Question

If ${\displaystyle \frac{x}{a+b-c}=\frac{y}{b+c-a}=\frac{z}{c+a-b}=5}$ and a + b + c = 7; the value of x + y + z is ______.

पर्याय

• 35

• ${\displaystyle \frac{7}{5}}$

• ${\displaystyle \frac{5}{7}}$

• 42

Answer 1

If ${\displaystyle \frac{x}{a+b-c}=\frac{y}{b+c-a}=\frac{z}{c+a-b}=5}$ and a + b + c = 7; the value of x + y + z is 35.

Explanation:

Given

${\displaystyle \frac{x}{a+b-c}=\frac{y}{b+c-a}=\frac{z}{c+a-b}=5}$

∴ Each fraction

= ${\displaystyle \frac{x+y+z}{a+b-c+b+c-a+c+a-b}}$

${\displaystyle \Rightarrow \frac{x+y+z}{a+b+c}=5}$

${\displaystyle \Rightarrow \frac{x+y+z}{7}=5}$  ...[Given a + b + c = 7]

${\displaystyle \Rightarrow }$ x + y + z = 35