# 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) 391 | B) 421 |

C) 481 | D) 511 |

Explanation:

Numbers are (2^{3} - 1), (3^{3} - 1), (4^{3} - 1), (5^{3} - 1), (6^{3} - 1), (7^{3} - 1) etc.

So, the next number is (8^{3} - 1) = (512 - 1) = 511.

A) 76 | B) 79 |

C) 85 | D) 87 |

Explanation:

Average = total runs / no.of innings = 32

So, total = Average x no.of innings = 32 x 10 = 320.

Now increase in avg = 4runs. So, new avg = 32+4 = 36runs

Total runs = new avg x new no. of innings = 36 x 11 = 396

Runs made in the 11th inning = 396 - 320 = 76

A) 25200 | B) 52000 |

C) 120 | D) 24400 |

Explanation:

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4) = ($7{C}_{3}$*$4{C}_{2}$)

= 210.

Number of groups, each having 3 consonants and 2 vowels = 210.

Each group contains 5 letters.

Number of ways of arranging 5 letters among themselves = 5! = 120

Required number of ways = (210 x 120) = 25200.

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) 6 % | B) 8 % |

C) 10 % | D) 12 % |

Explanation:

Let the total income be x.

Then, income left = (100 -80)% of [100 - (35 + 25)] % of x = 20% of 40% of x = 20/100 * 40/100 * 100) % of x = 8 % of x.

A) 1 | B) 2 |

C) 3 | D) 4 |

Explanation:

Let the numbers 13a and 13b.

Then, 13a x 13b = 2028

=>ab = 12.

Now, the co-primes with product 12 are (1, 12) and (3, 4).

[Note: Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]

So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4).

Clearly, there are 2 such pairs.