Question

58. 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

Step-by-step explanation:

let the unit digit be x and tens digit be y

thus the no is 10y+x

we have that x+y= x+5

from here we get that y=5

no formed by reversing the digits is 10x+y

original no is 10y+x

sum of the digits is x+5

10y+x-(x+5)= 10+ 10x+y

we know that value of y is 5

putting value of y we get

50 + x -x -5 = 10 + 10x +5

=50 -10-5-5 = 10x

= 10x = 50- 20

=10x = 30

= x= 3

y= 5

thus the number is 53

plz plz mark as brainliest

Answer 2

$Given:-$

• $Sum \: \: of \: \: digits \: \: is \: \: 5 \: \: more \: \: that \: \: the \: \: unit's \: \: digit.$

$The \: \: difference \: \: between</p><p>\: \: the \: \: original \: \: number \: \: and \: \: the \: \: sum \: \: of \: \: digits \: \: is \: \: 10 \: \: more \: \: than \: \: the \: \: number \: \: formed \: \: by \: \: reversing \: \: the \: \: digits.$

$To \: \: Find:-$

• $Different \: \: between \: \: the \: \: digits.$

$Solutions:-$

• $Let \: \: the \: \: ten's \: \: place \: \: be \: \: x$

• $Let \: \: the \: \: unit's \: \: place \: \: be \: \: y$

$Then, \: \: the \: \: number \: \: {10x} \: + {y}$

$Sum \: \: of \: \: digits \: \: ⇢ \: \: {x} \: + \: {y} \: \: = \: \: {5} \: + \: {y}$

$\: \: \: \: \: ⇢ {x} \: \: = \: \: {5} \: + \: {y} \: - \: {y}$

$\: \: \: \: \: ⇢ {x} \: \: = \: \: {5} \: \: ......(1).$

$So, \: \: ten's \: \: place \: \: digit \: \: = \: \: 5$

$After \: \: reversing \: \: the \: \: digit:-$

• $Number \: \: formed \: \: = \: \: {10y} \: + \: {x}$
• $Original \: \: number \: \: = \: \: {10x} \: + \: {y}$

$According \: \: To \: \: Questions$

$\: \: \: \: \: The \: \: difference \: \: between \: \: the \: \: original \: \: number \: \: and \: \: the \: \: sum \: \: of \: \: digits \: \: is \: \: 10 \: \: more \: \: than \: \: the \: \: number \: \: formed \: \: by \: \: reversing \: \: the \: \: digits.$

$\: \: \: \: \: ⇢(10x \: + \: y) \: - \: (y \: - \: 5) \: = \: \: (10y \: + \: x) \: + \: 10$

$\: \: \: \: \: ⇢ 10x \: + \: y \: - \: y \: - \: 5 \: - \: 10y \: + \: x \: + \: 10$

$\: \: \: \: \: ⇢ 10x \: - \: 5 \: = \: 10y \: + \: x \: + \: 10 \: \: .......(2).$

$Putting \: \: value \: \: of \: \: x \: \: = \: \: 5 \: \: from \: \: (1) \: \: in \: \: Eq. \: \: (2).$

$\: \: \: \: \: ⇢ 10 \: × 5 \: - \: 5 \: = \: 10y \: + \: 5 \: + \: 10$

$\: \: \: \: \: ⇢ 50 \: - \: 5 \: = \: 10y \: + \: 15$

$\: \: \: \: \: ⇢ 45 \: = \: 10y \: + \: 15$

$\: \: \: \: \: ⇢ 10y \: = \: 45 \: - \: 15$

$\: \: \: \: \: ⇢ 10y \: = \: 30$

$\: \: \: \: \: ⇢ y \: = \: 3$

$So, \: \: x \: = \: \: 5 \: \: and \: \: y \: \: = \: \: 3$

$\: \: \: \: \: \: \: Original \: \: number \: \: = \: \: 53$

$Difference \: \: between \: \: the \: \: digits \: \: = \: \: 5 \: - \: 3 \: \: = \: \: 2$

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