Are you smarter than a fifth grader- Part 5

In the time it takes A to run 1000 metres, B runs 960 metres, and C runs 930 metres.

By the time B will have run 1000 metres, C will have run $\large\frac{1000}{960}\times930$ metres.

Therefore, the distance by which B beats C is $\large1000-\frac{1000}{960}\times930=\large\boxed{\frac{125}{4}}$

Therefore, $\large x\times y=125\times4=500$

$Let\quad their\quad speeds\quad be\quad A,B\quad and\quad C\\ Let\quad time\quad taken\quad by\quad A\quad to\quad finish\quad the\quad race={ t }_{ 1 }=\frac { 1000 }{ A } \\ Then,\quad \frac { 1000 }{ A } \left( A-B \right) =40m,\quad Thus\quad B=\frac { 96A }{ 100 } (1)\\ Similarly,\quad \frac { 1000 }{ A } (A-C)\quad implies\quad C=\frac { 93A }{ 100 } (2)\\ Let\quad time\quad taken\quad by\quad B\quad to\quad finish\quad the\quad race=T=\frac { 1000 }{ B } \\ Thus\quad we\quad need\quad to\quad compute:\\ \frac { 1000 }{ B } \left( B-C \right) =\quad \frac { 25,000 }{ 24A } \left( \frac { 3A }{ 100 } \right) =\frac { 125 }{ 4 } m\\ Thus\quad required\quad answer=\quad 125\times 4=\quad \boxed { 500 }$

Answer: At first glance, it might seem as if we are missing some crucial information like the speed of each runner. So, lets start by labeling the speed of the first runner, Va m/s. When A crosses the finish line, A has travelled 1000m. The time it took for A to do so can be expressed as (1000m/Va) seconds ( Remember the Distance/Speed=Time equality)

By this same amount of time (1000m/Va), B had ran 960 meters and C, 930 meters. From this information, we can calculate the speed of B. B's speed is simply the distance of 960 meters covered in (1000m/Va) seconds, or 960m/(1000m/Va) meter/second =(960/1000) Va meter/second or simplify further, (24/25) Va meter/second.

At the speed of (24/25)Va meter/second , B would have finished the 1000m race in (1000meter)/(24/25)Va meter/second, or (3125/3Va) seconds.

Similarly, as A crosses the line in a time of 1000/Va , C would have covered a distance of 930 meters and hence, C's speed is 930meters/(1000/Va) seconds or 0.93Va meter/second. So how much more did C travelled when it took B the time of (3125)/(3Va) seconds to run 30 more meters to the finish line? Thus, C would have travelled $0.93Va meter/second \times(3125)/(3Va)seconds$ or 968.75 meters. ( Note the terms Va cancels off)

Therefore, B beat C by 1000-968.75 or 31.25 meters or 3125/100 or 125/4. x=125 and y=4. xy= $\boxed {500}$