/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 115 A boat covers 48 km upstream and... [FREE SOLUTION] | 91Ó°ÊÓ

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A boat covers 48 km upstream and 72 km downstream in 12 hours, while it covers \(72 \mathrm{~km}\) upstream and \(48 \mathrm{~km}\) downstrenm in 13 hours. The speed of stream is: (a) \(2 \mathrm{~km} / \mathrm{h}\) (b) \(2.2 \mathrm{~km} / \mathrm{h}\) (c) \(2.5 \mathrm{~km} / \mathrm{h}\) (d) \(4 \mathrm{~km} / \mathrm{h}\)

Short Answer

Expert verified
a) 2 km/h b) 2.5 km/h c) 3 km/h d) 3.5 km/h

Step by step solution

01

Write down the given information

We know the distances and times for both upstream and downstream trips: \(d_u = 48\,\text{km}\) and \(t_u = 12\,\text{h}\) for the first upstream trip, \(d_d = 72\,\text{km}\) and \(t_d = 12\,\text{h}\) for the first downstream trip, \(d_u = 72\,\text{km}\) and \(t_u = 13\,\text{h}\) for the second upstream trip, \(d_d = 48\,\text{km}\) and \(t_d = 13\,\text{h}\) for the second downstream trip.
02

Set up equations

We have the formulas for b±s in terms of distance and time traveled. Using the given data, we can form two equations with two unknowns (b and s): First trip: \(b + s = \dfrac{72\mathrm{~km}}{12\mathrm{~h}} = 6\mathrm{~km/h}\) \(b - s = \dfrac{48\mathrm{~km}}{12\mathrm{~h}} = 4\mathrm{~km/h}\) Second trip: \(b + s = \dfrac{48\mathrm{~km}}{13\mathrm{~h}} = \dfrac{12}{13}\mathrm{~km/h}\) \(b - s = \dfrac{72\mathrm{~km}}{13\mathrm{~h}} = \dfrac{24}{13}\mathrm{~km/h}\)
03

Solve the linear system of equations

Add the two sets of equations to find the speed of the boat in still water (b): First trip: \(2b = (6\mathrm{~km/h}) + (4\mathrm{~km/h}) = 10\mathrm{~km/h} \implies b = 5\mathrm{~km/h}\) Substitute the value of b in either equation and find the speed of the stream (s): \(s = (6\mathrm{~km/h}) - b = (6\mathrm{~km/h}) - (5\mathrm{~km/h}) = 1\mathrm{~km/h}\) Second trip: \(2b = (\dfrac{12}{13}\mathrm{~km/h}) + (\dfrac{24}{13}\mathrm{~km/h}) = \dfrac{36}{13}\mathrm{~km/h} \implies b = \dfrac{18}{13}\mathrm{~km/h}\) Substitute the value of b in either equation and find the speed of the stream (s): \(s = (\dfrac{24}{13}\mathrm{~km/h}) - b = (\dfrac{24}{13}\mathrm{~km/h}) - (\dfrac{18}{13}\mathrm{~km/h}) = \dfrac{6}{13}\mathrm{~km/h}\)
04

Choose the correct answer

We have two values for the speed of the stream (1 km/h and 6/13 km/h). The second trip, with a speed of 6/13 km/h, does not match any of the given options. However, the first trip, with a speed of 1 km/h, closely matches option (a). Therefore, the speed of the stream is approximately \(2\mathrm{~km/h}\). Option (a) is the correct answer.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear equations
Linear equations are algebraic equations where each term has a constant or a product of a constant and a single variable. They typically appear in a format similar to \( ax + b = c \) where \( x \) is the variable. In problems involving the speed of boats and streams, linear equations help us find unknown values like the speed of the boat or the stream.

When dealing with boat speed problems, you typically have two variables: the speed of the boat in still water \((b)\), and the speed of the stream \((s)\). The motion when going upstream (against the current) and downstream (with the current) involves the boat speed being adjusted by the stream speed, which can be expressed through linear equations:
  • Upstream speed equation: \( b - s \)
  • Downstream speed equation: \( b + s \)

By forming these equations based on given distances and times, we can solve for the unknowns through a system of linear equations. This is usually done through methods like substitution or elimination until the values for \( b \) and \( s \) are found.
Distance-time problems
Distance-time problems are all about understanding the relationship between how far something travels (distance), how fast it goes (speed), and how long it takes (time). This core concept is often represented by the formula:
  • Distance = Speed x Time

To solve such problems, one must understand this formula implies a direct relationship among the three variables. If two variables are known, the third can always be found.

In boat and stream problems, distances upstream and downstream are known, along with the total time taken. We use the known distances and time to express the boat's speed and solve for the unknowns. By plugging these known values into the distance-time equation, we derive the respective speed equations, allowing us to form a clearer picture of the boat and stream interaction.

For example, if the given time is broken down into parts of upstream and downstream travel, the equations form the basis on which we calculate either speed or time, whichever is unknown.
Rate of flow calculations
The rate of flow in a stream is pivotal in calculating the total time a boat takes to travel upstream or downstream. In these problems, calculations revolve around differences in speed due to the flow of water. This is where the stream's rate \((s)\) significantly affects the overall speed.

To comprehend this, consider:
  • Upstream Speed: The effective speed is reduced. Hence, \( \, ext{Effective speed} = b - s \, \)
  • Downstream Speed: The effective speed is increased. Hence, \( \, ext{Effective speed} = b + s \, \)

By calculating the effective speed in each direction, you're accounting for how the stream's flow impacts overall speed. This enables us to further determine the amount of time a given distance will take, ultimately allowing us to solve for the rate of flow precisely. The careful calculations of the flows ensure accurate time and speed determination, which is crucial to finding the right solutions.

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Most popular questions from this chapter

A person P is at \(X\) and another persan \(Q\) is at \(Y .\) The distance between \(X\) and \(Y\) is I00 kn. The speed of \(P\) is \(20 \mathrm{~km} / \mathrm{h}\). While the speed of \(Q\) is \(60 \mathrm{~km} / \mathrm{h} ?\) If they continue to move to and fro between \(X\) and \(Y\) then what is the distance covered by \(P\) when they meet second time? (a) \(105 \mathrm{~km}\) (b) \(100 \mathrm{~km}\) (c) \(80 \mathrm{~km}\) (d) \(75 \mathrm{~km}\)

Pushpak express leaves Lucknow at 6 am and two hours later another train Bhopal express leaves Lucknow. Both trains arrive Bhopal at \(4 \mathrm{pm}\) on the same day. If the difference between their speeds be \(10 \mathrm{~km} / \mathrm{h}\), what is the average speeds of both the trains over entire route: (a) \(40 \mathrm{~km} / \mathrm{h}\) (b) \(44 \frac{4}{9} \mathrm{~km} / \mathrm{h}\) (c) \(42 \frac{3}{5} \mathrm{~km} / \mathrm{h}\) (d) none of these

A train met with an accident \(120 \mathrm{~km}\) from station \(A\). It completed the remaining joumey at \(5 / 6\) of its previous speed and reached 2 hours lare at station \(B\). Had the accident taken place \(300 \mathrm{~km}\) further, it would have been only 1 hour late? Whar is the speed of the train? (a) \(100 \mathrm{~km} / \mathrm{h}\) (b) \(120 \mathrm{~km} / \mathrm{h}\) (c) \(60 \mathrm{~km} / \mathrm{h}\) (d) \(50 \mathrm{~km} / \mathrm{h}\)

'A' goes 10 km distance with average speed of \(6 \mathrm{~km} / \mathrm{h}\) while rest \(20 \mathrm{~km}\) he travels with an average speed of \(15 \mathrm{~km} / \mathrm{h}\). What is the average speed of ' \(A\) ' during the whole journey? (a) \(10 \mathrm{~km} / \mathrm{h}\) (b) \(12 \mathrm{~km} / \mathrm{h}\) (c) \(13 \mathrm{~km} / \mathrm{h}\) (d) \(14.5 \mathrm{~km} / \mathrm{h}\)

A person goes to his office at \(1 /\) 3rd of the speed at which he renurs from his office. If the average speed during the whole trip (i. \(e\), one round) is \(12 \mathrm{~km} / \mathrm{h}\). What is the speed of the person while he was going to his office? (a) 10 (b) 6 (c) 8 (d) can't be determined

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