# Aptitude and Reasoning Questions

A) 5, 3, 4, 1, 2 | B) 3, 5, 4, 2, 1 |

C) 3, 5, 1, 4, 2 | D) 5, 3, 1, 2, 4 |

Explanation:

The correct order is :

Arrival Introduction Presentation Discussion Recommendation

3 5 1 4 2

A) Friday | B) Saturday |

C) Sunday | D) Thursday |

Explanation:

**15 Aug, 1947** = (1946 years + Period from 1.1.1947 to 15.8.1947)

**Odd days** in 1600 years = 0

**Odd days** in 300 years = 1

**46 years** = (35 ordinary years + 11 leap years) = (35 x 1 + 11 x 2)= 57 (8 weeks + 1 day) = **1 odd day **

**Jan. Feb. Mar. Apr. May. Jun. Jul. Aug **

( **31 + 28 + 31 + 30 + 31 + 30 + 31 + 15** ) = 227 days = (32 weeks + 3 days) = **3 odd days.**

**Total number of odd days = (0 + 1 + 1 + 3) = 5 odd days. **

Hence, as the number of **odd days = 5 , given day is Friday.**

A) Tuesday | B) Monday |

C) Sunday | D) Saturday |

Explanation:

Each day of the week is repeated after 7 days. So, after 63 days, it will be Monday.

After 61 days, it will be Saturday.

A) 2500 | B) 2700 |

C) 2900 | D) 3100 |

Explanation:

Total number of votes = 7500

Given that 20% of Percentage votes were invalid

=> Valid votes = 80%

Total valid votes = 7500*(80/100)

1st candidate got 55% of the total valid votes.

Hence the 2nd candidate should have got 45% of the total valid votes

=> Valid votes that 2nd candidate got = total valid votes x (45/100)

7500*(80/100)*(45/100) = 2700

A) 360, 160, 200 | B) 160, 360, 200 |

C) 200, 360,160 | D) 200,160,300 |

Explanation:

let ratio be x.

Hence no. of coins be 5x ,9x , 4x respectively

Now given total amount = Rs.206

=> (.50)(5x) + (.25)(9x) + (.10)(4x) = 206

we get x = 40

=> No. of 50p coins = 200

=> No. of 25p coins = 360

=> No. of 10p coins = 160

A) 65.25 | B) 56.25 |

C) 65 | D) 56 |

Explanation:

let each side of the square be a , then area = ${a}^{2}$

As given that The side is increased by 25%, then

New side = 125a/100 = 5a/4

New area = ${\left(\frac{5a}{4}\right)}^{2}$

Increased area= $\frac{25{a}^{2}}{16}-{a}^{2}$

Increase %=$\frac{\left[9{a}^{2}/16\right]}{{a}^{2}}*100$ % = 56.25%

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) 1/4 | B) 1/2 |

C) 3/4 | D) 7/12 |

Explanation:

Let A, B, C be the respective events of solving the problem and $\overline{)A},\overline{)B},\overline{)C}$ be the respective events of not solving the problem. Then A, B, C are independent event

$\therefore \overline{)A},\overline{)B},\overline{)C}$ are independent events

Now, P(A) = 1/2 , P(B) = 1/3 and P(C)=1/4

$P\left(\overline{)A}\right)=\frac{1}{2},P\left(\overline{)B}\right)=\frac{2}{3},P\left(\overline{)C}\right)=\frac{3}{4}$

$\therefore $ P( none solves the problem) = P(not A) and (not B) and (not C)

= $P\left(\overline{)A}\cap \overline{)B}\cap \overline{)C}\right)$

= $P\left(\overline{)A}\right)P\left(\overline{)B}\right)P\left(\overline{)C}\right)$ $\left[\because \overline{)A},\overline{)B},\overline{)C}areIndependent\right]$

= $\frac{1}{2}\times \frac{2}{3}\times \frac{3}{4}$

= $\frac{1}{4}$

Hence, P(the problem will be solved) = 1 - P(none solves the problem)

= $1-\frac{1}{4}$= **3/4**