If K is the sum of reciprocals of the consecutive integers

[M7MBA]

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

The OA is C .

How can I solve this PS solving? I don't have it clear.

[Brent@GMATPrepNow]

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

We want the approximate sum of 1/43 + 1/44 + 1/45 + . . . + 1/48

Let's make the following observations about the upper and lower values:

Upper values: If all 6 fractions were 1/43, the sum would be (6)(1/43) = 6/43 ~ 6/42 = 1/7

Lower values: If all 6 fractions were 1/48, the sum would be (6)(1/48) = 6/48 = 1/8

From this we can conclude that 1/8 < K < 1/7

If K is between 1/8 and 1/7, then K must be closer to 1/8 than it is to 1/6

Answer = C

Cheers,
Brent

[Scott@TargetTestPrep]

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

The OA is C .

How can I solve this PS solving? I don't have it clear.

We see that K = 1/43 + 1/44 + 1/45 + 1/46 + 1/47 + 1/48. We see that K is a sum of 6 numbers. Since each number is less than 1/42 and greater than or equal to 1/48, a lower estimate for K is 6 x 1/48 = 6/48 = 1/8 and an upper estimate for K is 6 x 1/42 = 6/42 = 1/7. In other words, K is between 1/7 and 1/8. Therefore, K is closest to 1/8.

Answer: C