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Bank Exams Questions

A) 36 | B) 232 |

C) 216 | D) 484 |

A) 49 | B) 154 |

C) 378 | D) 1078 |

Explanation:

The diameter is equal to the shortest side of the rectangle.

So radius= 14/2 = 7cm.

Therefore, $areaofcircle={\mathrm{\pi r}}^{2}=\frac{22}{7}\times 49=154{\mathrm{cm}}^{2}$

A) 20 | B) 25 |

C) 30 | D) 35 |

Explanation:

Let breadth = x m

Then, length = (x+5)m

x(x+5) = 750

x² + 5x - 750= 0

(x+30)(x-25)= 0

x = 25

A) 1 | B) 2 |

C) 3 | D) 4 |

Explanation:

Area of the park = (60 x 40) = 2400${m}^{2}$

Area of the lawn = 2109${m}^{2}$

Area of the crossroads = (2400 - 2109) = 291${m}^{2}$

Let the width of the road be x metres. Then,

$60x+40x-{X}^{2}=291$

${x}^{2}-100x+291=0$

(x - 97)(x - 3) = 0

x = 3.

A) Closed source software | B) Open source software |

C) Horizontal market software | D) vertical market software |

A) 4360 | B) 4460 |

C) 4560 | D) 4660 |

Explanation:

Let the sum be Rs.P.then

P(1+R/100)^3=6690…(i) and P(1+R/100)^6=10035…(ii)

On dividing,we get (1+R/100)^3=10025/6690=3/2.

Substituting this value in (i),we get:

P*(3/2)=6690 or P=(6690*2/3)=4460

Hence,the sum is rs.4460.

A) Rs.24.64 | B) Rs.21.85 |

C) Rs.16 | D) Rs.16.80 |

Explanation:

For 1st year S.I =C.I.

Thus, Rs.16 is the S.I. on S.I. for 1 year, which at 8% is thus Rs.200

i.e S.I on the principal for 1 year is Rs.200

Principle = $Rs.\frac{100*200}{8*1}$ = Rs.2500

Amount for 2 years, compounded half-yearly

$Rs.\left[2500*{\left(1+\frac{4}{100}\right)}^{4}\right]=Rs.2924.4$

C.I = Rs.424.64

Also, $S.I=Rs.\left(\frac{2500*8*2}{100}\right)=Rs.400$

Hence, [(C.I) - (S.I)] = Rs. (424.64 - 400) = Rs.24.64

A) 1:1 | B) 1:2 |

C) 1:3 | D) 1:4 |

Explanation:

Let the side of the square be x. Then, its diagonal = $\sqrt{2{x}^{2}}=\sqrt{2x}$

Radius of incircle = $\frac{x}{2}$

Radius of circum circle= $\sqrt{2}\times \frac{x}{2}=\frac{x}{\sqrt{2}}$

Required ratio = $\frac{{\mathrm{\pi x}}^{2}}{4}:\frac{{\displaystyle {\mathrm{\pi x}}^{2}}}{{\displaystyle 2}}=\frac{1}{4}:\frac{1}{2}=1:2$