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The average of 63, 37 and x is greater than the average of the

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The average of 63, 37 and x is greater than the average of the  [#permalink]

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New post 25 Oct 2018, 01:36
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A
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D
E

Difficulty:

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Question Stats:

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The average of 63, 37 and x is greater than the average of the multiples of 3 between 1 and 111, inclusive. Which of the following is a possible value of x?
I. 64 II. 71 III. 75 IV. 85
A) I, II, Ill, and IV
B) IV only
C) II, Ill and IV
D) III and IV
E). None of these

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Re: The average of 63, 37 and x is greater than the average of the  [#permalink]

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New post 25 Oct 2018, 01:56
1
Hi,

It is a simple question testing the concept of Average.

Average = Total / Number of elements

So, average of three numbers 63,37 and x is

(100+x)/3

Average of multiples of 3 between 1 and 111(inclusive) is,

Multiples of 3 are,

3 ,6, 9, …111

We can find the average of multiples of 3 in using different ways,

One such way is,

Average = (3+111)/2

Rule: If a list of numbers were evenly spaced (common difference) then Average of the entire list is equal to the average of first element and last element of the set.

So average = 57

Given,

(100+x)/3 > 57

100 + x > 171

x > 71

So the answers are 75 and 85.

So the answer is D.

Hope this helps.
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Re: The average of 63, 37 and x is greater than the average of the  [#permalink]

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New post 30 Jul 2019, 06:55
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Abhi077 wrote:
The average of 63, 37 and x is greater than the average of the multiples of 3 between 1 and 111, inclusive. Which of the following is a possible value of x?
I. 64 II. 71 III. 75 IV. 85
A) I, II, Ill, and IV
B) IV only
C) II, Ill and IV
D) III and IV
E). None of these


rule: avg=sum•n
rule: multiple(x) between range: greatestm(x)-leastm(x)/x+1
rule: average of an arithmetic progression: greatestm(x)+leastm(x)/2
rule: any number is divisible by 3 if the sum of its digits is divisibly by 3

avg m(3) from 1 to 111 inclusive: greatestm(3)=111, leastm(3)=3, avg=111+3/2=57

given:
[63+37+x]/3>57
100+x>171
x>71

Answer (D).
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Re: The average of 63, 37 and x is greater than the average of the   [#permalink] 30 Jul 2019, 06:55
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