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Asifpirlo
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The possible arguments of the functions f(x,y) and g(x,y) are x=0 or 1 06 Aug 2013, 15:29
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The possible arguments of the functions f(x,y) and g(x,y) are x=0 or 1 and y = 0 or 1 and are defined by the following tables (Attachment):

What is the value of f(a,b) + g(a,b) ?

(1) f(a,b) = 0
(2) g(a,b)= 0


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The possible arguments of the functions f(x,y) and g(x,y) are x=0 or 1 24 Apr 2014, 03:30
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qlx
Asifpirlo
The possible arguments of the functions f(x,y) and g(x,y) are x=0 or 1 and y = 0 or 1 and are defined by the following tables (Attachment):

What is the value of f(a,b) + g(a,b) ?

(1) f(a,b) = 0
(2) g(a,b)= 0


Can some one please explain the table to me . As of now I am not able to comprehend the two tables of f(a,b ) and g(a, b).
in a 2*2 table , rows stand for something and column stand for something. In this sum what does row stand for and what does column stand for?

Thank you members.
Show more


Yellow rows are the values of x, green columns are the values of y, and the red boxes are the outcomes of the functions f and g.

The possible arguments of the functions f(x,y) and g(x,y) are x=0 or 1 and y = 0 or 1 and are defined by the following tables:

What is the value of f(a,b) + g(a,b) ?

(1) f(a,b) = 0. According to the second table we can have the following cases:
a=0, b=0 --> f(a,b) + g(a,b) = 0 + 0 = 0;
a=1, b=0 --> f(a,b) + g(a,b) = 0 + 1 = 1;
a=0, b=1 --> f(a,b) + g(a,b) = 0 + 1 = 1.

Not sufficient.


(2) g(a,b) = 0. According to the first table we can have only the following case:
a=0, b=0 --> f(a,b) + g(a,b) = 0 + 0 = 0.

Sufficient.

Answer: B.

Hope it's clear.

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Zarrolou
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The possible arguments of the functions f(x,y) and g(x,y) are x=0 or 1 06 Aug 2013, 23:16
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What is the value of f(a,b) + g(a,b) ?

(1) f(a,b) = 0
This could be \(a,b=0\) and \(f(a,b) + g(a,b)=0\); or could be \(a=0\) and \(b=1\) and \(f(a,b) + g(a,b)=1\).
Not sufficient

(2) g(a,b)= 0
This works only for \(a,b=0\) so \(f(a,b) + g(a,b)=0\).
Sufficient
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The possible arguments of the functions f(x,y) and g(x,y) are x=0 or 1 24 Apr 2014, 03:13
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Asifpirlo
The possible arguments of the functions f(x,y) and g(x,y) are x=0 or 1 and y = 0 or 1 and are defined by the following tables (Attachment) :

What is the value of f(a,b) + g(a,b) ?

(1) f(a,b) = 0
(2) g(a,b)= 0
Show more


Can some one please explain the table to me . As of now I am not able to comprehend the two tables of f(a,b ) and g(a, b).
in a 2*2 table , rows stand for something and column stand for something. In this sum what does row stand for and what does column stand for?

Thank you members.
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qlx
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The possible arguments of the functions f(x,y) and g(x,y) are x=0 or 1 24 Apr 2014, 04:07
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Asifpirlo
The possible arguments of the functions f(x,y) and g(x,y) are x=0 or 1 and y = 0 or 1 and are defined by the following tables (Attachment) :

What is the value of f(a,b) + g(a,b) ?

(1) f(a,b) = 0
(2) g(a,b)= 0


Can some one please explain the table to me . As of now I am not able to comprehend the two tables of f(a,b ) and g(a, b).
in a 2*2 table , rows stand for something and column stand for something. In this sum what does row stand for and what does column stand for?

Thank you members.

Attachment:
Untitled.png
Yellow rows are the values of x, green columns are the values of y, and the red boxes are the outcomes of the functions f and g.

What is the value of f(a,b) + g(a,b) ?

(1) f(a,b) = 0. According to the second table we can have the following cases:
a=0, b=0 --> f(a,b) + g(a,b) = 0 + 0 = 0;
a=1, b=0 --> f(a,b) + g(a,b) = 0 + 1 = 1;
a=0, b=1 --> f(a,b) + g(a,b) = 0 + 1 = 1.

Not sufficient.


(2) g(a,b) = 0. According to the first table we can have only the following case:
a=0, b=0 --> f(a,b) + g(a,b) = 0 + 0 = 0.

Sufficient.

Answer: B.

Hope it's clear.
Show more

Great that clears it! I guess a real GMAT question could have mentioned horizontal row is x and vertical column is y.
Thank you.
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The possible arguments of the functions f(x,y) and g(x,y) are x=0 or 1 10 Mar 2016, 19:15
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first time seeing a question with such notations. unless I see similar official question, I would rather disregard it...
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The possible arguments of the functions f(x,y) and g(x,y) are x=0 or 1 05 Mar 2019, 23:30
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Asifpirlo
The possible arguments of the functions f(x,y) and g(x,y) are x=0 or 1 and y = 0 or 1 and are defined by the following tables (Attachment):

What is the value of f(a,b) + g(a,b) ?

(1) f(a,b) = 0
(2) g(a,b)= 0


Can some one please explain the table to me . As of now I am not able to comprehend the two tables of f(a,b ) and g(a, b).
in a 2*2 table , rows stand for something and column stand for something. In this sum what does row stand for and what does column stand for?

Thank you members.


Yellow rows are the values of x, green columns are the values of y, and the red boxes are the outcomes of the functions f and g.

The possible arguments of the functions f(x,y) and g(x,y) are x=0 or 1 and y = 0 or 1 and are defined by the following tables:

What is the value of f(a,b) + g(a,b) ?

(1) f(a,b) = 0. According to the second table we can have the following cases:
a=0, b=0 --> f(a,b) + g(a,b) = 0 + 0 = 0;
a=1, b=0 --> f(a,b) + g(a,b) = 0 + 1 = 1;
a=0, b=1 --> f(a,b) + g(a,b) = 0 + 1 = 1.

Not sufficient.


(2) g(a,b) = 0. According to the first table we can have only the following case:
a=0, b=0 --> f(a,b) + g(a,b) = 0 + 0 = 0.

Sufficient.

Answer: B.

Hope it's clear.
Show more


hi bunuel , can't a=1 and b=1 also be a possible domain for f(a,b) as you have mentioned only 3 possible cases
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The possible arguments of the functions f(x,y) and g(x,y) are x=0 or 1 05 Mar 2019, 23:46
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Asifpirlo
The possible arguments of the functions f(x,y) and g(x,y) are x=0 or 1 and y = 0 or 1 and are defined by the following tables (Attachment):

What is the value of f(a,b) + g(a,b) ?

(1) f(a,b) = 0
(2) g(a,b)= 0



Yellow rows are the values of x, green columns are the values of y, and the red boxes are the outcomes of the functions f and g.

The possible arguments of the functions f(x,y) and g(x,y) are x=0 or 1 and y = 0 or 1 and are defined by the following tables:

What is the value of f(a,b) + g(a,b) ?

(1) f(a,b) = 0. According to the second table we can have the following cases:
a=0, b=0 --> f(a,b) + g(a,b) = 0 + 0 = 0;
a=1, b=0 --> f(a,b) + g(a,b) = 0 + 1 = 1;
a=0, b=1 --> f(a,b) + g(a,b) = 0 + 1 = 1.

Not sufficient.


(2) g(a,b) = 0. According to the first table we can have only the following case:
a=0, b=0 --> f(a,b) + g(a,b) = 0 + 0 = 0.

Sufficient.

Answer: B.

Hope it's clear.


hi bunuel , can't a=1 and b=1 also be a possible domain for f(a,b) as you have mentioned only 3 possible cases
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(1) says that f(a,b) = 0.

If, as you say, a = 1 and b = 1, then f(a,b) = f(1,1) = 1, not 0 (bottom right corner of table for f).
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The possible arguments of the functions f(x,y) and g(x,y) are x=0 or 1 15 May 2024, 13:24
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