albany09 wrote:

Each of 20 parents chose one of five days from Monday through Friday to attend parent-teacher conferences. If more parents chose Monday than Tuesday, did at least one of the parents choose Friday?

(1) None of the five days was chosen by more than 5 parents.

(2) More parents chose Monday than Wednesday.

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

D. EACH statement ALONE is sufficient.

E. Statements (1) and (2) TOGETHER are NOT sufficient.

Key to the solution: Try to allocate as many parents to Mon, Tue, Wed, and Thu and as possible. Or, try to minimize parents in Fri.

(1) Since the maximum parents allowed is 5 parents per day.

Mon = 5

Tue = 4 ---------- Since Mon > Tue

Try to minimize parents in Friday.

Wed = 5

Thu = 5

The minimum parents in Fri = 1.

Even though we try our best to make Fri = 0, the condition given here make Fri = 1.

So, at least one of the parents must choose Friday. Sufficient.

(2) Mon > Wed

Mon > Tue ---------- Given from the main question.

Case 1:

Mon = 17

Tue = 2

Wed = 1

Thu, Fri = 0.

Case 2:

Mon = 16

Tue = 2

Wed = 1

Thu = 0

Fri = 1.

Two answers. Insufficient.

The answer is A.