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 6 of 8.
Deependra singh said:
9 years ago
Take length and breadth each 10 then area = 100.
Increased length and breadth will be 12 each then 12*12 = 144 increased area is 44 so 44%.
Increased length and breadth will be 12 each then 12*12 = 144 increased area is 44 so 44%.
Sarvesh agarwal said:
8 years ago
A = L*B.
A = (1+1/5)(1+1/5).
A = 36/25LB.
A = 144%.
i.e 44%.
A = (1+1/5)(1+1/5).
A = 36/25LB.
A = 144%.
i.e 44%.
Ravali yadav said:
8 years ago
Assume it as 100.
increase 20 for 100 then it will become 120,
again increase 20% for 120, then it will become 144,
100----------20%-----120-------20%(24)------120+24 = 144.
increase 20 for 100 then it will become 120,
again increase 20% for 120, then it will become 144,
100----------20%-----120-------20%(24)------120+24 = 144.
Viki said:
8 years ago
Let us keep.
l = 100.
b = 50.
The area will be 5000
If changed be,
l = 120
b = 60
Area = 7200.
In original area, 1% of 5000 is 50.
Then changed area is 7200-5000 = 2200
Now (2200/50 = 44).
Then changed area is 44%.
l = 100.
b = 50.
The area will be 5000
If changed be,
l = 120
b = 60
Area = 7200.
In original area, 1% of 5000 is 50.
Then changed area is 7200-5000 = 2200
Now (2200/50 = 44).
Then changed area is 44%.
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
Mero baduwal said:
8 years ago
Please solve the question that what percent changes in area of rectangular if there is mistake in length and breadth by one percent.
Madhu said:
8 years ago
Increases 20% so (120/100) * (120/100) =14400/10000=144%.
44% increases.
44% increases.
Veer said:
8 years ago
This can be solved in very simple steps. It is as follow.
The area of rectangle = l * b,
Consider initial is 100 and given that 20% increase,
100----->20%--->120----->20%---->144,
(144-100) = 44.
The area of rectangle = l * b,
Consider initial is 100 and given that 20% increase,
100----->20%--->120----->20%---->144,
(144-100) = 44.
UDAY KINGS said:
8 years ago
Take side = a.
So, a increased by 20 so a = 120/100.
Area of square = a^2.
120/100*120/100 = 44.
So, a increased by 20 so a = 120/100.
Area of square = a^2.
120/100*120/100 = 44.
ER RAJKUMAR ARYA said:
8 years ago
Let, l = b = 1 unit.
a1 = 1*1 = 1 sq.unit.
20% increase in both sides, new area a2 = 1.2*1.2 = 1.44.
i.e, % increase in area = 1.44 - 1 = .44 or 44%.
a1 = 1*1 = 1 sq.unit.
20% increase in both sides, new area a2 = 1.2*1.2 = 1.44.
i.e, % increase in area = 1.44 - 1 = .44 or 44%.
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