UPSC CDS 2 2022 Math Paper

Let the length and breadth of two rectangles be l1b1 and l2b2 respectively.
According to the question
⇒11b1=480
⇒12b2=480
They differ in length by 6‌cm and breadth by 4‌cm
Then, length of second rectangle (l2)=(l1+6)‌cm
Then, breadth of second rectangle (b2)=(b1−4)‌cm
Perimeter of first rectangle =2(l1+b1)
Perimeter of second rectangle =2(l1+6+b1−4)
⇒2(11+b1)+4
The difference in their perimeters is
⇒2(l1+b1)−2(l1+b1)+4
⇒4‌cm