Question

If $a:b=c:d=e:f$ then $(pa+qc+re):(pb+qd+rf)$ is equal to

• A. $p:(q+r)$
• B. $(p+q):r$
• C. $a:b$
• D. $1:2$

Answer 1

The correct option is C $a:b$

Let $a:b=c:d=e:f$

$\therefore a=bk$

$\therefore c=dk$

$\therefore e=fk$

Now $\frac{pa+qc+re}{pb+qd+rf}=\frac{pbk+qdk+rfk}{pb+qd+rf}=\frac{k(pb+qd+rf)}{(pb+qd+rf)}\therefore (pa+qc+re):(pb+qd+rf)=k:1=a:b=c:d=e:f$