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

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Manager
Joined: 24 Feb 2020
Posts: 165
Location: Italy
WE: Analyst (Investment Banking)
Integers x and y are chosen at random so that -5<=x<=5 and  [#permalink]

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17 Apr 2020, 11:19
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Difficulty:

15% (low)

Question Stats:

83% (02:11) correct 17% (02:21) wrong based on 53 sessions

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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) $$\frac{19}{35}$$
DS Forum Moderator
Joined: 19 Oct 2018
Posts: 1998
Location: India
Re: Integers x and y are chosen at random so that -5<=x<=5 and  [#permalink]

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