Question

In a two digit number, the unit digit is 3 less than the ten's digit. On reversing the
digit the number obtained is 9 more than 3 times the sum of the digits, find the
numbers​

Answer 1

Answer:

63 will be answer..

Step-by-step explanation:

see attachment.

hope it helpes.......

mark brainliest

Answer 2

The required number is 63, and that obtained on reversing its digits is 36.

Given:

The unit digit of a two-digit unknown number is 3 less than the ten's digit. On reversing the digits, the number obtained is 9 more than 3 times the sum of the digits

To Find:

The required number, and that obtained on reversing its digits.

Solution:

Let us consider that the tens place and the unit place of the unknown number are 't' and 'u' respectively.

Hence the unknown number = 10t + u.

Now, we are given that the unit digit of this unknown number is 3 less than the ten's digit, i.e.

u = t-3

or, u - t = -3  ........................(I)

On reversing the digits (i.e. reversing the tens and unit places) the new number obtained is = 10u + t.

Now this number is 9 more than 3 times the sum of the digits, i.e.

10u + t = 9 + 3(u + t)

⇒ 10u + t = 9 + 3u + 3t

⇒ 7u -2t = 9    ........................(II)

On multiplying (I) by 2 and subtracting it from (II), we get:

(7u -2t) - 2(u-t) = 9 - 2(-3)

⇒ 7u -2t -2u +2t = 9+6 = 15

⇒ 5u = 15

⇒ u = 3.

Substituting the value of 'u' in (I), we get:

3-t = -3

⇒ 3+3 = t

⇒ t = 6.

Hence the required number is = 10t + u = 10(6) + 3 = 63.

The number obtained on reversing its digits is 36.

∴ The required number is 63, and that obtained on reversing its digits is 36.

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