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Ay + B = C

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Ay + B = C

by rahulvsd » Fri Apr 20, 2012 7:42 am
How many different numbers y are there such that Ay + B = C (A, B, and C are known) ?

1. C > B
2. A >1

[spoiler]OA: B. I found this question weird. Please explain. [/spoiler]
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by aneesh.kg » Fri Apr 20, 2012 11:03 am
Given:
y = (C - B)/A

Statement 1: C - B > 0
which only tells that the numerator of 'y' is positive. Depending on the value of A, y can hold many values.

Statement 2: A > 0
this talks about only the denominator. Based on the values of B and C, y can hold many values.
There is no way (B) could be the correct option.

Let's try to combine the two statements.
Now, y = (positive quantity)/(A quantity greater than 1)

Again, 'y' can hold many values. The question doesn't even say that 'y' has to be an integer.

No conclusive answer. (E) should be correct.
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by neelgandham » Fri Apr 20, 2012 12:24 pm
How many different numbers y are there such that Ay + B = C (A, B, and C are known) ?

The question states that the value of y = (C-B)/A. So, if the value of A is not equal to zero, we can say that y =(C-B)/A, a single value. If the value of A is zero, the value of y is undefined and there are infinite/No numbers in the number system that satisfy the equation.
1. C > B
The statement doesn't speak of A or its value, hence the statement is insufficient to answer the question.
2. A > 1
If the value of A is greater than 1, then we know that A is definitely not equal to 0. So, The value of y = (C-B)/A and y has just one value. Therefore the statement is sufficient to answer the question.

IMO B
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by aneesh.kg » Fri Apr 20, 2012 12:33 pm
I agree with the analysis above.
I didn't note that the values of A,B and C are known, and that's the catch.
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