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Given: 3 pieces of fabric with a total length of 105m. After cutting \(\frac{2}{5}\) the first, \(\frac{4}{7}\) of the second, and \(\frac{2}{3}\) of the third, the rest of the three pieces are equal in length.

Asked: The original lengths of each piece respectively are:

A. 35m; 25m; 45m: The rest of first piece = (1-2/5)*35 = 3/5*35 = 21; The rest of second piece = (1-4/7)*25 = 3/7*25 = 75/7; The rest of third piece = (1-2/3)*45 = 1/3*45 = 15: The rest of the three pieces are not equal in length; Incorrect

B. 35m; 35m; 35m; The rest of first piece = (1-2/5)*35 = 3/5*35 = 21; The rest of second piece = (1-4/7)*35 = 3/7*35 = 15; The rest of third piece = (1-2/3)*35 = 1/3*35 = 35/3: The rest of the three pieces are not equal in length; Incorrect

C. 25m; 35m; 45m; The rest of first piece = (1-2/5)*25 = 3/5*25 = 15; The rest of second piece = (1-4/7)*35 = 3/7*35 = 15; The rest of third piece = (1-2/3)*45 = 1/3*45 = 15: The rest of the three pieces are equal in length; Correct

D. 45m; 25m; 35m; The rest of first piece = (1-2/5)*45 = 3/5*45 = 27; The rest of second piece = (1-4/7)*25 = 3/7*25 = 75/7; The rest of third piece = (1-2/3)*35 = 1/3*35 = 35/3: The rest of the three pieces are not equal in length; Incorrect

E. 45m; 35m; 25m; The rest of first piece = (1-2/5)*45 = 3/5*45 = 27; The rest of second piece = (1-4/7)*35 = 3/7*35 = 15; The rest of third piece = (1-2/3)*25 = 1/3*25 = 25/3: The rest of the three pieces are not equal in length; Incorrect

IMO C
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3 pieces of fabric with a total length of 105m. After cutting \(\frac{2}{5}\) of the first, \(\frac{4}{7}\) of the second, and \(\frac{2}{3}\) of the third, the rest of the three pieces are equal in length. The original lengths of each piece respectively are:

A. 35m; 25m; 45m
B. 35m; 35m; 35m
C. 25m; 35m; 45m
D. 45m; 25m; 35m
E. 45m; 35m; 25m

let the length of the fabrics be x,y, 105-x-y

=>3/5 *x = 3/7 *y = 1/3 * (105 - x-y)

Taking each equations individually and solving through substitution we get
x=25 , y=35 , 105-35-25 = 45

Therefore IMO C
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\(Let\) \(the\) \(lengths\) \(of\) \(three\) \(fabrics\) \(be\) = \(x,y,z\)

\(x\)-\(\frac{2x}{5}\) = \(\frac{3x}{5}\)

\(y\)-\(\frac{4y}{7}\) = \(\frac{3y}{7}\)

\(z\)-\(\frac{2z}{3}\) = \(\frac{1z}{3}\)

\(Given,\)

\(\frac{3x}{5}\) = \(\frac{3y}{7}\) => \(\frac{x}{y}\) = \(\frac{5}{7}\)

\(\frac{3x}{5}\) = \(\frac{1z}{3}\) => \(\frac{x}{z}\) = \(\frac{5}{9}\)

\(x:y:z = 5:7:9\)

\(5k+7k+9k = 105\)
\(21k = 105\)
\(k = 5\)

\(x = 5k = 25\)
\(y = 7k = 35\)
\(z = 9k = 45 (C)\)
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So this is just a logic trick for this particular problem and answer set:

If you cut off a smaller portion of a ribbon and the remainder is still equal to the remainder of a larger portion cut from another ribbon,
that would mean that the original length of the ribbon in which the smaller portion was cut has to be smaller than the ribbon in which
the larger portion was cut.

If you recognize that the fractions are in order from smallest to largest; 2/5, 4/7, 2/3 Then the answer choice
has to be the one that's also ordered smallest to largest.

This works as a very quick way to solve this particular problem because only one answer choice is arranged in order from smallest to largest.
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