Question : If $\frac{P}{2}=\frac{Q}{5}=\frac{R}{6}$, then what is the value of $P^2:(P+Q)^2:(Q+R)^2$?
Option 1: 10 : 56 : 169
Option 2: 4 : 49 : 121
Option 3: 9 : 64 : 144
Option 4: 16 : 10 : 125
Correct Answer: 4 : 49 : 121
Solution :
Given: The expression is $\frac{P}{2}=\frac{Q}{5}=\frac{R}{6}$.
Let $\frac{P}{2}=\frac{Q}{5}=\frac{R}{6}=k$.
⇒ $P=2k, Q=5k, R=6k$
The value of $P^2:(P+Q)^2:(Q+R)^2$
$=(2k)^2:(2k+5k)^2:(5k+6k)^2$
$=4k^2:(7k)^2:(11k)^2$
$=4k^2:49k^2:121k^2$
$=4:49:121$
Hence, the correct answer is 4 : 49 : 121.
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