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If the operation x&y is defined by the equation x&y=x·y/(x+y) for all

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If the operation x&y is defined by the equation x&y=x·y/(x+y) for all [#permalink]

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22 Sep 2016, 03:37
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If the operation x&y is defined by the equation x&y=x·y/(x+y) for all x≠−y, then m&m must be equal to

A. 1/2
B. m/2
C. 1
D. m^2/2
E. m
[Reveal] Spoiler: OA

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Re: If the operation x&y is defined by the equation x&y=x·y/(x+y) for all [#permalink]

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22 Sep 2016, 04:02
its B m*m/2m = m/2

I choose B
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Re: If the operation x&y is defined by the equation x&y=x·y/(x+y) for all [#permalink]

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25 Aug 2017, 01:09
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If the operation x&y is defined by the equation x&y=x·y/(x+y) for all [#permalink]

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25 Aug 2017, 12:48
Bunuel wrote:
If the operation x&y is defined by the equation x&y=x·y/(x+y) for all x≠−y, then m&m must be equal to

A. 1/2
B. m/2
C. 1
D. m^2/2
E. m

deepthit wrote:
its B m*m/2m = m/2
I choose B

karishmajain06 wrote:

karishmajain06 , I can't tell which part of the answer doesn't make sense. So I have discussed beginning to end.

I put the symbol & in bold to emphasize that it does NOT mean "and."

1. Treat the strange symbol as the abbreviation for a rule or "recipe" that you have to follow. (Some people find it easier to think of & as a function.)

The strange symbol & means "do these steps."

This rule says do these steps: if you have two numbers or variables, x and y, with that & symbol between them, make a particular kind of fraction using x and y -- plugging them in.

The numerator is x * y

The denominator is x + y -->

$$\frac{(x * y)}{(x + y)}$$

2. m&m with the rule

We're given m&m. There are still two variables. They simply are equal. So you could also say the rule is

(LHS variable)&(RHS variable) =

$$\frac{(LHS * RHS)}{(LHS + RHS)}$$

We just plug in m, and then m again. The first m corresponds to x in the rule; the second m corresponds to y.

m&m = $$\frac{m * m}{m + m}$$, or $$\frac{(m)(m)}{(2)(m)}$$

3. Finish the arithmetic to get the answer

Remove one factor of m from the numerator and the denominator ("cancel").

$$\frac{(m)(m)}{(2)(m)}$$ =

$$\frac{m}{2}$$

OR: $$\frac{m^2}{2m}$$, factor out one m, =

$$\frac{m}{2}$$

Hope that helps!
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Re: If the operation x&y is defined by the equation x&y=x·y/(x+y) for all [#permalink]

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29 Aug 2017, 16:27
Bunuel wrote:
If the operation x&y is defined by the equation x&y=x·y/(x+y) for all x≠−y, then m&m must be equal to

A. 1/2
B. m/2
C. 1
D. m^2/2
E. m

m & m = m·m/(m + m) = m^2/(2m) = m/2

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Re: If the operation x&y is defined by the equation x&y=x·y/(x+y) for all   [#permalink] 29 Aug 2017, 16:27
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