Question

the sum of the digit of two digit number is 11. The number is less than the number obtained by interchanging the digit by 9. find the number​

Answer 1

Answer:

Question

• ✳️Let x be the unit digit of the number
• ✳️Then the ten's digit =11-x
• ✳️Therefore, number=10(11-x)+x

✍️110-10x+x=100-9

• ✳️ when digit are interchanged unit digit=11-1 and ten's digit =x
• ✳️ So number obtained by interchanging the digits = 10x+(11-x)=9x+11

As per the question

✍️9x+11-(110-9x)=9

✍️9x+11-110+9x=9

✍️18x+11-110=9

✍️18x=9-11+110

✍️18x=108

$= = > x = \frac{108}{18} = \frac{6}{1} = 6$

✍️ Hence,unit digit = 6

✍️And ten's digit=11-6=5

➡️ Therefore, the required number is 56

Step-by-step explanation:

hope you appreciate this ans

Answer 2

Let X be The Unit digit of Number.

Then,

$The \: Tens \: Digit = 11 - x$

$\huge\mathfrak\red{☟ \: \: \: answer \: \: \: \: ✎}$

Therefore,

Number

$= 10(11 - x) + x = 110 - 10x + x = 110 - 9x$

When digits are interchanged,

$unit \: digit = 11 - x$

‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎And

$tens \: digit = x$

So Number Obtained by interchanged the digit

$= 10x + (11 - x) = 9x + 11$

$\sf \colorbox{pink} {\red(As Per the Question-}$

$9x + 11 - (110 - 9x) = 9$

$⇒9x + 11 - 110 + 9x = 9$

$⇒18x + 11 - 110 = 9$

$⇒18x = 9 - 11 + 110$

$⇒18x = 108$

$⇒x = 6$

Hence,

$units \: digit = 6$

‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎And

$tens \: digit = 11 - 6$

$= 5$

Therefore the Required Number is 56

$\\ \\ \\ \\ \sf \colorbox{gold} {\red(ANSWER ᵇʸ ⁿᵃʷᵃᵇ⁰⁰⁰⁸} \\ and\\ \sf \colorbox{lightgreen} {\red❤ANSWER ᵇʸ ᶠˡⁱʳᵗʸ ᵇᵒʸ}$