Question

the sum of digits of a two digits number is 11 . on reversing the digits the new number so formed is 63 more original number . find the original number​

Answer 1

Answer:

Sum of digits=11

Let the unit's digit be 'x'

Hence ,the ten's digit = (11-x)

i.e. The number =

$10(11 - x) + 1 \times x \\ = 110 - 10x + x \\ = 110 - 9x$

On reversing the digits,

Unit's digit = (11-x)

Ten's digit = x

i.e. The number =

$10 \times x + 1 \times (11 - x) \\ = 10x + 11 - x \\ = 11 + 9x$

According to the question ,

$11 + 9x = (110 - 9x) + 63 \\ = > 9x + 9 x = 110 + 63 - 11 \\ = >18x = 162 \\ = > x = \frac{162}{18} = 9$

Therefore , (11-x) = 11-9 =2

HENCE , THE ORIGINAL NUMBER =

$110 - 9x \\ = 110 - (9 \times 9) \\ = 110 - 81 = 29$

ANSWER : 29

Step-by-step explanation:

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