If x, y, z are three numbers such that $x+y=13, y+z=15$ and $z+x=16$, the value of $\frac{x y+x z}{x y z}$ is:
Answer & explanation
Correct answer: option 4
$x+y=13, y+z=15$ and $z+x=16$
By adding all of these we get,
x + y + y + z + z + x = 13 + 15 + 16
2(x + y + z) = 44
x + y + z = 22
x = 22 - 15 = 7
y = 22 - 16 = 6
x = 22 - 13 = 9
The value of $\frac{x y+x z}{x y z}$ = $\frac{7 × 6+7 × 9}{7 × 6 × 9}$
The value of $\frac{x y+x z}{x y z}$ = $\frac{105}{378}$
The value of $\frac{x y+x z}{x y z}$ = $\frac{5}{18}$