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from latex2sympy2_extended import NormalizationConfig
from math_verify import LatexExtractionConfig, parse, verify
from sympy.parsing.latex.errors import LaTeXParsingError
content = """Let's denote the original number of workers and the original production amount by \( W \) and \( P \), respectively. We'll use these to find out how many workers there are in the plant.
When 40 workers switch to the Stakhanovite method, the production increases by 20%, so the new production amount is \( 1.2W \).
Then, when 60% of the workers (which is \( 1.2W \times 0.6 = 0.72W \)) switch to the new method, the production increases to two and a half times the original production:
\[ W + 0.72W = 2.5W \]
Let \( N \) be the total number of workers in the plant, and \( P \) be the original production amount. After all the workers switch to the new method, the production becomes \( N \times P \), and we know from the previous equation that \( N \times P = 2.5W \).
Since 40 workers switch to the Stakhanovite method, we have:
\[ W = N \times (1 - 0.4) = 0.6N \]
And since 60% of the workers have switched, we get:
\[ W \times 0.6 = N \times (1 - 0.4 - 0.32) = 0.28N \]
Using the equation \( W = 0.6N \), we can substitute \( W \) with \( 0.6N \) in the second equation:
\[ 0.6N \times 0.6 = 2.5N \]
Reducing the left side:
\[ 0.36N = 2.5N \]
\[ N = 6.25N \]
When all workers switch, the production becomes:
\[ N \times N = N^2 = (2.5N)^2 = 6.25W \]
Thus, the production increases by \( \frac{6.25W}{W} = 6.25 \) times."""
answer_parsed = parse(
content,
extraction_config=[
LatexExtractionConfig(
normalization_config=NormalizationConfig(
nits=False,
malformed_operators=False,
basic_latex=True,
equations=True,
boxed="all",
units=True,
),
# Ensures that boxed is tried first
boxed_match_priority=0,
try_extract_without_anchor=True,
)
],
extraction_mode="first_match",
)
print (answer_parsed)
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