Question

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In a two-digit number, the sum of the digits is 5 more than the units digit. The difference between the original number and the sum of digits is 10 more than the number formed by reversing the digits. Then find the difference between the digits.

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Answer 1

Let :-

• The ones digit of the number = x
• The tens digit of the number = y
• The original number = 10y + x
• The reversing number = 10x + y

To Find :-

• The difference between ones digit and tens digit = ?

Solution :-

• To calculate the difference between its digit at first we have to set up equation by helping clue in question.

Calculation for 1st equation :-

⇒ Sum of the digits = 5 more then ones digit

⇒ x + y = x + 5

⇒ y = 5----------(i)

Calculation for 2nd equation :-

⇒ Orginal number - (sum of digits) = reversing no + 10

⇒ 10y + x - (x + y) = 10x + y + 10

⇒ 10y + x - x - y = 10x + y + 10

⇒ 9y = 10x + y + 10

⇒ 10x + y - 9y = - 10

⇒ 10x - 8y = - 10 ----------(ii)

• Putting y = 5 in eq (ii) here we get :-

⇒ 10x - 8(5) = - 10

⇒ 10x - 40 = - 10

⇒ 10x = - 10 + 40

⇒ 10x = 30

⇒ x = 3

• Now calculate difference here :-

⇒ Difference (tens - one's) digit

⇒ y - x

⇒ 5 - 3

⇒ 2

Hence, the required difference is 2 :-

Answer 2

Given : In a two-digit number, the sum of the digits is 5 more than the units digit. The difference between the original number and the sum of digits is 10 more than the number formed by reversing the digits. Then find the difference between the digits.

_______________________________

$\frak{Let \: us \: assume} \rightarrow \begin{cases} \frak{The \: units \: digits \: be \: \green{a}} \\ \\ \frak{The \: tens \: digit \: be \: \green{b}}\end{cases}$

Hence, we get

$\implies \frak{ a + b = 5 + b} \\ \\ \implies \frak{ \pink{a = 5}}$

After which the original number is

$\dashrightarrow \frak{ \frak{Original \: Number : \underline{10a + b}}}$

Reversing the digit we will get the new number

$\dashrightarrow \frak{New \: Number = \underline{10b + a}}$

Now, according to the question putting all the required values to get the equation

$\rightharpoondown \frak{10a + b - (a + b) = 10 + 10b + a} \\ \\ \rightharpoonup \frak{10a + b - a - b = 10 + 10b + a} \\ \\ \rightharpoondown \frak{9a + \cancel b - \cancel b = 10 + 10b + a} \\ \\ \rightharpoonup \frak{9a - a = 10 + 10b} \\ \\ \rightharpoondown \frak{8(5) - 10 = 10b} \\ \\ \rightharpoonup \frak{40 - 10 = 10b} \\ \\ \rightharpoondown \frak{10b = 30} \\ \\ \star \quad \underline{ \boxed{ \frak{ \purple{b = 3}}}}$

$\dag \: \: \underline{ \frak{As \: we \: know \: that} }:$

$\frak{ \blue{a = 5 } \: and\: \pink{b = 3}} \\ \\ \therefore \sf Difference \: between \: the \: numbers = \frak{5 - 3 =\red{2}}$