Question

the sum of the digits of a two-digit number is 11 the number obtained by interchanging the digits exceeds the original number by 27. find the number​

Answer 1

$\mathfrak{\large{\underline{\underline{Answer:-}}}}$

The required number is 47.

$\mathfrak{\large{\underline{\underline{Explanation:-}}}}$

Given :- Sum of digits of a number = 11

Number obtained by interchanging digits exceeds the original number by 27

To find : Required number

Solution :-

Let the digits of a two digit number be x and y

Sum of the digits of a two digit number = 11

⇒ x + y  = 11

⇒ x = 11 - y ------(1)

Original number = 10(x) + y = 10(11  - y) + y = 110 - 10y + y = 110 - 9y

New number when digits are interchanged = 10(y) + x = 10y + (11 - y) = 10y + 11 - y = 9y + 11

Given that Number obtained by interchanging digits exceeds the original number by 27

It means difference between number obtained by interchanging digits is equal to 27

According to the question :-

Equation formed :-

⇒ 9y + 11 - (110 - 9y) = 27

⇒ 9y + 11 - 110 + 9y  = 27

⇒ 18y - 99 = 27

⇒ 18y = 27 + 99

⇒ 18y = 126

⇒ y = 126/18

⇒ y = (126 ÷ 18) / (18 ÷ 18)

⇒ y = 7

Now substitute y = 7 in eq(1)

⇒ x = 11 - 7

⇒ x = 4

Now to find the number substitute value of x and y in 10x + y

⇒ 10(4) + 7

⇒ 40 + 7

⇒ 47

$\mathfrak{\large{\underline{\underline{Verification:-}}}}$

⇒ 74 - 47 = 27

⇒ 27 = 27

Therefore the required number is 47.

Answer 2

Answer:

that would be the area of the inner cylinder, plus the area of the outer cylinder, plus the area of the two rings at the end. So, if

R = outer radius

r = inner radius

h = length of pipe

the total area is

2πrh + 2πRh + π(R^2-r^2)

= 2πh(r+R) + π(R+r)(R-r)

= π(R+r)(2h+R-r)