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Intern  B
Joined: 22 Feb 2018
Posts: 10
In a parallelogram ABCD, P is the midpoint of AB. Diagonal AC inter...  [#permalink]

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Difficulty:   35% (medium)

Question Stats: 64% (01:35) correct 36% (02:02) wrong based on 47 sessions

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In a parallelogram ABCD, P is the midpoint of AB. Diagonal AC intersects PD at Q. What proportion of AC is AQ?
(A) 1/2
(B) 1/3
(C) 1/4
(D) 1/5
(E) 1/6

Attachment: Screen Shot 2019-01-25 at 3.50.29 PM.png [ 78.65 KiB | Viewed 726 times ]

Thank you in advance!

Source: Manhattan Review

Originally posted by jpfg259 on 25 Jan 2019, 07:54.
Last edited by jpfg259 on 30 Jan 2019, 08:14, edited 1 time in total.
Math Expert V
Joined: 02 Aug 2009
Posts: 7954
Re: In a parallelogram ABCD, P is the midpoint of AB. Diagonal AC inter...  [#permalink]

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jpfg259 wrote:
In a parallelogram ABCD, P is the midpoint of AB. Diagonal AC intersects PD at Q. What proportion of AC is AQ?
(A) 1/2
(B) 1/3
(C) 1/4
(D) 1/5
(E) 1/6

Attachment:
The attachment Screen Shot 2019-01-25 at 3.50.29 PM.png is no longer available

Thank you in advance!

Triangles APQ and CDQ are similar triangles as AP is parallel to CD...
The corresponding sides are in same ratio...
so $$\frac{AP}{CD}=\frac{PQ}{DQ}=\frac{AQ}{CQ}$$ means $$\frac{AP}{CD}=\frac{AQ}{CQ}... =>\frac{x}{2x}=\frac{AQ}{CQ}....=>CQ=2AQ$$..
We are looking for $$\frac{AQ}{AC}=\frac{AQ}{AQ+QC}=\frac{AQ}{AQ+2AQ}=\frac{AQ}{3AQ}=\frac{1}{3}$$
Attachments Screen Shot 2019-01-25 at 3.50.29 PM.png [ 61.58 KiB | Viewed 674 times ]

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Intern  B
Joined: 22 Feb 2018
Posts: 10
Re: In a parallelogram ABCD, P is the midpoint of AB. Diagonal AC inter...  [#permalink]

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chetan2u wrote:
jpfg259 wrote:
In a parallelogram ABCD, P is the midpoint of AB. Diagonal AC intersects PD at Q. What proportion of AC is AQ?
(A) 1/2
(B) 1/3
(C) 1/4
(D) 1/5
(E) 1/6

Attachment:
Screen Shot 2019-01-25 at 3.50.29 PM.png

Thank you in advance!

Triangles APQ and CDQ are similar triangles as AP is parallel to CD...
The corresponding sides are in same ratio...
so $$\frac{AP}{CD}=\frac{PQ}{DQ}=\frac{AQ}{CQ}$$ means $$\frac{AP}{CD}=\frac{AQ}{CQ}... =>\frac{x}{2x}=\frac{AQ}{CQ}....=>CQ=2AQ$$..
We are looking for $$\frac{AQ}{AC}=\frac{AQ}{AQ+QC}=\frac{AQ}{AQ+2AQ}=\frac{AQ}{3AQ}=\frac{1}{3}$$

That was a great explanation, thank you so much!
Director  V
Joined: 12 Feb 2015
Posts: 915
Re: In a parallelogram ABCD, P is the midpoint of AB. Diagonal AC inter...  [#permalink]

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jpfg259 wrote:
In a parallelogram ABCD, P is the midpoint of AB. Diagonal AC intersects PD at Q. What proportion of AC is AQ?
(A) 1/2
(B) 1/3
(C) 1/4
(D) 1/5
(E) 1/6

Attachment:
Screen Shot 2019-01-25 at 3.50.29 PM.png

Thank you in advance!

Source: Manhattan Review

Important thing to note in this question is that triangles QDC & QPA are similar. Since P is the mid point; AP:DC is 1:2; Hence AQ:QC is 1:2 or AQ:AC is 1:3

Corect ans is Option B
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Manish "Only I can change my life. No one can do it for me" Re: In a parallelogram ABCD, P is the midpoint of AB. Diagonal AC inter...   [#permalink] 05 Feb 2019, 09:48
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