Question

The digit at the tens place of a two digit number is three times the digit at the units place. Il
reversed, the new number will be 36 less than the original number. Find the number​

Answer 1

Answer:

jkdmdkjsskskskkdkks

Answer 2

Let the digit at unit place be m

And, the digit at tens place be n

Then,

• The number is 10n+m

Here,

→ n = 3m

→ n - 3m = 0 -----(i)

On reversing the digits,

• The new number is 10m+n

Given that,

• New number = Old number - 36

i.e.

$\to \rm{}10m + n = 10n + m - 36 \\ \to \rm{}10m - m + n - 10n = - 36 \\ \to \rm{ \: \: }9m - 9n = - 36 \\ \to \rm \: \: \: 9(m - n) = - 36 \\ \\ \to \rm \: \: \: (m - n) = \frac{ - 36}{9} \\ \\ \to \rm \: \: \: ( m- n) = - 4 \: \: \: \: \: \: \: \: \: \cdots(ii)$

Now,

Adding equation (i) and equation (ii), we get,

$\to \sf{}\cancel{n} - 3m + m - \cancel{n} = 0 + ( - 4) \\ \to \sf{} \: \: \: \: \: \: \: \: \: \cancel- 2m = \cancel - 4 \\ \\ \to \sf{} \: \: \: \: \: \: \: \frac{2m}{2} = \frac{4}{2} \\ \\ \to \boxed{ \sf\: \: \: \: \: \: \: \: \: \: m = 2 \: \: \: }$

Thus,

$\leadsto \rm{}n = 3m \: \: \: \: \: \: \: \: \{using \: \: {eq}^{n} \: \: \: \: (i) \} \\\leadsto \rm{}n = 3(2) \\\leadsto \boxed{ \: \: \: \: \rm{} n= 6 \: \: \: }$

Hence,

• Original number (10n+m) = 62