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Andrew and Maria each collected six rocks, and the masses of the rocks are shown in the table above. The...

GMAT Problem-Solving and Data Analysis : (PS_DA) Questions

Source: Official
Problem-Solving and Data Analysis
One-variable data: distributions and measures of center and spread
HARD
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Andrew and Maria each collected six rocks, and the masses of the rocks are shown in the table above. The mean of the masses of the rocks Maria collected is \(\mathrm{0.1}\) kilogram greater than the mean of the masses of the rocks Andrew collected. What is the value of \(\mathrm{x}\)?

Masses (kilograms)
Andrew2.4, 2.5, 3.6, 3.1, 2.5, 2.7
Maria\(\mathrm{x}\), 3.1, 2.7, 2.9, 3.3, 2.8
Enter your answer here
Solution

1. TRANSLATE the problem information

  • Given information:
    • Andrew's masses: \(2.4, 2.5, 3.6, 3.1, 2.5, 2.7\) kg
    • Maria's masses: \(\mathrm{x}, 3.1, 2.7, 2.9, 3.3, 2.8\) kg
    • Maria's mean is \(0.1\) kg greater than Andrew's mean
  • What this tells us: We need to find Andrew's mean first, then use it to determine what Maria's mean should be, then solve for \(\mathrm{x}\).

2. SIMPLIFY to find Andrew's mean

  • Calculate the sum: \(2.4 + 2.5 + 3.6 + 3.1 + 2.5 + 2.7 = 16.8\) kg
  • Find the mean: \(16.8 \div 6 = 2.8\) kg

3. TRANSLATE the mean relationship

  • Since Maria's mean is \(0.1\) kg greater than Andrew's mean:
  • Maria's mean = \(2.8 + 0.1 = 2.9\) kg

4. SIMPLIFY to set up Maria's equation

  • Sum of Maria's known masses: \(3.1 + 2.7 + 2.9 + 3.3 + 2.8 = 14.8\) kg
  • Total sum including \(\mathrm{x}\): \(\mathrm{x} + 14.8\) kg
  • Mean equation: \((\mathrm{x} + 14.8) \div 6 = 2.9\)

5. SIMPLIFY to solve for x

  • Multiply both sides by \(6\): \(\mathrm{x} + 14.8 = 17.4\)
  • Subtract \(14.8\) from both sides: \(\mathrm{x} = 17.4 - 14.8 = 2.6\)

Answer: \(2.6\) (also acceptable: \(\frac{13}{5}\) or \(2.60\))




Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY skill: Students make arithmetic errors when adding Andrew's masses or Maria's known masses, leading to incorrect mean calculations.

For example, if they miscalculate Andrew's sum as \(16.2\) instead of \(16.8\), they get Andrew's mean as \(2.7\), making Maria's mean \(2.8\), and ultimately \(\mathrm{x} = 2.0\). This creates confusion since none of the typical multiple choice answers would match, leading to guessing.

Second Most Common Error:

Poor TRANSLATE reasoning: Students misinterpret "\(0.1\) kilogram greater" and subtract instead of add, thinking Maria's mean should be \(2.8 - 0.1 = 2.7\) kg.

This leads them to set up the equation \((\mathrm{x} + 14.8) \div 6 = 2.7\), giving \(\mathrm{x} = 1.4\), which again doesn't match expected answer patterns and causes confusion.

The Bottom Line:

This problem requires careful arithmetic at multiple steps combined with clear translation of the comparative relationship. Students who rush through the calculations or misread the comparison direction will struggle to reach the correct answer.

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