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

Let the CE be X

Let the ME be Y

Let the EE be Z

the average age of the CE's is c years

the average age of the ME's is m years

the average age of the EE's is e years

The Sum of the total ages is 2160: Equation:

 (1)

And there total average age is 36 so:

 (2)

And the average of CE and ME is 39 so:

 (3)

the average age of the ME’s and EE’s is 32 8/11 ---->360/11

 (4)

the average age of the CE's and EE's is 36 2/3 --->110/3

 (5)

"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"

SO the equation would be:

 (6)

We have 6 equations in our case:

---> Replace equation 2: --> SImplfy that:

X+Y+Z=60 (7)

ANd SInce X+Y+Z=60 replace it with equation 6.





 

And from equation number 1 We can replace Xc+Ym+Ze by 2160:

2160+X+6Y+7Z=2460 

X+6Y+7Z=300(8)

Cross multiply the equations 3 4 and 5.

Xc + Ym = 39(X + Y) (set 1)

Ym + Ze = (360/11)(Y + Z) (set 2)

Xc + Ze = (110/3)(X + Z) (set 3)

Add all the sets to get:

2Xc + 2Ym + 2Ze = (227/3)X + (789/11)Y + (2290/33)Z

Since 33 is the lowest multiple; multiply the whole equation by 33.

 and simplify

66Xc + 66Ym + 66Ze = 2497X + 2367Y + 2290Z

Apply the common factor then subsitute with the first equation:

66(2160)=2497X + 2367Y + 2290Z (9)

So after simplfying we have 7,8,9 unknowns

X+Y+Z=60 (7)

X+6Y+7Z=300  (8)

142560=2497X + 2367Y + 2290Z (9)

Subtract 8 from 7 to get:

5Y+6Z=240

Solve for Y:

y=48-1.2Z (subsitution)

Simplfy throughout using substution and elimination method to get:

X=16, Y=24, Z=20

Plug that throughout the quation and solve for the rest as averages:

If you still need help e-mail me.

Paul.