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For all nonzero numbers x and y, operation x#y is equal to y#x. Operat

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For all nonzero numbers x and y, operation x#y is equal to y#x. Operat  [#permalink]

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New post Updated on: 29 Jul 2019, 08:44
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For all nonzero numbers x and y, operation x#y is equal to y#x. Operation # could be defined by which of the following operations?

I. x#y = y/x + x/y
II. x#y = x/y - y/x
III. x#y = x^2 - 2xy + y^2

(A) I only
(B) II only
(C) III only
(D) I and II
(E) I and III

Originally posted by GOHOSATRAJIT on 29 Jul 2019, 08:40.
Last edited by Bunuel on 29 Jul 2019, 08:44, edited 1 time in total.
Renamed the topic, edited the question and moved to PS forum..
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Re: For all nonzero numbers x and y, operation x#y is equal to y#x. Operat  [#permalink]

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New post 30 Jul 2019, 02:18
How to solve this in less than a minute?

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For all nonzero numbers x and y, operation x#y is equal to y#x. Operat  [#permalink]

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New post 30 Jul 2019, 02:41
Since x#y =y#x ,we can infer that # is commutative
x+y =y+x and x•y =y•x ( addition and multiplication)

I. x#y = y/x + x/y
x#y = (y/y(y/x) + (x/x)(x/y)
x# y =( x^2+y^2)/xy
Numerator (adding)/denominator(multiplying)
So satisfies #

II. x#y = x/y - y/x
x#y = (x^2-y^2)/xy
Here numberator (subtracting)/denominator (multiplying)
Does NOT satisfy #

III. x#y = x^2 - 2xy + y^2
x#y = (x-y)^2
Though (subtracting ) since it’s a square will always be +ve regardless of the change of x and y
Eg. 1#2 = (1-2)^2=1
2#1 = (2-1)^2 =1
So satisfies #
Answer E

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Re: For all nonzero numbers x and y, operation x#y is equal to y#x. Operat  [#permalink]

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New post 06 Aug 2019, 17:55
GOHOSATRAJIT wrote:
For all nonzero numbers x and y, operation x#y is equal to y#x. Operation # could be defined by which of the following operations?

I. x#y = y/x + x/y
II. x#y = x/y - y/x
III. x#y = x^2 - 2xy + y^2

(A) I only
(B) II only
(C) III only
(D) I and II
(E) I and III


Now let’s verify each Roman numeral for y#x and see if it is same as x#y.

I. y#x = x/y + y/x = y/x + x/y → This is the same as x#y.

II. y#x = y/x - x/y → This is NOT the same as x#y.

III. y#x = y^2 - 2yx + x^2 = x^2 - 2xy + y^2 → This is the same as x#y.

Solution: E
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Re: For all nonzero numbers x and y, operation x#y is equal to y#x. Operat   [#permalink] 06 Aug 2019, 17:55
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