 Home
 Quantitative
 English
 Reasoning
 IT Officer
 Programming

Computer
 Computer Awareness Questions Answers  Set 1
 Computer Awareness Questions Answers  Set 2
 Important Abbreviations Computer Awareness Questions Answers
 Important File Extensions Questions Answers
 Computer System Architecture Questions Answers
 MS Office Questions Answers
 MS Excel Questions Answers
 MS PowerPoint Questions Answers

GK
 Geography Questions Answers
 Indian History Questions Answers
 World History Questions Answers
 Indian Economy Questions Answers
 Indian Polity and Constitution
 Physics Questions Answers
 Chemistry Questions Answers
 Biology Questions Answers
 First In India
 First In World
 Longest and Largest
 Books and Authors
 Important Days of year
 Countries and Capitals
 Inventions and Inventors
 Current Affairs
 Govt Jobs
 You are here
 Home
 Quantitative Aptitude
 Arithmetic Aptitude Questions Answers
 Area Questions Answers
 Aptitude Question
 Current Affairs 2015
 Current Affairs 2014
 Current Affairs Jan 2014
 Current Affairs Feb 2014
 Current Affairs Mar 2014
 Current Affairs April 2014
 Current Affairs May 2014
 Current Affairs June 2014
 Current Affairs July 2014
 Current Affairs August 2014
 Current Affairs September 2014
 Current Affairs October 2014
 Current Affairs November 2014
 Current Affairs December 2014
 Current Affairs 2014
Question Detail
The percentage increase in the area of a rectangle, if each of its sides is increased by 20% is:
 32%
 34%
 42%
 44%
Answer: Option D
Explanation:
Let original length = x metres and original breadth = y metres.
\begin{aligned}
\text{Original area } = \text{xy } m^2 \\
\text{New Length }= \frac{120}{100}x = \frac{6}{5}x \\
\text{New Breadth }= \frac{120}{100}y = \frac{6}{5}y \\
=>\text{New Area }= \frac{6}{5}x * \frac{6}{5}y \\
=>\text{New Area }= \frac{36}{25}xy \\
\text{Area Difference} = \frac{36}{25}xy  xy \\
= \frac{11}{25}xy \\
Increase \% = \frac{Differnce}{Actual}*100 \\
= \frac{11xy}{25}*\frac{1}{xy}*100 = 44\%
\end{aligned}
1. The length of a rectangle is three times of its width. If the length of the diagonal is \begin{aligned}8\sqrt{10}\end{aligned}, then find the perimeter of the rectangle.
 60 cm
 62 cm
 64 cm
 66 cm
Answer: Option C
Explanation:
Let Breadth = x cm,
then, Length = 3x cm
\begin{aligned}
x^2+{(3x)}^2 = {(8\sqrt{10})}^2 \\
=> 10x^2 = 640 \\
=> x = 8 \\
\end{aligned}
So, length = 24 cm and breadth = 8 cm
Perimeter = 2(l+b)
= 2(24+8) = 64 cm
2. What are the least number of square tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad ?
 714
 814
 850
 866
Answer: Option B
Explanation:
In this type of questions, first we need to calculate the area of tiles. With we can get by obtaining the length of largest tile.
Length of largest tile can be obtained from HCF of length and breadth.
So lets solve this,
Length of largest tile = HCF of (1517 cm and 902 cm)
= 41 cm
Required number of tiles =
\begin{aligned}
\frac{\text{Area of floor}}{\text{Area of tile}} \\
= \left(\frac{1517\times902}{41 \times 41}\right)\\
= 814
\end{aligned}
3. The perimeters of 5 squares are 24 cm, 32 cm, 40 cm, 76 cm and 80 cm respectively. The perimeter of another square equal in area to the sum of the area of these square is:
 124 cm
 120 cm
 64 cm
 56 cm
Answer: Option A
Explanation:
Clearly first we need to find the areas of the given squares, for that we need its side.
Side of sqaure = Perimeter/4
So sides are,
\begin{aligned}
\left(\frac{24}{4}\right),\left(\frac{32}{4}\right),\left(\frac{40}{4}\right),\left(\frac{76}{4}\right),\left(\frac{80}{4}\right) \\
= 6,8,10,19,20 \\
\text{Area of new square will be }\\
= [6^2+8^2+10^2+19^2+20^2] \\
= 36+64+100+361+400 \\
= 961 cm^2 \\
\text{Side of new Sqaure =}\sqrt{961} \\
= 31 cm \\
\text{Required perimeter =}(4\times31) \\
= 124 cm
\end{aligned}
4. A courtyard is 25 meter long and 16 meter board is to be paved with bricks of dimensions 20 cm by 10 cm. The total number of bricks required is :
 16000
 18000
 20000
 22000
Answer: Option C
Explanation:
\begin{aligned}
\text{Number of bricks =}\frac{\text{Courtyard area}}{\text{1 brick area}} \\
= \left( \frac{2500 \times 1600}{20 \times 10} \right) \\
= 20000
\end{aligned}
5. If the circumference of a circle increases from 4pi to 8 pi, what change occurs in the area ?
 Area is quadrupled
 Area is tripled
 Area is doubles
 Area become half
Answer: Option A
Explanation:
\begin{aligned}
2\pi R1 = 4 \pi \\
=> R1 = 2 \\
2\pi R2 = 8 \pi \\
=> R2 = 4 \\
\text{Original Area =} 4\pi * 2^2 \\
= 16 \pi \\
\text{New Area =} 4\pi * 4^2 \\
= 64 \pi
\end{aligned}
So the area quadruples.
 Copyright 2014  All rights reserved
 Terms Of Use & Privacy Policy
 Copyright
 Contact Us