# Bank Exams Questions

Q:

MS-Word is an example of _____

 A) An operating system B) A processing device C) Application software D) An input device

Answer & Explanation Answer: C) Application software

Explanation:

Filed Under: Computer
Exam Prep: Bank Exams

82 21018
Q:

If log 2 = 0.3010 and log 3 = 0.4771, the values of log5 512 is

 A) 2.875 B) 3.875 C) 4.875 D) 5.875

Explanation:

ANS:      log5512 = ${\color{Black}&space;\frac{\log&space;512}{\log&space;5}}$  =  ${\color{Black}&space;\frac{\log&space;2^{9}}{\log&space;(\frac{10}{2})}}$  =${\color{Black}&space;\frac{9\log&space;2}{\log10-\log&space;2&space;}}$ =${\color{Black}&space;\frac{(9\times&space;0.3010)}{1-0.3010&space;}}$ =${\color{Black}&space;\frac{2.709}{0.699&space;}}$ =${\color{Black}&space;\frac{2709}{699&space;}}$ =3.876

71 20312
Q:

At what time between 4 and 5 o'clock will the hands of a watch point in opposite directions?

 A) 54 past 4 B) (53 + 7/11) past 4 C) (54 + 8/11) past 4 D) (54 + 6/11) past 4

Answer & Explanation Answer: D) (54 + 6/11) past 4

Explanation:

4 o'clock, the hands of the watch are 20 min. spaces apart.
To be in opposite directions, they must be 30 min. spaces apart.
$\inline \therefore$Minute hand will have to gain 50 min. spaces.
55 min. spaces are gained in 60 min

50 min. spaces are gained in $\inline (\frac{60}{55}\times 50)$ min. or $\inline 54\frac{6}{11}$

$\inline \therefore$ Required time = $\inline 54\frac{6}{11}$ min. past 4.

78 19169
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

27 18181
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:

Let original radius = R.

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%