Six years ago, the average storage life of 8 rare botanical samples in a seed bank was 27 weeks. Today, the average storage life of the 3 longest-lasting samples is 65 weeks, and the average storage life of another 3 samples is 20 weeks. Among the remaining two samples, one lasts 3 weeks longer than the other. Find the storage life of the longer-lasting sample of these remaining two?

The correct answer is 6 weeks.

Let's work through this carefully step by step.

Step 1: Establish the total storage life six years ago.

Six years ago, the average storage life of all 8 samples was 27 weeks.

\text{Total storage life (6 years ago)}=8\times 27=216\text{weeks}

Step 2: Find today's total storage life.

Since 6 years have passed, every sample has aged by 6 more weeks. Therefore, the average storage life of all 8 samples today is:

\text{Average today}=27+6=33\text{weeks}

\text{Total storage life (today)}=8\times 33=264\text{weeks}

Step 3: Calculate the total storage life of the 6 known samples today.

The 3 longest-lasting samples have an average of 65 weeks:

\text{Total (3 longest)}=3\times 65=195\text{weeks}

Another 3 samples have an average of 20 weeks:

\text{Total (next 3)}=3\times 20=60\text{weeks}

Step 4: Find the combined storage life of the remaining two samples.

\text{Combined (remaining 2)}=264-195-60=9\text{weeks}

Step 5: Solve for the individual storage lives of the two remaining samples.

Let the shorter-lasting sample have a storage life of x weeks. Then the longer-lasting sample has a storage life of (x + 3) weeks (since one lasts 3 weeks longer than the other).

x+(x+3)=9

2x+3=9

2x=6

x=3\text{weeks}

So the shorter-lasting sample has a storage life of 3 weeks, and the longer-lasting sample has a storage life of:

x+3=3+3=6\text{weeks}

Therefore, the storage life of the longer-lasting of the two remaining samples is 6 weeks.