# Bank Exams Questions

Q:

A sum of money amounts to Rs.6690 after 3 years and to Rs.10,035 after 6 years on compound interest.find the sum.

 A) 4360 B) 4460 C) 4560 D) 4660

Explanation:

Let the sum be Rs.P.then
P(1+R/100)^3=6690…(i) and P(1+R/100)^6=10035…(ii)
On dividing,we get (1+R/100)^3=10025/6690=3/2.
Substituting this value in (i),we get:
P*(3/2)=6690 or P=(6690*2/3)=4460
Hence,the sum is rs.4460.

33 10793
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 = $\inline \fn_cm \frac{\prod x^{2}}{4}:\frac{\prod x^{2}}{2}=\frac{1}{4}:\frac{1}{2}=1:2$

29 10518
Q:

Find compound interest on Rs. 8000 at 15% per annum for 2 years 4 months, compounded annually

 A) 2109 B) 3109 C) 4109 D) 6109

Explanation:

Time = 2 years 4 months = 2(4/12) years = 2(1/3) years.
Amount = Rs'. [8000 X (1+(15/100))^2 X (1+((1/3)*15)/100)]
=Rs. [8000 * (23/20) * (23/20) * (21/20)]
= Rs. 11109. .
:. C.I. = Rs. (11109 - 8000) = Rs. 3109.

26 10484
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.

11 10346
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