(a) If \((y - 1)\log_{10}4 = y\log_{10}16\), without using Mathematics tables or calculator, find the value of y.
(b) When I walk from my house at 4km/h, I will get to my office 30mins later than when I walk at 5km/h. Calculate the distance between my house and office.
(b) When I walk from my house at 4km/h, I will get to my office 30mins later than when I walk at 5km/h. Calculate the distance between my house and office.
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option n
Explanation:
(a) \((y - 1)\log_{10} 4 = y\log_{10} 16\)
\((y - 1)\log_{10} 4 = y \log_{10} 4^{2}\)
\((y - 1)\log_{10} 4 = 2y\log_{10} 4\)
Equating both sides, we have
\(y - 1 = 2y \implies -1 = 2y - y\)
\(\therefore y = -1\)
(b) Let the distance from my house to the office = c.
At 4km/h, the time taken to get to the office from the house = \(\frac{c}{4} hr\)
At 5km/h, the time taken to get to the office from the house = \(\frac{c}{5} hr\)
\(\frac{c}{4} = \frac{c}{5} + \frac{30}{60}\)
\(\frac{c}{4} - \frac{c}{5} = \frac{1}{2}\)
\(\frac{c}{20} = \frac{1}{2} \implies c = 10km\)
(a) \((y - 1)\log_{10} 4 = y\log_{10} 16\)
\((y - 1)\log_{10} 4 = y \log_{10} 4^{2}\)
\((y - 1)\log_{10} 4 = 2y\log_{10} 4\)
Equating both sides, we have
\(y - 1 = 2y \implies -1 = 2y - y\)
\(\therefore y = -1\)
(b) Let the distance from my house to the office = c.
At 4km/h, the time taken to get to the office from the house = \(\frac{c}{4} hr\)
At 5km/h, the time taken to get to the office from the house = \(\frac{c}{5} hr\)
\(\frac{c}{4} = \frac{c}{5} + \frac{30}{60}\)
\(\frac{c}{4} - \frac{c}{5} = \frac{1}{2}\)
\(\frac{c}{20} = \frac{1}{2} \implies c = 10km\)