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A certain bakery has 6 employees. It pays annual salaries of [#permalink]
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I came across the following GMATPrep® question: A certain bakery has 6 employees. It pays annual salaries of $14,000 to each of 2 employees, $16,000 to 1 employee, and $17,000 to each of the remaining 3 employees. The average (arithmetic mean) annual salary of these employees is closest to which of the following? (A) $15,200 (B) $15,500 (C) $15,800 (D) $16,000 (E) $16,400 Lets try to solve this problem using some unconventional method, to save time. First and foremost, lets not consider the three zeros, we'll add them afterwards
The generic method to solve this question would be adding the salaries and dividing by 6  (14 x 2 + 16 + 17 x 3) x1000/6 This would give the right answer.
The other way, to solve this average problem, using lesser calculations is:
Since the salaries are 14, 16, 17, it would mean that the average is going to be more than 14. We have, 14 x 2 and 17 x 3.
Had it been, 14 x 3 and 17 x 3, we could have concluded, the average would be (14 + 17) / 2 = 15.5
But we have 14 x 2 and 16 x 1, which would mean, the average should increase by a small amount. This would eliminate most of the choices  A B E
We are left with 15800 and 16000.
An increase of 2 in the salary, will increase the average by 2 / 6 = .36
Thus you'll get the answer 15.86
We can avoid the last step calculation too, if we observe that the increase is from 14 to 16, and will that result in an increase of .5 ?
If you have some other good and faster method to solve this problem, please share.
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Originally posted by UmangMathur on 05 Mar 2013, 04:31.
Last edited by Bunuel on 05 Mar 2013, 04:38, edited 1 time in total.
Renamed the topic, edited the question and the tags.



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Re: A certain bakery has 6 employees. It pays annual salaries of [#permalink]
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UmangMathur wrote: I came across the following GMATPrep® question:
A certain bakery has 6 employees. It pays annual salaries of $14,000 to each of 2 employees, $16,000 to 1 employee, and $17,000 to each of the remaining 3 employees. The average (arithmetic mean) annual salary of these employees is closest to which of the following?
(A) $15,200 (B) $15,500 (C) $15,800 (D) $16,000 (E) $16,400
Let the salary for each person be 14. Thus, the bakery ends up getting 2 + 3*3 = 11 as extra. This again has to be divided among 6 people. 12/6 = 2 which makes it 14+2 =16. Thus, as it is 11/6, it will be somewhere close to 16 and less than that. Also, 1/6 = 0.166. Thus 10/6 = 1.66 which gives the final salary after redistribution as 15.66. Thus, the answer is C.
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Re: A certain bakery has 6 employees. It pays annual salaries of [#permalink]
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10 Mar 2013, 03:52
İ am not sure that my method is ideal. So, I will be happy to hear experts' opinion on this  we have a set 14 14 16 17 17 17 lets derive from it two sets  1) 12 12 18 18 18 18 2) +2 +2 2 3 (+2 means we need +2 to get 14 from 12; 2 means we need 2 to get 16 from 18 etc) 12*2/6 +18*4/6 =4+12=16 ; 16 is the mean of the 1st set 1/6 is the mean of the 2nd set 160.1666= 15.8333 ans is C
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Re: A certain bakery has 6 employees. It pays annual salaries of [#permalink]
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LalaB wrote: İ am not sure that my method is ideal. So, I will be happy to hear experts' opinion on this 
we have a set 14 14 16 17 17 17 lets derive from it two sets  1) 12 12 18 18 18 18 2) +2 +2 2 3 (+2 means we need +2 to get 14 from 12; 2 means we need 2 to get 16 from 18 etc)
12*2/6 +18*4/6 =4+12=16 ; 16 is the mean of the 1st set
1/6 is the mean of the 2nd set
160.1666= 15.8333
ans is C I see what you are doing and it is fine but the logic behind the approach is not very clear to me. (By the way, good call on dropping the '000s. They add no value and just make the solution cumbersome) If I were to split it into two sets, they would be: 16, 16, 16, 16, 16, 16  so that I know that the mean here is certainly none other than 16. 16 is the middle number and the means is expected to be something around it. 2, 2, 0, +1, +1, +1  because actually the first term is 14 i.e. 2 less and so on. This adds up to 1 giving an average of 1/6 So average = 16  1/6 = 15.8333 (Essentially the same thing)
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Re: A certain bakery has 6 employees. It pays annual salaries of [#permalink]
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11 Mar 2013, 00:42
Thoughtosphere wrote: I came across the following GMATPrep® question: A certain bakery has 6 employees. It pays annual salaries of $14,000 to each of 2 employees, $16,000 to 1 employee, and $17,000 to each of the remaining 3 employees. The average (arithmetic mean) annual salary of these employees is closest to which of the following? (A) $15,200 (B) $15,500 (C) $15,800 (D) $16,000 (E) $16,400 Lets try to solve this problem using some unconventional method, to save time. First and foremost, lets not consider the three zeros, we'll add them afterwards
The generic method to solve this question would be adding the salaries and dividing by 6  (14 x 2 + 16 + 17 x 3) x1000/6 This would give the right answer.
The other way, to solve this average problem, using lesser calculations is:
Since the salaries are 14, 16, 17, it would mean that the average is going to be more than 14. We have, 14 x 2 and 17 x 3.
Had it been, 14 x 3 and 17 x 3, we could have concluded, the average would be (14 + 17) / 2 = 15.5
But we have 14 x 2 and 16 x 1, which would mean, the average should increase by a small amount. This would eliminate most of the choices  A B E
We are left with 15800 and 16000.
An increase of 2 in the salary, will increase the average by 2 / 6 = .36
Thus you'll get the answer 15.86
We can avoid the last step calculation too, if we observe that the increase is from 14 to 16, and will that result in an increase of .5 ?
If you have some other good and faster method to solve this problem, please share. I am wondering why we cannot just take the average of the numbers the conventional way instead of looking at novel ways to solve this simple problem We can ignore the zeroes and just go ahead and compute (14*2 + 16 + 17*3) / 6....which approximately comes to a little less than 16 and we have our answer. Took me less than 30 secs



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Re: A certain bakery has 6 employees. It pays annual salaries of [#permalink]
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11 Mar 2013, 01:26
Dipankar6435 wrote: I am wondering why we cannot just take the average of the numbers the conventional way instead of looking at novel ways to solve this simple problem We can ignore the zeroes and just go ahead and compute (14*2 + 16 + 17*3) / 6....which approximately comes to a little less than 16 and we have our answer. Took me less than 30 secs We most certainly can and it is quite easy to do in this question. Nevertheless, people like to keep a repertoire of various methods that can be used to solve questions of a particular type. Depending on the values given in the question, innovative methods can be much faster than conventional approaches.
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Re: A certain bakery has 6 employees. It pays annual salaries of [#permalink]
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11 Mar 2013, 09:05
Dipankar6435 wrote: I am wondering why we cannot just take the average of the numbers the conventional way instead of looking at novel ways to solve this simple problem We can ignore the zeroes and just go ahead and compute (14*2 + 16 + 17*3) / 6....which approximately comes to a little less than 16 and we have our answer. Took me less than 30 secs Read the article "Want a 750+? Do this question in 30 seconds", and u will get the answer http://www.manhattangmat.com/blog/index.php/2013/03/04/wanta750dothisquestionin30seconds/to VeritasPrepKarishma Thanks for checking my approach and giving good advice!
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Re: A certain bakery has 6 employees. It pays annual salaries of [#permalink]
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25 Jan 2014, 21:27
VeritasPrepKarishma wrote: LalaB wrote: İ am not sure that my method is ideal. So, I will be happy to hear experts' opinion on this 
we have a set 14 14 16 17 17 17 lets derive from it two sets  1) 12 12 18 18 18 18 2) +2 +2 2 3 (+2 means we need +2 to get 14 from 12; 2 means we need 2 to get 16 from 18 etc)
12*2/6 +18*4/6 =4+12=16 ; 16 is the mean of the 1st set
1/6 is the mean of the 2nd set
160.1666= 15.8333
ans is C I see what you are doing and it is fine but the logic behind the approach is not very clear to me. (By the way, good call on dropping the '000s. They add no value and just make the solution cumbersome) If I were to split it into two sets, they would be: 16, 16, 16, 16, 16, 16  so that I know that the mean here is certainly none other than 16. 16 is the middle number and the means is expected to be something around it. 2, 2, 0, +1, +1, +1  because actually the first term is 14 i.e. 2 less and so on. This adds up to 1 giving an average of 1/6 So average = 16  1/6 = 15.8333 (Essentially the same thing) I like this method but I have 2 questions. 1. When is it more time saving to use it? As numbers in these problems allow us to solve this problem very fast with the traditional method. 2. 16  1/6= 15.83 it takes me more time to calculate this than 95/6... Is there any faster way to calculate 161/6 other then (6*16+1)/6? Thank you so much.



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Re: A certain bakery has 6 employees. It pays annual salaries of [#permalink]
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27 Jan 2014, 03:04
bytatia wrote: I like this method but I have 2 questions. 1. When is it more time saving to use it? As numbers in these problems allow us to solve this problem very fast with the traditional method. 2. 16  1/6= 15.83 it takes me more time to calculate this than 95/6... Is there any faster way to calculate 161/6 other then (6*16+1)/6?
Thank you so much.
How about: 147, 156, 160, 143, 180, 167 You might appreciate the deviations method more here. Also, check out this link for more on this method: http://www.veritasprep.com/blog/2012/05 ... eviations/You should know that 1/6 = .166666... or approximately .17 You must memorize some important fractionpercentage equivalents. Check here: http://www.veritasprep.com/blog/2011/02 ... rcentages/1/6 is 16.7% which is .167 in decimal form.
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Re: A certain bakery has 6 employees. It pays annual salaries of [#permalink]
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Thoughtosphere wrote: I came across the following GMATPrep® question: A certain bakery has 6 employees. It pays annual salaries of $14,000 to each of 2 employees, $16,000 to 1 employee, and $17,000 to each of the remaining 3 employees. The average (arithmetic mean) annual salary of these employees is closest to which of the following? (A) $15,200 (B) $15,500 (C) $15,800 (D) $16,000 (E) $16,400 Lets try to solve this problem using some unconventional method, to save time. First and foremost, lets not consider the three zeros, we'll add them afterwards
The generic method to solve this question would be adding the salaries and dividing by 6  (14 x 2 + 16 + 17 x 3) x1000/6 This would give the right answer.
The other way, to solve this average problem, using lesser calculations is:
Since the salaries are 14, 16, 17, it would mean that the average is going to be more than 14. We have, 14 x 2 and 17 x 3.
Had it been, 14 x 3 and 17 x 3, we could have concluded, the average would be (14 + 17) / 2 = 15.5
But we have 14 x 2 and 16 x 1, which would mean, the average should increase by a small amount. This would eliminate most of the choices  A B E
We are left with 15800 and 16000.
An increase of 2 in the salary, will increase the average by 2 / 6 = .36
Thus you'll get the answer 15.86
We can avoid the last step calculation too, if we observe that the increase is from 14 to 16, and will that result in an increase of .5 ?
If you have some other good and faster method to solve this problem, please share. Dropping the zero's \(\frac{14*2 + 16*1 + 17*3}{6} = \frac{95}{6}\) \(\frac{96}{6} = 16\); so \(\frac{95}{6}\) would be just below 16 15800 fits in best Answer = C (Solved in 15 seconds)
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Re: A certain bakery has 6 employees. It pays annual salaries of [#permalink]
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10 Apr 2016, 12:59
Someone posted the solution below to the question  I understood the approach up until the last part of adding 10. Why does adding 10 works / what's the conceptual reason? Thank you! SOLUTION: As we are very good at Tables up to 10×10… 4*2=8 6*1=6 7*3=21 Total…35 Divide by 6=5.8333 Add 10…15.8333 …That’s all…
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Re: A certain bakery has 6 employees. It pays annual salaries of [#permalink]
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24 Nov 2016, 05:58
I mistook this problem to be a weighted average problem. As a result, I chose (E) 16,400 since there are 3 people getting paid 17,000$. Can someone please point out the error in this reasoning?



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Re: A certain bakery has 6 employees. It pays annual salaries of [#permalink]
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24 Nov 2016, 05:59
VeritasPrepKarishma wrote: LalaB wrote: İ am not sure that my method is ideal. So, I will be happy to hear experts' opinion on this 
we have a set 14 14 16 17 17 17 lets derive from it two sets  1) 12 12 18 18 18 18 2) +2 +2 2 3 (+2 means we need +2 to get 14 from 12; 2 means we need 2 to get 16 from 18 etc)
12*2/6 +18*4/6 =4+12=16 ; 16 is the mean of the 1st set
1/6 is the mean of the 2nd set
160.1666= 15.8333
ans is C I see what you are doing and it is fine but the logic behind the approach is not very clear to me. (By the way, good call on dropping the '000s. They add no value and just make the solution cumbersome) If I were to split it into two sets, they would be: 16, 16, 16, 16, 16, 16  so that I know that the mean here is certainly none other than 16. 16 is the middle number and the means is expected to be something around it. 2, 2, 0, +1, +1, +1  because actually the first term is 14 i.e. 2 less and so on. This adds up to 1 giving an average of 1/6 So average = 16  1/6 = 15.8333 (Essentially the same thing) Hello, I mistook this problem to be a weighted average problem. As a result, I chose (E) 16,400 since there are 3 people getting paid 17,000$. Can someone please point out the error in this reasoning?



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Re: A certain bakery has 6 employees. It pays annual salaries of [#permalink]
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24 Nov 2016, 22:53
gatreya14 wrote: VeritasPrepKarishma wrote: LalaB wrote: İ am not sure that my method is ideal. So, I will be happy to hear experts' opinion on this 
we have a set 14 14 16 17 17 17 lets derive from it two sets  1) 12 12 18 18 18 18 2) +2 +2 2 3 (+2 means we need +2 to get 14 from 12; 2 means we need 2 to get 16 from 18 etc)
12*2/6 +18*4/6 =4+12=16 ; 16 is the mean of the 1st set
1/6 is the mean of the 2nd set
160.1666= 15.8333
ans is C I see what you are doing and it is fine but the logic behind the approach is not very clear to me. (By the way, good call on dropping the '000s. They add no value and just make the solution cumbersome) If I were to split it into two sets, they would be: 16, 16, 16, 16, 16, 16  so that I know that the mean here is certainly none other than 16. 16 is the middle number and the means is expected to be something around it. 2, 2, 0, +1, +1, +1  because actually the first term is 14 i.e. 2 less and so on. This adds up to 1 giving an average of 1/6 So average = 16  1/6 = 15.8333 (Essentially the same thing) Hello, I mistook this problem to be a weighted average problem. As a result, I chose (E) 16,400 since there are 3 people getting paid 17,000$. Can someone please point out the error in this reasoning? It is a weighted average problem. C1 = 14k, w1 = 2 C2 = 16K, w2 = 1 C3 = 17K, w3 = 3 Cavg = (C1*w1 + C2*w2 + C3*w3)/(w1 + w2 + w3) Cavg = (2*14 + 1*16 + 3*17)/(2 + 1 + 3) Cavg = 15.833
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Re: A certain bakery has 6 employees. It pays annual salaries of [#permalink]
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29 Jan 2018, 11:03
UmangMathur wrote: I came across the following GMATPrep® question:
A certain bakery has 6 employees. It pays annual salaries of $14,000 to each of 2 employees, $16,000 to 1 employee, and $17,000 to each of the remaining 3 employees. The average (arithmetic mean) annual salary of these employees is closest to which of the following?
(A) $15,200 (B) $15,500 (C) $15,800 (D) $16,000 (E) $16,400 We can set up the weighted average equation: [2(14,000) + 16,000 + 3(17,000)]/6 (28,0000 + 16,000 + 51,000)/6 = 95,000/6 ≈15,800 Answer: C
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