# Bank Exams Questions

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 6927
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 6902
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 6895
Q:

How many cubes of 3cm edge can be cut out of a cube of 18cm edge

 A) 36 B) 232 C) 216 D) 484

Explanation:

number of cubes=(18 x 18 x 18) / (3 x 3 x 3) = 216

41 6715
Q:

A watch which gains 5 seconds in 3 minutes was set right at 7 a.m. In the afternoon of the same day, when the watch indicated quarter past 4 o'clock, the true time is:

 A) 4pm B) 5pm C) 6pm D) 7pm

Explanation:

Time from 7 am to 4.15 pm  = 9 hrs 15 min = $\inline \frac{37}{4}$ hrs

3 min. 5 sec. of this clock = 3 min. of the correct clock.

Now 3min.5sec. is$\inline \frac{37}{720}$ hrs of this clock = 3 min.is$\inline \frac{1}{20}$ hrs of the correct clock

$\inline \frac{37}{4}$ hrs of clock = $\inline (\frac{1}{20}\times \frac{720}{37}\times \frac{37}{4})$ hrs of the correct clock.

= 9 hrs of the correct clock.

$\inline \therefore$ The correct time is 9 hrs after 7 am. i.e., 4 pm.