If K is the sum of the reciprocals of the consecutive integers from 43

Question

If K is the sum of the 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

PS00573

How do we decide between 1/6 and 1/8

Bunuel · Accepted Answer

Given that K=\frac{1}{43}+\frac{1}{44}+\frac{1}{45}+\frac{1}{46}+\frac{1}{47}+\frac{1}{48} . Notice that 1/43 is the larges term and 1/48 is the smallest term. If all 6 terms were equal to 1/43, then the sum would be 6/43=~1/7, but since actual sum is less than that, then we have that K<1/7. If all 6 terms were equal to 1/48, then the sum would be 6/48=1/8, but since actual sum is more than that, then we have that K>1/8. Therefore, 1/8m-is-the-sum-of-the-reciprocals-of-the-consecutive-integers-143703.html Hope it helps.

MacFauz · Answer

I believe a good approximation would be to take the mean, reciprocal of that and multiply by 6 (No of numbers being added)

=  \frac{6}{45.5}  which is closest to  \frac{6}{48}  (\frac{1}{6} would be  \frac{6}{36}  and  \frac{1}{10} would be  \frac{6}{60} ) and hence  \frac{1}{8}

Marcab · Answer

The numbers are  1/43 + 1/44+ 1/45 + 1/46 + 1/47 + 1/48 .
The easiest method is to find smart numbers.
If you consider each of the numbers as  1/42 , then there sum will be  6/42  or  1/7 . Remember that since we chose a higher number than those given, hence the actual sum will be smaller than  1/7 .
Now consider each of the numbers  1/48 . Then in such case, the sum will be  6/48  or  1/8 . Remember that since we chose a smaller number than those given, hence the actual sum will be greater than  1/8 .
Therefore the sum lies between  1/7  and  1/8 . Hence among teh answer choices, the sum is closest to  1/8 .
+1C

egmat · Answer

Hi, Well all other approaches are correct. Here is one more. A little less calculation intensive. From 1/43 to 1/48, there are 6 #. => We can infer that sum of 6 # of 1/40s > Sum of (1/43+/1/44+......+1/48) > Sum of 6 # of 1/50s =>So, 6/40 > Sum of (1/43+/1/44+......+1/48) > 6/50 => 1/6.66 > Sum of (1/43+/1/44+......+1/48) > 1/8.33 => 1/6.66 > 1/ (6.66< Denominator < 8.33) > 1/8.33 Only option available is C. Answer is 1/8. -Shalabh Jain

Zarrolou · Answer

What is the sum of  \frac{1}{43}+ ... +\frac{1}{48} ?

\frac{1}{43}(1+\frac{43}{44}+\frac{43}{45}+\frac{43}{46}+\frac{43}{47}+\frac{43}{48}) 
we can rewrite as:  \frac{1}{43}(1+1+1+1+1+1)=\frac{6}{43}

6/43 is something more than 7, so is colse to 8
 \frac{6}{43}=(almost)\frac{1}{8} 
 C

danzig · Answer

Bunuel, I understand your method. However, how can we know that the distance between k and 1/8 is shorter than the distance between k and 1/6. For example, if k were almost 1/7, we would have to calculate the distance between 1/8 and 1/7 and also the distance between 1/7 and 1/6.
I make this comment because the GMAT Prep explains that point, but it does that in a complex way.
Thanks!

Bunuel · Answer

Even if K=1/7, still the distance between 1/8 and 1/7 is less than the distance between 1/7 and 1/6.

PareshGmat · Answer

I did in this way
 \frac{1}{43} + \frac{1}{44} + \frac{1}{45} + \frac{1}{46} + \frac{1}{47} + \frac{1}{48}

= (\frac{1}{43} + \frac{1}{48}) + (\frac{1}{44} + \frac{1}{47}) + (\frac{1}{45} + \frac{1}{46})  .... Grouping the denominator's whose addition is same (91)

= \frac{1}{24} + \frac{1}{24} + \frac{1}{24}  (Approx)

= \frac{3}{24}  (Approx)

= \frac{1}{8} 
Answer = C

swanidhi · Answer

One way would be to find the middle terms.
Since total terms is 5. Middle term will be 3rd term. i.e. 1/45. Which should be the approximate (but less) than original mean.
1/45 * 5 = 1/9. So you know that the sum will be very lose to 1/9 but just a little more. 1/8 is the closest and also the correct answer.

minwoswoh · Answer

There are actually 6 terms. Anyway, your approach may work in this case.

BrentGMATPrepNow · Answer

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

saukrit · Answer

all 1/42...would have led to 6/42=1/7

all 1/48 would have led to 6/48=1/8

Clearly it will be nearer to 1/8. than anything else.  Hence C

SchruteDwight · Answer

total numbers = 48-43+1=6
Average (48+43)/2=45.5
Sum of recriproal = 6*10/455 which is approximately 6/45 which is 1/7.5 –> closest is 1/8

ScottTargetTestPrep · Answer

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 = ⅛, 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

Basshead · Answer

The reciprocals of consecutive integers from 43 to 48, inclusive, are:  \frac{1}{43} + \frac{1}{44} + \frac{1}{45} + \frac{1}{46} + \frac{1}{47} + \frac{1}{47}

Notice that if all 6 numbers were the reciprocal of 43, we would have  \frac{6}{43} .

If all 6 numbers were 48, we would have  \frac{6}{48} .

Simply to get:  \frac{1}{7}  (approximately) and  \frac{1}{8}

\frac{1}{7}  < K <  \frac{1}{8}

K is closest to  \frac{1}{8} .

Answer is C.

DanTheGMATMan · Answer

GMAT test writers absolutely love this problem, and so do we:

https://www.youtube.com/watch?v=28u-vstFgVw

bumpbot · Answer

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If K is the sum of the reciprocals of the consecutive integers from 43

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