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Difficulty:
45%
(medium)
Question Stats:
69% (02:09) correct
31%
(02:24)
wrong
based on 683
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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
A) 5
B) 4
C) 3
D) 2
E) 1
17 Mar 2013, 01:23
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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.
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.
27 Oct 2016, 12:21
Kudos
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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
B is the correct answer.
[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
B is the correct answer.
