Question

The sum of the digits of a 2 digit number is 9. On reversing its digits, the new number obtained is 45 more than the original number. Find the number.
$the \: sum \: of \: the \: digits \: of \: a \: 2 \: digit \: number \: is \: 9. \: on \: reversing \: its \: digits \: te \: new \: number \: obtained \: is \: 45 \: more \: than \: the \: original \: number. \: find \: the \: number$

Answer 1

x+y=9(1 eq)
10x+y=10y+x+45(2 eq)
A/Q
9Y-9x=45
x-y=5

x+y=9
x-y=5
_____
2x = 14
x=7
7+y=9
y=2
the new number
20+7=27

Answer 2

☆ Question ☆

The sum of the digits of a 2 digit number is 9. On reversing its digits, the new number obtained is 45 more than the original number. Find the number.

☆ Solution ☆

☆ Given ☆

• The sum of the digits of a 2 digit number is 9. On reversing its digits, the new number obtained is 45 more than the original number.

☆ To Find ☆

• The Number

☆ Step-by-Step-Explaination ☆

Let the number be of the form of 10x + y

Now, Sum of digits is 9.

So, we will write x + y = 9.

Now on reversing the new digit will be 10y + x .

So, 10y + x = 10x + y + 45

9y - 9x = 45

y - x = 5

We have got two equation

x + y = 9 and y - x = 5

On solving these equation we get x = 2 and y = 7.

Hence, the number is 10 × 2 + 7 = 27.