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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
Are You Up For the Challenge: 700 Level Questions
A. 39
B. 34
C. 30
D. 29
E. 24
Are You Up For the Challenge: 700 Level Questions
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
(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
Updated on: 09 Dec 2019, 09:57
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.
Last edited by Archit3110 on 09 Dec 2019, 09:57, edited 1 time in total.
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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
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
