Question

the sum of the digits of 2 digit number is 9. On reversing its digits the new number obtained is 45 more than the original number find the number​

Answer 1

Answer:

The required two-digit number is 27

Step-by-step explanation:

Let "x" be the digit in the 10s place of the original number

Let "y" be the digits in the 1s place of the original number

Hence, the number is 10x + y

ATQ,

x + y = 9 ....Eqn1

The number obtained by reversing the digits is 10y + x.

ATQ, reversed number is 45 more than original number.

Hence,

10y + x = (10x + y) + 45

10y + x = 10x + y + 45

9y - 9x = 45 ....Eqn 2

Substituting for y from Eqn 1 in Eqn 2, we get:

9(9 - x) -  9x = 45

(9 - x) - x = 5 ......Dividing by 9 throughout

9 - 2x = 5

-2x = -4

x = 2

Using Eqn1, y = 7

x = 2 and y = 7

Original number is 10x + y = 27

Check:

Original number = 27

Sum of digits = 9 .....verified

Number obtained by reversing digits = 72

72 is 45 more than 27 ........verified

Answer 2

Answer:

Required original number is 27.

Step-by-step explanation:

Let "x" be the digit in the 10s place of the original number

Let "y" be the digits in the 1s place of the

original number Hence, the number is 10x + y

According to question,

$x + y = 9....(1)$

The number obtained by reversing the digits is 10y + x

According to question, reversed number is 45 more than original number.

Hence,

$10y + x = 10x + y + 45 \\ 10y - y + x - 10x = 45 \\ 9y - 9x = 45 \\ y - x = \frac{45}{9} \\ y - x = 5....(2)$

We are adding equation (1) and (2),

$x + y + y - x = 5 + 9 \\ 2y = 14 \\ y = \frac{14}{2} \\ y = 7$

We are putting value of y in equation (2),

$7 - x = 5 \\ x = 7 - 5 \\ x = 2$

So,original number is

$(10x + y) = (10 \times 2 + 7) = 27$

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