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 3 of 8.
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 %.
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%.
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%.
Lexinah said:
1 decade ago
What if it says each side of a square not rectangle is increased by 15%?
What is the percentage area increase in its area?
What is the percentage area increase in its area?
Mhbkhb said:
1 decade ago
Taking 100*100 is not correct method. Because if we take 100*100 it will bcom a square but not rectangle.
Rozenelle said:
1 decade ago
If the answer is 44% it is a square not a rectangle, if it is a rectangle the correct answer is 32%.
Deepa ezhil. said:
1 decade ago
I can't understand Sandeepk logic. In the given problem it is given as rectangle but, he took the values of length and breadth as same value. I think the logic he used is wrong.
Pagalbuoy said:
1 decade ago
Let,
Length = 10, breadth = 20 unit.
So the area is 200 sq.unit.
Each side increased by 20%.
So length = 10+(10*20/100) = 12 unit.
Breadth = 20+(20*20/100) = 24 unit.
So area = (12*24) = 288 sq.unit.
So percentage area increase = (288-200)/200*100 = 44%.
I think this is the simplest way to do this sum. Hope you will enjoy it.
Length = 10, breadth = 20 unit.
So the area is 200 sq.unit.
Each side increased by 20%.
So length = 10+(10*20/100) = 12 unit.
Breadth = 20+(20*20/100) = 24 unit.
So area = (12*24) = 288 sq.unit.
So percentage area increase = (288-200)/200*100 = 44%.
I think this is the simplest way to do this sum. Hope you will enjoy it.
Hema said:
1 decade ago
Why here you take length = (120/100)x and breadth = (120/100)y?
Michelle said:
1 decade ago
Let length 40 and breadth is =20. After increasing length is 48 and breadth is 24. So the area is 1152. The difference between new are and old area is 352. So (352/800*100) = 44%.
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