Customers were asked to rate

Question

Customers were asked to rate their satisfaction on an integer scale from 0 to 10, with 0 being very dissatisfied and 10 being very satisfied. The average (mean) rating was 7.25, the median rating was 5, and exactly four customers gave a rating of 0.

What is the smallest possible number of customers who participated in the survey?

(A) 70
(B) 72
(C) 75
(D) 76
(E) 100

What is the smallest possible number of customers who participated in the survey?

(A) 70
(B) 72
(C) 75
(D) 76
(E) 100

paragw · Answer

Here's my solution:

MehakKaushal · Answer

Mean =7.25=4/29, Mean= total score/no of customers, so the number of customers must be a multiple of 4.
Possible choices: 72, 76, 100.
To maximize the total score for a given n:
4 customers rate 0.
Both middle ratings are 5 (median = 5).
All other lower-half ratings are 5.
All remaining upper-half ratings are 10.
Maximum possible total score:
5(n/2−4)+5+10(n/2−1)=15n/2 −25.

Check the options:

n=72: max =515, required =7.25×72=522 ❌
n=76: max =545, required =551 ❌
n=100: max =725, required =725 ✅
Answer: (E) 100

MistyNova · Answer

why should the terms preceding to median be 5?

similiqueamet · Answer

Since the mean is a terminating decimal we can be sure that the number of participants must be a multiple of 2^n * 5^n. Since only 100 satisfies this, we can quickly choose E for this.

sujoykrdatta · Answer

Since the mean is 7.25 = 29/4, the number of people must be a multiple of 4, i.e. 72, 76 or 100
Let the number of terms be 4n => Sum of the numbers = 29/4 * 4n = 29n
4 of them are zero => total = 0
The two middle values m1, m2 have average of 5 (since median is 5) => Sum = 10

If 4n = 72: 72 numbers add up to 522
0,0,0,0, a1,a2,a3... m1,m2, b1,b2,b3... = 522
Best possible case is if all numbers after 0 till 35th are 5, and 36th + 37th is 10, adding up to 0*4 + 31*5 + 10 = 165
So, we need to make 522-165 = 367 from the remaining 35 numbers which is not possible since max value can be 10

If 4n = 76: 76 numbers add up to 551
 0,0,0,0, a1,a2,a3... m1,m2, b1,b2,b3... = 551
Best possible case is if all numbers after 0 till 37th are 5, and 38th + 39th is 10, adding up to 0*4 + 33*5 + 10 = 175
So, we need to make 551-175 = 376 from the remaining 37 numbers which is not possible since max value can be 10
 
By elimination, it must be 100:

If 4n = 100: 100 numbers add up to 725
 0,0,0,0, a1,a2,a3... m1,m2, b1,b2,b3... = 725
Best possible case is if all numbers after 0 till 49th are 5, and 50th + 51st is 10, adding up to 0*4 + 49*5 + 10 = 255
So, we need to make 725-255 = 470 from the remaining 49 numbers which is easily possible (since max value can be 10)
 
 Ans E

kevincan · Answer

Be careful ! We can say that the number of participants is a multiple of 4, but need not be a multiple of 5. Nor can you rule out the possibility that the number of participants is a multiple of any other prime number .

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