GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 27 May 2020, 23:13 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # If K is the sum of reciprocals of the consecutive integers from 43 to

Author Message
TAGS:

### Hide Tags

Director  Joined: 29 Nov 2012
Posts: 673
If K is the sum of reciprocals of the consecutive integers from 43 to  [#permalink]

### Show Tags

12
1
111 00:00

Difficulty:   45% (medium)

Question Stats: 65% (01:38) correct 35% (01:51) wrong based on 1546 sessions

### HideShow timer Statistics

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

How do we decide between 1/6 and 1/8
Math Expert V
Joined: 02 Sep 2009
Posts: 64175
Re: If K is the sum of reciprocals of the consecutive integers from 43 to  [#permalink]

### Show Tags

35
40
fozzzy wrote:
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

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

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, 1/8<K<1/7. So, K must be closer to 1/8 than it is to 1/6.

Similar question to practice from OG: m-is-the-sum-of-the-reciprocals-of-the-consecutive-integers-143703.html

Hope it helps.
_________________
VP  Joined: 02 Jul 2012
Posts: 1091
Location: India
Concentration: Strategy
GMAT 1: 740 Q49 V42
GPA: 3.8
WE: Engineering (Energy and Utilities)
Re: If K is the sum of reciprocals of the consecutive integers from 43 to  [#permalink]

### Show Tags

6
fozzzy wrote:
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

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

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 $$\frac{1}{8}$$
##### General Discussion
Director  Status: Been a long time guys...
Joined: 03 Feb 2011
Posts: 964
Location: United States (NY)
Concentration: Finance, Marketing
GPA: 3.75
Re: If K is the sum of reciprocals of the consecutive integers from 43 to  [#permalink]

### Show Tags

4
1
fozzzy wrote:
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

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

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
_________________
e-GMAT Representative P
Joined: 02 Nov 2011
Posts: 2993
Re: If K is the sum of reciprocals of the consecutive integers from 43 to  [#permalink]

### Show Tags

3
5
fozzzy wrote:
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

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

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
_________________
Director  Status: Far, far away!
Joined: 02 Sep 2012
Posts: 998
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8
Re: If K is the sum of reciprocals of the consecutive integers from 43 to  [#permalink]

### Show Tags

5
1
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
Manager  Joined: 11 Aug 2012
Posts: 107
Schools: HBS '16, Stanford '16
Re: If K is the sum of reciprocals of the consecutive integers from 43 to  [#permalink]

### Show Tags

Bunuel wrote:
fozzzy wrote:
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

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

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, 1/8<K<1/7. So, K must be closer to 1/8 than it is to 1/6.

Similar question to practice from OG: m-is-the-sum-of-the-reciprocals-of-the-consecutive-integers-143703.html

Hope it helps.

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!
Math Expert V
Joined: 02 Sep 2009
Posts: 64175
Re: If K is the sum of reciprocals of the consecutive integers from 43 to  [#permalink]

### Show Tags

1
1
danzig wrote:
Bunuel wrote:
fozzzy wrote:
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

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

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, 1/8<K<1/7. So, K must be closer to 1/8 than it is to 1/6.

Similar question to practice from OG: m-is-the-sum-of-the-reciprocals-of-the-consecutive-integers-143703.html

Hope it helps.

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!

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.
_________________
SVP  Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1709
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: If K is the sum of reciprocals of the consecutive integers from 43 to  [#permalink]

### Show Tags

3
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)

$$= \frac{1}{8}$$

Originally posted by PareshGmat on 29 Apr 2013, 00:55.
Last edited by PareshGmat on 03 Aug 2014, 22:01, edited 1 time in total.
Manager  Joined: 11 Oct 2013
Posts: 95
Concentration: Marketing, General Management
GMAT 1: 600 Q41 V31
Re: If K is the sum of reciprocals of the consecutive integers from 43 to  [#permalink]

### Show Tags

2
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.
_________________
Its not over..
Manager  Joined: 10 May 2014
Posts: 134
Re: If K is the sum of reciprocals of the consecutive integers from 43 to  [#permalink]

### Show Tags

swanidhi wrote:
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.

There are actually 6 terms. Anyway, your approach may work in this case.
GMAT Club Legend  V
Joined: 11 Sep 2015
Posts: 4876
GMAT 1: 770 Q49 V46
Re: If K is the sum of reciprocals of the consecutive integers from 43 to  [#permalink]

### Show Tags

1
Top Contributor
1
fozzzy wrote:
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

Cheers,
Brent
_________________
Senior Manager  D
Status: Current student at IIMB
Affiliations: IIM Bangalore
Joined: 05 Jul 2018
Posts: 438
Location: India
Concentration: General Management, Technology
Schools: IIM (A)
GMAT 1: 600 Q47 V26 GRE 1: Q162 V149 GPA: 3.6
WE: Information Technology (Consulting)
Re: If K is the sum of reciprocals of the consecutive integers from 43 to  [#permalink]

### Show Tags

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
Senior Manager  S
Joined: 03 Sep 2018
Posts: 259
Location: Netherlands
Schools: HEC MiM "22 (S)
GPA: 4
Re: If K is the sum of reciprocals of the consecutive integers from 43 to  [#permalink]

### Show Tags

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
_________________
Good luck to you. Retired from this forum.
Target Test Prep Representative V
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 10569
Location: United States (CA)
Re: If K is the sum of reciprocals of the consecutive integers from 43 to  [#permalink]

### Show Tags

fozzzy wrote:
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

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

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.

_________________

# Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Re: If K is the sum of reciprocals of the consecutive integers from 43 to   [#permalink] 17 Oct 2019, 19:31

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