If the length of a rectangle is increased by 120 and its width is decreased by 20

Given : The length of a rectangle is greater than the breadth by 18cm. If both length and breadth are increased by 6cm , then area is increases by 168cm2 

To do: Find the breadth of the rectangle 

Solution Let the breadth of the rectangle = x cm

Length of the rectangle = $x + 18$ cm

Area of the rectangle = $L \times B = (x + 18) . x$

=$ x(x+ 18)$

length and breadth increased by 6cm each and area increases by 168 sq cm

$(x + 18 + 6)(x + 6) = x(x + 18) + 168$

$(x + 24)(x + 6)=  x^2 + 18x + 168$

$x^2 + 30x + 144 = x^2 + 18x + 168$

$30x - 18x = 12x = 168 - 144 = 24$

$12x = 24$

$x = \frac{24}{12} = 2; x = 2$

So length and breadth of the rectangle are

2 + 18, 2 or 20 cm and 2 cm respectively

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A) increase by 2%

B) decrease by 2%.

C) increase by 4%

D) decrease by 4%

Description for Correct answer:
Let 'I' be the length lncrease = 20%

New length = \( \Large \frac{120}{100}l=\frac{6}{5}l \)

Let b be the width. Decrease = 20%

New width = \( \Large \frac{80}{100}b=\frac{4}{5}b \)

Original area = lb

New area = \( \Large \frac{6}{5}l \times \frac{4}{5}b = \frac{24}{25}lb \)

Change in area = \( \Large lb-\frac{24}{25}lb=\frac{1}{25}lb \)

Percentage in decrease = \( \Large 100 \times \frac{1}{25} \)= 4%

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