Question

please solve question number 11 ​

Answer 1

$\huge\cal\blue{UR\: ANSWER❤}$

$\sf\red{\fbox{\fbox\pink{11th \:\:\:Solution}}}$

$\sf\purple{Given\:In \:a\: Rectangle:}$

‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎length = (6+x) cm

‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎breadth = x cm

‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎Perimeter = 80 cm

$\sf\purple{Perimeter \:of\: Rectangle:}$$\large\red{2(l+b)}$

$\red\Rightarrow$ 2(6+x+x) = 80

$\red\Rightarrow$ 2(6+2x) = 80

$\red\Rightarrow$ 12x+4x = 80

$\red\Rightarrow$ 16x = 80

$\red\Rightarrow$ x = $\large\frac{80}{16}$

$\red\Rightarrow$ x = 5

$\therefore$ The length of rectangle = (6+x) cm

‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎= (6+5) cm

‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎= 11 cm

The breadth of rectangle = 5cm

★$\bf\purple{Length=11cm}$

★$\bf\purple{Breadth=5cm}$

‎

‎_______________

~$\cal\red{HOPE\: IT\: HELPS\:\: UH❗}$

Answer 2

Let the number be "X"

⠀⠀⠀Sum of terms= "S"

⠀⠀⠀No. of terms = "N"

⠀⠀⠀⠀⠀Average = "A"

$average \: = \frac{sum \:of \:the \:terms}{no. \:of \:terms}$

⠀⠀⠀⠀⠀⠀3rd no. =$\frac{x}{3}$

⠀⠀⠀⠀⠀quarter =$\frac{x}{4}$

⠀⠀⠀⠀12th no. =$\frac{x}{12}$

Sum of the terms = one third of X + quarter of X + one twelfth of X

$\color {black} \looparrowright s = \frac{x}{3} + \frac{x}{4} + \frac{x}{12}$

$\color{black} \looparrowright s = \frac{4x + 3x + x}{12} = \frac{8x}{12}$

$\color{black}\looparrowright \: s = \cancel{ \frac{8x}{12} } = \frac{2x}{3}$

$\color{black}\looparrowright$n = 3

$\color{black} \looparrowright a = \frac{2x}{3} \times \frac{1}{3}$

$\color{black}\looparrowright \: a = \frac{2x}{9}$

∴ A.T.Q no. is 56 more than average

i.e.

$\color{black}{\rightarrow} \: 56 + \frac{2x}{9}$

$\color{black}\rightarrow \: x - \frac{2x}{9} = 56$

$\color{black}\rightarrow \: \frac{9x - 2x}{9} = 56$

$\color{black}\rightarrow \: \frac{7x}{9} = 56$

$\color{black}\rightarrow \: 7x = 504$

$\color{red}\rightarrow \: x = 72$