Math, asked by hemantlah07, 7 months ago

Sum of the digits of a two-digit number
is 5. When we interchange the digits, it is
found that the resulting new number is
less than the original number by 27.
What is the two-digit number?

Answers

Answered by ButterFliee
23

GIVEN:

  • Sum of the digits of a two-digit number is 5.
  • When we interchange the digits, the resulting new number is less than the original number by 27.

TO FIND:

  • What is the original number ?

SOLUTION:

Let the digit at ten's place be 'x' and the digit at unit's place be 'y'

NUMBER = 10x + y

CASE:- 1)

Sum of the digits of a two-digit number is 5.

According to question:-

x + y = 5....

x = 5 –y

CASE:- 2)

When we interchange the digits, the resulting new number is less than the original number by 27.

Reversed Number = 10y + x

Reversed Number = Original Number 27

According to question:-

10y + x = 10x + y –27

27 = 10x + y –10y –x

27 = 9x –9y

Take common 9 from both sides

3 = x y.... 

Put the value of 'x' from equation 1) in equation 2)

3 = 5 –y –y

3 = 5 –2y

3 –5 = –2y

–2 = –2y

y = \sf{\cancel\dfrac{-2}{-2}}

y = 1

Put the value of 'y' in equation 1)

x + 1 = 5

x = 5 –1

x = 4

NUMBER = 10x + y

NUMBER = 10(4) + 1

NUMBER = 40 + 1

NUMBER = 41

Hence, the number formed is 41

______________________


VishalSharma01: Awesome :)
Answered by Anonymous
90

Given : Sum of the digits of a two-digit number is 5. When we interchange the digits, it is found that the resulting new number is less than the original number by 27.

\rule{130}{1}

Solution :

Let the digits at tens place and ones place be x and 5 - x respectively.

\therefore\:{\underline{\sf Original\: number\::}}\\\\\\:\implies\sf 10x + (5 - x)\\\\\\:\implies\sf 50 - 9x\\\\\\:\implies\sf 9x + 5

\rule{130}{1}

Now interchange the digits: digit at one's place and tens place be x and 5 - x respectively.

\therefore\:{\underline{\sf New\: number\::}}\\\\\:\implies\sf 10(5 - x) + x\\\\\\:\implies\sf 50 - 10x + x\\\\\\:\implies\sf 50 - 9x

\rule{130}{1}

\underline{\boldsymbol{According\: to \:the\: Question\:now :}}

:\implies\sf  New\: number = Original\: number + 27\\\\\\:\implies\sf 50 - 9x = 9x + 5 + 27\\\\\\:\implies\sf 50 - 9x = 9x + 32\\\\\\:\implies\sf 18x = 18\\\\\\:\implies\sf  x = 1

Hence the digit at tens place and ones place of the number are 1 and 5 - x = 5 - 1 = 4.

\therefore\:\underline{\textsf{The two digit number is  50 - 9x = 50 - 9 (1) = \textbf{41}}}.

Peace :)!

\rule{185}{2}


VishalSharma01: Nice :)
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