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Richard works as a waiter in an elegant downtown restaurant. His avera

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New post 01 Apr 2018, 10:28
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Question Stats:

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Richard works as a waiter in an elegant downtown restaurant. His average (arithmetic mean) gratuity per day over the past 10 workdays is $200. If Richard’s average gratuity per day for the first six of the past 10 workdays was $180, what was his average gratuity per day for the last four of the past 10 workdays?

(A) $210
(B) $220
(C) $225
(D) $230
(E) $240

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Re: Richard works as a waiter in an elegant downtown restaurant. His avera  [#permalink]

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New post 01 Apr 2018, 11:02
Bunuel wrote:
Richard works as a waiter in an elegant downtown restaurant. His average (arithmetic mean) gratuity per day over the past 10 workdays is $200. If Richard’s average gratuity per day for the first six of the past 10 workdays was $180, what was his average gratuity per day for the last four of the past 10 workdays?

(A) $210
(B) $220
(C) $225
(D) $230
(E) $240


average gratuity for 10days=(x+y)/10=200

average gratuity for 6days=x/6=180

Putting in equation 6*180+y=2000,y= 920

average gratuity for 4days=y/4=z=920/4=230

option D
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Richard works as a waiter in an elegant downtown restaurant. His avera  [#permalink]

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New post 01 Apr 2018, 13:45
Bunuel wrote:
Richard works as a waiter in an elegant downtown restaurant. His average (arithmetic mean) gratuity per day over the past 10 workdays is $200. If Richard’s average gratuity per day for the first six of the past 10 workdays was $180, what was his average gratuity per day for the last four of the past 10 workdays?

(A) $210
(B) $220
(C) $225
(D) $230
(E) $240

Weight the averages.
Let 4 days' average = \(x\)
6-day average = $180
Total 10-day average = $200

\(4(x) + 6($180) = 10($200)\)
\(4x + $1,080 = $2,000\)
\(4x = $920\)
\(x = $230\)


Answer D

Similar method - a little less direct

\(A * n = S\)
(10-day Sum) - (6-day Sum) = (4-day Sum)

Divide result by 4 for last four days' average, because (n=4): \(A= \frac{S}{n}\)

Sum for all 10 days: ($200 * 10) = $2,000
Sum for 6 days: ($180 * 6) = $1,080

Sum for 4 days: ($2,000 - $1,080) = $920
Average for 4 days: \(\frac{$920}{4} = $230\) per day

Answer D
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Re: Richard works as a waiter in an elegant downtown restaurant. His avera  [#permalink]

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New post 02 Apr 2018, 22:20
Bunuel wrote:
Richard works as a waiter in an elegant downtown restaurant. His average (arithmetic mean) gratuity per day over the past 10 workdays is $200. If Richard’s average gratuity per day for the first six of the past 10 workdays was $180, what was his average gratuity per day for the last four of the past 10 workdays?

(A) $210
(B) $220
(C) $225
(D) $230
(E) $240

IMO D avg of 10days= 200 total= 200*10=2000
perday for 1st 6 days =180
total of 6 days= 180*6=1080
now total of last 4 days= 2000-1080=920
now perday grtuity= 920/4= 230
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Re: Richard works as a waiter in an elegant downtown restaurant. His avera   [#permalink] 02 Apr 2018, 22:20
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