Question

Sum of the digits of a two-digit number is 9 by interchanging the place of the digits the number reduced by 63 find the number

Answer 1

☞ To Find :

➝ The original number...

☞ Taken :

Let the digits of number be a and b.

So, the two digit number formed is (10a + b) and the number obtained on reversing the digits is (10b + a).

☞ Given :

The sum of digits is 9.

so ,the equation formed is :

$\bold{\mathtt{\therefore a + b = 9}}$(Equation...i)

• Difference of the the original no. and the no. obtained on reversing the digits is 63.

☞ Concept :

ATQ ,

In the question , it says that when the digits of original no. are interchanged, the Original no. is reduced by 63.

Hence ,

• Original no. = 10 a + b.
• New no. = 10b + a.

10b + a = 10a + b - 63

By solving ,the equation, we get :

$\mathtt{10b + a = 10a + b - 63}$

$\mathtt{\Rightarrow 10a + b - (10b + a) = 63}$

$\mathtt{\Rightarrow 10a + b - 10b - a = 63}$

$\mathtt{\Rightarrow 9a - 9b = 63}$

Taking the common numeral 9 ,we get the equation as ,

$\mathtt{\Rightarrow 9(a - b) = 63}$

$\mathtt{\Rightarrow a - b = \dfrac{63}{9}}$

$\mathtt{\Rightarrow a - b = \dfrac{\cancel{63}}{\cancel{9}}}$

$\mathtt{\Rightarrow a - b = 7}$

$\bold{\mathtt{\therefore a - b = 7}}$(Equation...ii)

☞ Solution :

Equation (i)...

$\mathtt{a + b = 9}$

Equation (ii)...

$\mathtt{a - b = 7}$

By solving the two equations linearly , we get :

$\mathtt{a + \cancel{b} = 9}$

$\mathtt{a - \cancel{b} = 7}$

_______(By adding)

$\mathtt{2a = 16}$

By solving it , we get :

$\mathtt{\Rightarrow 2a = 16}$

$\mathtt{\Rightarrow a = \dfrac{16}{2}}$

$\mathtt{\Rightarrow a = \dfrac{\cancel{16}}{\cancel{2}}}$

$\mathtt{\Rightarrow a = 8}$

Putting the value of a in the equation...i ,we get :

$\mathtt{a + b = 9}$

$\mathtt{\Rightarrow 8 + b = 9}$

$\mathtt{\Rightarrow b = 9 - 8}$

$\mathtt{\Rightarrow b = 1}$

Hence , the value of a is 8 and the value of b is 1 .

We know the original no. i.e 10a + b.

Putting the value of a and b in the equation , we can find the original number.

$\mathtt{10a + b}$

$\mathtt{\Rightarrow 10(8) + 1}$

$\mathtt{\Rightarrow 80 + 1}$

$\mathtt{\Rightarrow 81}$

Hence ,the original number is 81.

☞ Verification :

• Sum of the digits of the no. is 9

Original no .= 81

Sum of digits = 8 + 1 → 9

Proved :

• Difference of original and no. obtained on reversing the digits is 63.

Original no = 81

No. obtained on reversing the digits = 18

ATQ

$\mathtt{\Rightarrow 10a + b - (10b + a) = 63}$

Putting the value of (10a + b) and (10b + a) in the equation , we get:

$\mathtt{\Rightarrow 81 - 18 = 63}$

$\mathtt{\Rightarrow 63 = 63}$

LHS = RHS

Proved :

Answer 2

$\huge{\sf{\underline{\red{Answer :- }}}}$

$\looparrowright$

$\small{\sf{\underline{\orange{According to Question:}}}}$

• Find the original number.

• sum of the digits of the original number = 09

• after interchanging the digits , number is reduced by = 63

$\small{\sf{\underline{\pink{Considering :-}}}}$

➜ let the original number be XY

➜ hence , number after reversing the digits : YX

$\small{\sf{\underline{\orange{\boxed{Equations\: formed}}}}}$

• 1) X + Y = 09

• 2) 10Y + X = 10X + Y

➜ rearranging the equation 2, we get....

• 10X + Y - 10Y - X = 63..

➜ by solving equation 2 we get .....

• 9x - 9y = 63

• x - y = 7 (after dividing the equation by 9)
• x = 08

➜put the x in equation 1 we get...

• y = 9 - 1 = 8..

➜ put the value of X and Y

• 10 × 8 + 1 = 80 + 1 = 81

Original number = 81..

Verification :-

➜ ATQ, difference between original number and number after reversing it's digits is = 63

• original number = 81

• number after reversing the digits = 18

➜ equation = 81 - 18 = 63...

• 81 - 18 = 63

• 63 = 63

➜ LHS = RHS...

➜HENCE, VERIFIED.....