`(a)/(b)=(c )/(d)=(e )/(f)=(7)/(11)`
Find `(3a+5c+5e)/(2a-3d+5f)=?`

Answer

To solve the problem, we start with the given ratios:

\[ \frac{a}{b} = \frac{c}{d} = \frac{e}{f} = \frac{7}{11} \]

This implies that we can express \(a\), \(b\), \(c\), \(d\), \(e\), and \(f\) in terms of a common variable \(k\):

\[ a = 7k, \quad b = 11k, \quad c = 7m, \quad d = 11m, \quad e = 7n, \quad f = 11n \]

Since all ratios are equal, we can set \(m = n = k\). Thus, we have:

\[ a = 7k, \quad b = 11k, \quad c = 7k, \quad d = 11k, \quad e = 7k, \quad f = 11k \]

Now, we need to find the value of the expression:

\[ \frac{3a + 5c + 5e}{2a - 3d + 5f} \]

Substituting the values of \(a\), \(c\), \(e\), \(d\), and \(f\):

**Numerator:**

\[ 3a + 5c + 5e = 3(7k) + 5(7k) + 5(7k) \]

Calculating this gives:

\[ = 21k + 35k + 35k = 91k \]

**Denominator:**

\[ 2a - 3d + 5f = 2(7k) - 3(11k) + 5(11k) \]

Calculating this gives:

\[ = 14k - 33k + 55k = 14k + 22k = 36k \]

Now we can substitute the values of the numerator and denominator back into the expression:

\[ \frac{3a + 5c + 5e}{2a - 3d + 5f} = \frac{91k}{36k} \]

Cancelling \(k\) from the numerator and denominator:

\[ = \frac{91}{36} \]

Thus, the final answer is:

\[ \frac{91}{36} \]