# Bank PO Questions

Q:

Find the ratio of the areas of the incircle and circumcircle of a square.

 A) 1:1 B) 1:2 C) 1:3 D) 1:4

Explanation:

Let the side of the square be x. Then, its diagonal = ${\color{Black}\sqrt{2x^{2}}=\sqrt{2}x}$

Radius of incircle = $\inline \fn_cm \frac{x}{2}$

Radius of circum circle= ${\color{Black}\sqrt{{2}}\times&space;\frac{x}{2}=\frac{x}{\sqrt{2}}}$

Required ratio = $\inline \fn_cm \frac{\prod x^{2}}{4}:\frac{\prod x^{2}}{2}=\frac{1}{4}:\frac{1}{2}=1:2$

24 6488
Q:

If the diagonal of a rectangle is 17cm long and its perimeter is 46 cm. Find the area of the rectangle.

 A) 110 B) 120 C) 130 D) 140

Explanation:

let length = x and breadth = y then

2(x+y) = 46  $\inline \fn_cm \Rightarrow$  x+y = 23

x²+y² = 17² = 289

now (x+y)² = 23²

$\inline \fn_cm \Rightarrow$x²+y²+2xy= 529

289+ 2xy = 529

$\inline \fn_cm \Rightarrow$ xy = 120

area = xy = 120 sq.cm

20 6343
Q:

The length of a rectangular hall is 5m more than its breadth. The area of the hall is 750 m. The length of the hall is

 A) 20 B) 25 C) 30 D) 35

Explanation:

Then, length = (x+5)m

x(x+5) = 750

x² + 5x - 750= 0

(x+30)(x-25)= 0

x = 25

15 6284
Q:

A parallelogram has sides 30m and 14m and one of its diagonals is 40m long. Then its area is

 A) 136 B) 236 C) 336 D) 436

Explanation:

let ABCD be the given parallelogram

area of parallelogram ABCD = 2 x (area of triangle ABC)

now a = 30m, b = 14m and c = 40m

$\inline \fn_jvn s=\frac{1}{2}\times \left ( 30+14+40 \right ) = 42$

Area of triangle ABC = $\inline&space;{\color{Black}\sqrt{s(s-a)(s-b)(s-c)}}$

= $\inline&space;{\color{Black}\sqrt{42(12)(28)(2)}}$= 168sq m

area of parallelogram ABCD = 2 x 168 = 336 sq m

11 6129
Q:

A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. If the area of the lawn is 2109 sq. m, then what is the width of the road?

 A) 1 B) 2 C) 3 D) 4

Explanation:

Area of the park = (60 x 40) = 2400$\inline m^{2}$

Area of the lawn = 2109$\inline m^{2}$

Area of the crossroads = (2400 - 2109) = 291$\inline m^{2}$

Let the width of the road be x metres. Then,

$\inline 60x+40x-x^{2}=291$

$\inline x^{2}-100x+291=0$

(x - 97)(x - 3) = 0
x = 3.