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 7 of 8.
G@ni said:
1 decade ago
I had a small and sweet solution for this problem.
The formula for % increase is x+y+xy/100.
And for %decrease is x+y-xy/100.
Now x=20.
y=20.
Therefore substituting in the above formula .
20+20+(20*20/100)=40+(400/100).
=40+4.
=44%.
The formula for % increase is x+y+xy/100.
And for %decrease is x+y-xy/100.
Now x=20.
y=20.
Therefore substituting in the above formula .
20+20+(20*20/100)=40+(400/100).
=40+4.
=44%.
Deepi said:
1 decade ago
Hi guys i have a easy step to this kind of sum,
if any one side is increased a% mean what is the increment of area can be calculated by the following formula
=2a+(a^2/100)
similarly if a% decrease means area also decreased by
=2a-(a^2/100)
In the above sum side is increased 20 while applying it in the formula
=2*20+(20*20/100)
=40+4
=44
if any one side is increased a% mean what is the increment of area can be calculated by the following formula
=2a+(a^2/100)
similarly if a% decrease means area also decreased by
=2a-(a^2/100)
In the above sum side is increased 20 while applying it in the formula
=2*20+(20*20/100)
=40+4
=44
Rahul said:
1 decade ago
Let us supoose l=10, b=10 area=l*b=10*10=100
n aftr inc l=12 , b=12 area=l*b=12*12=144
144-100=44%ans
n aftr inc l=12 , b=12 area=l*b=12*12=144
144-100=44%ans
AKASH SONI said:
1 decade ago
Its a lengthy approach
let length=L Breadth=b
intial area =L*B
increased length and breadth are 1.2L and 1.2B respectively
new area =1.44LB
increase area =.44lb
increased percentage area =.44LB/(LB)*100 =44%
let length=L Breadth=b
intial area =L*B
increased length and breadth are 1.2L and 1.2B respectively
new area =1.44LB
increase area =.44lb
increased percentage area =.44LB/(LB)*100 =44%
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 %
Pratheep Santho said:
1 decade ago
Very Good Chandra
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%
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 !
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