Question

A beats b by 100 m in a race of 1200 m and b beats c by 200 m in race of 1600 m. Approximately by how many metres can a beat c in a race of 9600 m?

Answer 1

Given:

• A beats B by $100m$ in a race of $1200 m$
• B beats C by $200 m$ in race of $1600 m$

To Find:

• By how many meters can a beat C in a race of $960m$ ?

Solution:

    When A runs $1200m$, B runs $1100m$.

    Hence, when A runs $600m$, B runs $550m$.

   Again, when B runs $1600m$, C runs $1400m$.

   And, when B runs $800m$,

        C runs  $= 1400 \times \frac{550}{1600}$

                     $= 481.25m.$

Required distance $= 960 - 481.25 = 498.75m$

That means when A runs $600m$ then B can run $550m$ then C runs $498.75m$

Answer 2

Ans .          Let t1 be the time of completion of the race between A & B where A beats B . Then ,

$\left \{ {{D_{A} =D_{B}+100m} \atop {1200m=1100m+100m}} \right\\So ,\\\frac{V_{A} }{V_{B} } =\frac{12}{11}$

            Let t2 be the time of completion of the race between B & C where B beats C. Then ,

$\left \{ {{D_{B} =D_{C}+200m} \atop {1600m=1400m+200m}} \right\\So ,\\\frac{V_{B} }{V_{C} } =\frac{8}{7}$

            Let t3 be the time of completion of the race between A & C where A beats C . Then ,

$\left \{ {{D_{A} =D_{C}+?} \atop {9600m = x+?}} \right\\So ,\\\frac{V_{A} }{V_{C} } =\frac{9600}{x}$

          Solving all the above equations , we get x = 7700m , and ? = 1900m .