Question

the area of the classroom is 10,125 meters
example: an electrician has a piece of wire that is 4 and 3/8 centimetres long. she cuts the wire into pieces that are 1 and 2/3 centimetres long. how many pieces does she have?

Explanation:

Step1: Convert mixed numbers to improper fractions

First, convert \(1\frac{2}{3}\) to an improper fraction: \(1\frac{2}{3}=\frac{1\times3 + 2}{3}=\frac{5}{3}\)
Then, convert \(4\frac{3}{8}\) to an improper fraction: \(4\frac{3}{8}=\frac{4\times8+3}{8}=\frac{35}{8}\)

Step2: Divide the total length by the length of each piece

To find the number of pieces, we divide the total length of the wire by the length of each piece. Let the total length be \(L = \frac{35}{8}\) cm and the length of each piece be \(l=\frac{5}{3}\) cm. The number of pieces \(n\) is given by \(n=\frac{L}{l}\)
\[n=\frac{\frac{35}{8}}{\frac{5}{3}}\]
When dividing fractions, we multiply by the reciprocal of the divisor:
\[n=\frac{35}{8}\times\frac{3}{5}\]
Simplify the fraction:
\[n=\frac{35\times3}{8\times5}=\frac{7\times3}{8\times1}=\frac{21}{8} = 2\frac{5}{8}\]
Wait, this seems incorrect. Wait, maybe the total length is \(10.125\) meters? Wait, the text is a bit unclear. Wait, maybe the wire length is \(10.125\) meters? Let's re - examine.

Wait, maybe the wire length is \(10.125\) meters, and we need to cut it into pieces of length \(1\frac{2}{3}\) meters and \(4\frac{3}{8}\) meters? No, the original problem: "An electrician has a piece of wire that is \(4\frac{3}{8}\) centimetres long. She cuts the wire into pieces that are \(1\frac{2}{3}\) centimetres long. How many pieces does she have?"

Wait, let's recalculate:

\(4\frac{3}{8}=\frac{35}{8}\) cm, \(1\frac{2}{3}=\frac{5}{3}\) cm

\(n=\frac{35}{8}\div\frac{5}{3}=\frac{35}{8}\times\frac{3}{5}=\frac{21}{8}=2.625\)

But we can't have a fraction of a piece in a practical sense, but mathematically, the number of pieces is \(\frac{21}{8}\) or \(2\frac{5}{8}\). But maybe there was a mistake in the problem statement, maybe the total length is \(10.125\) meters (which is \(1012.5\) cm) and the piece length is \(1\frac{2}{3}\) cm? Let's try that.

\(10.125\) meters \( = 1012.5\) cm \(=\frac{10125}{10}=\frac{2025}{2}\) cm

\(1\frac{2}{3}=\frac{5}{3}\) cm

Number of pieces \(n=\frac{2025}{2}\div\frac{5}{3}=\frac{2025}{2}\times\frac{3}{5}=\frac{405\times3}{2}=\frac{1215}{2} = 607.5\)

This is also odd. Wait, maybe the total length is \(10.125\) meters and the piece length is \(4\frac{3}{8}\) meters?

\(10.125\) meters \(=\frac{81}{8}\) meters, \(4\frac{3}{8}=\frac{35}{8}\) meters

Number of pieces \(n=\frac{81}{8}\div\frac{35}{8}=\frac{81}{35}\approx2.31\)

Alternatively, maybe the problem is: The area of the classroom is \(10.125\) square meters, and we have tiles of area \(1\frac{2}{3}\) square meters and \(4\frac{3}{8}\) square meters? No, the original text is about an electrician and wire.

Wait, let's assume the correct problem is: An electrician has a piece of wire that is \(10.125\) meters long. She cuts the wire into pieces that are \(1\frac{2}{3}\) meters long. How many pieces does she have?

\(10.125=\frac{81}{8}\) meters, \(1\frac{2}{3}=\frac{5}{3}\) meters

\(n=\frac{81}{8}\div\frac{5}{3}=\frac{81}{8}\times\frac{3}{5}=\frac{243}{40}=6.075\)

This is still odd. Maybe the piece length is \(1\frac{2}{3}\) centimeters and the total length is \(4\frac{3}{8}\) centimeters. Then:

\(4\frac{3}{8}=\frac{35}{8}\) cm, \(1\frac{2}{3}=\frac{5}{3}\) cm

\(n = \frac{35}{8}\div\frac{5}{3}=\frac{35\times3}{8\times5}=\frac{21}{8}=2.625\)

Since we can't have a fraction of a piece, maybe we take the whole number part, but mathematically, the number of pieces is \(\frac{21}{8}\) or \(2\frac{5}{8}\). But perhaps there was a typo and the total length is \(10.125\) centimeters.

\(10.125=\frac{81}{8}\) cm

\(n=\frac{81}{8}\div\frac{5}{3}=\frac{81\times3}{8\times5}=\frac{243}{40}=6.075\)

This is also not a whole number. Maybe the piece length is \(4\frac{3}{8}\) centimeters and the total length is \(10.125\) centimeters.

\(10.125=\frac{81}{8}\) cm, \(4\frac{3}{8}=\frac{35}{8}\) cm

\(n=\frac{81}{8}\div\frac{35}{8}=\frac{81}{35}\approx2.31\)

This is very confusing. Maybe the original problem has a different set of numbers. Let's assume that the total length of the wire is \(10.125\) meters (\(1012.5\) cm) and the length of each piece is \(1.625\) cm (\(1\frac{5}{8}\) cm) or something else. But based on the given numbers, let's proceed with the first interpretation where the total length is \(4\frac{3}{8}\) cm and the piece length is \(1\frac{2}{3}\) cm.

The number of pieces is \(\frac{4\frac{3}{8}}{1\frac{2}{3}}=\frac{\frac{35}{8}}{\frac{5}{3}}=\frac{35\times3}{8\times5}=\frac{21}{8}=2\frac{5}{8}\)

But if we consider that we can only have whole pieces, we can say 2 pieces with a remainder. But mathematically, the division gives \(\frac{21}{8}\) or \(2.625\)

Answer:

\(\frac{21}{8}\) (or \(2\frac{5}{8}\))