Question

The sum of the digits of a two-digit number is 9. If the digits are interchanged, the number obtained exceeds the original number by 27. Find the number.
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Answer 1

EXPLANATION.

• GIVEN

sum of the digit of the two digit number = 9

the digit are interchange the number obtained

exceeds the original number by 27

=> To find the number,

Let the tens place = x

Let the unit place = y

original number = 10x + y

reversing number = 10y + x

according to the question,

sum of Digit of two digit number = 9

x + y = 9 ...... (1)

the digit are interchange the number obtained

exceeds the original number by 27

10y + x - ( 10x + y) = 27

10y + x - 10x - y = 27

9y - 9x = 27

y - x = 3 ...... (2)

From equation (1) and (2) we get,

2y = 12

y = 6

put the value of y = 6 in equation (1)

we get,

x + 6 =9

x = 3

Therefore,

original number = 10x + y

10(3) + 6 = 36

original number = 36

Answer 2

$\large{\red{\underline{\underline{\bold{Given:-}}}}}$

• The sum of digits in a two digit number is 9

• The number obtained by interchanging
the digits exceeds the original number by 27.

$\large{\blue{\underline{\underline{\bold{To\:Find:-}}}}}$

• The original number

$\large{\green{\underline{\underline{\bold{Solution:-}}}}}$

Let the digit at the ones place be x

Then digit at tens place = (9 - x)

The original number = 10(9 - x) + x

By interchanging the digits

The number obtained = 10x + (9 - x)

According to condition 2:-

$\rightarrow \sf \{10x + (9- x )\} - \{10(9 - x) + x \} = 27$

$\rightarrow \sf (10x + 9 - x )- (90 - 10x+ x ) = 27$

$\rightarrow \sf (9x + 9 )- (90- 9x ) = 27$

$\rightarrow \sf 9x + 9 - 90 + 9x = 27$

$\rightarrow \sf 18x - 81= 27$

$\rightarrow \sf 18x = 27+81$

$\rightarrow \sf 18x = 108$

$\rightarrow \sf x = \dfrac{108}{18}$

$\rightarrow \sf x = 6$

Therefore:-

The ones digit of the number = x = 6

The tens digit of the number = 9-x = 9 - 6 = 3

The number = 10(3) + 6 = 36
Hence:-

$\large\boxed { \sf \purple{ \therefore The \: original \: number \: = 36}}$