SOLUTION: In an organization there are CE’s, ME’s and EE’s. The sum of their ages is 2160; the average age is 36; the average age of the CE’s and ME’s is 39; the average age of the ME’s and

In an organization there are CE’s, ME’s and EE’s. The sum of

their ages is 2160; the average age is 36; the average age of

the CE’s and ME’s is 39; the average age of the ME’s and EE’s

is 32 8/11; the average age of the CE’s and EE’s is 36 2/3.

If each CE had been 1 year older, each ME 6 years and each EE

7 years older, their average age would have been greater by 5

years. Find the number of each group and their average ages.

Suppose there are C CE's,

Suppose there are M ME's,

Suppose there are E EE's,

Suppose the average age of the CE's is x years

Suppose the average age of the ME's is y years

Suppose the average age of the EE's is z years

>>...The sum of their ages is 2160...<<

Cx + My + Ez = 2160

>>...the average age is 36...<<

(Cx + My + Ez)/(C + M + E) = 36

>>...the average age of the CE’s and ME’s is 39...<<

(Cx + My)/(C + M) = 39

>>...the average age of the ME’s and EE’s is 32 8/11...<<

(My + Ez)/(M + E) = 32 8/11

>>...the average age of the CE’s and EE’s is 36 2/3...<<

(Cx + Ez)/(C + E) = 36 2/3

>>...If each CE had been 1 year older, each ME 6 years and

each EE 7 years older, their average age would have been

greater by 5 years...<<

[C(x+1)+M(y+6)+E(z+7)]/(C + M + E) = 36 + 5

So we have 6 equations in 6 unknowns:

(1): Cx + My + Ez = 2160

(2): (Cx + My + Ez)/(C + M + E) = 36

(3): (Cx + My)/(C + M) = 39

(4): (My + Ez)/(M + E) = 32 8/11

(5): (Cx + Ez)/(C + E) = 36 2/3

(6): [C(x+1)+M(y+6)+E(z+7)]/(C + M + E) = 36 + 5

Replacing the numerator of (2) by 2160 from (1)

 2160/(C + M + E) = 36

             2160 = (C + M + E)(36)

               60 = C + M + E

(7)     C + M + E = 60

Using (7), we can put 60 for the denominator in (6)

     [C(x+1)+M(y+6)+E(z+7)]/60 = 36 + 5

     [C(x+1)+M(y+6)+E(z+7)]/60 = 41

          C(x+1)+M(y+6)+E(z+7) = 2460

          Cx+C + My+6M + Ez+7E = 2460 

        Cx+My+Ez + C + 6M + 7E = 2460

Now from (1) we can replace Cx+My+Ez by 2160

           2160 + C + 6M + 7E = 2460

(8)               C + 6M + 7E = 300

Cross-multiplying (3), (4), and (5)

(9)            Cx + My = 39(C + M)

(10)           My + Ez = (32 8/11)(M + E)

(11)           Cx + Ez = (36 2/3)(C + E) 

Adding the left sides and the right sides, we get

       2Cx + 2My + 2Ez = (227/3)C + (789/11)M + (2290/33)E

Multiplying thru by 33 to clear of fractions:

             66Cx + 66My + 66Ez = 2497C + 2367M + 2290E

               66(Cx + My + Ez) = 2497C + 2367M + 2290E

By (1) we can replace the parenthetical expression by 2160.

                       66(2160) = 2497C + 2367M + 2290E 

or

(12)       2497C + 2367M + 2290E = 142560

So with (7), (8), and (12), we have three equations in three

unknowns:

     C +     M +     E = 60

     C +    6M +    7E = 300

 2497C + 2367M + 2290E = 142560

Solving this system gives C=16, M=24, E=20 

Now (1) becomes

       16x + 24y + 20z = 2160

or dividing thru by 4

(13)      4x + 6y + 5z = 540

Now (9) becomes:

               16x + 24y = 39(16 + 24)

               16x + 24y = 1560

Dividing thru by 8

(14)             2x + 3y = 195

Now (11) becomes

           16x + 20z = (36 2/3)(16 + 20)

or         16x + 20z = 1320

Dividing thru by 4

(15)         4x + 5z = 330 

Now (13),(14), and (15) form a system of 3 equations in 3

unknowns

        4x + 6y + 5z = 540 

        2x + 3y      = 195

        4x      + 5z = 330

Solving that system gives x=45, y=35, z=30

Answer: 

Average age of the 16 CE's is 45

Average age of the 24 ME's is 35

Average age of the 20 EE's is 30

Edwin