The length and breadth of a rectangular field are increased by 15% and

1.	The length and breadth of a rectangular field are increased by 15% and 10% respectively. What will be the effect on its area?1). 26.5% increase2). 28.5% increase3). 21.5% increase4). 26.5% decrease
Answer» Let the length and BREADTH of the rectangular field are ‘l’ and ‘b’ RESPECTIVELY. Then, According to the question, Length of the field is increased by 15%- ⇒ New length = 115% of l = (23/20)l, Breadth of the field is increased by 10%- ⇒ New breadth = 110% of b = (11/10)b We KNOW that, Area of a rectangle = Length × Breadth ∴ New area of the rectangle = (23/20)l × (11/10)b = (253/200) × l × b Percent increase in the area- $(= \frac{{\frac{{253}}{{200}}lb - lb}}{{lb}} \times 100\%= \left( {\frac{{253}}{{200}} - 1} \right) \times 100\%= \frac{{53}}{{200}} \times 100\%= 26.5\%)$ Hence, the required change in the area of the field is 26.5% increase.

The length and breadth of a rectangular field are increased by 15% and 10% respectively. What will be the effect on its area?1). 26.5% increase2). 28.5% increase3). 21.5% increase4). 26.5% decrease

Answer»

Let the length and BREADTH of the rectangular field are ‘l’ and ‘b’ RESPECTIVELY. Then,

According to the question,

Length of the field is increased by 15%-

⇒ New length = 115% of l = (23/20)l,

Breadth of the field is increased by 10%-
⇒ New breadth = 110% of b = (11/10)b

We KNOW that,

Area of a rectangle = Length × Breadth

∴ New area of the rectangle = (23/20)l × (11/10)b

= (253/200) × l × b

Percent increase in the area-

$(= \frac{{\frac{{253}}{{200}}lb - lb}}{{lb}} \times 100\%= \left( {\frac{{253}}{{200}} - 1} \right) \times 100\%= \frac{{53}}{{200}} \times 100\%= 26.5\%)$

Hence, the required change in the area of the field is 26.5% increase.

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