Question

The digits of a two-digit number differ by 3. If digits are interchanged  and the resulting number is added to the original number, we get 121.  Find the original number.​

Answer 1

• To find : Original number

Solution

• Given Conditions

★ The digits of a two-digit number differ by 3.

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

• According to the first condition

Consider the two numbers be x and y. These two numbers are differ by 3

Eqⁿ (1) : x - y = 3

• According to the second condition

Original number = 10x + y

Reversed or interchanged number = 10y + x

→ Interchanged number + original number = 121

→ 10y + x + 10x + y = 121

→ 11y + 11x = 121

→ 11(x + y) = 121

→ x + y = 11

Eqⁿ (2) : x + y = 11.

• Add both the equation

→ x + y + x - y = 11 + 3

→ 2x = 14

→ x = 7

• Substitute the value of x in eqⁿ (1)

→ x - y = 3

→ 7 - y = 3

→ y = 7 - 3

→ y = 4

•°• Original number = 10x + y = 74

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Answer 2

• $\Large\boxed{\sf{\blue{Original\:number = 74}}}$

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Explanation :

$\underline{\bf\dag{\underline{\green{Given}}}:}$

• The digits of a two-digit number differ by 3.
• If digits are interchanged and the resulting number is added to the original number, we get 121.

$\underline{\bf\dag{\underline{\green{To\:Find}}}:}$

• Original number = ?

$\underline{\bf\dag{\underline{\green{Solution}}}:}$

• Let the two numbers be m and n.

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$\underline{\bf\dag{\underline{\green{In\:1^{st}\:case}}}:}$

Atq,

$\qquad\leadsto\quad\sf m - n = 3\qquad\qquad \bigg\lgroup eq^{n}\:(1) \bigg\rgroup$

$\underline{\bf\dag{\underline{\green{In\:2^{nd}\:case}}}:}$

$\qquad\bf\dag\:\bigg\lgroup \sf{ Original\:number\:=\:10m + n} \bigg\rgroup$

$\qquad\bf\dag\:\bigg\lgroup \sf{ Interchanged\:number\:=\:10n + m} \bigg\rgroup$

Atq,

$\qquad\leadsto\quad\small\sf Original\:number + Interchanged\:number = 121$

$\qquad\leadsto\quad\sf 10m + n + 10n + m = 121$

$\qquad\leadsto\quad\sf 11m + 11n = 121$

$\qquad\leadsto\quad\sf 11(m + n) = 121$

$\qquad\leadsto\quad\sf m + n = \dfrac{121}{11}$

$\qquad\leadsto\quad\sf m + n = \dfrac{\cancel{121}}{\cancel{11}}$

$\qquad\leadsto\quad\sf m + n = 11\qquad\qquad \bigg\lgroup eq^{n}\:(2) \bigg\rgroup$

Now,

★ Adding [eqⁿ (1)] and [eqⁿ (2)] :-

$\qquad\leadsto\quad\sf (m - n) + (m + n) = 3 + 11$

$\qquad\leadsto\quad\sf m - n + m + n = 14$

$\qquad\leadsto\quad\sf m - \cancel{n} + m + \cancel{n} = 14$

$\qquad\leadsto\quad\sf m + m = 14$

$\qquad\leadsto\quad\sf 2m = 14$

$\qquad\leadsto\quad\sf m = \dfrac{14}{2}$

$\qquad\leadsto\quad\sf m = \dfrac{\cancel{14}}{\cancel{2}}$

$\qquad\leadsto\quad\bf{ m = \red{7}}$

★ Putting value of "m" in [eqⁿ (2)] :-

$\qquad\leadsto\quad\sf 7 + n = 11$

$\qquad\leadsto\quad\sf n = 11 - 7$

$\qquad\leadsto\quad\bf{ n = \red{4}}$

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$\small\therefore\:{\underline{\sf{Hence,\:original\:number\:=\:\bf{74}\:\sf{respectively.}}}}$

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