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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

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Answer:

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

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

User Justin Vallely
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