# Bank Exams Questions

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.

68 21438
Q:

A wire can be bent in the form of a circle of radius 56cm. If it is bent in the form of a square, then its area will be

 A) 7744 B) 8844 C) 5544 D) 4444

Explanation:

length of wire = $2\mathrm{\pi r}$= 2 x (22/7 ) x 56 = 352 cm
side of the square = 352/4 = 88cm
area of the square = 88 x 88 = 7744sq cm

45 21251
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 = $\sqrt{2x}$ = (1.41x) m.

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

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

34 20856
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

s=1/2 x (30+14+40) = 42

Area of triangle ABC = $\sqrt{\left[s\left(s-a\right)\left(s-b\right)\left(s-c\right)\right]}$

= $\sqrt{\left[42\left(12\right)\left(28\right)\left(2\right)\right]}$= 168sq m

area of parallelogram ABCD = 2 x 168 = 336 sq m

42 20787
Q:

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

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

Explanation:

$\mathrm{log}\left({2}^{56}\right)$ =56*0.30103 =16.85768.

Its characteristics is 16.

Hence, the number of digits in ${2}^{56}$ is 17.

56 20192
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 $5+5\frac{48}{60}$ min in a time of  (7×24)+12 = 180 hours.

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

$\frac{54}{5}min\to 180hours$

5min ->

$\left(5}{\frac{54}{2}}×180\right)$

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

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

49 19902
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= $31\frac{5}{11}$ 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

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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