Question

The sum of the digits of a two-digit number is 15. The number obtained by interchanging its
digits exceeds the given number by 9. Find the original number ​

Answer 1

Answer:

Given :-

• The sum of the digits of a two-digit number is 15.
• The number obtained by interchanging its digits exceeds the given number by 9.

To Find :-

• What is the original number.

Solution :-

Let,

➲ Units digit = x

➲ Tens digit = y

Hence, the original number will be :

⇒ 10x + y

By interchanging its digits we get,

⇒ 10y + x

According to the question,

★ Sum of the two digits of a two-digit number is 15.

↦ x + y = 15 ------ (Equation No ①)

★ The number obtained by interchanging its digits exceeds the given number by 9.

↦ (10y + x) - (10x + y) = 9

↦ 10y + x - 10x - y = 9

↦ 10y - y + x - 10x = 9

↦ 9y - 9x = 9

↦ 9(y - x) = 9

↦ y - x = 9/9

↦ y - x = 1 ------ (Equation No ②)

By adding the equation no 1 and 2 we get,

↦ x + y + y - x = 15 + 1

↦ x - x + y + y = 16

↦ 2y = 16

↦ y = 16/2

↦ y = 8/1

➠ y = 8

Again, by putting y = 8 in the equation no 1 we get,

↦ x + y = 15

↦ x + 8 = 15

↦ x = 15 - 8

➠ x = 7

Hence, the required original number is :

⇒ Original Number = 10x + y

⇒ Original Number = 10(7) + 8

⇒ Original Number = (10 × 7) + 8

⇒ Original Number = 70 + 8

➦ Original Number = 78

∴ The original number is 78.

Answer 2

$\Huge{\texttt{{{\color{Magenta}{⛄A}}{\red{N}}{\purple{S}}{\pink{W}}{\blue{E}}{\green{R}}{\red{♡}}{\purple{࿐⛄}}{\color{pink}{:}}}}}$

$\sf{let \: the \: digit \: at \: unit \: place \: to \: be \: x}$

$\sf{then \: digit \: at \: tens \: place =15-x}$

$\sf{so \: original \: no=10(15-x)+x}$

$\sf \red{ \longrightarrow150-10x+x}$

$\boxed{ \tt{ \red{=150-9x}}}$

$\sf {if \: digit \: get \: interchange \: then;}$

$\sf{and \: at \: tens \: place=x}$

$\sf{so \: new \: no=10(x)+15-x}$

$\sf \pink{ \longrightarrow10x+15-x}$

$\boxed{ \sf \pink{ = 9x+15}}$

$\sf{so \: the \: no \: is : original \: no -new \: no=9}$

$\sf \purple{ \longrightarrow150-9x -(9x+15)=9}$

$\sf \purple{ \longrightarrow150-9x-9x-15=9}$

$\sf \purple{ \longrightarrow135-18x=9}$

$\sf \purple{ \longrightarrow-18x=-135+9}$

$\sf \purple{ \longrightarrow-18x=-126}$

$\boxed{ \sf \purple{x=7}}$

$\tt \red{so \: digit \: at \: unit \: place \: is \: 7}$

$\sf \red{and \: at \: tens \: place \: is \: 8 \: so \: the \: digit \: formed \: is \: 87}$

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$\huge\red{\boxed{\orange{\mathcal{{{\fcolorbox{red}{i}{{\red{@Master}}}}}}}}}$