Question

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

Answer 1

Answer:

47

Step-by-step explanation:

$\bf{let \: one \: digit \: number \: be \: x} \\ \\ \sf{so \: numbers \: are \: x \: and \: x + 3} \\ \\ \bf{number \: is \: represented \: as \: 10x + y} \\ 10x + x + 3 + 10(x + 3) + x = 121 \\ \\ 11x + 3 + 11x + 30 = 121 \\ 22x + 33 = 121$

$22x=121-33</p><p>22x=88</p><p>x=4$

$\sf{number \: is \: 47}$

$★\sf{4 \: and \: (4+3)}$

Answer 2

Answer:

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

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.

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

• Original number = ?

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

Let the two numbers be m and n.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

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

Atq,

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

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

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

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

ATQ,

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

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

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

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

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

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

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

Now,

★ Adding [eqⁿ (1)] and [eqⁿ (2)] :-

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

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

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

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

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

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

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

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

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

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

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

$\qquad\leadsto\quad\bf{ n = \bold{ \gray{4}}}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

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