Mathematics Questions and Answers

\(\frac{6.3 \times 10^5}{8.1 \times 10^3}\)

\(\frac{7}{9} \times 10^{2}\)

= \(0.77778 \times 10^{2}\)

= \(77.778 \approxeq 77.80\)

\(\begin{array}{c|c} 2 & 25\\2 & 12 R 1\\2 & 6 R 0\\2 & 3 R 0\\2 & 1 R 1\end{array}\)

= (11001)

\(\frac{6^{2n + 1} \times 9^n \times 4^{2n}}{18^n \times 2^n \times 12^{2n}}\) = \(\frac{(2 \times 3^{2n + 1} \times 3^{2n}) \times 2^{4n}}{(2 \times 9)^n \times 2^n \times (6 \times 2^{2n})}\)

= \(\frac{(2^{2n + 1} \times 3^{2n + 1}) \times 3^{2n} \times 2^{4n}}{2^n \times 3^{2n} \times 2^n \times 2^{4n} \times 3^{2n}}\)

= \(\frac{2^{2n} + 1 + 4^n \times 3^{2n} + 1 + 2^n}{2^{n + n + 4n} \times 3^{2n + 4n} \times 3^{2n + 2n}}\)

= \(\frac{2^{6n + 1} \times 3^{4n + 1}}{2^{6n} x 3^{4n}}\)

= 2^{6n + 1 - 6n} x 3^{4n + 1 - 4n}

2 x 3 = 6

Distance travelled by both x and y = 150 km.

Speed of y = 60km per hour, Time = \(\frac{dist.}{Speed}\)

Time spent by y = \(\frac{150}{60}\)

= 2\(\frac{3}{6}\)hrs

= \(\frac{1}{2}\) hrs

if x arrived at p 25 minutes earlier than y, then time spent by x = 2 hour 5 mins

= 125 mins

x average speed = \(\frac{150}{125}\) x \(\frac{60}{1}\)

= \(\frac{9000}{125}\)

= 72 km\hr

Given that log₁₀2 = 0.3010, If 10^x = 0.2

log₁₀10^x = log₁₀ 0.2 = log₁₀

xlog₁₀ 10 = log₁₀ 2 - log₁₀10 since log₁₀ 10 = 1

x = 0.3010 - 1

x = -0.6990