Was the median height of the 25 children in a certain class at least 2 percent greater than the average...

Understanding the Question

We need to determine: Is the median height of 25 children at least 2% greater than their average height?

In other words: Is \(\mathrm{Median} \geq 1.02 \times \mathrm{Average}\)?

This is a yes/no question. Since we have 25 children (odd number), the median is the height of the 13th child when heights are arranged in order.

Key insight: To answer YES or NO definitively, we need to know the exact relationship between median and average.

Analyzing Statement 1

Statement 1 tells us: The median is exactly 2 cm greater than the average.

So: \(\mathrm{Median} = \mathrm{Average} + 2\text{ cm}\)

But here's the crucial question: Does 2 cm represent at least 2% of the average?

The answer depends entirely on the average height:

• If \(\mathrm{average} = 80\text{ cm}\): \(\mathrm{Median} = 82\text{ cm}\). Is \(82 \geq 1.02 \times 80 = 81.6\)? YES ✓
• If \(\mathrm{average} = 120\text{ cm}\): \(\mathrm{Median} = 122\text{ cm}\). Is \(122 \geq 1.02 \times 120 = 122.4\)? NO ✗

Since we get different answers depending on the average height, Statement 1 alone is NOT sufficient.

[STOP - Not Sufficient!] → Eliminate choices A and D

Analyzing Statement 2

Important: We now analyze Statement 2 independently, forgetting Statement 1 completely.

Statement 2 tells us: The sum of all 25 heights < 2,550 cm

This means: \(\mathrm{Average\,height} < \frac{2550}{25} = 102\text{ cm}\)

But this reveals nothing about how the median compares to the average. The median could be:

• Higher than average (many short children, few very tall ones)
• Equal to average (heights symmetrically distributed)
• Lower than average (many tall children, few very short ones)

Without knowing the median-average relationship, we cannot answer our question.

Statement 2 alone is NOT sufficient.

[STOP - Not Sufficient!] → Eliminate choice B

Combining Both Statements

Now we use both statements together:

• From Statement 1: \(\mathrm{Median} = \mathrm{Average} + 2\text{ cm}\)
• From Statement 2: \(\mathrm{Average} < 102\text{ cm}\)

The critical question remains: Is 2 cm at least 2% of the average?

Here's the key insight:

• 2% of 100 cm = 2 cm (exactly at the boundary!)
• If \(\mathrm{Average} \leq 100\text{ cm}\) → 2 cm ≥ 2% of average → Answer is YES
• If \(\mathrm{Average} > 100\text{ cm}\) → 2 cm < 2% of average → Answer is NO

But Statement 2 only tells us \(\mathrm{Average} < 102\), which includes values both above and below 100:

Example 1: \(\mathrm{Average} = 99\text{ cm}\)

• \(\mathrm{Median} = 101\text{ cm}\)
• Is \(101 \geq 1.02 \times 99 = 100.98\)? YES ✓

Example 2: \(\mathrm{Average} = 101\text{ cm}\)

• \(\mathrm{Median} = 103\text{ cm}\)
• Is \(103 \geq 1.02 \times 101 = 103.02\)? NO ✗

Since we still get different answers, the statements together are NOT sufficient.

[STOP - Not Sufficient!] → Eliminate choice C

The Answer: E

The statements together are not sufficient because the critical threshold (\(\mathrm{Average} = 100\text{ cm}\)) falls within the possible range allowed by both statements. We cannot determine a definitive YES or NO.

Answer: E - Both statements together are not sufficient.