# Aptitude and Reasoning Questions

A) 10000 | B) 12000 |

C) 14000 | D) 16000 |

Explanation:

Purchase price = $Rs.\left[\frac{8748}{{{}^{\left(1-{\displaystyle \frac{10}{100}}\right)}}^{3}}\right]$ = Rs. [8748 * (10/9) * (10/9 )* (10/9)] = Rs.12000

A) 4991 | B) 5467 |

C) 5987 | D) 6453 |

Explanation:

Total sale for 5 months = Rs. (6435 + 6927 + 6855 + 7230 + 6562) = Rs. 34009.

Required sale = Rs.[(6500 x 6) - 34009]

= Rs. (39000 - 34009)

= Rs. 4991.

A) 1/2 | B) 3/5 |

C) 9/20 | D) 8/15 |

Explanation:

Here, S = {1, 2, 3, 4, ...., 19, 20}.

Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.

P(E) = n(E)/n(S) = 9/20.

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) 125% | B) 150% |

C) 175% | D) 110% |

Explanation:

Let the edge = a cm

So increase by 50 % = a + a/2 = 3a/2

Total surface Area of original cube = $6{a}^{2}$

TSA of new cube = $6{\left(\frac{3a}{2}\right)}^{2}$ =$6\left(\frac{9{a}^{2}}{4}\right)$= $13.5{a}^{2}$

Increase in area = $13.5{a}^{2}-6{a}^{2}$ =$7.5{a}^{2}$

$7.5{a}^{2}$ Increase % =$\frac{7.5{a}^{2}}{6{a}^{2}}\times 100$ = 125%

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) 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) 18 litres | B) 24 litres |

C) 32 litres | D) 42 litres |

Explanation:

Let the quantity of the wine in the cask originally be x litres

Then, quantity of wine left in cask after 4 operations =$\left[x{\left(1-\frac{8}{x}\right)}^{4}\right]$litres

$\therefore \left[\frac{x{\left(1-{\displaystyle \frac{8}{x}}\right)}^{4}}{x}\right]=\frac{16}{81}$

$\Rightarrow {\left[1-\frac{8}{x}\right]}^{4}={\left(\frac{2}{3}\right)}^{4}$

$\Rightarrow x=24$