Question

At a certain regatta the number of races on each day varies jointly as the number of days from the beginning and end of the regatta up to and including the day in question. On three successive days there were respectively \[6,5\] and \[3\] races. Which days were there, and how long did the regatta last?

Answer 1

Hint: With the given conditions we will find three equations and by solving we get the required answer. To form the equation we will initially take the three consecutive days as variables.

Complete step-by-step answer:
It is given that at a certain regatta the number of races on each day varies jointly as the number of days from the beginning and end of the regatta up to and including the day in question. On three successive days there were respectively $6$, $5$ and \[3\] races.
Let us consider, the regatta lasted \[N\] days and the days in question are \[(x - 1) th, x th, (x + 1) th\]. Then,
Number of races on \[(x - 1) th\] days \[ = (x - 1)(N - (x - 2)) = 6k\]… (1)
Number of races on \[x th\] days \[ = x(N - (x - 1)) = 5k\]… (2)
Number of races on \[(x - 1) th\] days \[ = (x + 1)(N - x) = 3k\]… (3)
Now, let us subtract (2) from (1), we get,
\[2x - 2 - N = k\]… (4)
Again, by subtracting (3) from (2), we get,
\[2x - N = 2k\]… (5)
Let us solve equation (4) and (5) by subtracting (4) and (5) we get,
\[k = 2\]
From (5), on substituting k we get,
\[2x - N = 4\]
By simplifying for N we get,
\[N = 2x - 4\]
Let us substitute the value of \[k\] and \[N\] in (2), we get,
\[x(x - 3) = 10\]
By rearranging the equation we get,
\[{x^2} - 3x - 10 = 0\]
By simplifying the quadratic equation we get,
\[(x - 5)(x + 2) = 0\]
Equating the product to zero we get,
\[x = 5\] as we won’t accept negative numbers for x,
So, \[N = 2(5) - 4 = 6\]
Hence, the regatta lasted for \[6\] days and the \[3\] successive days were \[4 th, 5 th, 6 th\].

Note: From the quadratic equation we get, two values of number of days. We will take the positive value since the number of days cannot be negative.