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At a certain restaurant, the average (arithmetic mean) number of custo

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At a certain restaurant, the average (arithmetic mean) number of custo  [#permalink]

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New post 21 Aug 2015, 05:15
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Re: At a certain restaurant, the average (arithmetic mean) number of custo  [#permalink]

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New post 21 Aug 2015, 05:20
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Bunuel wrote:
At a certain restaurant, the average (arithmetic mean) number of customers served for the past x days was 75. If the restaurant serves 120 customers today, raising the average to 90 customers per day, what is the value of x?

A. 2
B. 5
C. 9
D. 15
E. 30

Kudos for a correct solution.


Per the question,

\(\frac{75x+120}{x+1}\) =90 ---> x =2. A is the correct answer.
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Re: At a certain restaurant, the average (arithmetic mean) number of custo  [#permalink]

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New post 21 Aug 2015, 06:14
1
Bunuel wrote:
At a certain restaurant, the average (arithmetic mean) number of customers served for the past x days was 75. If the restaurant serves 120 customers today, raising the average to 90 customers per day, what is the value of x?

A. 2
B. 5
C. 9
D. 15
E. 30

Kudos for a correct solution.



No of customers/No of days=75
C/x=75
C=75x
Now, C+120/x+1=90
Put C=75x from the earlier equation,
15x=30
x=2
Answer A
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Re: At a certain restaurant, the average (arithmetic mean) number of custo  [#permalink]

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New post 21 Aug 2015, 21:13
Bunuel wrote:
At a certain restaurant, the average (arithmetic mean) number of customers served for the past x days was 75. If the restaurant serves 120 customers today, raising the average to 90 customers per day, what is the value of x?

A. 2
B. 5
C. 9
D. 15
E. 30


Ans: A
Solution: from the given info we can say if N is the total number of people served in x says then avg 75=N/x = N=75x
and after todayavg 90= (N+120)/(x+1); puttine the value of N we get the x=2
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At a certain restaurant, the average (arithmetic mean) number of custo  [#permalink]

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New post 21 Aug 2015, 23:11
1
Algebraic:

y/x=75
(y+120)/(x+1)=90
75x+120=90x+90
30=15x => x=2

Backsolving:

take 9 (C) 75*9=675+120=795/10=79.5, too low average so should take less days
take 5 (B) 75*5=375+120=495/6=82.5, low again but going up, so must be 2

A
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Re: At a certain restaurant, the average (arithmetic mean) number of custo  [#permalink]

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New post 22 Aug 2015, 01:09
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Equating total number of customers served
75x+120 = 90(x+1)
=> 15x = 30
x=2

Answer A.
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Re: At a certain restaurant, the average (arithmetic mean) number of custo  [#permalink]

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New post 23 Aug 2015, 03:20
1
Total Customers Served in X days = 75x

Total customers served after today 75x+120

New Average (75x+120)/x+1 = 90

Solving X =2 ( Option A)
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Re: At a certain restaurant, the average (arithmetic mean) number of custo  [#permalink]

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New post 23 Aug 2015, 04:40
1
Bunuel wrote:
At a certain restaurant, the average (arithmetic mean) number of customers served for the past x days was 75. If the restaurant serves 120 customers today, raising the average to 90 customers per day, what is the value of x?

A. 2
B. 5
C. 9
D. 15
E. 30

Kudos for a correct solution.


the increase in average is 15
the increase in customers from the mean is 45
45/15 = 3 is x+1
Therefore, x=2
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Re: At a certain restaurant, the average (arithmetic mean) number of custo  [#permalink]

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New post 23 Aug 2015, 12:22
1
1
Bunuel wrote:
At a certain restaurant, the average (arithmetic mean) number of customers served for the past x days was 75. If the restaurant serves 120 customers today, raising the average to 90 customers per day, what is the value of x?

A. 2
B. 5
C. 9
D. 15
E. 30

Kudos for a correct solution.


GROCKIT OFFICIAL SOLUTION:

WITHOUT using the formula, we can see that today the restaurant served 30 customers above the average. The total amount ABOVE the average must equal total amount BELOW the average. This additional 30 customers must offset the “deficit” below the average of 90 created on the x days the restaurant served only 75 customers per day.

30/15 = 2 days. Choice (A).

WITH the formula, we can set up the following:

90 = (75x + 120)/(x + 1)

90x + 90 = 75x + 120

15x = 30

x = 2 Answer Choice (A)

Use whichever makes more sense to you!
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Re: At a certain restaurant, the average (arithmetic mean) number of custo  [#permalink]

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New post 24 Aug 2015, 09:18
HI Bunnel,

If you could you explain more elaborately, that would be very helpful.
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Re: At a certain restaurant, the average (arithmetic mean) number of custo  [#permalink]

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New post 24 Aug 2015, 18:20
riteshgmat wrote:
HI Bunnel,

If you could you explain more elaborately, that would be very helpful.



Hi riteshgmat

Perhaps I can be of assistance. He means, whatever the residual is, it must be spread across the days equally, one part below and another above. The residual here is 30 days (\(120 - 90 = 30\)) and so to have the new average be 90, it must have been split into two \(15\)s and therefore, the number of days is \(2\).

Kr,
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Re: At a certain restaurant, the average (arithmetic mean) number of custo  [#permalink]

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New post 16 Dec 2016, 17:03
Nice Question
Here is my solution to this one =>

\(Using Mean = \frac{Sum}{#}\)

Sum(x)=75x

Sum(x+1)=75x+120

As pre Question => 75x+120/x+1=90
Hence x=2


Alternately,
75+15(x+1)=120
x=2


Hence A

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Re: At a certain restaurant, the average (arithmetic mean) number of custo  [#permalink]

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Re: At a certain restaurant, the average (arithmetic mean) number of custo   [#permalink] 05 Oct 2018, 05:06
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