# Bank Exams Questions

Q:

A watch which gains uniformly ,is 5 min,slow at 8 o'clock in the morning on sunday and it is 5 min 48 sec.fast at 8 p.m on following sunday. when was it correct?

 A) 7pm on wednesday B) 20 min past 7pm on wednesday C) 15min past 7pm on wednesday D) 8pm on wednesday

Explanation:

This sunday morning at 8:00 AM, the watch is 5 min. Slow, and the next sunday at 8:00PM it becomes 5 min 48 sec fast.  The watch gains $\inline \fn_jvn \small 5+5\tfrac{48}{60}$ min in a time of  (7×24)+12 = 180 hours.

To show the correct time, it has to gain 5 min.

$\inline \fn_jvn \frac{54}{5}min\rightarrow 180 \; hours$

$\inline \fn_jvn 5\; min\rightarrow ?$

$\inline \fn_jvn \Rightarrow \frac{5}{\frac{54}{5}}\times 180$

$\inline \fn_jvn 83\frac{1}{3}hrs =72hrs+11\frac{1}{3}hrs=3days+11hrs+20min$

So the correct time will be shown on wednesday at 7:20 PM

42 15261
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.

39 15255
Q:

If log 2 = 0.30103, Find the number of digits in 256 is

 A) 17 B) 19 C) 23 D) 25

Explanation:

${\color{Black}\log&space;(2^{56})=(56\times0.30103)&space;}$ =16.85768.

Its characteristics is 16.

Hence, the number of digits in ${\color{Black}2^{56}&space;}$ is 17.

41 14841
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

26 14362
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