No ..wrong answer but would like to know your approach . What's your approach?Team BV--Pratik Gauri
Haan realised a little after posting that I had made a cancelling mistake (not sure about concept mistake) by what I did should have got 750ish....
1st case A running 1000 B running 960 gap 19 s for A 2nd case A running 960 B running 1000 gap 30 s for A
I converted the second case to A running 1000, B running 1000*25/24 i.e 1041.66 m and time gap will be 30 * 25/24 = 31.25 for A. (Idea being to remove A from the story)
So B runs the extra 1041.66 - 960 = 81.66 in 31.25 - 19 = 12.25 s so he will run 5000 in 5000*12.25 /81.66 = 5000 * 49/4 * 3/245 = 750 s? regards scrabbler
Il check once i reach home at 8 dont worry One ques from my side : In a km race, if A gives a start of 40m to B, A wins by 19s. If A gives B a 30 sec start , B wins by 40 m. Find time taken by B to run 5000m .Do post your approach . Let's see who solves in the shortest way Team BV--Pratik Gauri
Il check once i reach home at 8 dont worry One ques from my side : In a km race, if A gives a start of 40m to B, A wins by 19s. If A gives B a 30 sec start , B wins by 40 m. Find time taken by B to run 5000m .Do post your approach . Let's see who solves in the shortest way Team BV--Pratik Gauri
method (960-19b)/b = 1000/a (1000-30b)/b = 960/a divide both the eqs 1000^2 - 960^2 = (30*1000- 19*960)b speed b= 6.666
Another way of stating the above thing....summation from 1 to 22 implies woh jo sigma sign use karte they....with 1 below and 22 above...regardsscrabbler
I was expecting a number, myself bashing my head for properties of prime number factorials if any .