Aptitude - Area - Discussion
Discussion Forum : Area - General Questions (Q.No. 4)
4.
The percentage increase in the area of a rectangle, if each of its sides is increased by 20% is:
Answer: Option
Explanation:
Let original length = x metres and original breadth = y metres.
Original area = (xy) m2.
| 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 | m2 |
= | |
36 | xy | m2. |
| 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
Discussion:
74 comments Page 7 of 8.
Hussain said:
8 years ago
a*(200+a)/100.
so, 20*(200+20)/100.
=44%.
so, 20*(200+20)/100.
=44%.
Puja said:
8 years ago
20 +20+20*20/100=44%.
Harmanpreet said:
7 years ago
Each side 20% increased thus 120*120/100=144,
then 144-100 =44.
then 144-100 =44.
MAHESHKRISHNAN D said:
7 years ago
100*100 = 10000(1)
120*120 = 14400(2)
1-2 = 4400/100.
= 44%.
120*120 = 14400(2)
1-2 = 4400/100.
= 44%.
(3)
Kumar Gowda said:
6 years ago
= 20+20(20*20/100).
= 40+(400/100),
= 40+(4),
= 40+4,
= 44%.
= 40+(400/100),
= 40+(4),
= 40+4,
= 44%.
(5)
Tridib Samanta said:
6 years ago
Let,
Original length of rectangle be, x = 100m,
Original breadth of rectangle be, y = 10m,
Original area of the rectangle is, A = 1000 sq. m.
Now, each side is incresed by 20%.
New length of the rectangle is, x1 = 120m,
New breadth of the rectangle is, y1 = 12m,
New area of the rectangle is, A1 = 1440 sq. m.
Therefore, total increase in area = (A1 - A) = (1440 - 1000)m = 440m.
% increase in area;
= (total increase in area / original area) x 100.
= {( A1- A) / A } x 100,
= (440/1000) x 100,
= .44 x 100,
= 44 % (Answer).
Thanks!
Original length of rectangle be, x = 100m,
Original breadth of rectangle be, y = 10m,
Original area of the rectangle is, A = 1000 sq. m.
Now, each side is incresed by 20%.
New length of the rectangle is, x1 = 120m,
New breadth of the rectangle is, y1 = 12m,
New area of the rectangle is, A1 = 1440 sq. m.
Therefore, total increase in area = (A1 - A) = (1440 - 1000)m = 440m.
% increase in area;
= (total increase in area / original area) x 100.
= {( A1- A) / A } x 100,
= (440/1000) x 100,
= .44 x 100,
= 44 % (Answer).
Thanks!
(4)
Bhavik said:
6 years ago
Say length =100.
But it increased to 120.
Therefore the area of rectangle = 120*120 = 14400.
Now divide by 100-->144. So 144-100 = 44 is the answer.
But it increased to 120.
Therefore the area of rectangle = 120*120 = 14400.
Now divide by 100-->144. So 144-100 = 44 is the answer.
(5)
Vaibhav said:
5 years ago
Simple trick for such problems is. If increase by x% we can calculate by the formula.
x(x+200)%100.
x(x+200)%100.
(2)
Subbu said:
5 years ago
Can anyone show how it can be solved by using derivatives? please.
(1)
GANESH SHAM TANDEL said:
5 years ago
Let's assume that length N width of this rectangle will be 4 and 2 respectively ( you can take any number).
So as per the question, It is increase by 20% by its side.
That means length N width increase to (4+ 4*20%=4.8) & (2+2*20%=2.40) respectively.
Old Sq feet was 4 * 2 = 8
New sq feet is 4.8 * 2.4 = 11.52
Area increased by 11.52-8= 3.52 sq feet.
In percentage,
= area increase/old area * 100.
= 3.52 /8= 44%
So as per the question, It is increase by 20% by its side.
That means length N width increase to (4+ 4*20%=4.8) & (2+2*20%=2.40) respectively.
Old Sq feet was 4 * 2 = 8
New sq feet is 4.8 * 2.4 = 11.52
Area increased by 11.52-8= 3.52 sq feet.
In percentage,
= area increase/old area * 100.
= 3.52 /8= 44%
(1)
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