Question

The sum of digits of two digit Number is
7. If the digits are reversed and
resulting number is decreased by 2
twice the original number is obtained
find the original no.​

Answer 1

Answer:

83

Step-by-step explanation:

Let the tens place be x

let the unit place be x-7

A/Q

No. 10*x+(7-x)

=10x +7-x= 9x+7

R.no=10*(7-x)+x

=70-9x+x

=9x+7 =70-9x-2

=18x=70-8

=18x=2

therefore,=18÷2 =9

No.= 9x+7=9×9+2

=81+2 =83Ans..

I hope it is useful for you...

Thanks...

Answer 2

Ans

Let 10th digit no be x

and unit digit be y .

so original no will be=(10x+y) reversed no will be=(10y+ x)

Case 1

x+y=7

x=7-y-------------eq.1

Case 2

(10y+x) - 2 = 2(10x+y)

10y+x - 2=20x +2y

10y- 2y + x - 20x -2 =0

8y- 19x - 2=0

Put the Value of x

8y - 19(7- y) - 2=0

8y - 133 +19y - 2=0

27y -135=0

27y=135

y=135/27

y=5

Put the Value of y in eq.1

x= 7- y

x= 7-5

x= 2

Original no= 10x+y

= 10×2+5

=25

Reversed no =10y+x

=10×5+2

= 52