Question

6.
The sum of the digits of a 2-digit number is 10. The number obtained by
interchanging the digits exceeds the original number by 36. Find the original
number.​

Answer 1

Let the number at unit's place be x and that of ten's place be y.

Therefore, The Original number=10y+x

From First condition,

x+y=10............ (1)

From Second Condition,

10x+y=10y+x+36

10x-x+y-10y=36

9x-9y=36

x-y=4......... (2)

Adding (1) and (2) we get,

2x=14

x=14/2

x=7

Substituting the value of x in equation (1)

7+y=10...........(1)

y=10-7

y=3

Therefore, the original number,

10y+x

=10(3) +7

=30+7

=37

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Answer 2

${\huge{\bold{\purple{\bigstar{\boxed{\boxed{\bf{Question}}}}}}}}$

The sum of the digits of a 2-digit number is 10. The number obtained by

interchanging the digits exceeds the original number by 36. Find the original

number.

${\huge{\bold{\purple{\bigstar{\boxed{\boxed{\bf{solution}}}}}}}}$

let the no. be 10x + y

${\bold{\blue{\boxed{\bf{Given }}}}}$

• x + y = 10
• 10x + y - 10y + x = 36

${\bold{\blue{\boxed{\bf{Find:- }}}}}$

• the correct number

${\bold{\blue{\boxed{\bf{solution }}}}}$

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

10x + y - (10y +x) = 36

10x + y -10y -x = 36

9x - 9y = 36

9(x-y)=36

x-y = 4 ------------(ii)

• add eq (i) and eq (ii)

x + y + x - y = 10 + 4

2x = 14

x = 7

• putting value of x in eq (i)

x+y = 10

7 + y = 10

y = 3

correct no. = 10x + y = 10*7+3 = 70 + 3 = 73

${\bold{\blue{\boxed{\bf{correct \: no. = 73}}}}}$

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