Carcass wrote:
If k is the sum of the reciprocals of the consecutive integers from 41 to 60 inclusive, which of the following are less than k?
Indicate
all such statements.
❑ \(\frac{1}{4}\)
❑ \(\frac{1}{3}\)
❑ \(\frac{1}{2}\)
Approximation works to solve this question.
We know that:
\(\frac{1}{40} > \frac{1}{41}, \frac{1}{42}, .... ,\frac{1}{49}\)
And:
\(\frac{1}{50} > \frac{1}{51}, \frac{1}{52}, .... ,\frac{1}{59}\)
So we can approximate \(k\):
\(k = \frac{1}{41} + \frac{1}{42} + ... + \frac{1}{51} + \frac{1}{52} + .... + \frac{1}{60}\)
\( k < \frac{1}{40} + \frac{1}{40} + .... + \frac{1}{50} + \frac{1}{50} + .... + \frac{1}{60}\)
There are 9 numbers between 41 and 49, and ten between 50 and 59, with the lone 60 at the end, so:
\( k < 9 * \frac{1}{40} + 10 * \frac{1}{50} + \frac{1}{60}\)
This simplifies to:
\( k < \frac{9}{40} + \frac{10}{50} + \frac{1}{60}\)
\( k < \frac{9}{40} + \frac{1}{5} + \frac{1}{60}\)
\(\frac{9}{40}\) is slightly less than \(\frac{10}{40}\) or \(\frac{1}{4}\), and \(\frac{1}{60}\) is less than \(\frac{1}{50}\), so essentially what we have is:
\( k < \frac{1}{4} + \frac{1}{5} + \frac{1}{50}\)
\( k < 0.25 + 0.2 + 0.02 \)
\( k < 0.47 \)
Giving us the answers A and B