \frac{3}{2} = 1.5\) -> 1.5, 2, 2.5 triangle.
Case 2: 2x - 1 is the hypotenuse. Then \(x^2 + (x + 1)^2 = (2x - 1)^2\). After simplifying we get \(2x^2 - 6x = 0\), then x = 3 is the meaningful solution, which gives us the 3, 4, 5 triangle.
Again we have two viable solutions, one integer and one not so insufficient.
Combined:
Only x = 3 is in both, sufficient.
Ans: C
