Question

.
The digits of a two-digits numben differ by 3
digits are interaged and the resulting number
is added to the original number we get 121 find
the original number​

Answer 1

ANSWER:

• Original number is 74 and 47.

GIVEN:

• Digits of a two-digits numben differ by 3.
• Sum of interchanged digit and original number = 121

TO FIND:

• The original number

SOLUTION:

Let the digit at tens place be 'x'

Let the digit at once place be 'y'

Original number = 10x+y

Reversed number = 10y+x

According to Question:

Case 1

=> 10x+y+10y+x = 121

=> 11x+11y = 121

=> 11(x+y) = 121

=> x+y = 121/11

=> x +y = 11 ....(i)

Case 2

When digit at tens place is greater than once place then:

=> x-y = 3. ....(ii)

Adding eq (i) and (ii) we get:

=> x+y+x-y = 11+3

=> 2x = 14

=> x = 14/2

=> x = 7

Putting x = 7 in eq(i) we get:

=> 7+y=11

=> y = 11-7

=> y = 4

Original number = 10*(7)+4

=70+4

=74

Case 3

when the digit at once place is greater than tens place:

=> y-x = 3 ....(iii)

Adding eq (I) and (iii) we get;

=> 2y = 14

=> y = 7

putting y = 7 in eq(I)

=> x+7 = 11

=> x = 4

Original number = 10*(4)+7

= 40+7

=47