Answer: Option B. -> 60

Answer: Option B. -> 20 m
$$\eqalign{
& A:B = 100:90 \cr
& A:C = 100:72 \cr
& B:C = {B \over A} \times {A \over C} = {{90} \over {100}} \times {{100} \over {72}} = {{90} \over {72}} \cr} $$
When B runs 90 m, C runs 72 m.
When B runs 100m, C run
$$\left( {{{72} \over {90}} \times 100} \right)m = 80\,m$$
∴ B can give C 20 m

Answer: Option B. -> 4.14 kmph
$$\eqalign{
& {\text{A's}}{\kern 1pt} {\kern 1pt} {\text{speed}} = \left( {5 \times \frac{5}{{18}}} \right){\text{m/sec}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{25}}{{18}}{\text{m/sec}} \cr
& {\text{Time}}{\kern 1pt} {\kern 1pt} {\text{taken}}{\kern 1pt} {\kern 1pt} {\text{by}}{\kern 1pt} {\kern 1pt} {\text{A}}{\kern 1pt} {\kern 1pt} {\text{to}}{\kern 1pt} {\kern 1pt} {\text{cover 100 m}} \cr
& = \left( {100 \times \frac{{18}}{{25}}} \right)\sec \cr
& = 72\sec \cr
& \therefore {\text{Time}}{\kern 1pt} {\kern 1pt} {\text{taken}}{\kern 1pt} {\kern 1pt} {\text{by}}{\kern 1pt} {\kern 1pt} {\text{B}}{\kern 1pt} {\kern 1pt} {\text{to}}{\kern 1pt} {\kern 1pt} {\text{cover 92 m}} \cr
& = \left( {72 + 8} \right) = 80\sec \cr
& \therefore {\text{B's}}{\kern 1pt} {\kern 1pt} {\text{speed}} = \left( {\frac{{92}}{{80}} \times \frac{{18}}{5}} \right){\text{kmph}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4.14{\text{ kmph}} \cr} $$

Answer: Option C. -> 10 points
$$\eqalign{
& A:B = 60:45 \cr
& A:C = 60:40 \cr
& \therefore \frac{B}{C} = {\frac{B}{A} \times \frac{A}{C}} = {\frac{{45}}{{60}} \times \frac{{60}}{{40}}} \cr
& = \frac{{45}}{{40}} = \frac{{90}}{{80}} = 90:80 \cr
& \therefore {\text{B}}\,{\text{can}}\,{\text{give}}\,{\text{C}}\,{\text{10}}\,{\text{points}}\,{\text{in}}\,{\text{a}}\,{\text{game}}\,{\text{of}}\,{\text{90}} \cr} $$

Answer: Option C. -> 20 m
To reach the winning post A will have to cover a distance of (500 - 140)m, i.e., 360 m.
While A covers 3 m, B covers 4 m.
While A covers 360 m, B covers $$\left( {\frac{4}{3} \times 360} \right)$$  m = 480 m.
Thus, when A reaches the winning post, B covers 480 m and therefore remains 20 m behind.
∴ A wins by 20 m.

Answer: Option D. -> 6 m
$$\eqalign{
& A:B = 100:90 \cr
& A:C = 100:87 \cr
& \frac{B}{C} = \frac{B}{A} \times \frac{A}{C} \cr
& \,\,\,\,\,\,\,\, = \frac{{90}}{{100}} \times \frac{{100}}{{87}} \cr
& \,\,\,\,\,\,\,\, = \frac{{30}}{{29}} \cr} $$
When B runs180m, C runs $$ \left( {\frac{{29}}{{30}} \times 180} \right){\text{m}} = 174{\text{m}}$$
∴ B beats C by (180 - 174) m = 6 m

Answer: Option B. -> 80 sec
$$\eqalign{
& {\text{B runs}}\frac{{45}}{2}{\text{m in 6 sec}} \cr
& \therefore {\text{B covers 300m in}} \cr
& = \left( {6 \times \frac{2}{{45}} \times 300} \right)\sec \cr
& = 80\,{\text{sec}} \cr} $$

Answer: Option B. -> 205 sec
In a km race, suppose A takes t sec,
Then, B takes (t + 30) sec, and C takes (t + 45) sec
180 m is covered by C in 45 sec
∴ 1000 m is covered by C in
= $$\left( {\frac{{45}}{{180}} \times 1000} \right)$$   sec = 250 sec
∴ A covers 1000 m in= (250 - 45) sec = 205 sec

Answer: Option B. -> 29 m
B : A = 200 : 190C : B = 200 : 180
$$\eqalign{
& = \frac{C}{A} \cr
& = {\frac{C}{B} \times \frac{B}{A}} \cr
& = {\frac{{200}}{{180}} \times \frac{{200}}{{190}}} \cr
& = \frac{{200}}{{171}} \cr} $$
∴ C can give to A, a start of (200 - 171) m = 29 m

Answer: Option A. -> 200 m
Ratio of speeds of A and B $$ = \frac{5}{3}:1 = 5:3$$
Thus, in race of 5m, A gains 2m over B
2m are gained by A in a race of 5m
80m will be gained by A in a race of
$$\left( {\frac{5}{2} \times 80} \right){\text{m}} = 200{\text{ m}}$$
∴ Winning post is 200 m away from the starting point