If k is the sum of the reciprocals of the consecutive integers from 41

Question

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?

I.  \frac{1}{4}

II.  \frac{1}{3}

III.  \frac{1}{2}

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III

LeoN88 · Accepted Answer

K= (1/41) + (1/42) + (1/43).....(1/60) Total terms= (60-41)+1 => 20, so there are 20 terms. Out of these 20 terms the term with maximum value is 1/41 again, 1/40 > 1/41 Let k' = 20* (1/40) = 1/2, so 1/2 definitely more than original k. Let's compare 20/41 _ 1/4 we can say 20/41 is > 1/4. Similarly, 20/41 > 1/3 How to find which fraction is greater: check here-> https://gmatclub.com/forum/which-of-the ... l#p2280549 So Option D (I and II) are less than k.

Bunuel · Answer

k = 1/41 + 1/42 + 1/43 + ... + 1/60

We have the sum of 20 terms, which are reciprocals of consecutive integers from 41 to 60, inclusive. Notice that the largest term among the 20 is  \frac{1}{41} , and the smallest term is  \frac{1}{60} . 
 
If all 20 terms were equal to the largest term  \frac{1}{41} , the sum would be  \frac{20}{41} . Since the actual sum must be less than that, then  k < \frac{20}{41} < \frac{1}{2}   
 
If all 20 terms were equal to the smallest term  \frac{1}{60} , the sum would be  \frac{20}{60}=\frac{1}{3} . Since the actual sum must be more than that, then  k > \frac{1}{3}   
 
Therefore,   \frac{1}{3} < k < \frac{1}{2}  .

Answer: D.

nick1816 · Answer

(20/60) < (1/41) + (1/42).....+(1/60) < 20/41
1/3< (1/41) + (1/42).....+(1/60) < 1/2

Hence k will always greater than 1/3 and 1/4

ScottTargetTestPrep · Answer

We can create a range of the sum of the 20 numbers. The low end of the range is 20 x 1/60 = 1/3, and the high end of the range is 20 x 1/40 = 1/2.

Thus, k must be less than 1/2, so 1/3 and 1/4 are less than k.

Answer: D

Aliprep · Answer

Can we assume that sum of reciprocals consecutive integers equal to approx? = 1/((41+60)2 and this all multiplied by * 20 items
Then we have around 2/5 and I got the right answer.

As if I am finding the average and then multiplying it by the n items. Is that right assumption or I cannot do that with reciprocals?
Don't know whether I was lucky or not.

Ic0830de · Answer

Hi, I looked at the solutions posted but I am a little confused. Why did we multiply the max and min fraction with 20? Is this a concept in fractions?

Adarsh_24 · Answer

basically we replace every number with 1/80, 1/60, 1/40 respectively.

20*1/80=1/4
20*1/60=1/3
20*1/40=1/2

and for example we know 1/80 smaller than even smallest in given question 1/60.

so first answer is smaller than k.

ADisHere · Answer

I am confused with the methodology here.
Range of sum of reciprocals is no. of terms * highest value< K < no. of terms * lowest value2
20/41<K < 20/60
~1/2 < k < 1/3
I get I but how II?? wouldn't it be within the range and not less than k?

shubhim20 · Answer

how did you decode the actual sum should be less than or more than

Bunuel · Answer

If you replace every term with the  biggest  term, you create a sum that is  as large as possible . The real sum uses many  smaller  terms, so it must be  less  than that maximum.

If you replace every term with the  smallest  term, you create a sum that is  as small as possible . The real sum uses many  bigger  terms, so it must be  greater  than that minimum.

That’s why those two fake sums form the upper and lower limits.

If k is the sum of the reciprocals of the consecutive integers from 41

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If k is the sum of the reciprocals of the consecutive integers from 41

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