I could be wrong but I would say b)1.
This is because here three cases can be considered
1. Both x and y are integers
2. Both x and y have o.5 as the decimal part.
3. One of these is an integer and the other has a decimal part of 0.5
In each of these cases the difference between 2x and 2y will have to be covered by the difference between [x] and [y], which is not possible even by taking minimum difference between x any y in each of the cases mentioned above.

I have not given the correct reasoning in my first response to this problem.This is what I mean.
Firstly, we assign values to x any y to get a values of a and b.Now we have to find a pair of x and y for which the values of a and b are the same.for this four cases are possible
1. If x,y are integers and the next values taken are integers with a min. difference of 1.
In this case.
[x] + 2y =a
If both x and y are being integers are increased or decreased by 1, a cannot be obtained again.
2.If x,y are both have 0.5 as the decimal part.
Two cases are possible here
a)If x and y are increased here lets say by 0.5 Then [x] and [y] will increase to the next integer and so will the value of 2x and 2y, so again obtaining a and b from the equations will not be possible.
b) If decreased by 0.5 the [x],[y] remain same but again 2x and 2y change.
3.if one of x and y has o.5 as the decimal part and the other is an integer, in this case as well the integer part will increase or decrease by 1 and will be doubled on multiplying by2.and the [x],[y]will be increased to the next integer or remain same depending on the next value chosen is greater or less.
Hence there b)1

I would proceed like this..
say [x] = greatest integer <=x
and {x} = fractional part of x.

For positive integer
So x = [x] + {x}.

Please see its not very difficult to observe that for negative x.
x = [x] + 1 + {x}.

Consider the case where both x and y are positive.
then 1st equation can be written as:-
[x] + 2[y] + 2{y} = a …….. (Equ 1′)
and second equation can be rewritten as :-
[y] + 2[x] + 2{x} = b …………(Equ 2′)

Now 1′ + 2′ gives:-
3[x] + 3[y] + 2({x} + {y}) = a + b ……. Equ 3′

and 1′ – 2′ gives
[y] – [x] + 2({y} – {x}) = a – b ……….Equ 4′

since {x} + {y} can take values 0, 0.5, 1.5 in order to satisfy 3′
and {x} – {y} can take values 0, 0.5 in order to satisy 4′

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I could be wrong but I would say b)1.

This is because here three cases can be considered

1. Both x and y are integers

2. Both x and y have o.5 as the decimal part.

3. One of these is an integer and the other has a decimal part of 0.5

In each of these cases the difference between 2x and 2y will have to be covered by the difference between [x] and [y], which is not possible even by taking minimum difference between x any y in each of the cases mentioned above.

DeepshikhaAugust 28, 2008 at 2:04 am

I have not given the correct reasoning in my first response to this problem.This is what I mean.

Firstly, we assign values to x any y to get a values of a and b.Now we have to find a pair of x and y for which the values of a and b are the same.for this four cases are possible

1. If x,y are integers and the next values taken are integers with a min. difference of 1.

In this case.

[x] + 2y =a

If both x and y are being integers are increased or decreased by 1, a cannot be obtained again.

2.If x,y are both have 0.5 as the decimal part.

Two cases are possible here

a)If x and y are increased here lets say by 0.5 Then [x] and [y] will increase to the next integer and so will the value of 2x and 2y, so again obtaining a and b from the equations will not be possible.

b) If decreased by 0.5 the [x],[y] remain same but again 2x and 2y change.

3.if one of x and y has o.5 as the decimal part and the other is an integer, in this case as well the integer part will increase or decrease by 1 and will be doubled on multiplying by2.and the [x],[y]will be increased to the next integer or remain same depending on the next value chosen is greater or less.

Hence there b)1

Pardon me for my mistakes. I am fallible!!

DeepshikhaAugust 28, 2008 at 4:45 am

Nice attempt, but you have misunderstood the problem !

outtimedAugust 28, 2008 at 10:33 am

I would proceed like this..

say [x] = greatest integer <=x

and {x} = fractional part of x.

For positive integer

So x = [x] + {x}.

Please see its not very difficult to observe that for negative x.

x = [x] + 1 + {x}.

Consider the case where both x and y are positive.

then 1st equation can be written as:-

[x] + 2[y] + 2{y} = a …….. (Equ 1′)

and second equation can be rewritten as :-

[y] + 2[x] + 2{x} = b …………(Equ 2′)

Now 1′ + 2′ gives:-

3[x] + 3[y] + 2({x} + {y}) = a + b ……. Equ 3′

and 1′ – 2′ gives

[y] – [x] + 2({y} – {x}) = a – b ……….Equ 4′

since {x} + {y} can take values 0, 0.5, 1.5 in order to satisfy 3′

and {x} – {y} can take values 0, 0.5 in order to satisy 4′

Now there are fe

DufferOctober 8, 2008 at 7:11 am