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Anarel [89]
3 years ago
10

Together two ferries can transport a day's cargo in 7 hours. The larger ferry can transport cargo three times faster than the sm

aller ferry. How long does it take the larger ferry to transport a day's cargo working alone?
Mathematics
1 answer:
Savatey [412]3 years ago
5 0

Answer:

It takes the larger ferry 9 hours and 20 minutes to work alone.

Step-by-step explanation:

If the two ferries work together and transport a day's cargo in 7 hours

And if the larger ferry can transport cargo three times faster than the smaller ferry,

This implies that the speed ratio of the larger ferry to the smaller ferry is equal to 3 : 1

Then the sum of the ratio = 4 (3 + 1)

From the above, it will take the smaller ferry  28 hours to work alone (7 x 4), given its speed proportion of 1: 3.

Therefore, when the larger ferry works alone, it will take it (7 x 4)/3 = 9.33 hours.

9.33 hours = 9 hours, 20 minutes.

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For the first question please find the attached diagram.

As per the diagram, P is the upstream point and Q is the downstream point. The distance between P and Q is 22.5 miles.

Let the speed of the boat in the still waters of the lake be represented by S.

Then, when the boat travels upstream, the net speed of the boat will be (S-6) miles per hour because the river flows downstream and thus the speed of the boat will have to be subtracted from the speed of the river.

Now, we know that the relationship between the net speed, distance and time of travel is give as:

Distance = Net Speed x Time of travel

For the upstream ride of the board we know that Distance is 22.5 miles and Net Speed is (S-6). Therefore, the above equation will become:

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We can rearrange the above equation to be:

T_{1}=\frac{22.5 }{S-6}......................(Equation 1)

By similar arguments we know that the downstream speed of the boat is S+6 and the distance travelled is the same and so the time taken to travel downstream (represented by T_{2}) will be:

T_{2}=\frac{22.5}{S+6}................(Equation 2)

Now, we know that the total time of travel should be 9 hours.

This means that: T_{1}+T_{2}=9............(Equation 3)

Plugging in the values of T_{1} and T_{2} from (Equation 1) and (Equation 2) into (Equation 3), we get:

\frac{22.5 }{S-6} +\frac{22.5 }{S+6}=9

Simplifying the above we will get a quadratic equation:

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The roots of this quadratic equation are:

S=-1 and S=6

Since, speed cannot be negative, S=-1 is out of consideration.

The speed of the boat in the lake is thus S=6 miles per hour.

But we have a problem with S=6 too. The problem is that if S=6, then the boat will not be able to move upstream.

Let us solve problem 2

We are given that: \frac{x-2}{x+3}+\frac{10x}{x^2-9}

We can rewrite it as:

\frac{x-2}{x+3}+\frac{10x}{(x-3)(x+3)}

\frac{(x-3)(x-2)+10x}{(x-3)(x+3)} =\frac{x^2-5x+6+10x}{(x-3)(x+3)}

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\frac{x^2+10x+6}{(x-3)(x+3)} =\frac{(x+3)(x+2)}{(x-3)(x+3)} =\frac{x+2}{x-3}

Thus, our final simplified answer is:

\frac{x+2}{x-3}

The restriction on the variable x is that it cannot be equal to either +3 or -3 as that would make the denominator of the original question equal to zero.

Thus, the restriction is x\neq \pm 3

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