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 1 of 8.
Vijay said:
2 decades ago
How did you get 11/25 ?
(1)
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%
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
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.
Krishna said:
1 decade ago
Bhupendra has solved this correctly !
Stephan said:
1 decade ago
Very good Ajmal.
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....
Krishna said:
1 decade ago
Old area:
length=l
breadth=b
area=lb
new area:
length=(120/100)l = 6l/5
breadth=(120/100)b=6b/5
area=36lb/25
error% = (n.a - o.a)*100/o.a
=(((36lb/25)-lb)*100)/lb
=44%
length=l
breadth=b
area=lb
new area:
length=(120/100)l = 6l/5
breadth=(120/100)b=6b/5
area=36lb/25
error% = (n.a - o.a)*100/o.a
=(((36lb/25)-lb)*100)/lb
=44%
Pratheep Santho said:
1 decade ago
Very Good Chandra
Santhosh said:
1 decade ago
Let the original length be 100
so the increment on all sides by 20% i.e. length = 120
now calculate the area of original rectangle and new increased length's rectangle
i.e. A1 = 100*100= 10000 ; A2 = 120 * 120 = 14400
A2-A1 = 4400
so % increase in the area = 4400/100 = 44 %
so the increment on all sides by 20% i.e. length = 120
now calculate the area of original rectangle and new increased length's rectangle
i.e. A1 = 100*100= 10000 ; A2 = 120 * 120 = 14400
A2-A1 = 4400
so % increase in the area = 4400/100 = 44 %
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