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# In circle K shown, chords AB and CD intersect at point E within the

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Math Expert
Joined: 02 Sep 2009
Posts: 52390
In circle K shown, chords AB and CD intersect at point E within the  [#permalink]

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30 Apr 2018, 08:11
00:00

Difficulty:

25% (medium)

Question Stats:

57% (01:32) correct 43% (02:23) wrong based on 32 sessions

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In circle K shown, chords AB and CD intersect at point E within the circle. What is the value of x?

(A) 3
(B) 5
(C) 10
(D) 21
(E) 35

Attachment:

2018-04-30_2009.png [ 25.87 KiB | Viewed 641 times ]

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Senior Manager
Joined: 29 Dec 2017
Posts: 386
Location: United States
Concentration: Marketing, Technology
GMAT 1: 630 Q44 V33
GMAT 2: 690 Q47 V37
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GPA: 3.25
WE: Marketing (Telecommunications)
In circle K shown, chords AB and CD intersect at point E within the  [#permalink]

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30 Apr 2018, 08:51
3*(6x+5) = 5 (4x+1), x=5, Answer (C)
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Manager
Joined: 15 Nov 2017
Posts: 115
Location: India
Concentration: Operations, Marketing
GPA: 4
WE: Operations (Retail)
Re: In circle K shown, chords AB and CD intersect at point E within the  [#permalink]

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30 Apr 2018, 09:03
1
Bunuel wrote:

In circle K shown, chords AB and CD intersect at point E within the circle. What is the value of x?

(A) 3
(B) 5
(C) 10
(D) 21
(E) 35

Attachment:
2018-04-30_2009.png

CE*ED = AE*EB
X = 5 (option B)
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Manager
Joined: 31 Jan 2018
Posts: 69
Re: In circle K shown, chords AB and CD intersect at point E within the  [#permalink]

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30 Apr 2018, 11:56
Bunuel wrote:

In circle K shown, chords AB and CD intersect at point E within the circle. What is the value of x?

(A) 3
(B) 5
(C) 10
(D) 21
(E) 35

Attachment:
2018-04-30_2009.png

3 * (6x+5) = 5(4x +1)
==>18x + 15 = 20x + 5
==>20x - 18x =15 -5
==>2x = 10
therefore x = 5(B)
Re: In circle K shown, chords AB and CD intersect at point E within the &nbs [#permalink] 30 Apr 2018, 11:56
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