(y+15) men can do a piece of work in (z-30) days. (y-15)

Solution

(y+15) men can do a piece of work in (z-30) days.

Total work = Men_1 x Days_1

= (y+15)x(z-30)    Eq.(i)

(y-15) men can do the same piece of work in (z+45) days.

Total work = Men_2 x Days_2

= (y-15)x(z+45)    Eq.(ii)

If (y+25) men can do the same work in (z-45) days.

Total work = Men_3 x Days_3

= (y+25)x(z-45)    Eq.(iii)

So Eq.(i) = Eq.(ii).

(y+15)x(z-30) = (y-15)x(z+45)

yz-30y+15z-450 = yz+45y-15z-675

-30y+15z-450 = 45y-15z-675

45y-15z-675+30y-15z+450 = 0

75y-30z = 675-450

75y-30z = 225

5y-2z = 15    Eq.(iv)

So Eq.(i) = Eq.(iii).

(y+15)x(z-30) = (y+25)x(z-45)

yz-30y+15z-450 = yz-45y+25z-1125

-30y+15z-450 = -45y+25z-1125

45y-25z+1125-30y+15z-450 = 0

15y-10z+675 = 0

3y-2z = -135    Eq.(v)

So Eq.(iv)-Eq.(v).

5y-2z-(3y-2z) = 15-(-135)

5y-2z-3y+2z = 15+135

2y = 150

y = 75

Put the value of ‘y’ in Eq.(v)

3x75-2z = -135

225-2z = -135

225+135 = 2z

2z = 360

z = 180

So the values of ‘y’ and ‘z’ are 75 and 180 respectively.