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In the diagram given below, ∠ABD = ∠CDB = ∠PQD = 90°. If AB: CD = 3: 1

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In the diagram given below, ∠ABD = ∠CDB = ∠PQD = 90°. If AB: CD = 3: 1  [#permalink]

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New post 10 Mar 2017, 12:39
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54% (02:53) correct 46% (02:23) wrong based on 49 sessions

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In the diagram given below, ∠ABD = ∠CDB = ∠PQD = 90°. If AB: CD = 3: 1, the ratio of CD: PQ

a). 1: 0.69
b). 1: 0.75
c). 1: 0.72
d). 1: 0.50
e). None of these

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Re: In the diagram given below, ∠ABD = ∠CDB = ∠PQD = 90°. If AB: CD = 3: 1  [#permalink]

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New post 10 Mar 2017, 19:28
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quantumliner wrote:
In the diagram given below, ∠ABD = ∠CDB = ∠PQD = 90°. If AB: CD = 3: 1, the ratio of CD: PQ

a). 1: 0.69
b). 1: 0.75
c). 1: 0.72
d). 1: 0.50
e). None of these



triangles ABD & PQD && CDB &PQB are similar
and AB/CD=3/1=AB=3CD-------((a)

AB/PQ = BD/QD-----(b)
CD/PQ = BD/BQ----(c)

substituting value BD from (c) to (b)
CD*BQ=AB*QD
substituting value AB from (a)
BQ/QD=3/1
thus BD/BQ= 4/3 (as BD=BQ+QD)

or CD/PQ= 4/3 =1/0.75
Ans B
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Re: In the diagram given below, ∠ABD = ∠CDB = ∠PQD = 90°. If AB: CD = 3: 1  [#permalink]

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New post 11 Mar 2017, 22:57
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Okay I completely don't understand the proportionality theorem- kudos for help
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Re: In the diagram given below, ∠ABD = ∠CDB = ∠PQD = 90°. If AB: CD = 3: 1  [#permalink]

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New post 12 Mar 2017, 01:30
In the figure, Let BQ = a and DQ = b

Triangle ABD and triangle PQD are similar. Therefore, PQ/AB = b/(a+b)

Also triangle CBD and triangle PBQ are similar. Therefore, PQ/CD = a/(a+b)

Dividing the second equality by the first, we get. AB/CD = a/b = 3

Therefore, CD/PQ = (a+b)/a = 4/3 = 1 : 0.75
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Re: In the diagram given below, ∠ABD = ∠CDB = ∠PQD = 90°. If AB: CD = 3: 1  [#permalink]

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New post 23 Sep 2018, 12:05
Hello from the GMAT Club BumpBot!

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Re: In the diagram given below, ∠ABD = ∠CDB = ∠PQD = 90°. If AB: CD = 3: 1 &nbs [#permalink] 23 Sep 2018, 12:05
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In the diagram given below, ∠ABD = ∠CDB = ∠PQD = 90°. If AB: CD = 3: 1

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