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), (5), 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