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# If B and D are centers of the circles shown in the figure above, and

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Math Expert
Joined: 02 Sep 2009
Posts: 42618

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If B and D are centers of the circles shown in the figure above, and [#permalink]

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21 Nov 2017, 22:11
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79% (01:13) correct 21% (02:58) wrong based on 14 sessions

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If B and D are centers of the circles shown in the figure above, and if BD = 12, BC = 2 and DE = 3, then AB =

(A) 12
(B) 14
(C) 21
(D) 22
(E) 24

[Reveal] Spoiler:
Attachment:

2017-11-21_1032.png [ 8.73 KiB | Viewed 163 times ]
[Reveal] Spoiler: OA

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Kudos [?]: 135771 [0], given: 12708

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Joined: 18 Aug 2016
Posts: 598

Kudos [?]: 181 [0], given: 138

GMAT 1: 630 Q47 V29
GMAT 2: 740 Q51 V38
Re: If B and D are centers of the circles shown in the figure above, and [#permalink]

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21 Nov 2017, 22:55
Bunuel wrote:

If B and D are centers of the circles shown in the figure above, and if BD = 12, BC = 2 and DE = 3, then AB =

(A) 12
(B) 14
(C) 21
(D) 22
(E) 24

[Reveal] Spoiler:
Attachment:
2017-11-21_1032.png

Let AB = x..Consider two triangles
A and a Diameter passing through B
A and a Diameter passing through D

We know Diameter of smaller circle = 4 and Diameter of larger circle = 6
Hence
$$AB/AD = 4/6$$
$$x/x+12 = 4/6$$
x= 24
E
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Kudos [?]: 181 [0], given: 138

Manager
Joined: 06 Aug 2017
Posts: 74

Kudos [?]: 15 [0], given: 30

GMAT 1: 610 Q48 V24
If B and D are centers of the circles shown in the figure above, and [#permalink]

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22 Nov 2017, 00:18
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Answer is E as follows.

Attachment:
File comment: Image

1966360.jpg [ 9.26 KiB | Viewed 100 times ]

We have to use properties of similar triangle here.

Triangle ACB is similar to AED
$$\frac{AD}{AB} = \frac{DE}{BC}$$ => $$\frac{x+12}{x} = \frac{3}{2}$$ => 2x+24 = 3x => x=24

Hence E is the answer
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Kudos [?]: 15 [0], given: 30

If B and D are centers of the circles shown in the figure above, and   [#permalink] 22 Nov 2017, 00:18
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# If B and D are centers of the circles shown in the figure above, and

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