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 8 of 8.
Manan said:
1 decade ago
There is a much simpler method.
Suppose the length of rectangle = 100 cm & breadth = 50 cm.
So Area equals to 100*50 = 5000.
Now if 20% of both sides are increased then,
New length = 20% of 100 = 120 cm & 20% of 50 = 60 cm.
So, new area = 120*60 = 7200.
Difference = 7200-5000 = 2200.
Percentage increase = (2200\5000)*100 = 44%.
Suppose the length of rectangle = 100 cm & breadth = 50 cm.
So Area equals to 100*50 = 5000.
Now if 20% of both sides are increased then,
New length = 20% of 100 = 120 cm & 20% of 50 = 60 cm.
So, new area = 120*60 = 7200.
Difference = 7200-5000 = 2200.
Percentage increase = (2200\5000)*100 = 44%.
Lipu said:
1 decade ago
Let true area = (100)^2.
Error area = (120)^2.
So error %= (error area-true area)/true area.
= (120)^2-(100)^2/(100)^2 *100.
= 220*20/100.
= 44%.
Error area = (120)^2.
So error %= (error area-true area)/true area.
= (120)^2-(100)^2/(100)^2 *100.
= 220*20/100.
= 44%.
Hanny Gupta said:
1 decade ago
A1 = 100*100 = 10000.
A2 = 120*120 = 14400.
A2-A1 = 14400-10000 = 4400.
(A2-A1)% = 44 %.
A2 = 120*120 = 14400.
A2-A1 = 14400-10000 = 4400.
(A2-A1)% = 44 %.
Mani said:
1 decade ago
From where have we got new length and new breadth.
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