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# The average (arithmetic mean) of four different positive...

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Joined: 16 Mar 2013
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The average (arithmetic mean) of four different positive...  [#permalink]

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Updated on: 17 Mar 2013, 01:30
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45% (medium)

Question Stats:

69% (02:16) correct 31% (02:25) wrong based on 274 sessions

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The average (arithmetic mean) of four different positive integers is 9. If the first of these integers in 3 times the second integer and the second integer is 2 less than the third integer, what is the least possible value of the fourth integer?

A) 5
B) 4
C) 3
D) 2
E) 1

Originally posted by sheyla96 on 16 Mar 2013, 23:15.
Last edited by Bunuel on 17 Mar 2013, 01:30, edited 1 time in total.
Moved to PS forum.
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Re: The average (arithmetic mean) of four different positive...  [#permalink]

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17 Mar 2013, 01:23
Let the second integer be x and the fourth be a.

Then [3x + x + (x+2) + a]/4 = 9
=> 5x + 2 + a = 36
=> 5x + a = 34
=> a = 34 - 5x

From the above equation we can see that a is minimum when x is maximum, provided both are positive
The maximum value that x can take in the above equation while still keeping a positive is x=6
This gives us a= 34 - 30 = 4

Therefore the minimum value that the fourth integer can have is 4. Option B.
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Re: The average (arithmetic mean) of four different positive...  [#permalink]

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27 Oct 2016, 12:21
Starting out...

[w+x+y+z]/4 = 9

w=3x --> 3y-6
x = y-2

Plug back into main equation

3y-6+y-2+y+z=36
5y-8+z=36
5y+z=44

We know multiples of 5 give us either a unit digit of 0 or 5 --> In this case all of the values w,x,y,z can take on are positive, thus we are limited to a unit digit of 0, making z = 4 & y = 8

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Re: The average (arithmetic mean) of four different positive...  [#permalink]

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27 Oct 2016, 15:03
sheyla96 wrote:
The average (arithmetic mean) of four different positive integers is 9. If the first of these integers in 3 times the second integer and the second integer is 2 less than the third integer, what is the least possible value of the fourth integer?

A) 5
B) 4
C) 3
D) 2
E) 1

a+b+c+d=36
a+b+c=5b+2
greatest possible value of 5b+2<36=32
36-32=4=least possible value of d
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Re: The average (arithmetic mean) of four different positive...  [#permalink]

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18 Dec 2016, 14:08
1
Here is my solution to this one ->

Let the integers be-->
w1
w2
w3
w4

Mean = 9

Using -->

$$Mean = \frac{Sum}{#}$$

Hence sum(4)=9*4=36

As per Question-->
w1=3w3-6
w2=w3-2
w3
w4

Hence w4=44-5w3

To minimise w4 => w3=> 8

Hence w4 => 4

Hence => B

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Re: The average (arithmetic mean) of four different positive...  [#permalink]

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19 Dec 2016, 12:14
sheyla96 wrote:
The average (arithmetic mean) of four different positive integers is 9. If the first of these integers in 3 times the second integer and the second integer is 2 less than the third integer, what is the least possible value of the fourth integer?

A) 5
B) 4
C) 3
D) 2
E) 1

a + b + c + d = 36

(3c - 6) + (c - 2) + c + d = 36

Or, 5c + d = 44

Now, use Maximization/Minimization...

least possible value of the fourth integer ( ie, d ) is the maximum possible value of 5c

Here maximum possible value of 5c = 40 ; so d = 4

Hence, Correct answer will be (B) 4

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Re: The average (arithmetic mean) of four different positive...  [#permalink]

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03 Jan 2017, 17:49
1
sheyla96 wrote:
The average (arithmetic mean) of four different positive integers is 9. If the first of these integers in 3 times the second integer and the second integer is 2 less than the third integer, what is the least possible value of the fourth integer?

A) 5
B) 4
C) 3
D) 2
E) 1

We are given that the average of four different positive integers is 9. Thus, the sum of the integers is 4 x 9 = 36.

We can create the following equation in which a = the first integer, b = the second integer, c = the third integer, and d = the fourth integer:

a + b + c + d = 36

We are given that the first of these integers is 3 times the second integer. Thus:

a = 3b

We are also given that the second integer is 2 less than the third integer. Thus:

b = c - 2

b + 2 = c

We can substitute 3b for a and (b + 2) for c in the equation a + b + c + d = 36:

3b + b + b + 2 + d = 36

5b + d + 2 = 36

5b + d = 34

5b = 34 - d

b = (34 - d)/5

Since b is an integer, 34 - d must be a multiple of 5, and since the largest multiple of 5 less than 34 is 30, the smallest value for d is 4.

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The average (arithmetic mean) of four different positive...  [#permalink]

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04 Jan 2017, 09:40
sheyla96 wrote:
The average (arithmetic mean) of four different positive integers is 9. If the first of these integers in 3 times the second integer and the second integer is 2 less than the third integer, what is the least possible value of the fourth integer?

A) 5
B) 4
C) 3
D) 2
E) 1

SUM ($$a_1$$,...,$$a_5$$) = 36
$$a_1$$ = 3$$a_2$$
$$a_3$$ = $$a_2$$ + 2

SUM ($$a_1$$,...,$$a_5$$) = 3$$a_2$$ + $$a_2$$ + $$a_2$$ + 2 + $$a_4$$ = 36 --> 5$$a_2$$ + $$a_4$$ = 34 --> Min($$a_4$$) = 4
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Re: The average (arithmetic mean) of four different positive...  [#permalink]

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09 Feb 2019, 04:37
sheyla96 wrote:
The average (arithmetic mean) of four different positive integers is 9. If the first of these integers in 3 times the second integer and the second integer is 2 less than the third integer, what is the least possible value of the fourth integer?

A) 5
B) 4
C) 3
D) 2
E) 1

If i can get the 3 unknowns in one variable and add the 4th unknown variable to it, i should be good

a= 3b = 3c -6
b =c-2
c =c
d= d

Now a+b+c+d = 36
d = 44-5c
c = 8

d = 4

B
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Re: The average (arithmetic mean) of four different positive...   [#permalink] 09 Feb 2019, 04:37
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