Question

Q1.the length of a rectangular fields ifs increased by 50 percent and its breadth is decreased by 50 percent to form a new rectangular field.Find the percentage change in the area of the fields.

Answer 1

Answer:

Let the length of rectangle be a

Let the length of rectangle be aAnd width of rectangle be b

Let the length of rectangle be aAnd width of rectangle be bArea =ab

Let the length of rectangle be aAnd width of rectangle be bArea =abNow,

Let the length of rectangle be aAnd width of rectangle be bArea =abNow, New length =a+50 % of a

Let the length of rectangle be aAnd width of rectangle be bArea =abNow, New length =a+50 % of a =a+

Let the length of rectangle be aAnd width of rectangle be bArea =abNow, New length =a+50 % of a =a+ 100

Let the length of rectangle be aAnd width of rectangle be bArea =abNow, New length =a+50 % of a =a+ 10050a

=a+

=a+ 2

=a+ 2a

=a+ 2a

=a+ 2a

=a+ 2a =

=a+ 2a = 2

=a+ 2a = 23a

=a+ 2a = 23a

=a+ 2a = 23a

=a+ 2a = 23a New width =b−50% of b

=a+ 2a = 23a New width =b−50% of b =b−

=a+ 2a = 23a New width =b−50% of b =b− 2

=a+ 2a = 23a New width =b−50% of b =b− 2b

=a+ 2a = 23a New width =b−50% of b =b− 2b

=a+ 2a = 23a New width =b−50% of b =b− 2b

=a+ 2a = 23a New width =b−50% of b =b− 2b =

=a+ 2a = 23a New width =b−50% of b =b− 2b = 2

=a+ 2a = 23a New width =b−50% of b =b− 2b = 2b

=a+ 2a = 23a New width =b−50% of b =b− 2b = 2b

=a+ 2a = 23a New width =b−50% of b =b− 2b = 2b

=a+ 2a = 23a New width =b−50% of b =b− 2b = 2b New Area =

=a+ 2a = 23a New width =b−50% of b =b− 2b = 2b New Area = 2

=a+ 2a = 23a New width =b−50% of b =b− 2b = 2b New Area = 23a

=a+ 2a = 23a New width =b−50% of b =b− 2b = 2b New Area = 23a

=a+ 2a = 23a New width =b−50% of b =b− 2b = 2b New Area = 23a ×

=a+ 2a = 23a New width =b−50% of b =b− 2b = 2b New Area = 23a × 2

=a+ 2a = 23a New width =b−50% of b =b− 2b = 2b New Area = 23a × 2b

=a+ 2a = 23a New width =b−50% of b =b− 2b = 2b New Area = 23a × 2b

=a+ 2a = 23a New width =b−50% of b =b− 2b = 2b New Area = 23a × 2b

=a+ 2a = 23a New width =b−50% of b =b− 2b = 2b New Area = 23a × 2b =

=a+ 2a = 23a New width =b−50% of b =b− 2b = 2b New Area = 23a × 2b = 4

=a+ 2a = 23a New width =b−50% of b =b− 2b = 2b New Area = 23a × 2b = 43ab

=a+ 2a = 23a New width =b−50% of b =b− 2b = 2b New Area = 23a × 2b = 43ab

=a+ 2a = 23a New width =b−50% of b =b− 2b = 2b New Area = 23a × 2b = 43ab

=a+ 2a = 23a New width =b−50% of b =b− 2b = 2b New Area = 23a × 2b = 43ab Decrease in Area =ab−

=a+ 2a = 23a New width =b−50% of b =b− 2b = 2b New Area = 23a × 2b = 43ab Decrease in Area =ab− 4

=a+ 2a = 23a New width =b−50% of b =b− 2b = 2b New Area = 23a × 2b = 43ab Decrease in Area =ab− 43ab

=a+ 2a = 23a New width =b−50% of b =b− 2b = 2b New Area = 23a × 2b = 43ab Decrease in Area =ab− 43ab

=a+ 2a = 23a New width =b−50% of b =b− 2b = 2b New Area = 23a × 2b = 43ab Decrease in Area =ab− 43ab

=a+ 2a = 23a New width =b−50% of b =b− 2b = 2b New Area = 23a × 2b = 43ab Decrease in Area =ab− 43ab =

=a+ 2a = 23a New width =b−50% of b =b− 2b = 2b New Area = 23a × 2b = 43ab Decrease in Area =ab− 43ab = 4

=a+ 2a = 23a New width =b−50% of b =b− 2b = 2b New Area = 23a × 2b = 43ab Decrease in Area =ab− 43ab = 4ab

=a+ 2a = 23a New width =b−50% of b =b− 2b = 2b New Area = 23a × 2b = 43ab Decrease in Area =ab− 43ab = 4ab

=a+ 2a = 23a New width =b−50% of b =b− 2b = 2b New Area = 23a × 2b = 43ab Decrease in Area =ab− 43ab = 4ab

=a+ 2a = 23a New width =b−50% of b =b− 2b = 2b New Area = 23a × 2b = 43ab Decrease in Area =ab− 43ab = 4ab % of Decrease in Area =[

=a+ 2a = 23a New width =b−50% of b =b− 2b = 2b New Area = 23a × 2b = 43ab Decrease in Area =ab− 43ab = 4ab % of Decrease in Area =[ 4

=a+ 2a = 23a New width =b−50% of b =b− 2b = 2b New Area = 23a × 2b = 43ab Decrease in Area =ab− 43ab = 4ab % of Decrease in Area =[ 4ab

=a+ 2a = 23a New width =b−50% of b =b− 2b = 2b New Area = 23a × 2b = 43ab Decrease in Area =ab− 43ab = 4ab % of Decrease in Area =[ 4ab

=a+ 2a = 23a New width =b−50% of b =b− 2b = 2b New Area = 23a × 2b = 43ab Decrease in Area =ab− 43ab = 4ab % of Decrease in Area =[ 4ab ×100]/ab

=a+ 2a = 23a New width =b−50% of b =b− 2b = 2b New Area = 23a × 2b = 43ab Decrease in Area =ab− 43ab = 4ab % of Decrease in Area =[ 4ab ×100]/ab=

=a+ 2a = 23a New width =b−50% of b =b− 2b = 2b New Area = 23a × 2b = 43ab Decrease in Area =ab− 43ab = 4ab % of Decrease in Area =[ 4ab ×100]/ab= 4

=a+ 2a = 23a New width =b−50% of b =b− 2b = 2b New Area = 23a × 2b = 43ab Decrease in Area =ab− 43ab = 4ab % of Decrease in Area =[ 4ab ×100]/ab= 4100

=a+ 2a = 23a New width =b−50% of b =b− 2b = 2b New Area = 23a × 2b = 43ab Decrease in Area =ab− 43ab = 4ab % of Decrease in Area =[ 4ab ×100]/ab= 4100

=a+ 2a = 23a New width =b−50% of b =b− 2b = 2b New Area = 23a × 2b = 43ab Decrease in Area =ab− 43ab = 4ab % of Decrease in Area =[ 4ab ×100]/ab= 4100

=a+ 2a = 23a New width =b−50% of b =b− 2b = 2b New Area = 23a × 2b = 43ab Decrease in Area =ab− 43ab = 4ab % of Decrease in Area =[ 4ab ×100]/ab= 4100 = 25 %

Answer 2

Step-by-step explanation:

Answer:

Let the original length be 'l' and the breadth be 'b'.

Hence the Area of the rectangular field originally is:

→ Area = lb

Now it is given that the length is increased by 50% and the breadth is decreased by 50%.

Hence the new length and breadth are given as:

$</p><p>\begin{gathered}\implies l_{new} = l + \text{50 percent of l}\\\\\\\implies l_{new} = l + \dfrac{50\:l}{100}\\\\\\\implies l_{new} = l + \dfrac{l}{2} = \dfrac{3l}{2}\end{gathered} </p><p>$

Hence the new length is 1.5 l.

$\begin{gathered}\implies b_{new} = b - \text{50 percent of b}\\\\\\\implies b_{new} = b - \dfrac{50\:b}{100}\\\\\\\implies b_{new} = b - \dfrac{b}{2} = \dfrac{b}{2}\end{gathered} </p><p>$

Hence the new breadth is 0.5 b.

Calculating the new area we get:

⇒ New Area = New L × New B

⇒ New Area = 1.5 l × 0.5 b

⇒ New Area = 0.75 lb

Hence the change in the Area is given as:

⇒ Change = Old Area - New Area

⇒ Change = lb - 0.75 lb

⇒ Change = 0.25 lb

Percentage change in area is given as:

$\begin{gathered}\implies Change \:\% = \dfrac{\text{Change in Area}}{\text{Old Area}}\times 100\\\\\\\implies Change \:\% = \dfrac{0.25\:lb}{lb} \times 100\\\\\\\implies Change \:\% = 0.25 \times 100 = \boxed{ \bf{25 \:\%}}\end{gathered} </p><p>$

Hence the percentage change in area of the fields is 25%.