# Bank PO Questions

Q:

If log 27 = 1.431, then the value of log 9 is

 A) 0.754 B) 0.854 C) 0.954 D) 0.654

Explanation:

log 27 = 1.431
${\color{Black}&space;\Rightarrow&space;\log&space;(3^{3})=1.431}$
3 log 3 = 1.431
log 3 = 0.477
log 9 = ${\color{Black}&space;\log&space;(3^{2})}$ = 2 log 3 = (2 x 0.477) = 0.954

26 17335
Q:

If the radius of a circle is decreased by 50%, find the percentage decrease in its area.

 A) 55% B) 65% C) 75% D) 85%

Explanation:

New radius = $\inline \fn_cm \frac{50}{100}R$$\inline \fn_cm \frac{R}{2}$

Original area =$\inline \fn_cm \tiny \prod R^{2}$  and new area = $\inline \fn_cm \tiny \prod(\frac{R}{2})^{2}=\frac{\prod R^{2}}{4}$

Decrease in area = $\inline \fn_cm \tiny \frac{3\prod R^{2}}{4}\times \frac{1}{\prod R^{2}}\times 100$ = 75%

47 17179
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)

25 14362
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

60 13959
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.