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Last year Manfred received 26 paychecks. Each of his first 6 [#permalink]
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07 Mar 2014, 04:42
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The Official Guide For GMAT® Quantitative Review, 2ND EditionLast year Manfred received 26 paychecks. Each of his first 6 paychecks was $750; each of his remaining paychecks was $30 more than each of his first 6 paychecks. To the nearest dollar, what was the average (arithmetic mean) amount of his paychecks for the year? (A) $752 (B) $755 (C) $765 (D) $773 (E) $775 Problem Solving Question: 137 Category: Arithmetic Statistics Page: 79 Difficulty: 600 GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition  Quantitative Questions ProjectEach week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution. We'll be glad if you participate in development of this project: 1. Please provide your solutions to the questions; 2. Please vote for the best solutions by pressing Kudos button; 3. Please vote for the questions themselves by pressing Kudos button; 4. Please share your views on difficulty level of the questions, so that we have most precise evaluation. Thank you!
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Re: Last year Manfred received 26 paychecks. Each of his first 6 [#permalink]
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07 Mar 2014, 04:42



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Re: Last year Manfred received 26 paychecks. Each of his first 6 [#permalink]
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07 Mar 2014, 05:40
The first 6 paychecks are for 750 each and then next 20 paychecks are for 780 each. The average will be closer to 780. Eliminate options A, B and C.
How do we guesstimate the right option between D and E. Upon solving, we get 773 but the calculations are tedious. Any good way to eliminate 775?



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Re: Last year Manfred received 26 paychecks. Each of his first 6 [#permalink]
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07 Mar 2014, 18:25
Option D. A carafe paycheck=Total amt./total no. Of paychecks Total amt.=750*6+780*20=20100 Total no. Of paychecks=26 Avg.=20100/26=773 approx
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Re: Last year Manfred received 26 paychecks. Each of his first 6 [#permalink]
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08 Mar 2014, 05:12
looks like one of those questions where you just have to use your gut feeling to guess between 773 and 775. Otherwise, I can't see any easier way (apart from number crunching) to choose one over the other
Geniuses please, any suggestions



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Re: Last year Manfred received 26 paychecks. Each of his first 6 [#permalink]
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08 Mar 2014, 13:25



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Re: Last year Manfred received 26 paychecks. Each of his first 6 [#permalink]
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09 Mar 2014, 02:46
Donnie84 wrote: The first 6 paychecks are for 750 each and then next 20 paychecks are for 780 each. The average will be closer to 780. Eliminate options A, B and C.
How do we guesstimate the right option between D and E. Upon solving, we get 773 but the calculations are tedious. Any good way to eliminate 775? Hmm tough choice between D and E. A,B and C can be ruled out.... Actually Bunuel's way of calculating makes it less tedious.... But here is what I will do.....750....................................780... We know the average to be closer to 780... We can divide the distance of 30 in 26 parts and we get roughly as 1.15 (30/26) Now average will be 750+ 20*30/26 or 750+ 20*1.15= 773 or 7806*30/26 or 780 6*1.15 ~ 773
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Re: Last year Manfred received 26 paychecks. Each of his first 6 [#permalink]
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05 Dec 2014, 01:45
the solutions posted thus far are good but lots of computation, an area where i tend to make mistakes. here is how i broke it down.
6 checks are at 750. 20 checks are at 780.
difference between these checks is 30.
20 checks * 30 = 600 (easy math) when removing 600 from the latter checks, all checks are now 'evenly filled' at 750, as if they were glasses filled with 750ml of water, for example.
what's an even way to distribute the 600 among all of the 26 checks? ==> 600/26, but this is also a tedious calculation however, 600/30 is 20, since our actual denominator is smaller (and 600 numerator the same) we know that the resulting number will be just a little above 770 (750 + 20)
773 is more 'little' above 750 than 775.



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Last year Manfred received 26 paychecks. Each of his first 6 [#permalink]
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30 Jan 2015, 01:00
fhfuentes wrote: the solutions posted thus far are good but lots of computation, an area where i tend to make mistakes. here is how i broke it down.
6 checks are at 750. 20 checks are at 780.
difference between these checks is 30.
20 checks * 30 = 600 (easy math) when removing 600 from the latter checks, all checks are now 'evenly filled' at 750, as if they were glasses filled with 750ml of water, for example.
what's an even way to distribute the 600 among all of the 26 checks? ==> 600/26, but this is also a tedious calculation however, 600/30 is 20, since our actual denominator is smaller (and 600 numerator the same) we know that the resulting number will be just a little above 770 (750 + 20)
773 is more 'little' above 750 than 775. Alternative approach Since we are calculating average and all the nos are either at the lower end (750) or at the higher end (780), the points lost by the nos at the higher end (to the average) should be equal to the pints gained by the nos at the lower end. Points gained (Lower end * nos at lower end) = Points lost (upper end * nos at upper end ) For example, average of 2 and 12 is 7. Points gained by 2 = 5 which is = the no of points lost by 12 to reach the average. Strategically trying choice D first, as A and B are clearly ruled out. (773 750)*6 = (780 773)* 20 23*6 =7*20 130 ALMOST = 140. So, D is right choice.
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Re: Last year Manfred received 26 paychecks. Each of his first 6 [#permalink]
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04 Feb 2015, 01:04
Answer = D = 773 Average \(= \frac{750*6 + 780*20}{26} = \frac{7800+2250}{13} = \frac{10050}{13} \approx{773}\)
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Re: Last year Manfred received 26 paychecks. Each of his first 6 [#permalink]
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04 Feb 2015, 07:36
hi, there are various people in the thread asking for a short cut to estimate between 773 and 775 as answer. it is easy by seeing the spread of the number of cheques and of the value of cheques.. for eg the two values of cheques are 750 and 780.. the spread away from average if it is 775 is 25:5 or 5:1... the no of cheques should be in ratio 1:5 or x:5x... here if x=6, 5x=30.. this means you require 30 cheques of 780 if there are 6 cheques of 770 to get an avg of 775.. however we have only 20 cheques, therefore the avg has to be less than 775 .. leaving 773 as answer.. I hope it is slightly clear to people, earlier or now, in doubt of choosing between the two figures
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Re: Last year Manfred received 26 paychecks. Each of his first 6 [#permalink]
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22 Nov 2015, 13:09
My simple way:
Call 750 the starting point so = 0 780=30 6(0)+20(30) / 26 = 23 750+23 = 773



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Last year Manfred received 26 paychecks. Each of his first 6 [#permalink]
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23 Nov 2015, 17:47
IMO, the simplest way to deal with this type of questions is by calculating the weighted average. 750..................780 (diff is 30) 6......................20(total=26) So wted avg= 750+ (30/26)*20= 773
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Re: Last year Manfred received 26 paychecks. Each of his first 6 [#permalink]
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15 Dec 2015, 05:10
Donnie84 wrote: The first 6 paychecks are for 750 each and then next 20 paychecks are for 780 each. The average will be closer to 780. Eliminate options A, B and C.
How do we guesstimate the right option between D and E. Upon solving, we get 773 but the calculations are tedious. Any good way to eliminate 775? Any multiple of 6 will not give 5 at unit place. So D is the safe choice.



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Re: Last year Manfred received 26 paychecks. Each of his first 6 [#permalink]
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15 Dec 2015, 05:16
robu wrote: Donnie84 wrote: The first 6 paychecks are for 750 each and then next 20 paychecks are for 780 each. The average will be closer to 780. Eliminate options A, B and C.
How do we guesstimate the right option between D and E. Upon solving, we get 773 but the calculations are tedious. Any good way to eliminate 775? Any multiple of 6 will not give 5 at unit place. So D is the safe choice. Hi, if you follow that rule than you will have to negate the right answer too, which is 773.. as no multiple of 6 will have 3 in its units digit.. the Q asks us " to the nearest dollar", which clearly shows that it is mostly not an integer.. you will have to find ways to discard 775, and looking for multiple of 6 is not one of them..
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Re: Last year Manfred received 26 paychecks. Each of his first 6 [#permalink]
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25 May 2016, 06:57
Total no of paychecks = 26 6 paychecks = 750 20 paychecks = 780 Thus on an average 12 paychecks of (750+780)/2 = 765 and 14 paychecks of 780 Now if it had been equal quantities i.e. 13 paychecks of each, the simple average would be 772.5 Since there is an extra paycheck for 780 it will be slightly more than 772.5 Hence 773 Ans : D
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Re: Last year Manfred received 26 paychecks. Each of his first 6 [#permalink]
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21 Sep 2016, 00:30
Bunuel wrote: The Official Guide For GMAT® Quantitative Review, 2ND EditionLast year Manfred received 26 paychecks. Each of his first 6 paychecks was $750; each of his remaining paychecks was $30 more than each of his first 6 paychecks. To the nearest dollar, what was the average (arithmetic mean) amount of his paychecks for the year? (A) $752 (B) $755 (C) $765 (D) $773 (E) $775 Problem Solving Question: 137 Category: Arithmetic Statistics Page: 79 Difficulty: 600 GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition  Quantitative Questions ProjectEach week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution. We'll be glad if you participate in development of this project: 1. Please provide your solutions to the questions; 2. Please vote for the best solutions by pressing Kudos button; 3. Please vote for the questions themselves by pressing Kudos button; 4. Please share your views on difficulty level of the questions, so that we have most precise evaluation. Thank you! Another way would be to just focus on the amount which increased and needs to be added to the average. Take 750 as the original value for average. Now you know first 6 are at 750 so we are good on that part. However for remaining 20  30 more is added i.e 30*20 = 600. ok so now you need to distribute this 600 into 26 values... 300\26 = 23 approx so add this to 750 and this is your answer
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Re: Last year Manfred received 26 paychecks. Each of his first 6 [#permalink]
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06 Dec 2016, 05:59
Bunuel wrote: SOLUTION
Last year Manfred received 26 paychecks. Each of his first 6 paychecks was $750; each of his remaining paychecks was $30 more than each of his first 6 paychecks. To the nearest dollar, what was the average (arithmetic mean) amount of his paychecks for the year?
(A) $752 (B) $755 (C) $765 (D) $773 (E) $775
The average amount of the paychecks =
\(\frac{750*6+ (750 +30)*20}{26}=\frac{750*26+ 30*20}{26}=750+\frac{30*20}{26}=750+\frac{30*10}{13}\approx{750+23}=773\).
Answer: D. Hi Bunuel, on the first two steps, I'm not sure where that 26 came from? Would you be able to guide me in how you got to that computation?? Thanks!



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Re: Last year Manfred received 26 paychecks. Each of his first 6 [#permalink]
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06 Dec 2016, 09:58
alandizzle wrote: Bunuel wrote: SOLUTION
Last year Manfred received 26 paychecks. Each of his first 6 paychecks was $750; each of his remaining paychecks was $30 more than each of his first 6 paychecks. To the nearest dollar, what was the average (arithmetic mean) amount of his paychecks for the year?
(A) $752 (B) $755 (C) $765 (D) $773 (E) $775
The average amount of the paychecks =
\(\frac{750*6+ (750 +30)*20}{26}=\frac{750*26+ 30*20}{26}=750+\frac{30*20}{26}=750+\frac{30*10}{13}\approx{750+23}=773\).
Answer: D. Hi Bunuel, on the first two steps, I'm not sure where that 26 came from? Would you be able to guide me in how you got to that computation?? Thanks! The numerator: \(750*6+ (750 +30)*20 =750*6+ 750*20 +30*20 =750*(6+20) +30*20=750*26+30*20\)
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Re: Last year Manfred received 26 paychecks. Each of his first 6 [#permalink]
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08 Dec 2016, 10:10
Option D)Number of paychecks = 26 Value of first 6 Paychecks = 750 Value of next 20 Paychecks = 750 + 30 Mean value of Paychecks = \(750 + \frac{(30*20)}{26} = 750 + 23 (approx.) = 773 (approx.)\)
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