GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 22 Nov 2019, 13:54 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # In the diagram given below, ∠ABD = ∠CDB = ∠PQD = 90°. If AB: CD = 3:1

Author Message
TAGS:

### Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 59268
In the diagram given below, ∠ABD = ∠CDB = ∠PQD = 90°. If AB: CD = 3:1  [#permalink]

### Show Tags

1
17 00:00

Difficulty:   75% (hard)

Question Stats: 54% (02:11) correct 46% (02:49) wrong based on 65 sessions

### HideShow timer Statistics

Competition Mode Question 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

Attachment: diagram.PNG [ 3.01 KiB | Viewed 1351 times ]

_________________
Senior Manager  P
Status: Whatever it takes!
Joined: 10 Oct 2018
Posts: 383
GPA: 4
Re: In the diagram given below, ∠ABD = ∠CDB = ∠PQD = 90°. If AB: CD = 3:1  [#permalink]

### Show Tags

4
1
IMO answer is option B. I have done by two methods. Here's how I did:

Posted from my mobile device
Attachments IMG_20190920_102211.jpg [ 4.12 MiB | Viewed 1179 times ] IMG_20190920_102225.jpg [ 4.3 MiB | Viewed 1175 times ]

_________________

ALL ABOUT GMAT- $$https://exampal.com/gmat/blog/gmat-score-explained$$
##### General Discussion
Senior Manager  G
Joined: 01 Mar 2019
Posts: 297
Location: India
Concentration: Strategy, Social Entrepreneurship
Schools: Ross '22, ISB '20, NUS '20
GPA: 4
Re: In the diagram given below, ∠ABD = ∠CDB = ∠PQD = 90°. If AB: CD = 3:1  [#permalink]

### Show Tags

1
triangles, ABD, PQD are similar , AB=3CD , BD= BQ + QD

so AB/PQ = BD/QD

CDB,PQB are simialr

CD/PQ = BD/QB

SOLVING above we will get PQ/CD = 3/4
I.E CD/PQ = 1/0.75

OA:B
Senior Manager  P
Joined: 31 May 2018
Posts: 451
Location: United States
Concentration: Finance, Marketing
Re: In the diagram given below, ∠ABD = ∠CDB = ∠PQD = 90°. If AB: CD = 3:1  [#permalink]

### Show Tags

1
triangle BQP is similar to triangle BDC (AAA similarity)

$$\frac{BD}{BQ}$$ = $$\frac{CD}{PQ}$$

BD = $$\frac{CD*BQ}{PQ}$$---(1)

triangle DBA is similar to triangle DQP (AAA similarity)

$$\frac{BD}{QD}$$ = $$\frac{AB}{PQ}$$

BD = $$\frac{AB*QD}{PQ}$$---(2)

(1) = (2) we get

$$\frac{CD*BQ}{PQ}$$ = $$\frac{AB*QD}{PQ}$$

$$\frac{BD}{BQ}$$ = $$\frac{CD}{AB}$$

$$\frac{BD}{BQ}$$ = $$\frac{1}{3}$$ (given $$\frac{AB}{CD}$$=$$\frac{3}{1}$$)

from triangle BQP and triangle BDC

$$\frac{BD}{BQ}$$ = $$\frac{CD}{PQ}$$

so $$\frac{CD}{PQ}$$=$$\frac{1}{3}$$

Intern  B
Joined: 06 Jun 2019
Posts: 4
Re: In the diagram given below, ∠ABD = ∠CDB = ∠PQD = 90°. If AB: CD = 3:1  [#permalink]

### Show Tags

1
B is correct

According to Thales theorem
CD/PQ=BD/BQ=BQ+QD/BQ=4/3
CD=1then PQ=0.75

Posted from my mobile device
Senior Manager  P
Joined: 07 Mar 2019
Posts: 393
Location: India
GMAT 1: 580 Q43 V27 WE: Sales (Energy and Utilities)
Re: In the diagram given below, ∠ABD = ∠CDB = ∠PQD = 90°. If AB: CD = 3:1  [#permalink]

### Show Tags

1
In the given figure, since $$∠ABD = ∠CDB = ∠PQD = 90°$$, $$AB || PQ || CD$$.
BC and AD are two lines intersecting these lines. Thus, $$Δ ABP ˜ Δ DCP$$(AAA congruency) where $$∠A = ∠D, ∠B = ∠C$$ and ∠P is common.

Hence
$$\frac{AB}{CD} = \frac{BP}{CP} = \frac{AP}{PD} = \frac{3}{1}$$ → ①

Also $$Δ BQP ˜ Δ BDC$$(AAA congruency) where $$∠Q = ∠D, ∠P = ∠C$$ and ∠B is common.
Thus, $$\frac{BQ}{BD} = \frac{BP}{BC} = \frac{PQ}{CD}$$

 $$\frac{PQ}{CD} = \frac{BP}{(BP + PC)}$$
 $$= \frac{3PC}{(3PC + PC)}$$ [From ① BP = 3PC]
 $$= \frac{3PC}{4PC}$$
 $$= \frac{3}{4}$$
 $$= \frac{0.75}{1}$$

_________________
Ephemeral Epiphany..!

GMATPREP1 590(Q48,V23) March 6, 2019
GMATPREP2 610(Q44,V29) June 10, 2019
GMATPREPSoft1 680(Q48,V35) June 26, 2019
Director  P
Joined: 24 Nov 2016
Posts: 837
Location: United States
Re: In the diagram given below, ∠ABD = ∠CDB = ∠PQD = 90°. If AB: CD = 3:1  [#permalink]

### Show Tags

Quote: 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

ABD and CDB have the same base and their sides are in proportion; since, AB:CD=3:1, then let their base BD=4;
now, imagine this diagram in a plane, with AB at the origin;
point A(0,3) B(0,0) C(4,1) D(4,0), and point P is the place where both lines AD and BC intersect;
line BC: m=(1-0)/(4-0)=1/4…y=x/4+b…1=(4)/4+b…b=0…y=x/4
equate lines: y=x/4 and y=-3x/4+3… x/4=-3x/4+3…x=3…y=x/4=(3/4)/4=3/4
point P(3,3/4) so CD:PQ=1:3/4

Senior Manager  G
Joined: 25 Jul 2018
Posts: 352
Re: In the diagram given below, ∠ABD = ∠CDB = ∠PQD = 90°. If AB: CD = 3:1  [#permalink]

### Show Tags

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

∆ABD and ∆PQD are similar triangles
--> $$\frac{AB}{PQ}$$=$$\frac{BD}{QD}$$ --> AB=$$\frac{PQ*BD}{QD}$$

∆CDB and ∆PQB are similar triangles
--> $$\frac{CD}{PQ}$$=$$\frac{BD}{(BD-QD)}$$ --> CD=$$\frac{PQ*BD}{(BD-QD)}$$

$$\frac{AB}{CD}=\frac{3}{1}=(\frac{PQ*BD}{QD})/\frac{PQ*BD}{(BD-QD)}= \frac{(BD-QD)}{QD}$$
3=$$\frac{(BD-QD)}{QD}$$
--> BD=4*QD

CD=$$\frac{PQ*4*QD}{3*QD}$$=$$\frac{PQ*4}{3}$$
$$\frac{CD}{PQ}=\frac{4}{3}=\frac{1}{(3/4)}$$=1:0.75

Manager  S
Joined: 10 Jun 2019
Posts: 113
Re: In the diagram given below, ∠ABD = ∠CDB = ∠PQD = 90°. If AB: CD = 3:1  [#permalink]

### Show Tags

The triangles ABP and PCD are similar becasue they have the same sized angles. Their heights will be in the same ratio as AB and CD are in. They heights are BQ and QD.The ratio of BQ :QD =3:1 since the ratio of their bases AB:CD is 3:1. Also one can see that CBD is similar to PBQ,meaning their bases which are the heights of ABP and PCD will have to be in the ratio of 4 :3 This allows us to say that the ratio of CD to PQ is also 4:3.

SVP  D
Joined: 03 Jun 2019
Posts: 1854
Location: India
Re: In the diagram given below, ∠ABD = ∠CDB = ∠PQD = 90°. If AB: CD = 3:1  [#permalink]

### Show Tags

Bunuel wrote:

Competition Mode Question 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

Attachment:
diagram.PNG

Since triangles BCD and BPQ are similar
CD:PQ = BD:BQ ----------1

Since triangles ABD and PQD are similar
AB:PQ = BD:QD -----------2

(2)/(1)
AB:CD = BQ:QD = 3:1
BD:BQ = 4:3 = 1: .75

IMO B
_________________
"Success is not final; failure is not fatal: It is the courage to continue that counts."

Please provide kudos if you like my post. Kudos encourage active discussions.

My GMAT Resources: -

Efficient Learning
All you need to know about GMAT quant

Tele: +91-11-40396815
Mobile : +91-9910661622
E-mail : kinshook.chaturvedi@gmail.com Re: In the diagram given below, ∠ABD = ∠CDB = ∠PQD = 90°. If AB: CD = 3:1   [#permalink] 28 Oct 2019, 23:43
Display posts from previous: Sort by

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