In an organization there are CE’s, GE’s and ME’s. The sum of their ages is 2160

Let
C, G, M = number of engineers (Civil, Geodetic, Mechanical)
p, q, r = corresponding average age ((Civil, Geodetic, Mechanical)
 

The sum of their ages is 2160:
$pC + qG + rM = 2160$     ←   Equation (1)
 

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

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

The average age of CE’s and GE’s is 39:
$\dfrac{pC + qG}{C + G} = 39$

$pC + qG = 39(C + G)$     ←   Equation (3)
 

The average age of GE’s and ME’s is 32 and 8/11
$\dfrac{qG + rM}{G + M} = 32\frac{8}{11}$

$qG + rM = 32\frac{8}{11}(G + M)$     ←   Equation (4)
 

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

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

If each CE had been 1 year older, each GE 6 years and each ME 7 years older, their average age would have been greater by 5 years:
$\dfrac{(p + 1)C + (q + 6)G + (r + 7)M}{C + G + M} = 36 + 5$

$\dfrac{(pC + gG + rM) + (C + 6G + 7M)}{C + G + M} = 41$

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

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

Equation (3) + Equation (4) + Equation (5)
$2(pC + qG + rM) = 75\frac{2}{3}C + 71\frac{8}{11}G + 69\frac{13}{33}M$

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

From Equations (2), (5), and (7):
C = 16, G = 24, M = 20
 

Substitute C, G, and M to Equations (3), (4), and (5):
$16p + 24q = 39(16 + 24)$     ←   from Equation (3)

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

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

From the three equations above
p = 45, q = 35, r = 30
 

Number of Civil Engineers = 16
Average age of Civil Engineers = 45 years old