Question

the sum of digit of a two digit number is 15 if the new number formed by changing the place of the digit is greater than the original number by 9 find the original number and also verify the solution​

Answer 1

Assumption

Tens digit be p

Unit digit be y

Required Number = 10p + y

p + y = 15 ........ (1)

Now,

Reversing Number = 10y + p

10y + p = 10p + y + 9

10p + y + 9 = 10y + p

10p - p + 9 = 10y - y

9p + 9 = 9y

9p - 9y = - 9

9(p - y) = -9

$\tt{\rightarrow p-y=\dfrac{-9}{9}}$

p - y = -1  .......... (2)

Add (1) and (2) we get,

2p = 14

$\tt{\rightarrow p=\dfrac{12}{2}}$

p = 7

Put value of p in (2)

p - y = -1

7 - y = -1

-y = -1 - 7

-y = -8

y = 8

Tens digit p = 7

Unit digit = 8

Required Number

= 10p + y

= 10 × 7 + 8

= 78

Verification:

Sum of the digits is 15

7 + 8 = 15

Interchanging the digits then no. formed original number by 9

87 = 78 + 9

(Verified)

Answer 2

$\huge\star{\tt{\underline{\underline{\red{Answer}}}}}\star$

Given:

Sum of Digits =15

Sum of Digits when place has been changed

= Original number + 9

Solution:

Let The Ten's digit be x and one's digit be y

Number formed are

✰ 10x + y and 10y +x

So the Linear Equation made by his given information will be as follows

$\implies$ x + y = 15 _______✰[1]

Also,

When condition is applied we get,

$\implies$ 10x + y + 9= 10y + x

$\implies$ 9x - 9y = -9

Here,

When divided by 9 at both side

We get

$\implies$ x - y = -1__________✰[2]

From [1] and [2]

By substitution method,

x = 15 - y _________✰[3]

Now substituting x in [2]

(15-y)- y = -1

$\implies$ 15 - y - y = -1

$\implies$ 15 +1 = 2y

$\implies$y= $\frac{16}{2}$

$\implies$ $\large{\boxed{y=8}}$

In [3] substituting y

x= 15 - (8)

$\implies$ $\large{\boxed{x = 7}}$

Number formed= 10x + y

$\huge\implies$$\large{\boxed{\boxed{78}}}$

Verifying the solution

Sum of the digits = x + y

$\implies$ 7 + 8 = 15

When the digits interchange condition applied

87 = 78 +9

i.e True

So hence the Verified