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In ΔLMN, LM = |x-7|, MN = |x-4| and NL = x + 1, where x is a number wh

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In ΔLMN, LM = |x-7|, MN = |x-4| and NL = x + 1, where x is a number wh  [#permalink]

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New post 25 Feb 2020, 02:24
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D
E

Difficulty:

  55% (hard)

Question Stats:

69% (02:22) correct 31% (02:35) wrong based on 58 sessions

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In ΔLMN, LM = |x-7|, MN = |x-4| and NL = x + 1, where x is a number wh  [#permalink]

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New post 25 Feb 2020, 05:10
1
In ΔLMN, LM = |x - 7|, MN = |x - 4| and NL = x + 1, where x is a number whose value is not known. Is ΔLMN an acute triangle?

(Statement1): NM = 10
|x—4 | = 10
—> x—4 = 10 —> x= 14
—> x—4= —10 —> x= —6
x cannot be —6, because (x+1) must be greater than zero

The lengths of sides of ΔLMN:
—> MN= 10, LM= 7 and NL = 15
In order to be an acute triangle,
\(a^{2} < b^{2} + c^{2}\)

( a is the longest side of a triangle)
But, in our case
\(15^{2} > 7^2+ 10^{2}\)

That means ΔLMN is not an acute triangle ( always NO)
Sufficient

(Statement2): LM = 7
|x—7 | = 7
—> x—7 = 7 —> x= 14
—> x—7 = —7 —> x= 0
( X cannot be zero. If so, the lengths of sides of a triangle will be 1,4 and 7. —> it cannot be a triangle)

x= 14. In ΔLMN:
LM= 7, MN= 10 and NL= 15
It cannot be an acute triangle as proved above in statement1
\(15^{2} > 7^{2} + 10^{2}\)

Angle M is not acute —> ΔLMN is not an acute triangle ( always NO)
Sufficient

The answer is D

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Re: In ΔLMN, LM = |x-7|, MN = |x-4| and NL = x + 1, where x is a number wh  [#permalink]

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New post 28 Mar 2020, 03:52
Bunuel wrote:
In ΔLMN, LM = |x - 7|, MN = |x - 4| and NL = x + 1, where x is a number whose value is not known. Is ΔLMN an acute triangle (a triangle each of whose angles measures less than 90∘)?

(1) NM = 10
(2) LM = 7


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Asked: In ΔLMN, LM = |x - 7|, MN = |x - 4| and NL = x + 1, where x is a number whose value is not known. Is ΔLMN an acute triangle (a triangle each of whose angles measures less than 90∘)?

(1) NM = 10
NM = |x-4| = 10
x = 4 + -10 = 14 or -6
x = 14 since -6 is not feasible NL = x+1 = -5 is not feasible
LM = 7; MN = 10; NL = 15
SUFFICIENT

(2) LM = 7
LM = |x-7| = 7
x = 7 + -7
x = 14 or 0
If x=14
LM = 7; MN = 10; NL = 15
But if x=0
LM=7; MN = 4; NL=1; Since MN + NL = 5 < LM = 7; Triangle is not feasible
So x=14
LM = 7; MN = 10; NL = 15 is the only solution
SUFFICIENT

IMO D
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Re: In ΔLMN, LM = |x-7|, MN = |x-4| and NL = x + 1, where x is a number wh   [#permalink] 28 Mar 2020, 03:52

In ΔLMN, LM = |x-7|, MN = |x-4| and NL = x + 1, where x is a number wh

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