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.
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%.
N G GANESH said:
3 weeks ago
In simple way:
Let's take;
Length l = 10.
Breadth b = 10.
=> l × b = 100
20% increase in side length means 20% of 10 is 2.
L = 12.
B = 12.
L×b = 12 × 12 = 144, which means 44% increase in area.
Let's take;
Length l = 10.
Breadth b = 10.
=> l × b = 100
20% increase in side length means 20% of 10 is 2.
L = 12.
B = 12.
L×b = 12 × 12 = 144, which means 44% increase in area.
(3)
Bhupendra said:
1 decade ago
Let it be a squire then suppose side is x
then area =x^2
new side is= (6x/5)
then area= 36x^2/25
Now diff in area= 36x^2/25-x^2= (11x^2/25)
then % increase in area is = (11x^2/25)*100/x^2= 44 ans
then area =x^2
new side is= (6x/5)
then area= 36x^2/25
Now diff in area= 36x^2/25-x^2= (11x^2/25)
then % increase in area is = (11x^2/25)*100/x^2= 44 ans
Ramesh said:
9 years ago
Let original length and breadth be 100.
So new length and breadth are 120
New Area = 120 * 120 so 14400.
Old area = 100 * 100 = 10000.
So diff = 4400.
Error = 4400/(100 * 100) * 100.
So, 44%.
So new length and breadth are 120
New Area = 120 * 120 so 14400.
Old area = 100 * 100 = 10000.
So diff = 4400.
Error = 4400/(100 * 100) * 100.
So, 44%.
Omveer said:
3 years ago
Square Is also a Rectangle.
Assume both sides to be 100.
Actual Area = 10000.
With 20% increase in each side Area = 102*102=14400,Diffrence = 4400.
% Increase in Area = 4400/10000 * 100 = 44%.
Assume both sides to be 100.
Actual Area = 10000.
With 20% increase in each side Area = 102*102=14400,Diffrence = 4400.
% Increase in Area = 4400/10000 * 100 = 44%.
(13)
Sourav said:
1 decade ago
Easiest approach:
Length 10 cm.
Breadth 10 cm.
Area = 10*10 = 100 cm.
With 20% increase:
Length 12 cm.
Breadth 12 cm.
Area 144 cm.
Percentage increase (144-100) = 44.
Ans: 44%.
Length 10 cm.
Breadth 10 cm.
Area = 10*10 = 100 cm.
With 20% increase:
Length 12 cm.
Breadth 12 cm.
Area 144 cm.
Percentage increase (144-100) = 44.
Ans: 44%.
Sandeepk said:
1 decade ago
I found below method most easy :
Let l=b=1 units.
Area = 1 * 1 = 1.
New l = b = 1.2 (20% increase).
New Area = 1.2 * 1.2 = 1.44.
Thus, 0.44 is 44% increase in area is the answer.
Let l=b=1 units.
Area = 1 * 1 = 1.
New l = b = 1.2 (20% increase).
New Area = 1.2 * 1.2 = 1.44.
Thus, 0.44 is 44% increase in area is the answer.
Mr.issue said:
8 years ago
Let original L and B=100 Increased L and B = 120,
Difference in Area = (120)2-(100)2,
(120+100)(120-100) = 4400.
Error In Length And Breath =
4400*100
------------- =44%.
100*100
Difference in Area = (120)2-(100)2,
(120+100)(120-100) = 4400.
Error In Length And Breath =
4400*100
------------- =44%.
100*100
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%.
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.
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