Question

THE SUM OF THE DIGITS OF A TWO DIGIT NUMBER IS 7. THE NUMBER FORMED BY REVERSING THE DIGITS IS 45 MORE THAN THE ORIGINAL NUMBER. FIND THE ORIGINAL NUMBER....( WITH METHOD)

Answer 1

Let the digits be x and y... and so, the original number be 10x + y ( since x in tenth place and y in unit place)

so, given : sum of the digits = x + y = 7 ---------> (A)
                  
reversing the digits means y in tenth place and x in unit place.. so, the reversed number be 10y + x
         
   given: reversed number is 45 more than original number
 
          so, 10y + x = 45 + (10x + y) 
           simplifying :   -9x + 9y = 45       ------------> (B)

solving 2 eqns (A) & (B) ,
 we get x = 1 and y = 6

so the original number is 10x + y = 10(1) + 6 = 16
   & the reversed number is 10y + x = 10(6) + 1 = 61
    & the reversed number 61 is 45 more than original number 16 (61 = 45+16)

Ans: the original number is 16

Answer 2

Answer:

→ The original number is 16 .

Step-by-step explanation:

Let the unit's digit of the original number be x .

And, the ten's digit of the original number be y .

Now, A/Q,

→ Sum of the two digits number is 7 .

∵ x + y = 7 ............(1) .

Original number = 10x + y .

Number obtained on reversing the digits = 10y + x .

A/Q,

→ The number obtained on reversing the digit is 45 more than the original number .

∵ 10x + y + 45 = 10y + x  .

⇒ 10x - x + y - 10y = - 45 .

⇒ 9x - 9y = - 45 .

⇒ 9( x - y ) = - 45 .

⇒ x - y = - 45/9 .

∵ x - y = -5 ...........(2) .

On substracting equation (1) and (2), we get

x + y = 7 .

x - y = -5 .

-  +     +

________

⇒ 2y = 12 .

⇒ y = 12/2 .

∴ y = 6.

On putting the value of 'y' in equation (1),  we get

∵ x + y = 7 .

⇒ x + 6 = 7 .

⇒ x = 7 - 6 .

∴ x = 1 .

Therefore , the original number = 10x + y .

= 10 × 1 + 6 .

= 10 + 6 .

= 16 .

Hence, the original number is 16 .

THANKS

#BeBrainly .