Question

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 (it's linear equation in one variable)​

Answer 1

$\dag\:\underline{\sf AnsWer :}$

⌬ First assume that the unit digit number be M

⌬ And ten's digit number be N.

⌬ Original Number = 10M + N

⌬ Interchanged Number = 10N + M

Now we are given that, sum of the digits of a two-digit number is 11, mathematically :

$\longrightarrow\:\:\sf M + N = 11 \: \cdot\cdot\cdot\cdot(i)$

$\bigstar\:\underline{\textsf{ According to the Question Now :}}$

• When we interchange the digits it is found that the resulting new number is greater than the original number by 63. Mathematically :

$\longrightarrow\:\:\sf (10M + N) = (10 N +M ) + 63$

$\longrightarrow\:\:\sf (10M + N) - (10 N +M ) = 63$

$\longrightarrow\:\:\sf 10M + N - 10 N - M = 63$

$\longrightarrow\:\:\sf 9M - 9N = 63$

$\longrightarrow\:\:\sf 9(M - N) = 63$

$\longrightarrow\:\:\sf M - N = \frac{ 63}{9}$

$\longrightarrow\:\:\sf M - N = 7$

$\longrightarrow\:\:\sf M = 7 + N \cdot\cdot\cdot\cdot(ii)$

$\bigstar\:\underline{\sf{Substitue \: the \: value \: of \: M \: from \: eq^n \: (ii) \: to \: eq^n \: (i) : }}$

$\longrightarrow\:\:\sf 7 + N + N = 11$

$\longrightarrow\:\:\sf 7 + 2 N = 11$

$\longrightarrow\:\:\sf 2N = 11 - 7$

$\longrightarrow\:\:\sf 2N = 4$

$\longrightarrow\:\:\sf N = 4 \div 2$

$\longrightarrow\:\: \underline{ \boxed{\sf N = 2}}$

$\bigstar\:\underline{\sf{Substitue \: the \: value \: of \: N = 2 \: in \: eq^n \: (i) : }}$

$\longrightarrow\:\:\sf M + 2 = 11$

$\longrightarrow\:\:\sf M = 11 - 2$

$\longrightarrow\:\: \underline{ \boxed{\sf M = 9 }}$

$\bigstar\:\underline{\textsf{Original Number: }}$

$:\implies\sf Original Number = 10M + N$

$:\implies\sf Original Number = 10(9) + 2$

$:\implies\sf Original Number =90+ 2$

$:\implies \underline{ \boxed{\sf Original Number =92 }}$

Answer 2

Answer:

29

Step-by-step explanation:

let the two digits be x and y.

1.) When they are kept in the first number,x become 10x because it is in tens place and y would be same as y.

Then the number is 10x + y

Hence according to the question, sum of those two numbers is 11 :-

• (x + y) = 11-----------(1)

2.) Now,when they are interchanged they are 63 more than the original number.

• (10y + x) = (10x + y) + 63
• 10y + x - 10x - y = 63
• 9y - 9x = 63
• 9(y - x) = 63
• y - x = 7-----------(2)

Add equation 1 and 2

• x + y = 11
• -x +y = 7

------------

• 2y = 18
• y = 9

substitute"y" in equation (1)

• x + y = 11
• x + 9 = 11
• x = 11-9
• x = 2

we have assumed the number as 10x + y,

substitute the values of x and y in it. we get :

10(2)+9 = 29

verification =>

• 2+9 = 11.

reverse them,they must be 63 more than original number :-

• 29 --> 92 = 92-29 = 63.

Hence proved.

Hope it helps