More data sufficiency - pick numbers carefully!

When you pick numbers, the goal is is often to find a contradiction, i.e., to show that different numbers lead to different answers. In data sufficiency, this can be an extremely effective strategy, as in the following question.

Question
If y doesn't equal 0 and y doesn’t equal -1, which is greater, x/y or x/(y+1)?
(1) x doesn’t equal 0
(2) x > y

Solution
The best approach is to pick numbers with the goal of showing there is insufficient info.
Don’t forget that you can pick negative integers and fractions.
We will have to be careful about the signs of x and y, because signs will determine which is greater.

Consider (1).
If x is 2 and y is 1, then x/y = 2, and x/(y+1) = 1. So x/y is greater.
But if x is -2 and y is 1, then x/y = -2 and x/(y+1) = -1. So x/(y+1) is greater.
There is a contradiction, so there’s insufficient info.
Eliminate A and D.

Consider (2)
Again we could have x=2 and y=1, so x/y is greater
We could not x=-2 and y=1.
However both x and y could be negative, as long as x>y.
If x= -1 and y= -2, then x/y = 1/2 , and x/(y+1) = 1, so x/(y+1) is greater.
Contradiction.
Insufficient info.
Eliminate B.

Consider (1) and (2) together.
We have already done so and found contradiction.
Not enough info.
Answer is E.

Here are two more data sufficiency problems.
Question
1. What fractional part of the total surface area of cube C is red?
(1) Each of 3 faces of C is exactly 1/2 red.
(2) Each of 3 faces of C is entirely white.

Solution
Consider (1).
We don’t know what color the other faces are. They could be all red or not red at all.
So we do not know what proportion of the total surface area is red.
Eliminate A and D.
Consider (2).
Again, we don’t know what color the non-white faces are.
Eliminate B.
Consider (1) and (2) together.
Now we infer that half of three sides are red, and no part of the other three sides are white.
The fractional part that is read is half of half, a quarter.
Answer is C.


Question
S is a set of integers such that
i) if a is in S, then -a is in S, and
ii) if each of a and b is in S, then ab is in S
is -4 in S?

(1) 1 is in S
(2) 2 is in S

Solution
Consider (1).
1 is in S, so -1 is also in S.
Not enough info to determine whether -4 is in S.
Eliminate A and D.

Consider (2).
2 is in S, so -2 is also in S.
2 and -2 are both members of S, so their product,-4, must be in S according to (ii).
Sufficient info.
Answer is B.