### Discussion :: 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: |
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Answer: Option C Explanation: Let original length = Original area = (
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
Video Explanation: https://youtu.be/I3jLjLPn1W4 |

Vijay said: (Jun 22, 2010) | |

How did you get 11/25 ? |

Chandra said: (Aug 17, 2010) | |

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: (Aug 18, 2010) | |

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: (Oct 3, 2010) | |

@ 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: (Oct 3, 2010) | |

Bhupendra has solved this correctly ! |

Stephan said: (Oct 4, 2010) | |

Very good Ajmal. |

Sivasankari said: (Dec 10, 2010) | |

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: (Mar 20, 2011) | |

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: (Jul 19, 2011) | |

Very Good Chandra |

Santhosh said: (Aug 5, 2011) | |

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 % |

Akash Soni said: (Jan 19, 2012) | |

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% |

Rahul said: (Mar 4, 2012) | |

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 |

Deepi said: (May 2, 2012) | |

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 |

G@Ni said: (Aug 5, 2012) | |

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%. |

Abhishek Jain said: (Oct 16, 2012) | |

The percentage increase in the area of a rectangle, if each of its sides is increased by 20% is:::. 1. Suppose I have 10 of each side and then 20% increased of every side as a question now the 10 comes 12 so now, we have two sides one is new and one is old now we square both old and new sides and we receive 100 for old and 144 for new. 2. Now minus these numbers like. (144-100) =44. 3. Now we will do percentage of 44 like. Minus result/previous square nu. *100. 44/100*100=44 answer. . |

Sdasdasdad said: (Mar 21, 2013) | |

a=l*b=120l*120b. a=14400lb. a=100l*100b. a=10000lb. %=44/100. 44% answer. |

Sandeepk said: (May 16, 2013) | |

I found below method most easy : Let l=b=1 units. Area = 1 * 1 = 1. New l = b = 1.2 (20% increase). New Area = 1.2 * 1.2 = 1.44. Thus, 0.44 is 44% increase in area is the answer. |

Ram said: (Jul 25, 2013) | |

In a easy way. Assume l & b = 10, 10. Area = 10*10 = 100. New length with 20% increase = 10*120/100 = 12. So same as new breath with 20% increase = 12. New area = 12*12 = 144. Difference of both area = 44 %. |

T.Pratap said: (Aug 20, 2013) | |

As you know that a square is also a rectangle. We will suppose that its sides are l=100, b=100. Thus the area will be= l*b = 100*100 = 10000. Now if there. Is increment of 20 %. The new sides are l= 120 and b=120. Thus the area will be = l*b = 14400. Difference between the area before increment and after increment. = 14400-10000 = 4400. Now the % increase in area is = (4400/10000) *100 = 44%. |

Mani said: (Dec 15, 2013) | |

From where have we got new length and new breadth. |

Hanny Gupta said: (Feb 12, 2014) | |

A1 = 100*100 = 10000. A2 = 120*120 = 14400. A2-A1 = 14400-10000 = 4400. (A2-A1)% = 44 %. |

Lipu said: (Oct 10, 2014) | |

Let true area = (100)^2. Error area = (120)^2. So error %= (error area-true area)/true area. = (120)^2-(100)^2/(100)^2 *100. = 220*20/100. = 44%. |

Manan said: (Nov 14, 2014) | |

There is a much simpler method. Suppose the length of rectangle = 100 cm & breadth = 50 cm. So Area equals to 100*50 = 5000. Now if 20% of both sides are increased then, New length = 20% of 100 = 120 cm & 20% of 50 = 60 cm. So, new area = 120*60 = 7200. Difference = 7200-5000 = 2200. Percentage increase = (2200\5000)*100 = 44%. |

Lexinah said: (Nov 19, 2014) | |

What if it says each side of a square not rectangle is increased by 15%? What is the percentage area increase in its area? |

Mhbkhb said: (Nov 20, 2014) | |

Taking 100*100 is not correct method. Because if we take 100*100 it will bcom a square but not rectangle. |

Rozenelle said: (Dec 21, 2014) | |

If the answer is 44% it is a square not a rectangle, if it is a rectangle the correct answer is 32%. |

Deepa Ezhil. said: (Dec 28, 2014) | |

I can't understand Sandeepk logic. In the given problem it is given as rectangle but, he took the values of length and breadth as same value. I think the logic he used is wrong. |

Pagalbuoy said: (Feb 17, 2015) | |

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. |

Hema said: (Mar 10, 2015) | |

Why here you take length = (120/100)x and breadth = (120/100)y? |

Michelle said: (May 9, 2015) | |

Let length 40 and breadth is =20. After increasing length is 48 and breadth is 24. So the area is 1152. The difference between new are and old area is 352. So (352/800*100) = 44%. |

Sourav said: (May 14, 2015) | |

Easiest approach: Length 10 cm. Breadth 10 cm. Area = 10*10 = 100 cm. With 20% increase: Length 12 cm. Breadth 12 cm. Area 144 cm. Percentage increase (144-100) = 44. Ans: 44%. |

Gnit said: (Jun 9, 2015) | |

Let the area be 100. Since length and area are proportional 20% increase in length => area = 120 since breadth and area are proportional 20 % increase in breadth => area = 144. Therefore, 144-100 = 44. |

Sukumar Satyen said: (Jun 21, 2015) | |

Let's assume Length = L cm and Breadth = B cm. => Area (old) = L*B = LB....equation (1). When, Length is increased by 20%, Length = 1.2L. When, Breadth is increased by 20%, Breadth = 1.2B. => Area (increased) = 1.2L*1.2B = 1.44LB.....equation (2). Using equation (1) and (2). => LB (1+% increase/100) = 1.44LB. => % increase = (1.44-1)*100 = 0.44*100 = 44 per cent. |

Priyanka said: (Sep 17, 2015) | |

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%. |

Lork said: (Oct 10, 2015) | |

Old a b ==> ab. New 1.2 a 1.2 b ==> 1.44 ab. Increase = 1.44 ab - ab = 0.44 = 44%. |

Kumar said: (Jan 28, 2016) | |

I can't understand your answer. Please explain brief. |

Naveen said: (Feb 4, 2016) | |

Given sides of rectangle are increased by 20%. Let us take length L = 20. Breadth B = 10. Area of rectangle = LxB = 10x20 = 200. After 20% increase L = 24. B = 12. Area of rectangle = LxB = 24x12 = 288. Area increased to 288 from 200 = 44%. Same for square. |

Shailesh said: (Mar 8, 2016) | |

Make it simple yar! Parentage increase increase of rectangle or square will be same. Assume 10 x 10 original square Area will be 100 simple. Now increase side length by 20 % will make 12 x 12 square Area will be 144. So simple should I tell you the correct answer (C) 44. |

Ajit said: (Jun 1, 2016) | |

Assume that total area is 100. 20% of 100 = 120. 20% of 120 = 144. 144 - 100 = 44. |

Ramesh said: (Jul 28, 2016) | |

Let original length and breadth be 100. So new length and breadth are 120 New Area = 120 * 120 so 14400. Old area = 100 * 100 = 10000. So diff = 4400. Error = 4400/(100 * 100) * 100. So, 44%. |

Kasaiah said: (Aug 26, 2016) | |

Why all of them taking a length as 100 that and is not a square. Why don't you try it length and breadth with separate values like 100 & 200? |

Naveed Subhan said: (Sep 19, 2016) | |

What if you don't know what the percentage of increase or decrease is and what if one side is increasing and the other is decreasing? Let a = percentage change(increase or decrease)on one side and b= percentage change of other side then: If both sides as are increasing = (a + b) + ab/100. If both sides are decreasing = (a + b) - ab/100. If one side is increasing (positive, say a) and one side is decreasing (negative, say b) = a - b - ab/100. |

Bhaskar Chandra said: (Oct 4, 2016) | |

Trick: If increase in length of breadth & width is x% then the increase in area a' = 2x + (square x)/100. Quick method: if x=20% then a'. 20 * 2 + (20 * 20 = 4.00) = 44.00. If x = 4 then a' = 4 * 2 + 0.(4 * 4) = 8 + 0.16 = 8.16% If x = 5 then a' = 5 * 2 + 0.(5 * 5) = 10 + 0.25 = 10.25. If x = 6 then a' = 6 * 2 + 0.(6 * 6) = 12 + 0.36 = 12.36. If x = 7 then a' = 7 * 2 + 0.(7 * 7) = 14 + 0.49 = 14.49. If x = 8 then a' = 8 * 2 + 0.(8 * 8) = 16 + 0.64 = 16.64. If x = 9 then a' = 9 * 2 + 0.(9 * 9) = 18 + 0.81 = 18.81. If x = 10 then a' = 10 * 2 + 0.(10 * 10) = 20 + 1.00 = 21. If x = 11 then a' = 11 * 2 + 0.(11 * 11) = 22 + 1.21 = 23.21. Note : When the product is a number of more than three decimals, place a decimal point after two places from the right. |

Ravi said: (Oct 10, 2016) | |

The rational number with recurring decimal expansion 0.12 is __________. Can anyone answer the question? |

Bubai Mitra said: (Oct 14, 2016) | |

If let it 100. It will make more easy. |

Vasavi said: (Oct 18, 2016) | |

@Bhupendra. How did you get the new side as 6x/5? |

Shashank said: (Oct 26, 2016) | |

Increased percentage is 20%. that means 120%. For both sides, 120/100 * 120/100 = 144/100. i.e 44% more than 100. |

Gokul said: (Nov 5, 2016) | |

Explained well @Shashank. |

Ajay said: (Nov 5, 2016) | |

Shortcut : 20 + 20 + ((20 * 20)/100) = 44. |

Ratul Sinha said: (Jan 1, 2017) | |

Let each side 100. Area 100 * 100, Increased area 120 * 120, (14400-10000) ÷ 10000 = .44, 0.44 * 100 = 44%. |

Deependra Singh said: (Jan 25, 2017) | |

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%. |

Sarvesh Agarwal said: (Feb 18, 2017) | |

A = L*B. A = (1+1/5)(1+1/5). A = 36/25LB. A = 144%. i.e 44%. |

Ravali Yadav said: (Mar 2, 2017) | |

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. |

Viki said: (Mar 21, 2017) | |

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%. |

Mr.Issue said: (Apr 22, 2017) | |

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 |

Mero Baduwal said: (Jun 4, 2017) | |

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: (Jul 17, 2017) | |

Increases 20% so (120/100) * (120/100) =14400/10000=144%. 44% increases. |

Veer said: (Jul 19, 2017) | |

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. |

Uday Kings said: (Nov 4, 2017) | |

Take side = a. So, a increased by 20 so a = 120/100. Area of square = a^2. 120/100*120/100 = 44. |

Er Rajkumar Arya said: (Nov 18, 2017) | |

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%. |

Hussain said: (Jan 22, 2018) | |

a*(200+a)/100. so, 20*(200+20)/100. =44%. |

Puja said: (Jan 23, 2018) | |

20 +20+20*20/100=44%. |

Harmanpreet said: (Jun 21, 2018) | |

Each side 20% increased thus 120*120/100=144, then 144-100 =44. |

Maheshkrishnan D said: (Nov 22, 2018) | |

100*100 = 10000(1) 120*120 = 14400(2) 1-2 = 4400/100. = 44%. |

Kumar Gowda said: (Apr 15, 2019) | |

= 20+20(20*20/100). = 40+(400/100), = 40+(4), = 40+4, = 44%. |

Tridib Samanta said: (Jul 15, 2019) | |

Let, Original length of rectangle be, x = 100m, Original breadth of rectangle be, y = 10m, Original area of the rectangle is, A = 1000 sq. m. Now, each side is incresed by 20%. New length of the rectangle is, x1 = 120m, New breadth of the rectangle is, y1 = 12m, New area of the rectangle is, A1 = 1440 sq. m. Therefore, total increase in area = (A1 - A) = (1440 - 1000)m = 440m. % increase in area; = (total increase in area / original area) x 100. = {( A1- A) / A } x 100, = (440/1000) x 100, = .44 x 100, = 44 % (Answer). Thanks! |

Bhavik said: (Nov 4, 2019) | |

Say length =100. But it increased to 120. Therefore the area of rectangle = 120*120 = 14400. Now divide by 100-->144. So 144-100 = 44 is the answer. |

Vaibhav said: (Apr 23, 2020) | |

Simple trick for such problems is. If increase by x% we can calculate by the formula. x(x+200)%100. |

Subbu said: (May 17, 2020) | |

Can anyone show how it can be solved by using derivatives? please. |

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