Question

the sum of two digit number is 11 if the number obtained by reversing the order of digit in nine less than the original number find the original number​

Answer 1

Answer:

65

Step-by-step explanation:

Let m and n be the digits of the number. Then:

m+n=11 and

10m+n-9=10n+m

So:

n=11-m

10m+11-m-9=10(11-m)+m

9m+2=110–9m

18m=108

m=6

n=5

The number is 65…………….

Answer 2

${\small{\bold{ \underline{Let \ us \ assume :- }}}}$

• The unit digit of original no. = A
• And, the tens digit of original no. = B
• Then , the original no. = 10B + A

• The new no. = 10A + B

${\small{\bold{\underline{\underline{Given:}}}}}$

• B + A = 11 ___(1)
• And, 10A + B = 10B + A - 9 ___(2)

${\small{\bold{ \underline{ \underline{Solution:}}}}}$

${\small{\tt{ 10A + B = 10B + A - 9}}}$

${\small{\Rightarrow{\tt{ 10A - A = 10B - B -9}}}}$

${\small{\Rightarrow{\tt{9A = 9B - 9 }}}}$

${\small{\Rightarrow{\tt{9A = 9(B-1)}}}}$

${\small{\tt{\Rightarrow{\dfrac{ \cancel{9}A}{ \cancel{9}} = B-1}}}}$

${\small{\Rightarrow{\tt{A = B-1}}}}$

${\small{ \bold{\underline{Substituting \ the \ value \ of A = B -1 \: in \: equation (1) }}}}$

${\mapsto{\small{\sf{B + A = 11 }}}}$

${\mapsto{\small{\tt{B + (B-1) = 11}}}}$

${\mapsto{\small{\tt{B + B-1 = 11}}}}$

${\mapsto{\small{\tt{2B = 11+1 }}}}$

${\mapsto{\small{\tt{2B = 12}}}}$

${\small{\mapsto{\tt{B = \dfrac{ \cancel{12}}{ \cancel{2}}}}}}$

${\mapsto{\small{\tt{B = 6}}}}$

${\small{\bold{Substitute \ the \ value \ of \ B = 6 \ in \ equation \ A = B-1}}}$

${\Rightarrow{\small{\tt{A = 6 - 1 }}}}$

${\Rightarrow{\small{\tt{A = 5}}}}$

Now,

${\leadsto{\small{\sf{The \ original \ number = 10B + A}}}}$

${\small{\bold{ \underline{Substitute\ the \ value \ of\ A \ and \ B }}}}$

${\leadsto{\small{\sf{The \ original\ number = 10(6) + 5 }}}}$

${\leadsto{\small{\sf{The \ original\ number = 60+ 5 }}}}$

${\leadsto{\small{\sf{The \ original\ number = 65 }}}}$

${ \red{\leadsto{\small{\frak{\underline{\underline{ \therefore \ \ \ The \ original \ number \: is \: 65}}}}}}}$