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.
Sivasankari said:
1 decade ago
Just we take before increment area is 100%l * 100%b = lb
After 20% of increment area is 120%l * 120%b=1.44lb from this we get 0.44 that is 44% was increased....
After 20% of increment area is 120%l * 120%b=1.44lb from this we get 0.44 that is 44% was increased....
Stephan said:
1 decade ago
Very good Ajmal.
Krishna said:
1 decade ago
Bhupendra has solved this correctly !
Ajmal said:
1 decade ago
@ Vijay
The difference between the original area = xy and 36/25 xy is
= (36/25)xy - xy
= xy(36/25 - 1)
= xy(11/25)
or
= (11/25)xy.
The difference between the original area = xy and 36/25 xy is
= (36/25)xy - xy
= xy(36/25 - 1)
= xy(11/25)
or
= (11/25)xy.
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
Chandra said:
1 decade ago
When there is an increase in sides of a figure, the net increase in its area is x+y+xy/100 %
So here 20% +20% + 20x20/100% = 44%
So here 20% +20% + 20x20/100% = 44%
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.
Kumar said:
10 years ago
I can't understand your answer. Please explain brief.
Lork said:
10 years ago
Old a b ==> ab.
New 1.2 a 1.2 b ==> 1.44 ab.
Increase = 1.44 ab - ab = 0.44 = 44%.
New 1.2 a 1.2 b ==> 1.44 ab.
Increase = 1.44 ab - ab = 0.44 = 44%.
Priyanka said:
10 years ago
Let the length be 100 and breadth be 200.
So area = l*b = 20000.
If both the sides increases by 20% then,
The new length = 120 and new breadth = 240.
New area = 120*240 = 28800.
So % increase in area = (28800-20000)/20000*100 which is coming out to be 44%.
So area = l*b = 20000.
If both the sides increases by 20% then,
The new length = 120 and new breadth = 240.
New area = 120*240 = 28800.
So % increase in area = (28800-20000)/20000*100 which is coming out to be 44%.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers