# Bank Exams Questions

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)

20 7560
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.

7 7323
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

15 7321
Q:

The diagonal of a rectangle is $\inline \sqrt{41}$ cm and its area is 20 sq. cm. The perimeter of the rectangle must be:

 A) 18 B) 28 C) 38 D) 48

Explanation:

$\inline \sqrt{l^{2}+b^{2}}=\sqrt{41}$ (or)   ${\color{Black}&space;l^{2}+b^{2}=41}$

Also, $\inline lb=20$

${\color{Black}&space;(l+b)^{2}=l^{2}+b^{2}+2lb}$

= 41 + 40 = 81

(l + b) = 9.

Perimeter = 2(l + b) = 18 cm.

103 7023
Q:

At what time between 5 and 6 o' clock are the hands of a 3 minutes apart ?

 A) 24min B) 12min C) 13min D) 14min

Explanation:

In this type of problems the formuae is
(5*x+ or - t)*12/11
Here x is replaced by the first interval of given time. Here x is 5.
t is spaces apart

Case 1 : (5*x + t) * 12/11

(5*5 + 3) * 12/11

28 * 12/11 = 336/11= $\inline 31\frac{5}{11}$ min

therefore the hands will be 3 min apart at 31 5/11 min past 5.

Case 2 : (5*x - t) * 12/11

(5*5 -3 ) * 12/11

22 *12/11 = 24 min

therefore the hands will be 3 min apart at 24 min past 5