Number of Civil, Electrical, and Mechanical Engineers and Their Average Ages

Let
C, E, M = number of engineers (Civil, Electrical, Mechanical)
c, e, m = corresponding average age (Civil, Electrical, Mechanical)
 

The sum of their ages is 2160:
$cC + eE + mM = 2160$     ←   Equation (1)
 

The average age is 36:
$\dfrac{2160}{C + E + M} = 36$

$C + E + M = 60$     ←   Equation (2)
 

The average age of CE’s and EE’s is 39:
$\dfrac{cC + eE}{C + E} = 39$

$cC + eE = 39(C + E)$     ←   Equation (3)
 

The average age of EE’s and ME’s is 32 and 8/11
$\dfrac{eE + mM}{G + M} = 32\frac{8}{11}$

$eE + mM = 32\frac{8}{11}(E + M)$     ←   Equation (4)
 

the average age of the CE’s and ME’s is 36 and 2/3
$\dfrac{cC + mM}{C + M} = 36\frac{2}{3}$

$cC + mM = 36\frac{2}{3}(C + M)$     ←   Equation (5)
 

If each CE had been 1 year older, each EE 6 years and each ME 7 years older, their average age would have been greater by 5 years:
$\dfrac{(c + 1)C + (e + 6)E + (m + 7)M}{C + E + M} = 36 + 5$

$\dfrac{(cC + eE + mM) + (C + 6E + 7M)}{C + E + M} = 41$

$\dfrac{2160 + (C + 6E + 7M)}{60} = 41$

$C + 6E + 7M = 300$     ←   Equation (6)
 

Equation (3) + Equation (4) + Equation (5)
$2(cC + eE + mM) = 75\frac{2}{3}C + 71\frac{8}{11}E + 69\frac{13}{33}M$

$75\frac{2}{3}C + 71\frac{8}{11}E + 69\frac{13}{33}M = 2(2160)$     ←   Equation (7)
 

Number of CE, EE, and ME from Equations (2), (6), and (7):
C = 16 Civil Engineers
E = 24 Electrical Engineers
M = 20 Mechanical Engineers
 

Substitute C, E, and M to Equations (3), (4), and (5):
$16c + 24e = 39(16 + 24)$     ←   from Equation (3)

$24e + 20m = 32\frac{8}{11}(24 + 20)$     ←   from Equation (4)

$16c + 20m = 36\frac{2}{3}(16 + 20)$     ←   from Equation (5)
 

Average ages from the three equations above
c = 45 yrs old
e = 35 yrs old
m = 30 yrs old