Question

LLLLLL
. Sum of the digits of a two digit number is 11. When we interchange the digits, it is
found that the resulting new number is greater than the original number by 63. Find
the two digit number.
​

Answer 1

Answer:

let x be the unit digit number,then men's digit number=11-x(since sum of the digits of a two digits number is 11)

---two digit number(original)=10(11-x)

=10x+11-x

=9x+11

according to the condition in the question, we get

9x+11-(110-9x)=63

---18x=99+63

---18x=162

---x=9

therefore the required two digit number=110-9x

---110-9×9

---110-81

---29

Answer 2

$\huge\sf\pink{Answer}$

☞ The Number is 29

━━━━━━━━━━━━━

$\huge\sf\blue{Given}$

✭ Sum of a two digit number is 11

✭ When the number is interchanged we get a number that is 63 greater than the real one

━━━━━━━━━━━━━

$\huge\sf\gray{To \:Find}$

◈ The two digit number?

━━━━━━━━━━━━━

$\huge\sf\purple{Steps}$

$\large\underline{\underline{\sf Let }}$

◕ Original Number be 10x+y

◕ After Reversing it will be 10y+x

☯ $\underline{\textsf{As Per the Question}}$

Sum of the digits of the number is 11

⪼ $\sf x+y= 11 \:\:\: -eq(1)$

If the number is interchanged then the new number is 63 greater than the original number

➝ $\sf (10x+y)-(10y+x) = 63$

➝ $\sf 10x+y-10y-x = 63$

➝ $\sf 9x-9y = 63$

➝ $\sf 9(x-y) = 63$

➝ $\sf x-y = \dfrac{63}{9}$

➝ $\sf x-y = 7\:\:\: -eq(2)$

Subtracting eq(2) from eq(1)

»» $\sf (x+y)-(x-y) = 11-7$

»» $\sf x+y-x+y=4$

»» $\sf 2y = 4$

»» $\sf y = \dfrac{4}{2}$

»» $\sf \green{y = 2}$

Substituting the value of y in eq(1)

➢ $\sf x+y=11$

➢ $\sf x =11-y$

➢ $\sf x = 11-2$

➢ $\sf \red{x = 9}$

So the number will be,

➳ $\sf Original \ Number = 10x+y$

➳ $\sf Original \ Number = 10(9)+2$

➳ $\sf Original \ Number = 90+2$

➳ $\sf \orange{Original \ Number = 92}$

━━━━━━━━━━━━━━━━━━