We are given two problems. (a) Solve the equation $(y - 1) \log_{10}4 = y \log_{10}16$ for $y$, without using mathematical tables or a calculator. (b) Calculate the distance between house and office given that walking at 4 km/h leads to being 30 minutes late compared to walking at 5 km/h.

1. Problem Description

We are given two problems.

(a) Solve the equation $(y - 1) \log_{10}4 = y \log_{10}16$ for $y$, without using mathematical tables or a calculator.

(b) Calculate the distance between house and office given that walking at 4 km/h leads to being 30 minutes late compared to walking at 5 km/h.

2. Solution Steps

(a)

We have the equation $(y - 1) \log_{10}4 = y \log_{10}16$.

We know that $16 = 4^2$. Therefore, $\log_{10}16 = \log_{10}(4^2) = 2\log_{10}4$.

Substituting this into the equation gives $(y - 1) \log_{10}4 = y (2\log_{10}4)$.

Since $\log_{10}4 \ne 0$, we can divide both sides by $\log_{10}4$, giving

$y - 1 = 2y$.

Subtracting $y$ from both sides gives $-1 = y$.

Therefore, $y = -1$.

(b)

Let $d$ be the distance between the house and the office in km.

Let $t_1$ be the time taken to walk at 4 km/h, and $t_2$ be the time taken to walk at 5 km/h, both in hours.

Then $d = 4t_1$ and $d = 5t_2$. Also, $t_1 = t_2 + \frac{30}{60} = t_2 + \frac{1}{2}$.

Substituting $t_1 = t_2 + \frac{1}{2}$ into $d = 4t_1$ gives $d = 4(t_2 + \frac{1}{2}) = 4t_2 + 2$.

Also $d = 5t_2$.

So we have $5t_2 = 4t_2 + 2$.

Subtracting $4t_2$ from both sides gives $t_2 = 2$.

The distance $d = 5t_2 = 5(2) = 10$.

3. Final Answer

(a) $y = -1$

(b) The distance between the house and the office is 10 km.