#
Bank Exams Questions

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.

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

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)

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

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.

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(\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$\frac{54}{2}$}\right.\times 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

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

A) 36 | B) 232 |

C) 216 | D) 484 |