Question

If l, m, and n are real numbers such that \(l + m = 10lm, m + n = 12mn\), and \(n + l = 4nl\), then find the value of \(\frac{180lmn}{lm+mn+nl}\).

• A. 18
• B. 20
• C. 24
• D. 32
• E. 36

Answer 1

The Correct Option is B

\(l + m = 10lm ⇒ \frac{l}{lm} + \frac{m}{lm }= 10 ⇒ \frac{1}{m} + \frac{1}{l }= 10\) … (1)

Similarly, \(m + n = 12mn ⇒ \frac{1}{n} + \frac{1}{m} = 12\) … (2)

And, \(n + l = - 4nl ⇒ \frac{1}{l} + \frac{1}{n} = -4 \)… (3)

Adding equations \((1), (2)\), and \((3)\), we get the following:

\(2\bigg(\frac{1}{l} + \frac{1}{m} + \frac{1}{n}\bigg) = 18\)

\(⇒ \bigg(\frac{1}{l} + \frac{1}{m} + \frac{1}{n}\bigg) = 9\)

So, = \(\frac{180 lmn}{lm+mn+nl }= \frac{180 }{ \frac{1}{n}+\frac{1}{l}+\frac{1}{m}} = \frac{180}{9} = 20\)

Hence, option B is the correct answer.