Question

the sum of dighits of a 2 digit no is 12 if the no obtained by reversiong the order of digits is 18 no the the original no find the number
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Answer 1

ANSWER

Let the tens digit of the required number be x and the units digit be y. Then,

x+y=12 .........(1)

Required Number = (10x+y).

Number obtained on reversing the digits = (10y+x).

Therefore,

(10y+x)−(10x+y)=18

9y−9x=18

y−x=2 ..........(2)

On adding (1) and (2), we get,

2y=14⟹y=7

Therefore,

x=5

Hence, the required number is 57.

Answer 2

Correct Question :–

The sum of dighits of a 2 digit no is 12 if the no obtained by reversiong the order of digits exceeds is 18. Find the number.

Answer :–

• The number = 57.

Given :–

• The sum of digits of two numbers = 12.
• The number obtained by reversing the order of the digits exceeds = 18.

To Find :–

• The number.

Solution :–

Let,

The tens digit of the required number be x.

The ones digit be y.

According to the question,

Sum of the digits of two numbers is 12.

I.e.,

$\mapsto \sf x + y = 12$ ––––(1)

The number = 10x + y because we take tens digit of the required number be x.

By reversing the digits.

[tex] \mapsto [/tex] 10y + x

Now,

According to the question,

The number obtained by reversing the order of the digits exceeds is 8.

$\mapsto (10y + x) - (10x + y) = 18$

$\mapsto 10y + x - 10x - y = 18$

$\mapsto 10y - y + x - 10x = 18$

$\mapsto 9y - 9x = 18$

$\mapsto \sf y - x = 2$ ––––(2)

Now, add both the equation (1) and (2),

$\mapsto (x + y) + (y - x) = 12 + 2$

$\mapsto x + y + y - x = 14$

$\mapsto x - x + y + y = 14$

$\mapsto 0 + 2y = 14$

$\mapsto 2y = 14$

$\mapsto y = \cancel\dfrac{14}{2}$

$\mapsto y = 7$

Now, substitute the value of y in equation (1),

$\mapsto x + y = 12$

$\mapsto x + 7 = 12$

$\mapsto x = 12 - 7$

$\mapsto x = 5$

The number = 10x + y.

So,

• x = 5
• y = 7

$\mapsto 10(5) + 7$

$\mapsto 10 \times 5 + 7$

$\mapsto 50 + 7$

$\mapsto 57$

Hence,

The number is 57.