Question

The sum of the digits of a two digit number is 7.on reversing it's digits,the new number is 45 less than the original number.find the number.​

Answer 1

Answer:

Let the number be 10x+y also given that x+y = 7 so, x = 7-y ---(1)

also, 10y + x = 10x+y +45 9y = 45 +9x y = 5+x ----(2)

putting the value of (1) in (2)

so, y = 5+ (7-y) => y = 5+7-y => y = 6

=> y +y = 12

=> 2y = 12

so, x = 7-y = 7-6 => x = 1

Hence for original number we must put value of x and y that is 1 and respectively... Therefore 10x+y=10x1+6=16

so, the number is 16.

Hope that help you

Answer 2

Step-by-step explanation:

Solution :

Let,

• Ones digit = x

• Tens digit = 7 - x

★ Original number =

10 (7 - x) + x

70 - 10x + x

70 - 9x

On reversing it's digits :

10 (x) + 7 - x

10x + 7 - x

9x + 7

★ According to the Question :

On reversing it's digits, the new number is 45 less than the original number

(70 - 9x) - 45 = 9x + 7

70 - 45 - 9x = 9x + 7

25 - 9x = 9x + 7

25 - 7 = 9x + 9x

18 = 18x

x = 18/18

x = 1

Ones digit = 1

• Tens digit = 7 - x

7 - 1 = 6

Tens digit = 6

The number = 61

Therefore, The number is 61