A and B take part in 100 m race. A runs at 5 kmph. A gives B a start of 8 m and still beats him by 8 seconds. The speed of B is:

15.15 kmph
24.14 kmph
34.25 kmph
44.4 kmph

Answer:- 1

Explanation:-

Solution:

\begin{aligned} A^{'} s s p e e d = (5 \times \frac{5}{18}) m / \sec = \frac{25}{18} m / \sec \\ T i m e t a k e n b y A t o cover 100 m = (100 \times \frac{18}{25}) \sec = 72 \sec \\ ∴ T i m e t a k e n b y B t o cover 92 m = (72 + 8) = 80 \sec \\ ∴ B^{'} s s p e e d = (\frac{92}{80} \times \frac{18}{5}) k m p h = 4.14 k m p h \end{aligned}

$\eqalign{ & A's\,speed = \left( {5 \times \frac{5}{{18}}} \right)m/\sec = \frac{{25}}{{18}}m/\sec \cr & Time\,taken\,by\,A\,to\,\operatorname{cover} \,100m = \left( {100 \times \frac{{18}}{{25}}} \right)\sec = 72\sec \cr & \therefore Time\,taken\,by\,B\,to\,\operatorname{cover} \,92m = \left( {72 + 8} \right) = 80\sec \cr & \therefore B's\,speed = \left( {\frac{{92}}{{80}} \times \frac{{18}}{5}} \right)kmph = 4.14kmph \cr}$