(a) The product of the digits of a 2-digit number is 18. When 27 is subtracted from the number, the digits interchange their places. Find the number. OR (b) Two numbers are in the ratio 5 : 6. If 8 is subtracted from each of the numbers, the ratio becomes 4 : 5. Find the numbers.

Step 1: Set up equations for problem (a)

Let the two-digit number be represented as $10x + y$, where $x$ is the tens digit and $y$ is the units digit. From the problem statement, the product of the digits is 18, which gives us the equation $xy = 18$. When 27 is subtracted from the number, the digits interchange their places, meaning the new number is $10y + x$. This gives us the second equation $(10x + y) - 27 = 10y + x$.

Step 2: Solve the system of equations for problem (a)

Simplify the second equation: $10x + y - 27 = 10y + x \implies 9x - 9y = 27 \implies x - y = 3$. From this, we get $x = y + 3$. Substitute this into the first equation: $(y + 3)y = 18$, which simplifies to a quadratic equation $y^2 + 3y - 18 = 0$. Factoring this quadratic equation gives $(y + 6)(y - 3) = 0$. Since $y$ is a digit, it must be positive, so $y = 3$. Substituting $y = 3$ back into $x = y + 3$ gives $x = 6$.

Step 3: Determine the number for problem (a)

With $x = 6$ and $y = 3$, the two-digit number is $10(6) + 3 = 63$.

Step 4: Set up equations for problem (b)

Let the two numbers be $a$ and $b$. Their initial ratio is $5:6$, so $\frac{a}{b} = \frac{5}{6}$, which simplifies to $6a = 5b$. When 8 is subtracted from each number, the new ratio is $4:5$, so $\frac{a - 8}{b - 8} = \frac{4}{5}$. This cross-multiplies to $5(a - 8) = 4(b - 8)$.

Step 5: Solve the system of equations for problem (b)

Expand the second equation: $5a - 40 = 4b - 32$, which simplifies to $5a - 4b = 8$. From the first equation, $6a = 5b$, we can express $b$ as $b = \frac{6a}{5}$. Substitute this expression for $b$ into the simplified second equation: $5a - 4\left(\frac{6a}{5}\right) = 8$. Multiply by 5 to clear the denominator: $25a - 24a = 40$, which gives $a = 40$. Substitute $a = 40$ back into $b = \frac{6a}{5}$ to find $b = \frac{6(40)}{5} = 48$.

Step 6: Determine the numbers for problem (b)

The two numbers are $a = 40$ and $b = 48$.

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