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Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  In the right-angled ∆ ABC above, y/3 = 3/(x + y). What is the value of

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Math Expert V
Joined: 02 Sep 2009
Posts: 55758
In the right-angled ∆ ABC above, y/3 = 3/(x + y). What is the value of  [#permalink]

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Difficulty:   25% (medium)

Question Stats: 79% (01:36) correct 21% (02:52) wrong based on 17 sessions

HideShow timer Statistics In the right-angled ∆ABC above, y/3 = 3/(x + y). What is the value of y?

(A) 5/2
(B) 3
(C) 9/5
(D) 5/9
(E) 7/2

Attachment: 2017-08-08_2112_002.png [ 7.11 KiB | Viewed 462 times ]

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Director  V
Joined: 04 Dec 2015
Posts: 740
Location: India
Concentration: Technology, Strategy
WE: Information Technology (Consulting)
Re: In the right-angled ∆ ABC above, y/3 = 3/(x + y). What is the value of  [#permalink]

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Bunuel wrote: In the right-angled ∆ABC above, y/3 = 3/(x + y). What is the value of y?

(A) 5/2
(B) 3
(C) 9/5
(D) 5/9
(E) 7/2

Attachment:
2017-08-08_2112_002.png

Using Pythagorean Triplets Rule, $$\triangle$$ $$ABC$$ have sides in ratio $$= 3:4:5$$

Given $$BC = 3$$ and $$AB = 4$$

Therefore $$AC = 5$$

$$AC = x + y => 5$$

$$\frac{y}{3} = \frac{3}{(x + y)}$$

$$\frac{y}{3} = \frac{3}{5}$$

$$y = \frac{3*3}{5} = \frac{9}{5}$$ Re: In the right-angled ∆ ABC above, y/3 = 3/(x + y). What is the value of   [#permalink] 08 Aug 2017, 11:04
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In the right-angled ∆ ABC above, y/3 = 3/(x + y). What is the value of  