Question:D = m/VThe formula above gives the density D, in kilograms per liter, of a substance with mass m, in...
GMAT Algebra : (Alg) Questions
\(\mathrm{D = \frac{m}{V}}\)
The formula above gives the density \(\mathrm{D}\), in kilograms per liter, of a substance with mass \(\mathrm{m}\), in kilograms, and volume \(\mathrm{V}\), in liters. A sealed \(\mathrm{4}\)-liter container initially holds a solution with a mass of \(\mathrm{2.8}\) kilograms. You plan to dissolve \(\mathrm{n}\) identical packets into this solution, where each packet has a mass of \(\mathrm{150}\) grams and you may assume the volume remains \(\mathrm{4}\) liters after the packets are added. For shipping regulations, the final mixture's density must be no more than \(\mathrm{0.95}\) kilogram per liter. What is the greatest number \(\mathrm{n}\) of packets that can be added while meeting the density requirement? Enter your answer as an integer.
1. TRANSLATE the problem information
- Given information:
- Initial solution: 2.8 kg in 4 L container
- Each packet: \(\mathrm{150\ grams = 0.150\ kg}\)
- Volume stays 4 L after adding packets
- Maximum allowed density: 0.95 kg/L
- Need: greatest integer n
- What this tells us: We need to set up a density constraint using \(\mathrm{D = m/V}\)
2. INFER the mathematical approach
- Strategy: Express total mass in terms of n, then use density constraint
- The total mass will be: initial mass + (n packets × mass per packet)
- Then apply the density formula with the constraint \(\mathrm{D \leq 0.95}\)
3. TRANSLATE into mathematical expressions
- Total mass after adding n packets: \(\mathrm{2.8 + 0.150n\ kg}\)
- Volume remains: \(\mathrm{4\ L}\)
- Density constraint: \(\mathrm{(2.8 + 0.150n)/4 \leq 0.95}\)
4. SIMPLIFY the inequality
- Multiply both sides by 4: \(\mathrm{2.8 + 0.150n \leq 3.8}\)
- Subtract 2.8 from both sides: \(\mathrm{0.150n \leq 1.0}\)
- Divide by 0.150: \(\mathrm{n \leq 6.67}\) (use calculator)
5. APPLY CONSTRAINTS to get final answer
- Since n must be a whole number of packets: \(\mathrm{n \leq 6}\)
- Greatest integer value: \(\mathrm{n = 6}\)
Answer: 6
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak TRANSLATE skill: Students struggle with unit conversion and miss that \(\mathrm{150\ grams = 0.150\ kg}\), instead using 150 directly in their calculations.
This creates the incorrect constraint \(\mathrm{(2.8 + 150n)/4 \leq 0.95}\), leading to \(\mathrm{2.8 + 150n \leq 3.8}\), then \(\mathrm{150n \leq 1.0}\), and finally \(\mathrm{n \leq 0.0067}\). This nonsensical result leads to confusion and random guessing.
Second Most Common Error:
Inadequate APPLY CONSTRAINTS execution: Students correctly solve the inequality to get \(\mathrm{n \leq 6.67}\) but then round up to \(\mathrm{n = 7}\) instead of recognizing they need the greatest integer that satisfies the constraint.
Using \(\mathrm{n = 7}\) would give density = \(\mathrm{(2.8 + 0.150×7)/4 = 3.85/4 = 0.9625\ kg/L}\), which exceeds \(\mathrm{0.95\ kg/L}\). This leads them to enter 7 as their answer.
The Bottom Line:
This problem tests whether students can carefully track units through a real-world constraint problem and properly interpret "greatest integer that satisfies" rather than just rounding their final calculation.