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

28 8466
Q:

What is the rate of interest p.c.p.a.?

I. An amount doubles itself in 5 years on simple interest.

II. Difference between the compound interest and the simple interest earned on a certain amount in 2 years is Rs. 400.

III. Simple interest earned per annum is Rs. 2000

 A) I only B) II and III only C) All I, II and III D) I only or II and III only

Answer & Explanation Answer: D) I only or II and III only

Explanation:

$\inline \fn_jvn I. \frac{P\times R\times 5}{100}=P\Rightarrow R=20$

$\inline \fn_jvn II.P\left ( 1+\frac{R}{100} \right )^2-P-\frac{P\times R\times 2}{100}=400\Rightarrow pR^{2}=4000000$

$\inline \fn_jvn III.\frac{P\times R\times 1}{100}=2000\Rightarrow PR=200000$

$\inline \fn_jvn \therefore \frac{PR^{2}}{PR}=\frac{4000000}{200000}\Rightarrow R=20$

Thus I only or (II and III) give answer.

$\inline \fn_jvn \therefore$ Correct answer is (D)

21 8380
Q:

One side of a rectangular field is 15m and one of its diagonal is 17m. Find the area of field?

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

Explanation:

Other side = [(17 x 17) - (15 x 15)] = (289 - 225) = 8m
Area = 15 x 8 =120 sq. m

31 8349
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

16 8346
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.