3 pieces of fabric with a total length of 105m

Because each fabric is cut into a ratio of itself, then the correct answer choice will be the one where the first value is a multiple of 5, the second a multiple of 7 and the last will be a multiple of 3.

Only Answer choice C works.

To check:

when 2/5 (10m) of 25m are cut off 15m are left
when 4/7 (20m) of 35m are cut off then 15m are left
when 2/3 (30m) of 45m are cut off then 15m are left

Answer C

The best way to solve this question is to try to plug in the answers for the lengths of the three ropes.
Now, before we go and plug in, let's look at the fractions first.

• \(\frac{2}{5}th\) of the first wire is cut.
• \(\frac{4}{7}th\) of second wire is cut.
• \(\frac{2}{3}rd\) of third wire is cut.

Now GMAT generally doesn't provide answers that are in fractions, so since we are trying to plug in the answers and check, let's look at values that give whole numbers with fractions.

A. 35m; 25m; 45m
\(\frac{4}{7} * 25\) doesn't give a whole number, let's avoid this for now.

B. 35m; 35m; 35m
\(\frac{2}{3} * 45\) doesn't give a whole number, let's avoid this for now.

C. 25m; 35m; 45m
This gives whole numbers for all, and when we check the values we get for remaining lengths of wires, we get 15m for all three wires. This looks good. This seems to be the answer.

D. 45m; 25m; 35m
\(\frac{4}{7} * 25\) doesn't give a whole number, let's avoid this for now.

E. 45m; 35m; 25m
\(\frac{2}{3} * 25\) doesn't give a whole number, let's avoid this for now.

Since only C gave us whole numbers and upon checking it gave equal remaining lengths for all wires, C is the correct answer.

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
_________________

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

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

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

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

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

\(Given,\)

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

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

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

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

\(x = 5k = 25\)
\(y = 7k = 35\)
\(z = 9k = 45 (C)\)