This is a DS question:

If A is a positive # less than 20, is B greater than the average of A and 20?

1) On the number line, B is closer to 20 than it is to A

I don't understand why this is sufficient. Please help!! Thanks!!

If A is a positive # less than 20, is B greater than the average of A and 20?

1) On the number line, B is closer to 20 than it is to A

A <20; B> A+20/2 What is B or What is A?

from (1), B -> Closer to 20 which means B can be 19 (which is max).
19 > A+20/2; When you have any value for A, this is sufficient to answer YES or NO if B is greater than Avg of A and 20.

Hence it is sufficient enough to answer the question.

Thanks

Original poster - Please post complete questions.

I think A is Insuff.

e.g. A = 10
A+20/2 = 15.
B could be either 14 or 16 satisfying St1. But 14 is less than A+20/2 but 16 is not.

Pathik

nope, (1) is sufficient.
the flaw in the above argument is that one of your choices of illustrative #s just plain doesn't work. you set b = 14, but 14 is not closer to 20 than to 10.
try it again: with your choice of a = 10, the only choices of b that satisfy statement (1) are precisely those values that are greater than 15.

--

here's why (1) is sufficient:
** let's call the average of a & 20 'm', to streamline the following discussion.
imagine a number line, with 'a' to the left of 20. then 'm', because it is the average of 'a' and 20, is exactly halfway between 'a' and 20.
statement (1) tells us that 'b' is closer to 20 than to 'a'.
but this means that 'b' is to the right of the halfway point between those two numbers - which means precisely that it is to the right of 'm'.
sufficient.

Good question.

A is sufficient because if you rearrange the problem:

2B>20+A ; on further manipulation

(B-A)>(20-B); which is what the statement was asking to prove.