Question

the perimeter of an aluminium sheet is 120 CM if its length is reduced by 10% and its breadth is increased by 20% the perimeter does not change find the measure of the length and the breadth of the sheet​

Answer 1

Given :-

• The perimeter of sheet = 120cm

To Find :-

• The length of sheet = ?
• The breadth of sheet = ?

Solution :-

• To calculate the dimensions of sheet at first we have to set up equation with the help of given clue in the question. By applying formula we can easily calculate the length and breadth of sheet. Let, assume the length of sheet be L and breadth be B

Calculation begins :-

Given in case - (i) :-

⇒ Perimeter of sheet = 2(L + B)

⇒ 2(L + B) = 120

⇒ L + B = 60------(i)

Given in case - (ii) :-

• If its length is reduced by 10% and its breadth is increased by 20% the perimeter does not change.

⇒ Perimeter of new sheet = perimeter of Original sheet

⇒ 2(L - 10% of L + B + 20% of B) = 2(L + B)

⇒ 2(L - L × 10/100 + B + B × 20/100) = 2(L + B)

⇒ 2(L - L/10 + B + B/5) = 2(L + B)

⇒ 2(10L - L)/10 + (6B + B)/5) = 2(L + B)

⇒ 9L/10 + 7B/5 = L + B

⇒ 9L + 14B/10 = L + B

⇒ 9L + 14B = 10L + 10B

⇒ 10L - 9L + 10B - 14B = 0

⇒ L - 4B = 0----(ii)

• Eq (i) and (ii) solving we get here :-]

⇒ L + B = 60

⇒ L - 4B = 0

⇒ 5B = 60

⇒ B = 12

• Putting the value of B = 12 in eq (i)

⇒ L + B = 60

⇒ L + 12 = 60

⇒ L = 60 - 12

⇒ L = 48

Hence,

• The length of sheet (L) = 48cm
• The breadth of sheet (B) = 12cm

Answer 2

Answer:

Given :-

• The perimeter of an aluminium sheet is 120 cm, if its length is reduced by 10% and its breadth is increased by 20% the perimeter does not change.

To Find:-

• What is the measure of the length and the breadth of the sheet.

Formula Used :-

$\clubsuit$ Perimeter Of Rectangle Formula :

$\footnotesize\mapsto \sf\boxed{\bold{\pink{Perimeter_{(Rectangle)} =\: 2(Length + Breadth)}}}$

Solution:-

Let,

$\mapsto \bf Length_{(Sheet)} =\: x\: cm$

$\mapsto \bf Breadth_{(Sheet)} =\: y\: cm$

$\clubsuit\: \: \sf\bold{\purple{\underline{In\: the\: 1^{st}\: case\: :-}}}$

Given :

• Perimeter of an aluminium sheet = 120 cm

According to the question by using the formula we get,

$\implies \sf 120 =\: 2(x + y)$

$\implies \sf \dfrac{\cancel{120}}{\cancel{2}} =\: x + y$

$\implies \sf 60 =\: x + y$

$\implies \sf\bold{\green{x + y =\: 60\: ------\: (Equation\: No\: 1)}}$

$\clubsuit\: \: \sf\bold{\purple{\underline{In\: the\: 2^{nd}\: case\: :-}}}$

$\bigstar$ The length is reduced by 10% and its breadth is increased by 20%, the perimeter does not change.

According to the question,

$\implies \sf 2(x - 10\% \times x + y + 20\% \times y) =\: 2(x + y)$

$\implies \sf 2\bigg(x - \dfrac{1\cancel{0}}{10\cancel{0}} \times x + y + \dfrac{2\cancel{0}}{10\cancel{0}} \times y\bigg) =\: 2\bigg(x + y\bigg)$

$\implies \sf 2\bigg(x - x \times \dfrac{1}{10} + y + y \times \dfrac{\cancel{2}}{\cancel{10}}\bigg) =\: 2\bigg(x + y\bigg)$

$\implies \sf 2\bigg(x - \dfrac{x}{10} + y + y \times \dfrac{1}{5}\bigg) =\: 2\bigg(x + y\bigg)$

$\implies \sf {\cancel{2}}\bigg(x - \dfrac{x}{10} + y + \dfrac{y}{5}\bigg) =\: {\cancel{2}}\bigg(x + y\bigg)$

$\implies \sf \dfrac{9x}{10} + \dfrac{7y}{5} =\: x + y$

$\implies \sf \dfrac{9x + 14y}{10} =\: x + y$

By doing cross multiplication we get,

$\implies \sf 9x + 14y =\: 10(x + y)$

$\implies \sf 9x + 14y =\: 10x + 10y$

$\implies \sf 10x - 9x + 10y - 14y =\: 0$

$\implies \sf\bold{\green{x - 4y =\: 0\: ------\: (Equation\: No\: 2)}}$

By solving the equation no 1 and 2 we get,

$\implies \sf x + y - (x - 4y) =\: 60 - 0$

$\implies \sf x + y - x + 4y =\: 60$

$\implies \sf {\cancel{x}} {\cancel{- x}} + y + 4y =\: 60$

$\implies \sf y + 4y =\: 60$

$\implies \sf 5y =\: 60$

$\implies \sf y =\: \dfrac{\cancel{60}}{\cancel{5}}$

$\implies \sf\bold{\blue{y =\: 12}}$

By putting the value of y in the equation no 2 we get,

$\implies \sf x - 4y =\: 0$

$\implies \sf x - 4(12) =\: 0$

$\implies \sf x - 4 \times 12 =\: 0$

$\implies \sf x - 48 =\: 0$

$\implies \sf\bold{\blue{x =\: 48}}$

Hence, the required length and breadth of the sheet are :

$\longrightarrow \sf\bold{\red{Length_{(Sheet)} =\: 48\: cm}}$

$\longrightarrow \sf\bold{\red{Breadth_{(Sheet)} =\: 12\: cm}}$

${\small{\bold{\underline{\therefore\: The\: length\: and\: breadth\: of\: sheet\: is\: 48\: cm\: and\: 12\: cm\: respectively\: .}}}}$