Three positive integers a,b, and c, where c is larger than b and b is larger than a, such that their median is 11, their mean is 9, and their range is 10. Find a.
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simple:
(a+b+c)/3=9,
multiply both sides by three you get
a+b+c =27
median is b=11
so a+11+c=27,then a+c=16
since the range is 10,it means c-a=10,this gives us two simultaneous equations:
a+c=16
-a+c=10
solve them by elimination and you will get a=3,b=11 and c=13
(a+b+c)/3=9,
multiply both sides by three you get
a+b+c =27
median is b=11
so a+11+c=27,then a+c=16
since the range is 10,it means c-a=10,this gives us two simultaneous equations:
a+c=16
-a+c=10
solve them by elimination and you will get a=3,b=11 and c=13
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Sum of the 3 positive integers = N*Xbar = 3*9 = 27
Median is the second positive integer in the ascending order = b = 11
Therefore c+a = 27 - 11 = 16 --------I
Range = c-a = 10 -----------------------II
solve these two equations
on adding we get 2c = 26
c = 26/2 = 13
on substituting the value of c in the first equation we get 13+a = 16
a = 16-13 = 3
Therefore a = 3, b = 11 and c = 13
Median is the second positive integer in the ascending order = b = 11
Therefore c+a = 27 - 11 = 16 --------I
Range = c-a = 10 -----------------------II
solve these two equations
on adding we get 2c = 26
c = 26/2 = 13
on substituting the value of c in the first equation we get 13+a = 16
a = 16-13 = 3
Therefore a = 3, b = 11 and c = 13
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a = 3
b = 11
c = 13
b = 11
c = 13