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The average (arithmetic mean) of four different positive integers is 9

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Bunuel
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The average (arithmetic mean) of four different positive integers is 9 [#permalink] New post   09 Dec 2019, 02:47
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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 highest possible value of the fourth integer?

A. 39
B. 34
C. 30
D. 29
E. 24


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E
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The average (arithmetic mean) of four different positive integers is 9 [#permalink] New post   Updated on: 09 Dec 2019, 10:33
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 highest possible value of the fourth integer?

(A+B+c+D)/4=9
A+B+c+D=36
A=3B, B=C-2 therefore C=B+2
putting in equation
3B+B+B+2 +D=36
5B+D=34
D=34- 5B
To maximize D we need to minimize 5B
Now we know all are positive so Minimum B = 1
therefore Maximum D =29 (34-5(1)) but in this case A and C will have same value
so lets PUT B=2 so
D=24

Originally posted by OC2910 on 09 Dec 2019, 08:34.
Last edited by OC2910 on 09 Dec 2019, 10:33, edited 1 time in total.
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The average (arithmetic mean) of four different positive integers is 9 [#permalink] New post   Updated on: 09 Dec 2019, 09:57
given
a+b+c+d=36
a=3b
b=c-2
c=b+2
3b+b+b+2+d=36
d= 34-5b
b= 1;
d= 29

But since all have to be different integers and at b=1 we get c=a=3
so at b=3 d=24
IMO E

Bunuel 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 highest possible value of the fourth integer?

A. 39
B. 34
C. 30
D. 29
E. 24


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Originally posted by Archit3110 on 09 Dec 2019, 08:53.
Last edited by Archit3110 on 09 Dec 2019, 09:57, edited 1 time in total.
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Re: The average (arithmetic mean) of four different positive integers is 9 [#permalink] New post   09 Dec 2019, 09:43
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StrugglingGmat2910 Archit3110

At b=1, a=c=3.

It's mentioned in the question that 4 numbers are distinct.

At b=2, a=6; c=4; and d=24 (this case satisfies all the conditions)
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The average (arithmetic mean) of four different positive integers is 9 [#permalink] New post   09 Dec 2019, 09:50
nick1816 wrote:
StrugglingGmat2910 Archit3110

At b=1, a=c=3.

It's mentioned in the question that 4 numbers are distinct.

At b=2, a=6; c=4; and d=24 (this case satisfies all the conditions)


Nope this misses the condition of D to be maximum : please refer my solution
As per your numbers you have assumed B =2 in my final equation,which gives one of the many solutions not the maximum
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Re: The average (arithmetic mean) of four different positive integers is 9 [#permalink] New post   09 Dec 2019, 09:58
You can't have a case in which D greater than 24 and satisfies all the conditions mentioned in the questions. Anyways, you can wait till the OA gonna revealed.

StrugglingGmat2910 wrote:
nick1816 wrote:
StrugglingGmat2910 Archit3110

At b=1, a=c=3.

It's mentioned in the question that 4 numbers are distinct.

At b=2, a=6; c=4; and d=24 (this case satisfies all the conditions)


Nope this misses the condition of D to be maximum : please refer my solution
As per your numbers you have assumed B =2 in my final equation,which gives one of the many solutions not the maximum
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Re: The average (arithmetic mean) of four different positive integers is 9 [#permalink] New post   09 Dec 2019, 10:08
nick1816 wrote:
You can't have a case in which D greater than 24 and satisfies all the conditions mentioned in the questions. Anyways, you can wait till the OA gonna revealed.

StrugglingGmat2910 wrote:
nick1816 wrote:
StrugglingGmat2910 Archit3110

At b=1, a=c=3.

It's mentioned in the question that 4 numbers are distinct.

At b=2, a=6; c=4; and d=24 (this case satisfies all the conditions)


Nope this misses the condition of D to be maximum : please refer my solution
As per your numbers you have assumed B =2 in my final equation,which gives one of the many solutions not the maximum


absolutely pardon i missed that you perfectly correct
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Re: The average (arithmetic mean) of four different positive integers is 9 [#permalink] New post   09 Dec 2019, 10:12
StrugglingGmat2910 wrote:
nick1816 wrote:
StrugglingGmat2910 Archit3110

At b=1, a=c=3.

It's mentioned in the question that 4 numbers are distinct.

At b=2, a=6; c=4; and d=24 (this case satisfies all the conditions)


Nope this misses the condition of D to be maximum : please refer my solution
As per your numbers you have assumed B =2 in my final equation,which gives one of the many solutions not the maximum


I think nick1816 is right.
There is restriction on the selection of numbers.. i.e first the numbers should be different & secondly they should be +ve . The maximum value of D is to be arrived keeping these two restrictions in mind. You are ignoring "different numbers" restriction. Similarly, if we ignore +ve numbers restrictions, we may get even higher value of D like considering b as -ve
d= 34-5b
d= 34-5(-1) = 34+5 = 39.. even higher value of D.

So, i think both restrictions are to be followed--- DIFFERENT & +VE..
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Re: The average (arithmetic mean) of four different positive integers is 9 [#permalink] New post   09 Dec 2019, 11:11
X+1/3x+1/3x+2+y/4=9
34=5/3x+y
34-5/3x=y
If x=3 we have 34-5=29 every thing else is a non positive or non integer or smaller

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Re: The average (arithmetic mean) of four different positive integers is 9 [#permalink] New post   11 Dec 2019, 06:50
Bunuel 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 highest possible value of the fourth integer?

A. 39
B. 34
C. 30
D. 29
E. 24


a,b,c,d=9*4=36 (distinct positive integers)
a=3b, c=b+2;
3b+b+2+b+d=36
5b+d=34
b=34-d/5
d=4,9,…24,29

d=29: b=34-29/5=1, a=3b=3, c=b+2=3, a=c, invalid;
d=24: b=34-24/5=2, a=6, c=4, valid.

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Re: The average (arithmetic mean) of four different positive integers is 9 [#permalink] New post   12 Jan 2023, 08:52
3x+x+x+2+y=36
y=34-5x
now ,x=2 is satisfied the condition , 3x=6,x=2,x+2=4
so,y=34-5*2=24
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Re: The average (arithmetic mean) of four different positive integers is 9 [#permalink]
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