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elena55 [62]
1 year ago
12

Traveling with the current, a canoe went 30 miles in 3 hours. Traveling against the current, the canoe went 12 miles in 3 hours.

What is the rate of the canoe in calm water?
Mathematics
1 answer:
lora16 [44]1 year ago
6 0

Answer

Rate of canoe in calm water is 7 mph

Step-by-step explanation:

Let y be the rate of the canoe in calm water

Let z represent the rate of current

Traveling with the current, the canoe went 30miles in 3 hours

The total speed of the current and canoe will be (y + z) mph

Recall that,

speed = distance / time

Distance = speed x time

Distance = 30 miles

time = 3 hours

30 = 3( y + z)

divide through by 3

30/3 = 3/3 (y + z)

10 = y + z -------------------- equation 1

Against the current, the canoe went 12 miles in 3 hours

Let the speed = (y - z) mph

Distance = speed x time

12 = 3(y - z)

Divide through by = 3

4 = y - z ------------- equation 2

The two-equation can be solved simultaneously

y + z = 10 --------------- equation 1

y - z = 4 -------------- equation 2

Eliminate z by adding the two equations equation together

(y + y) + (z - z) = 10 + 4

2y + 0 = 14

2y = 14

y = 14/2

y = 7 mph

To find z, put y in equation 1

y + z = 10

z = 10 - y

z = 10 - 7

z = 3 mph

Hence, the rate of canoe in calm water is 7mph

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