# Aptitude - Area

#### Why should I learn to solve Aptitude questions and answers section on "Area"?

Learn and practise solving Aptitude questions and answers section on "Area" to enhance your skills so that you can clear interviews, competitive examinations, and various entrance tests (CAT, GATE, GRE, MAT, bank exams, railway exams, etc.) with full confidence.

#### Where can I get the Aptitude questions and answers section on "Area"?

IndiaBIX provides you with numerous Aptitude questions and answers based on "Area" along with fully solved examples and detailed explanations that will be easy to understand.

#### Where can I get the Aptitude section on "Area" MCQ-type interview questions and answers (objective type, multiple choice)?

Here you can find multiple-choice Aptitude questions and answers based on "Area" for your placement interviews and competitive exams. Objective-type and true-or-false-type questions are given too.

#### How do I download the Aptitude questions and answers section on "Area" in PDF format?

You can download the Aptitude quiz questions and answers section on "Area" as PDF files or eBooks.

#### How do I solve Aptitude quiz problems based on "Area"?

You can easily solve Aptitude quiz problems based on "Area" by practising the given exercises, including shortcuts and tricks.

- Area - Formulas
- Area - General Questions
- Area - Data Sufficiency 1
- Area - Data Sufficiency 2
- Area - Data Sufficiency 3

Perimeter = Distance covered in 8 min. = | 12000 | x 8 | m = 1600 m. | |

60 |

Let length = 3*x* metres and breadth = 2*x* metres.

Then, 2(3*x* + 2*x*) = 1600 or *x* = 160.

Length = 480 m and Breadth = 320 m.

Area = (480 x 320) m^{2} = 153600 m^{2}.

100 cm is read as 102 cm.

A_{1} = (100 x 100) cm^{2} and A_{2} (102 x 102) cm^{2}.

(A_{2} - A_{1}) = [(102)^{2} - (100)^{2}]

= (102 + 100) x (102 - 100)

= 404 cm^{2}.

Percentage error = | 404 | x 100 | % | = 4.04% | |

100 x 100 |

2(l + b) |
= | 5 |

b |
1 |

2*l* + 2*b* = 5*b*

3*b* = 2*l*

b = |
2 | l |

3 |

Then, Area = 216 cm^{2}

*l* x *b* = 216

l x |
2 | l |
= 216 |

3 |

*l*^{2} = 324

*l* = 18 cm.

Let original length = *x* metres and original breadth = *y* metres.

Original area = (*xy*) m^{2}.

New length = | 120 | x |
m | = | 6 | x |
m. | ||

100 | 5 |

New breadth = | 120 | y |
m | = | 6 | y |
m. | ||

100 | 5 |

New Area = | 6 | x x |
6 | y |
m^{2} |
= | 36 | xy |
m^{2}. |
||

5 | 5 | 25 |

The difference between the original area = xy and new-area 36/25 xy is

= (36/25)xy - xy

= xy(36/25 - 1)

= xy(11/25) or (11/25)xy

Increase % = | 11 | xy x |
1 | x 100 | % | = 44%. | |

25 | xy |

Video Explanation: https://youtu.be/I3jLjLPn1W4

Area of the park = (60 x 40) m^{2} = 2400 m^{2}.

Area of the lawn = 2109 m^{2}.

Area of the crossroads = (2400 - 2109) m^{2} = 291 m^{2}.

Let the width of the road be *x* metres. Then,

60*x* + 40*x* - *x*^{2} = 291

*x*^{2} - 100*x* + 291 = 0

(*x* - 97)(*x* - 3) = 0

*x* = 3.

Video Explanation: https://youtu.be/R3CtrAKGxkc