# Aptitude and Reasoning Questions

A) 52/221 | B) 55/190 |

C) 55/221 | D) 19/221 |

Explanation:

We have n(s) =$52{C}_{2}$ 52 = 52*51/2*1= 1326.

Let A = event of getting both black cards

B = event of getting both queens

A∩B = event of getting queen of black cards

n(A) = $\frac{52*51}{2*1}$ = $26{C}_{2}$ = 325, n(B)= $\frac{26*25}{2*1}$= 4*3/2*1= 6 and n(A∩B) = $4{C}_{2}$ = 1

P(A) = n(A)/n(S) = 325/1326;

P(B) = n(B)/n(S) = 6/1326 and

P(A∩B) = n(A∩B)/n(S) = 1/1326

P(A∪B) = P(A) + P(B) - P(A∩B) = (325+6-1) / 1326 = 330/1326 = 55/221

A) 2/91 | B) 1/22 |

C) 3/22 | D) 2/77 |

Explanation:

Let S be the sample space.

Then, n(S) = number of ways of drawing 3 balls out of 15 = $15{C}_{3}$ =$\frac{15*14*13}{3*2*1}$= 455.

Let E = event of getting all the 3 red balls.

n(E) = $5{C}_{3}$ = $\frac{5*4}{2*1}$ = 10.

=> P(E) = n(E)/n(S) = 10/455 = 2/91.

A) Monday | B) Friday |

C) Sunday | D) Tuesday |

Explanation:

On 31st December, 2005 it was Saturday.

Number of odd days from 2006 to 2009 = (1 + 1 + 2 + 1) = 5 days.

On 31st December 2009, it was Thursday.

Thus, on 1st Jan, 2010 it is Friday

A) 71 | B) 61 |

C) 79 | D) 69 |

Explanation:

Fiest series : 3, 5, 7, 9

Second series : 5, 19, 41, ?

Difference of Second series are 14, 22, 30 etc

Next term is 41+30 i.e equal to 71

A) 20% | B) 21% |

C) 22% | D) 23% |

Explanation:

Gain % = $\left(\frac{{\left(100+commongain\%\right)}^{2}}{100}-100\right)$%

=$\left(\frac{{\left(100+10\right)}^{2}}{100}-100\right)$

= 21%

A) 2/7 | B) 5/7 |

C) 1/5 | D) 1/2 |

Explanation:

Total number of outcomes possible, n(S) = 10 + 25 = 35

Total number of prizes, n(E) = 10

$P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{10}{35}=\frac{2}{7}$

A) Four | B) Three |

C) Six | D) Five |

Explanation:

7 4 2

7 4 6

7 4 1

7 4 6

Only at these placees 4 is preceded by 7 but not followed by 3

A) 28 | B) 21 |

C) 24 | D) 30 |