The ratio between a two-digit number and the sum of the digits of that number is $4 : 1$. If the digit in the unit's place is $3$ more than the digit in the ten's place, then the number is

### Concept & Logic Any two-digit number can be expressed algebraically as $10t + u$, where $t$ is the tens digit and $u$ is the units digit. This allows us to convert digit-based word problems into standard linear equations. ### Step-by-Step Solution * **Given:** Let the tens digit be $t$ and the units digit be $u$. The number itself is $10t + u$. The sum of the digits is $t + u$. Condition 1: The ratio of the number to the sum of its digits is $4 : 1$. Condition 2: The units digit is $3$ more than the tens digit ($u = t + 3$). * **Calculation:** Express Condition 1 as an equation: $$ \frac{10t + u}{t + u} = \frac{4}{1} $$ Cross-multiply to solve: $$ 10t + u = 4(t + u) $$ $$ 10t + u = 4t + 4u $$ Rearrange to group variables: $$ 10t - 4t = 4u - u $$ $$ 6t = 3u $$ Divide by $3$: $$ 2t = u $$ Now, substitute Condition 2 ($u = t + 3$) into this finding: $$ 2t = t + 3 $$ $$ 2t - t = 3 $$ $$ t = 3 $$ If $t = 3$, find $u$: $$ u = 3 + 3 = 6 $$ The tens digit is $3$ and units digit is $6$, so the number is $36$. ### Exam Strategy & Shortcut **Option Verification** is incredibly fast for two-digit number problems. Check the second condition first: "unit's place is $3$ more than ten's place". (a) $24$: $4 - 2 = 2$ (Fail) (b) $36$: $6 - 3 = 3$ (Pass) (c) $63$: $3 - 6 = -3$ (Fail) (d) $96$: $6 - 9 = -3$ (Fail) Only Option (b) passes the second condition. We found the answer in $5$ seconds without doing any algebra. You can verify it with the first condition: $36 / (3+6) = 36 / 9 = 4$, which matches the $4:1$ ratio perfectly. ### Common Pitfall Getting bogged down in the algebra and making a sign error during cross-multiplication. In competitive exams, algebraically solving two-digit number problems should be your *backup* method, not your primary one. ### Final Answer **Therefore, the correct answer is 36.**