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At a certain restaurant, the average

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At a certain restaurant, the average

by gmatdriller » Sat Feb 04, 2012 1:02 am
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
According to OA:
A quick way to answer this is to note that today, the restaurant served 30 customers more than they served on an average day. This 30 additional customers has to exactly offset the deficit created on the x days when the restaurant served only 75 customer per day.

I understand the algebraic approach, but please explain other methods...thanks
OA: A
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by Anurag@Gurome » Sat Feb 04, 2012 1:55 am
gmatdriller 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
According to OA:
A quick way to answer this is to note that today, the restaurant served 30 customers more than they served on an average day. This 30 additional customers has to exactly offset the deficit created on the x days when the restaurant served only 75 customer per day.

I understand the algebraic approach, but please explain other methods...thanks
OA: A
(75x + 120)/(1 + x) = 90
75x + 120 = 90x + 90
15x = 30
x = 2

The correct answer is A.
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by ronnie1985 » Sat Feb 04, 2012 2:31 am
The method u have described is shortcut method. If you are well conversant with the method it is good. The algebraic method is
(75x+120)/(x+1) = 90
Solve for x
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by bryan88 » Sat Feb 04, 2012 11:15 am
TNC/x =75 TNC=Total no of customers
TNC =75x (1)

(TNC+120)/x+1 =90 (2)

Solve.
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by gmatdriller » Sat Feb 04, 2012 1:48 pm
I wanted to understand the shortcut approach under the spoiler as it appears on
grockit.com. Some people might want to try the question naturally, so i hid the text.

Nevertheless, i think i got the explanations now...or so i think.

===========90=============
75, 75, 75... [+30] for 120
we want to know the number of 15's(90 less than 75) that will equal 30(90 less than 120)
15n = 30
n = 2..... only 2 days
OA is thus A
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by GMATGuruNY » Sat Feb 04, 2012 3:45 pm
gmatdriller 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
We can plug in the answer choices, which represent the number of days with an average of 75 customers per day.

The required average for all the days is 90, implying that the total number of customers is a multiple of 10.
Thus, the number of days with an average of 75 customers per day must be EVEN; otherwise, the total number of customers will be ODD (and thus NOT a multiple of 10).
Eliminate B, C and D.

Answer choice E implies a total of 31 days: VERY unlikely, since 31 is prime.

Answer choice A: 2 days at 75 per day.
Average for all 3 days = ((2*75) + 120)/3 = 90.
Success!

The correct answer is A.

Note that we had to try ONLY ONE answer choice -- a very efficient approach.
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by gmatdriller » Sun Feb 05, 2012 3:03 pm
Yes, GmatGuru,
testing the answer choices is another fast approach.
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by Scott@TargetTestPrep » Tue Oct 09, 2018 9:35 am
gmatdriller 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
Using the formula: Average * number = sum, we see that the total number of customers served over the past x days is 75*x, or 75x.

Using the formula: sum/number = average, we can create the equation:

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

75x + 120 = 90x + 90

30 = 15x

2 = x

Answer: A

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