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

Answer: (B)

20

Answer 2

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