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Last year Manfred received 26 paychecks. Each of his first 6 paychecks

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Last year Manfred received 26 paychecks. Each of his first 6 paychecks  [#permalink]

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New post 07 Mar 2014, 04:42
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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

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Re: Last year Manfred received 26 paychecks. Each of his first 6 paychecks  [#permalink]

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New post 07 Mar 2014, 04:42
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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.
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Re: Last year Manfred received 26 paychecks. Each of his first 6 paychecks  [#permalink]

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New post 22 Nov 2015, 13:09
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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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Re: Last year Manfred received 26 paychecks. Each of his first 6 paychecks  [#permalink]

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New post 07 Mar 2014, 05:40
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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 paychecks  [#permalink]

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New post 07 Mar 2014, 18:25
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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 paychecks  [#permalink]

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New post 05 Dec 2014, 01:45
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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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Re: Last year Manfred received 26 paychecks. Each of his first 6 paychecks  [#permalink]

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New post 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 paychecks  [#permalink]

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New post 23 Nov 2015, 17:47
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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 paychecks  [#permalink]

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New post 21 Jan 2018, 19:55
Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

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


Manfred received 6 paychecks of $750 each and 20 paychecks of $780 each.

The first 6 paychecks summed to 6(750) = 4,500.

The next 20 paychecks summed to 20(780) = 15,600.

So, the average of all the checks is:

(4,500 + 15,600)/26 = 20,100/26 ≈ 773

Answer: D
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Re: Last year Manfred received 26 paychecks. Each of his first 6 paychecks  [#permalink]

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New post 16 Aug 2018, 01:18
Hi,
We are given that there are 26 paychecks in all. 6 paychecks each pay $750, and rest of 20 paychecks each pay $30 more than $750.

New avg= 750+(20*30)/26.

Now lets assume there were 30 paychecks in total. 6 paychecks each pay $750, and rest of 24 paychecks each pay $30 more than $750. ( i just picked a number that would cancel the 30 in the numerator)


By 20th paychecks the average would have increased by 21. How {(14*30)/20}=21
By 30th paychecks the average would have increased by 24. How {(24*30)/20}=24

So the new avg would be between 771 to 774
We have 773 as option
Hence Choice D .

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Re: Last year Manfred received 26 paychecks. Each of his first 6 paychecks  [#permalink]

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New post 27 Sep 2019, 11:08
Distance between 750 and 780 is 30
We have 26 "units"
30/26 = 15/13, this is one unit
The distance from 780 to the average is 780 - (15/13)*6 --> 780 - 90/13 --> 780 - 7 = 773
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Re: Last year Manfred received 26 paychecks. Each of his first 6 paychecks   [#permalink] 27 Sep 2019, 11:08
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