# Aptitude and Reasoning Questions

A) 2:5 | B) 3:5 |

C) 4:5 | D) 5:4 |

Explanation:

Let the third number be x.

Then, first number = 120% of x =120x/100 = 6x/5

Second number =150% of x = 150x/100 = 3x/2

Ratio of first two numbers = 6x/5 : 3x/2 = 12x : 15x = 4 : 5

A) Monday | B) Wednesday |

C) Tuesday | D) Friday |

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) 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) 100% | B) 200% |

C) 300% | D) 400% |

Explanation:

Let the C.P be Rs.100 and S.P be Rs.x, Then

The profit is (x-100)

Now the S.P is doubled, then the new S.P is 2x

New profit is (2x-100)

Now as per the given condition;

=> 3(x-100) = 2x-100

By solving, we get

x = 200

Then the Profit percent = (200-100)/100 = 100

Hence the profit percentage is 100%

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) MFEDJJOE | B) EOJDEJFM |

C) MFEJDJOE | D) EOJDJEFM |

Explanation:

- There are 8 letters in the word.

- The coded word can be obtained by taking the immediately following letters of word, expect the first and the last letters of the given word but in the reverse order. That means, in the coded form the first and the last letters have been interchanged while the remaining letters are coded by taking their immediate next letters in the reverse order.

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.