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

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Bunuel
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In the right-angled ∆ABC above, y/3 = 3/(x + y). What is the value of [#permalink] New post   24 Oct 2019, 00:46
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A
B
C
D
E

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87% (01:26) correct 13% (02:35) wrong based on 47 sessions
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In the right-angled ∆ABC above, \(\frac{y}{3} = \frac{3}{x + y}\). What is the value of y?

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

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Mohammadmo
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Re: In the right-angled ∆ABC above, y/3 = 3/(x + y). What is the value of [#permalink] New post   24 Oct 2019, 02:36
4²+3²=(x+y)²=25
x+y=5
y/3=3/5
5y=9,y=9/5

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In the right-angled ∆ABC above, y/3 = 3/(x + y). What is the value of [#permalink] New post   24 Oct 2019, 03:50
Using Pythagoras theorem:

\(4^2 + 3^2 = (x+y)^2\)

x+y = +5 or -5 -> should be positive so +5

Given equation:

\(\frac{y}{3} = \frac{3}{(x + y)}\) => y(x+y) = 9 => y(5) = 9

\(y = \frac{9}{5}\)

IMO Option C it is!
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Re: In the right-angled ∆ABC above, y/3 = 3/(x + y). What is the value of [#permalink] New post   26 Oct 2019, 08:59
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Bunuel wrote:

In the right-angled ∆ABC above, \(\frac{y}{3} = \frac{3}{x + y}\). What is the value of y?

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

Show Spoiler
Attachment:
2017-08-08_2112_002.png


We see that triangle ABC is a 3-4-5 right triangle. Therefore, x + y = 5. Since we are given that y/3 = 3/(x + y), we have:

y/3 = 3/5

5y = 9

y = 9/5

Answer: C
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gmatclubot
Re: In the right-angled ∆ABC above, y/3 = 3/(x + y). What is the value of [#permalink]
26 Oct 2019, 08:59
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