Question

the digit of a tow digit number differ by 3 if digit are interchanged and the resulting number is added to the orinigal number we get 121find the original number​

Answer 1

Given :

• The digits of a two digit number differ by 3 .

• If the digits are interchanged and the resulting number is added to the original number we get 121 .

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To find :

• The original number.

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Concept :

In this question there are two criterias given. We would frame two equation following both the criteria . Then after solving the equations we would get our answers.

Thing to know here is :

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

• m is ten's digit

• n is one's digit

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Assumption :

• Let the digit at the ten's place be x

• Let the digit at the one's place be y

• Therefore the original number will be 10x + y

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

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Solution :

Let's frame up the equations -

Following Criteria 1 :

The digits of a two digit number differ by 3

$\sf\:Ten's~digit~-~One's~digit~=~3$

$\tt\color{navy}{\implies\:x-y=3}$ --(i)

Here we assumed x is greater than y [ x > y ]

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Following Criteria 2 :

If the digits are interchanged and the resulting number is added to the original number we get 121 .

$\sf\:Original~number~+~Interchanged~number~=~121$

$\tt\implies\:(10x+y)+(10y+x)=121$

Simplifying it

$\tt\implies\:10x+y+10y+x=121$

Arranging and proceeding with simple calculation

$\tt\implies\:10x+x+y+10y=121$

$\tt\implies\:11x+11y=121$

Taking 11 common from LHS

$\tt\implies\:11(x+y)=121$

Transposing 11 to RHS it goes to the denominator

$\tt\implies\:(x+y)=\frac{121}{11}$

Reducing the fraction in RHS to the lower terms

$\tt\implies\:(x+y)=\cancel{\frac{121}{11}}$

$\tt\color{navy}{\implies\:x+y=11}$ --(ii)

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Now adding the equation (i) and equation (ii) :

Equation (i) + Equation (ii)

• LHS + LHS = RHS + RHS

$\tt\implies\:(x-y)+(x+y)=3+11$

Removing the brackets

$\tt\implies\:x-y+x+y=3+11$

Arranging and proceeding with simple calculation

$\tt\implies\:x+x-y+y=14$

Terms with opposite signs gets cancelled out

$\tt\implies\:x+x-\cancel{y}+\cancel{y}=14$

$\tt\implies\:x+x=14$

$\tt\implies\:2x=14$

Transposing 2 to RHS it goes to the denominator

$\tt\implies\:x=\frac{14}{2}$

Reducing the fraction in RHS to the lower terms

$\tt\implies\:x=\cancel{\frac{14}{2}}$

$\small{\underline{\boxed{\mathrm{\implies~x~=~7}}}}$

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Substituting the value of x as 7 in equation (i) :

$\tt\:x-y=3$ --(i)

$\tt\implies\:7-y=3$

Transposing +7 from LHS to RHS it becomes -7

$\tt\implies\:-y=3-7$

$\tt\implies\:-y=-4$

Negative signs gets cancelled from both sides

$\tt\implies\:\cancel{-}y=\cancel{-}4$

$\small{\underline{\boxed{\mathrm{\implies~y~=~4}}}}$

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Therefore :

The original number = 10x + y

Plugging the values we got for x and y

$\tt\twoheadrightarrow\:Original~Number~=~10(7) +4$

$\tt\twoheadrightarrow\:Original~Number~=~70 +4$

$\small{\underline{\boxed{\mathrm\red{\twoheadrightarrow~Original~Number~=~74}}}}$ ★

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Answer 2

Answer:

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