386390 wrote:

fluke wrote:

So, the answer is C as we have proven that "a+b" is NOT divisible by 7 using both statements. What's the confusion or there isn't any?

Hi,

Please tell me if im thinking right:

generally the rule is: multiple of a number + or - multiple of a number = multiple of a number.

(1) says a = not a multiple of 7.

(2) a-b = multiple of 7

but since we don't know the exact values of the a or b, we cant say whether a+b is divisible by 7. Hence (c) gives us a definitive NO answer.?

The rule you stated is correct. But, we need to take that rule one step further.

Using 1 and 2, we can definitely say that "a+b" is NOT divisible by 7, irrespective of what value "a" or "b" bears.

Just because we can DEFINITELY answer the question as NO, the statements together become sufficient.

Question is: Is a+b divisible by 7?

Answer: No, "a+b" is not divisible by 7.

If we can answer the question asked in a stem as a definite YES or a definite NO, only then the statement(s) will be sufficient.

If the answer is: Maybe a+b is divisible by 7, then the statements become INSUFFICIENT.

In this case: a+b is definitely NOT divisible by 7. No matter what value you associate a or b with.

1. a is not divisible by 7.

a=5

b=2

a+b is divisible by 7.

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a=5

b=3

a+b is NOT divisible by 7.

Thus, NOT SUFFICIENT.

2. a-b is divisible by 7.

a=14

b=7

a+b=21 is divisible by 7.

a=9

b=2

a+b=11 is NOT divisible by 7.

Thus, NOT SUFFICIENT.

Together:

a is NOT divisible by 7.

BUT a-b is divisible by 7.

Means, b is also NOT divisible by 7.

a=9

b=2

a-b=7; divisible by 7.

But, a+b=11; NOT divisible by 7.

You can as many examples as you want that satisfy both statements and you will find that a+b is never divisible by 7.

a=23

b=2

a-b=21; divisible by 7.

a+b=25; NOT divisible by 7.

Thus, answer is "C"

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You can try this with other examples but you will never find a value for a or b, such that, a+b is divisible by 7. Just make sure that you don't violate any condition stated in the stem, statement 1 or statement 2.

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