# Bank Exams Questions

Q:

A sum of money lent at compound interest for 2 years at 20% per annum would fetch Rs.482 more, if the interest was payable half yearly than if it was payable annually . The sum is

 A) 10000 B) 20000 C) 40000 D) 50000

Explanation:

Let sum=Rs.x

C.I. when compounded half yearly = $\inline \fn_jvn \left [ x(1+\frac{10}{100})^{4}-x \right ]=\frac{4641}{10000}x$

C.I. when compounded annually =$\inline \fn_jvn \left [ x(\frac{20}{100})^{2}-x \right ]=\frac{11}{25}x$

$\inline \fn_jvn \therefore \frac{4641}{10000}x-\frac{11}{25}x=482$

=> x=20000

64 15025
Q:

A man walked diagonally across a square lot. Approximately, what was the percent saved by not walking along the edges?

 A) 30 B) 40 C) 50 D) 60

Explanation:

Let the side of the square(ABCD) be x meters.

Then, AB + BC = 2x metres.

AC = $\inline \fn_jvn \sqrt{2}x$ = (1.41x) m.

Saving on 2x metres = (0.59x) m.

Saving % =$\inline \fn_jvn \frac{0.59x}{2x}\times 100$ = 30% (approx)

27 15013
Q:

A clock is set right at 5 a.m. The clock loses 16 minutes in 24 hours.What will be the true time when the clock indicates 10 p.m. on 4th day?

 A) 11pm B) 12pm C) 1pm D) 2pm

Explanation:

Time from 5 am. on a day to 10 pm. on 4th day = 89 hours.

Now 23 hrs 44 min. of this clock = 24 hours of correct clock.

356/15 hrs of this clock = 24 hours of correct clock

89 hrs of this clock = (24 x 31556 x 89) hrs of correct clock.

= 90 hrs of correct clock.

So, the correct time is 11 p.m.

44 13729
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

36 13153
Q:

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

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

${\color{Black}\log&space;(2^{56})=(56\times0.30103)&space;}$ =16.85768.
Hence, the number of digits in ${\color{Black}2^{56}&space;}$ is 17.