Question

guys please solve this ​

Answer 1

Given :

• Sum of the digits of a two - digit number is 9 .

• If the digits are interchanged then the resulting new number is greater than the original number by 27 .

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To find :

• The original number.

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Concept :

In this question we are given two main points. Following those we would frame up two equations and solving these equations we would get our final answers.

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Important point to note in this question :

A two digit number is always of the form $\small\fbox\green{10m+n }$ where ,

• m is ten's digit

• n is unit's digit

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Assumption :

• Let the digit at ten's place be x

• Let the digit at unit's place be y

• Therefore the original number will be 10x + y

• The digits when interchanged , the number would be 10y + x

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Solution :

Lets frame up the equations -

Following statement 1 :

Sum of the digits of a two - digit number is 9

$\tt\:Ten's~digit~+~Unit's~digit~=9$

$\tt\color{navy}{\implies\:x+y=9}$ --- $\small\fbox{1}$

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According to the statement 2 :

$\tt\:New~Number~=~27~+~Original~Number$

Substituting the assumptions

$\tt\implies\:10y+x~=~27~+~(10x+y)$

Simplifying it

$\tt\implies\:10y+x~=~27~+~10x+y$

Transposing (+10x + y) to LHS it becomes (-10x+y)

$\tt\implies\:10y+x-(10x+y)=27$

Removing the brackets

$\tt\implies\:10y+x-10x-y=27$

Arranging the terms and proceeding with calculations

$\tt\implies\:10y-y+x-10x=27$

$\tt\implies\:9y+x-10x=27$

$\tt\implies\:9y-9x=27$

Taking 9 as common

$\tt\implies\:9(y-x)=27$

Transposing 9 to RHS it goes to the denominator

$\tt\implies\:y-x=\frac{27}{9}$

$\tt\implies\:y-x=\cancel{\frac{27}{9}}$

$\tt\color{navy}{\implies\:y-x=3}$ --- $\small\fbox{2}$

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Now adding equation (i) and equation (ii) -

$\tt\:Equation~(i) ~+~Equation~(ii)$

$\tt\implies\:(x+y) +(y-x) =9+3$

Removing the brackets

$\tt\implies\:x+y+y-x=9+3$

Arranging the terms

$\tt\implies\:x-x+y+y=9+3$

Terms with opposite signs gets cancelled

$\tt\implies\:\cancel{x}-\cancel{x}+y+y=9+3$

$\tt\implies\:y+y=12$

$\tt\implies\:2y=12$

Transposing 2 to RHS it goes to the denominator

$\tt\implies\:y=\frac{12}{2}$

$\tt\implies\:y=\cancel{\frac{12}{2}}$

$\small{\underline{\boxed{\mathrm{\implies\:y~=~6}}}}$ ★

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Plugging the value of y as 6 in equation (i) -

$\tt\:x+y=9$

$\tt\implies\:x+6=9$

Transposing +6 to RHS it becomes -6

$\tt\implies\:x=9-6$

$\small{\underline{\boxed{\mathrm{\implies\:x~=~3}}}}$ ★

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Therefore :

The original number = 10x + y

Plugging both the values of x and y

$\tt\leadsto\:Original~Number~=~10(3)+6$

$\tt\leadsto\:Original~Number~=~30+6$

$\small{\underline{\boxed{\mathrm\red{\leadsto\:Original~Number~=~36}}}}$ ⍟

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Henceforth :

⠀‎ ⠀⠀ ⠀⠀‎ ⠀❒ The original number is $\large{\mathfrak\red{36}}$‎

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Answer 2

Answer:

✌❤ I gave thanks but u didn't followed me