Quantity I: Area of square, given in figure, in `40%` of the area of rectangle. Valuw of percent y which length of rectangle is more than breadth.

Quantity II: A pair of opposite sides of a square when increased by 15 cm, the area of figure increased by 600 `cm^(2).` Value of percent by which area increased. (a) Quantity I > Quantity II (b) Quantity I < Quantity II (c) Quantity I >= Quantity II (d) Quantity I <= Quantity II

Options

A. Quantity I `gt` Quantity II
B. Quantity I `lt` Quantity II
C. Quantity I `le` Quantity II
D. Quantity I `ge` Quantity II

Answer

B

Explanation

Quantity I:
Let length of rectangle = P
Breadth of rectangle = Q
Diameter of square = Q
Now, `PxxQxx40/100=1/2xxQxxQ`
`4P=5Q`
`%implies((P-Q)/(Q))xx100=((5Q-4Q)/(5Q))xx100=(Q)/(5Q)xx100=20%`
Quantity II:
Square get change into the rectangle.
By increasing 15 cm in two opposote sides, Area increased `to600`
Side `to (600)/(15)=40` cm
Area of square `=40**40=1600` square cm.
`%` by which area increased `to (600)/(1600)xx100=37.5%`
Quantity `Ilt` Quantity II