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Re: |x| < 1 and y > 0 [#permalink]
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putting -0.1 in place of x and 0.1 in place of y will lead to correct answer, without putting values solving such questions becomes difficult
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Re: |x| < 1 and y > 0 [#permalink]
So, A is the answer.
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Re: |x| < 1 and y > 0 [#permalink]
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daina1323031 wrote:
So, A is the answer.


Yes Sir.

Give explanation whenever you can though

Thanks

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Re: |x| < 1 and y > 0 [#permalink]
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Firstly, we should consider that

Quantity A will include an integer and a fraction, on the other hand, Quantity B will reduce the value of the fraction.

For x it can be either a positive or negative fraction. But for all negative fractions, the value of B is always is negative whereas, the value of A is positive always. This term, A>B

Secondly, for Positive values of B will reach towards an integer, meanwhile, the values A will turn into a big integer.

So, A is always greater, there is no chance that A will be equal to B
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Re: |x| < 1 and y > 0 [#permalink]
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|x|<1 and y>0

|x| < 1, therefore x range must be within +/- 1. That is -1 < x < 1

Note -> Multiplication of decimal number within -1 to 1 and any other number yields output less than other number.

Example -> 0.5 * 2 = 1 [1 is less than 0.5]
-> 0.2 * 0.5 = 0.01 [0.01 is less than 0.5]

Therefore, QA > QB
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Re: |x| < 1 and y > 0 [#permalink]
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We Need to test the values as per conditions .
y = 1,2,3...
x = 1/2,1/3,1/5
Lets take x = 1/2, y = 1
A -> 3/2 b-> 1/2 { Qt A is greater}
Lets take x = 1/2, y = 2
A -> 5/4 , B -> 1 { Qt A is greater }
Hence Qty A > qty B
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