Problem Solving

Hi Guru(s),
If K is the sum of reciprocals of the consecutive integers from 43 to 48, inclusive, then K is closest in value to which of the following?

A. 1/12
B. 1/10
C. 1/8
D. 1/6
E. 1/4
OA is A
Not sure how to derive the final answer with approximation
Thanks
Dhiren

dhiren8182 wrote:Hi Guru(s),
If K is the sum of reciprocals of the consecutive integers from 43 to 48, inclusive, then K is closest in value to which of the following?

A. 1/12
B. 1/10
C. 1/8
D. 1/6
E. 1/4
OA is A
Not sure how to derive the final answer with approximation
Thanks
Dhiren

There are six integers in that set, 43, 44, 45, 46, 47 and 48. Now I was looking at the OA you gave and check this out.

The smallest possible reciprocal is 1/48. Even if all of them were 1/48, the total would be 6 x 1/48 = 6/48 = 1/8. Your OA is even less than 1/8. So it must be incorrect.

Anyway, we are on our way to solving this problem.

The answer has to be a little bigger than 1/8, since the other reciprocals, 1/43, 1/44 and so on, are bigger than 1/48. So let's see which answer is closest.

Hmm. I see a nice clean way to get the other end of the range.

If all of the integers were 42, then the reciprocals would all be 1/42 and the total would be 6/42 or 1/7. So the real answer is somewhere between 1/7 and 1/8.

As numbers get higher, the difference between reciprocals of numbers the same distance apart becomes less. The difference between 1/12 and 1/13 is much less than the difference between 1/2 and 1/3. So 1/7 is a little closer to 1/8 than it is to 1/6.

Do we even need to figure this out? Maybe not. Clearly this number is a fair amount smaller than 1/7.

In any case, the sum of the reciprocals is closer to 1/8 than it is to 1/6.

Choose C.


_________________
Marty Murray
GMAT Coach
[email protected]
https://infinitemindprep.com/

dhiren8182 wrote:Hi Guru(s),
If K is the sum of reciprocals of the consecutive integers from 43 to 48, inclusive, then K is closest in value to which of the following?

A. 1/12
B. 1/10
C. 1/8
D. 1/6
E. 1/4

I received a PM about this problem.
My approach would be similar to Marty's and Brent's.

Determine the RANGE of the sum.

If all of the fractions were 1/43, the sum = 6(1/43) ≈ 6/42 = 1/7.
If all of the fractions were 1/48, the sum = 6(1/48) = 6/48 = 1/8.
Thus, the sum must be between 1/8 and 1/7.

The correct answer is C.