Question

If a, b and c are positive real numbers, such that $a(b+c)=152,b(c+a)=162$ and $c(a+b)=110$ then which of the following is true

• A. $ab+bc+ca=424$
• B. $ab+bc+ca=212$
• C. $abc=672$
• D. $abc=720$

Answer 1

The correct option is B $ab+bc+ca=212$

Given:$a,b$ and $c$ are positive real numbers

$a(b+c)=152$ ........$\left(1\right)$

$b(c+a)=162$ ........$\left(2\right)$

$c(a+b)=110$ ........$\left(3\right)$

Adding the equations $\left(1\right),\left(2\right)$ and $\left(3\right)$ we get

$a(b+c)+b(c+a)+c(a+b)=152+162+110=424$

$\Rightarrow ab+ac+bc+ba+ca+cb=424$

$\Rightarrow 2(ab+bc+ca)=424$

$\Rightarrow ab+bc+ca=\frac{424}{2}=212$ ......$\left(4\right)$

Hence option $\left(b\right)$ is correct.

From $\left(4\right)$ we have $ab+ac+bc=212$

$\left(1\right)\Rightarrow 152+bc=212$

$\Rightarrow bc=212-152=60$

From $\left(4\right)$ we have $ab+bc+ac=212$

$\left(2\right)\Rightarrow 162+ac=212$

$\Rightarrow ac=212-162=50$

From $\left(4\right)$ we have $ab+bc+ac=212$

$\left(3\right)\Rightarrow ab+110=212$

$\Rightarrow ab=212-110=102$

Again $\left(ab\right)\left(bc\right)\left(ca\right)=102\times 60\times 50$

$\Rightarrow {\left(abc\right)}^2=306,000$

$\Rightarrow abc=553.172$

Hence only option $\left(b\right)$ is correct.