Question

given that 45 is a difference between a number and that formed by reversing its digit at the sum of the digit of the number is 13 then find the number​

Answer 1

$\mathfrak{\large{\underline{\underline{Answer :-}}}}$

The Original Number is 94.

$\mathfrak{\large{\underline{\underline{Explanation :-}}}}$

$\text{\underline{\underline{\purple{Given :}}}}$

Difference between the Original Number and the Reversed Number = 45

Sum of the digits of the number = 13

$\text{\underline{\underline{\purple{To Find :}}}}$

The Number

$\text{\underline{\underline{\purple{Solution :}}}}$

Original Number -

• Let the Units Place be x
• Tens Place be 10(13 - x)

The Original Number =

$\longrightarrow$ 10(13 - x) + x

$\longrightarrow$ 130 - 10x + x

$\longrightarrow$ 130 - 9x ........[ 1 ]

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Number with Reversed Digits -

• Units Place will be (13 - x)
• Tens place 10(x)

Number with Reversed Digits =

$\longrightarrow$10x + (13 - x)

$\longrightarrow$ 10x - x + 13

$\longrightarrow$ 9x + 13 ........[ 2 ]

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According to the Question ;

Difference between the Original Number and Reversed Number is 45.

★ $\boxed{\sf{130-9x-(9x+13)=45}}$

$\sf{\longrightarrow} \:130-9x-(9x+13)=45$

$\sf{\longrightarrow} \:130-9x-9x - 13=45$

$\sf{\longrightarrow} \:130 - 18x- 13=45$

$\sf{\longrightarrow} \: - 18x + 117=45$

$\sf{\longrightarrow} \: - 18x = 45 - 117$

$\sf{\longrightarrow} \: \cancel- 18x = \cancel- 72$

$\sf{\longrightarrow} \:x = \dfrac{72}{18}$

$\sf{\longrightarrow} \:x = 4$

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★ Value of 10(13 - x)

$\sf{\longrightarrow} \:10(13 - 4)$

$\sf{\longrightarrow} \:130 - 40$

$\sf{\longrightarrow} \:90$

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The Number =

$\sf{\longrightarrow} \:90 + 4$

$\sf{\longrightarrow} \:94$

$\therefore$ The Original Number is 94.

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$\mathfrak{\large{\underline{\underline{Verification :-}}}}$

By Reversing the Digits of 94 the Number is 49. If their difference is 45, then the answer is Correct.

$\sf{\longrightarrow} \:94 - 49 = 45$

$\sf{\longrightarrow} \:45 = 45$

$\therefore$ The Original Number is 94.

Answer 2

$\huge\red{Solution:-}$

According to the given question:-

• 45 is the sum of difference between a number and that formed by reversing its Digit.

• The number 13 is the sum of given digits.

Here, we need to find the number:-

• Let y be the Units Place.
• Let 10(3-y) be in Ten's Place.

Here, is the Original number :-

$= > 10(13 - y) + y$

$= > 130 - 10y + y$

$= 130 - 9y \: \: - - equation - (1)$

Reversing Digits:-

$= > 10y + (13 - y)$

$= > 10y - y + 13$

$= > 9y + 13 - - equation(2)$

Then:-

$= > 130 - 9y - (9y + 13) = 45$

$= > 130 - 9y - 9y(13 = 45)$

$= > 130 - 18 - 13 = 45$

$= > - 18y + 117 = 45$

$= > - 18y = 45 - 117$

$= > - 18y = - 72$

$= > y = \frac{72}{18}$

$= > y = 4$

Here, the value of 3:-

$= > 10(13 - 4)$

$= > 130 - 40$

$= > 90$

So, the number is:-

$= > 90 + 4 = 94$

Therefore, 94 is the original number