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Integers x and y are chosen at random so that -5<=x<=5 and

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Joined: 24 Feb 2020
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Integers x and y are chosen at random so that -5<=x<=5 and  [#permalink]

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New post 17 Apr 2020, 11:19
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Integers x and y are chosen at random so that \(-5\leq{x}\leq{5}\) and \(-5\leq{y}\leq{5}\). What is the probability that \((x+y)^{2}\) > \(x^{2}+y^{2}\)?


(A) \(\frac{73}{124}\)

(B) \(\frac{43}{59}\)

(C) \(\frac{50}{121}\)

(D) \(\frac{34}{73}\)

(E) \(\frac{19}{35}\)
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Re: Integers x and y are chosen at random so that -5<=x<=5 and  [#permalink]

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New post 17 Apr 2020, 11:35
Total number of cases = 11*11=121 (We can mark our answer right here, as there is only 1 option whose denominator is multiple of 11)

\((x+y)^{2}\) > \(x^{2}+y^{2}\)

xy>0

Case 1 - when x>0 and y>0

x and y can be 1,2,3,4 or 5

Total case= 5*5=25

Case 2- when x<0 and y<0

x and y can be -1,-2,-3,-4 or -5

Total case= 5*5=25

Favourable cases = 25+25=50

Probability = 50/121


Ravixxx wrote:
Integers x and y are chosen at random so that \(-5\leq{x}\leq{5}\) and \(-5\leq{y}\leq{5}\). What is the probability that \((x+y)^{2}\) > \(x^{2}+y^{2}\)?


(A) \(\frac{73}{124}\)

(B) \(\frac{43}{59}\)

(C) \(\frac{50}{121}\)

(D) \(\frac{34}{73}\)

(E) \(\frac{19}{35}\)
GMAT Club Bot
Re: Integers x and y are chosen at random so that -5<=x<=5 and   [#permalink] 17 Apr 2020, 11:35

Integers x and y are chosen at random so that -5<=x<=5 and

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