Q8. A train travels 360 km at a uniform speed. If, the speed had been 5 km/hr more, it would have taken 1 hour less for the same journey. Find the speed of the train.

Let the speed of the train = x km/hr

If, speed had been 5 km/hr more, train would have taken 1 hour less.

So, according to this condition

\({{360} \over {x}} = {{360} \over {x + 5}} + 1\)

=>\(360 ( {{1}\over{x}} - {{1}\over {x + 5}}) = 1\)

=> \(360 ( {{x + 5 - x}\over{x(x + 5)}}) = 1\)

=>\((360)(5) = x^2 + 5x\)

=>\(x^2 + 5x - 1800 = 0\)

Comparing equation \(x^2 + 5x - 1800 = 0\) with general equation \(ax^2 + bx + c = 0\),

We get a = 1, b = 5 and c = -1800

Applying quadratic formula = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)

=>\(x = {{-5 ± \sqrt{(5)^2 - 4(1)(-1800)}} \over {(2)(1)}}\)

=>\(x = {{-5 ± \sqrt{25 + 7200}} \over {2}} = {{-5 ± \sqrt{7225}} \over {2}}\)

=>\(x = {{-5 + 85} \over {2}} , {{-5 - 85} \over {2}}\)

=>\(x = 40,-45\)

Since speed of train cannot be in negative. Therefore, we discard x = -45

Therefore, speed of train = 40 km/hr