# Aptitude and Reasoning Questions

A) 48 min. past 12. | B) 46 min. past 12. |

C) 45 min. past 12. | D) 47 min. past 12. |

Explanation:

Time from 8 a.m. on a day to 1 p.m. on the following day = 29 hours.

24 hours 10 min. of this clock = 24 hours of the correct clock.

$\frac{145}{6}$ hrs of this clock = 24 hours of the correct clock.

29 hours of this clock = $24*\frac{6}{145}*29$ hrs of the correct clock

= 28 hrs 48 min of the correct clock.

Therefore, the correct time is 28 hrs 48 min. after 8 a.m.

This is 48 min. past 12.

A) No profit, no loss | B) 5% |

C) 8% | D) 10% |

Explanation:

C.P. of 56 kg rice = Rs. (26 x 20 + 30 x 36) = Rs. (520 + 1080) = Rs. 1600.

S.P. of 56 kg rice = Rs. (56 x 30) = Rs. 1680.

Gain =(80/1600*100) % = 5%

A) 1/2 | B) 7/15 |

C) 8/15 | D) 1/9 |

Explanation:

Let S be the sample space

Then n(S) = no of ways of drawing 2 balls out of (6+4) =$10{C}_{2}$ 10 =$\frac{10*9}{2*1}$ =45

Let E = event of getting both balls of same colour

Then,n(E) = no of ways (2 balls out of six) or (2 balls out of 4)

=$6{C}_{2}+4{C}_{2}$ = $\frac{6*5}{2*1}+\frac{4*3}{2*1}$= 15+6 = 21

Therefore, P(E) = n(E)/n(S) = 21/45 = 7/15

A) 11 days | B) 12 days |

C) 13 days | D) 14 days |

Explanation:

One day's work of A, B and C = (1/24 + 1/30 + 1/40) = 1/10.

C leaves 4 days before completion of the work, which means only A and B work during the last 4 days.

Work done by A and B together in the last 4 days = 4 (1/24 + 1/30) = 3/10.

Remaining Work = 7/10, which was done by A,B and C in the initial number of days.

Number of days required for this initial work = 7 days.

Thus, the total numbers of days required = 4 + 7 = 11 days.

A) 40 | B) 80 |

C) 120 | D) 200 |

Explanation:

Let the numbers be 3x, 4x and 5x.

Then, their L.C.M. = 60*x*.

So, 60*x* = 2400 or x = 40.

The numbers are (3 x 40), (4 x 40) and (5 x 40).

Hence, required H.C.F. = 40.

A) 48 | B) 50 |

C) 58 | D) 60 |

Explanation:

Take a Look at this series :

${2}^{3}$ - ${2}^{2}$ = 8 - 4 = 4

${3}^{3}$ - ${3}^{2}$ = 27 - 9 = 18

${4}^{3}$ - ${4}^{2}$ = 64 - 16 = 48

${5}^{3}$ - ${5}^{2}$ = 125 - 25 = 100

${6}^{3}$ - ${6}^{2}$ = 216 - 36 = 180

${7}^{3}$ - ${7}^{2}$ = 343 - 49 = 294

${8}^{3}$ - ${8}^{2}$ = 512 - 64 = 448

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) 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