Aptitude - Area - Discussion

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"The secret to creativity is knowing how to hide your sources."

- Albert Einstein

The percentage increase in the area of a rectangle, if each of its sides is increased by 20% is:

[A].	40%	[B].	42%
[C].	44%	[D].	46%

Answer: Option C

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

Chandra said: (Tue, Aug 17, 2010 06:46:39 AM)

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%

Bhupendra said: (Wed, Aug 18, 2010 03:19:26 AM)

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

Ajmal said: (Sun, Oct 3, 2010 12:02:54 AM)

@ 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.

Krishna said: (Sun, Oct 3, 2010 11:05:38 AM)

Bhupendra has solved this correctly !

Sivasankari said: (Fri, Dec 10, 2010 02:26:17 PM)

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....

Krishna said: (Sun, Mar 20, 2011 10:17:07 PM)

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%

Pratheep Santho said: (Tue, Jul 19, 2011 12:53:12 PM)

Very Good Chandra

Santhosh said: (Fri, Aug 5, 2011 07:08:31 PM)

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 = 100100= 10000 ; A2 = 120 120 = 14400 A2-A1 = 4400 so % increase in the area = 4400/100 = 44 %

Akash Soni said: (Thu, Jan 19, 2012 11:14:07 PM)

Its a lengthy approach let length=L Breadth=b intial area =LB 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%