During each shift at a certain factory, nails are manufactured at a constant rate of 60 nails per 5 seconds...
GMAT Word Problems : (WP) Questions
During each shift at a certain factory, nails are manufactured at a constant rate of \(\mathrm{60\ nails\ per\ 5\ seconds}\) and packaged at a constant rate of \(\mathrm{100\ nails\ per\ 9\ seconds}\). If there are \(\mathrm{1{,}000\ nails}\) that are not packaged at the start of a certain shift and if each shift lasts more than \(\mathrm{2\ hours}\), how many nails are there that are not packaged \(\mathrm{2\ hours}\) after the start of the shift?
- Translate the problem requirements: Convert the given rates to common units (nails per second), understand that we start with 1,000 unpackaged nails, and need to find how many unpackaged nails exist after exactly 2 hours
- Calculate the net accumulation rate: Determine how many more nails are manufactured than packaged each second, since this difference adds to the unpackaged pile
- Apply the time period: Multiply the net rate by 2 hours (7,200 seconds) to find total additional unpackaged nails created during this period
- Add to initial inventory: Combine the starting 1,000 unpackaged nails with the additional nails that accumulated over 2 hours
Execution of Strategic Approach
1. Translate the problem requirements
Let's start by understanding what's happening in plain English. We have a factory where nails are being made and packaged simultaneously, but at different speeds. Think of it like a conveyor belt where nails are being added at one rate and taken away at another rate.
First, let's convert the rates to the same time unit so we can compare them:
• Manufacturing rate: \(\mathrm{60\text{ nails per }5\text{ seconds} = 12\text{ nails per second}}\)
• Packaging rate: \(\mathrm{100\text{ nails per }9\text{ seconds} = 100/9\text{ nails per second} ≈ 11.11\text{ nails per second}}\)
We also need to convert 2 hours to seconds: \(\mathrm{2\text{ hours} = 2 × 60 × 60 = 7,200\text{ seconds}}\)
Starting condition: 1,000 nails are already waiting to be packaged
Goal: Find how many nails are unpackaged after exactly 2 hours
Process Skill: TRANSLATE - Converting different time units and rates into comparable forms
2. Calculate the net accumulation rate
Now let's think about what happens every second. Picture this scenario: every second, 12 new nails are manufactured and about 11.11 nails are packaged. This means that each second, slightly more nails are being made than are being packaged.
The difference between what's made and what's packaged gets added to our pile of unpackaged nails:
• Net accumulation per second = Manufacturing rate - Packaging rate
• \(\mathrm{Net\text{ accumulation per second} = 12 - 100/9 = 108/9 - 100/9 = 8/9\text{ nails per second}}\)
This means that every second, \(\mathrm{8/9}\) additional nails join the unpackaged pile. In other words, for every 9 seconds that pass, 8 more nails become unpackaged than were there before.
3. Apply the time period
Now we need to find out how many additional unpackaged nails accumulate over the entire 2-hour period.
Since \(\mathrm{8/9}\) nails accumulate each second, and we have 7,200 seconds:
• \(\mathrm{Total\text{ additional unpackaged nails} = (8/9) × 7,200}\)
• \(\mathrm{Total\text{ additional unpackaged nails} = 8 × 7,200 ÷ 9}\)
• \(\mathrm{Total\text{ additional unpackaged nails} = 8 × 800 = 6,400\text{ nails}}\)
So over the 2-hour shift, 6,400 more nails will join the unpackaged pile beyond what was already there.
4. Add to initial inventory
Finally, we need to account for the nails that were already unpackaged at the start of the shift.
\(\mathrm{Total\text{ unpackaged nails after }2\text{ hours} = Starting\text{ unpackaged nails} + Additional\text{ accumulated nails}}\)
\(\mathrm{Total\text{ unpackaged nails} = 1,000 + 6,400 = 7,400\text{ nails}}\)
Final Answer
After 2 hours, there will be 7,400 nails that are not packaged.
Looking at our answer choices, this matches choice (E) 7,400.
We can verify this makes sense: since we're manufacturing slightly faster than we're packaging (\(\mathrm{12\text{ vs }11.11\text{ nails per second}}\)), the unpackaged pile should grow significantly over 2 hours, which it does from 1,000 to 7,400 nails.
Common Faltering Points
Errors while devising the approach
1. Misunderstanding the problem setup
Students often confuse this as a sequential process where nails are first manufactured completely, then packaged, rather than understanding that manufacturing and packaging happen simultaneously. This leads them to calculate total manufactured nails first, then subtract packaged nails, missing the dynamic nature of the problem.
2. Inconsistent time unit conversion
Students frequently convert only some rates to common units but not others. For example, they might convert manufacturing to "nails per second" but leave packaging as "nails per 9 seconds," making it impossible to find the correct net accumulation rate.
3. Forgetting the initial inventory
Many students focus entirely on what happens during the 2-hour period and forget that there were already 1,000 unpackaged nails at the start. They solve for only the additional accumulation and select that as their final answer.
Errors while executing the approach
1. Fraction arithmetic errors with rates
When calculating the net rate of \(\mathrm{12 - 100/9}\), students often struggle with the fraction subtraction. Common errors include: calculating \(\mathrm{12 - 100/9\text{ as }(12-100)/9 = -88/9}\), or incorrectly finding a common denominator and getting \(\mathrm{(108-100)/9 = 8/9}\) wrong due to arithmetic mistakes.
2. Time conversion mistakes
Students frequently make errors when converting 2 hours to seconds. Common mistakes include: \(\mathrm{2 × 60 = 120\text{ seconds}}\) (forgetting to multiply by 60 again), or \(\mathrm{2 × 3600 = 7,200}\) but then using the wrong value in subsequent calculations.
3. Multiplication errors with large numbers
When calculating \(\mathrm{(8/9) × 7,200}\), students often make computational errors. They might incorrectly calculate \(\mathrm{8 × 7,200 ÷ 9}\) or make mistakes in simplifying \(\mathrm{7,200 ÷ 9 = 800}\), leading to wrong values for the additional accumulated nails.
Errors while selecting the answer
1. Selecting intermediate calculated values
Students who correctly calculate the additional accumulation as 6,400 nails might mistakenly select this value if it appears among the answer choices, forgetting to add the initial 1,000 unpackaged nails to get the final answer of 7,400.
2. Misreading what the question asks for
Some students might calculate correctly but then select an answer that represents something other than what's asked. For instance, if they calculated both "total nails manufactured" and "total nails not packaged," they might accidentally select the wrong final value from their work.