Question

If 44b - 3a = 22b + c = 8c - a and a + b + c = 11 , then find the value of $4(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca})$

• A. $\frac{11}{9}$
• B. $\frac{9}{11}$
• C. $\frac{6}{5}$
• D. $\frac{5}{6}$

Answer 1

Answer: (A)

$\frac{11}{9}$

Answer 2

44b-3a = 22b+c = 8c-a
22(4b-3a) = 22b+c = 23(c-a)
28b-6a = 22b+c = 23c-3a
28b-6a = 23c-3a
8b - 6a = 3c - 3a
8b = 3a + 3c ........ (1)
22b+c = 23c-3a
2b + c = 3c - 3a
3a + 2b = 2c ........ (2)
28b-6a = 22b+c
8b - 6a = 2b + c
6b = 6a + c ....... (3)
From (1) and (2):
8b = 2c - 2b + 3c => 10b = 5c => c = 2b =>b = c/2
From (1) and (3):
8b = 3a + 3 (6b - 6a) => 15a = 10b => 3a = 2b =>a = 2b/3 = c/3
a + b + c = 11
(c/3) + (c/2) + c = 11
c = 6, b = 3 and a = 2
Value of $4[\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}]$
$=4[\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}]$
$=4[\frac{1}{6}+\frac{1}{18}+\frac{1}{12}]$
$=4[\frac{6+2+3}{36}]$
$=\frac{11}{9}$
So, the correct option is (A) : $\frac{11}{9}$