A motor boat traveling with the current went 30mi in 3 h. Against the current, it took 5 h to travel the same distance. Find the rate of the boat in calm water and the rate of the current

Speed of the boat is 9.4mph and the current is 15.4mph

Step-by-step explanation:

Rate= distance/ time

With current; rate= boat+ current= b+c

Against current; rate= boat - current = b- c

30=3(b+c)••••••••••••••eqn(1)

30=5(b-c)•••••••••••••••••eqn(2)

30=3b +3c••••••••••••••eqn(3)

30=5b-5c•••••••••••••••eqn(4)

Solve simultaneously

b=12mph

From eqn(3)

30=3b +3c

Divide both sides by 3, we have

10= b+c

If we make "b" the subject of the formula we have

b= (10-c)••••••••eqn(5)

Divide both side of eqn(4) by 5 we have

6=b-c••••••••eqn(7)

Substitute eqn(5) into 7

6= (10-c)-c

6= 10c-c^2

C^2-10c+6

C= 9.4 or 0.64

Substitute the highest value 9.4 into equation 7

6=b-9.4

b=6+9.4

b= 15.4

Speed of the boat is 9.4mph and the current is 15.4mph