Practice · Spend downstream time first, bound upstream speed · Sensim Math

Variant 1 easy

A river flows at a steady $2$ km per hour. One point lies upstream and another lies downstream, $70$ km apart. A boat that moves at $18$ km per hour (its own speed, in still water) makes a round trip between the two points: it goes downstream at that speed, and for the trip back upstream it raises its own speed. To finish the round trip within $5$ hours, at least how many km per hour must the boat's own speed be on the way upstream? (The boat's speed is a whole number of km per hour.)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A boat goes $70$ km downstream at its own speed of $18$ km/h on a river flowing at $2$ km/h, then comes back upstream at a higher own speed. The whole round trip must take at most $5$ hours. Find the smallest whole-number own speed for the upstream trip.

Givens

The river flows at $2$ km/h.
The two points are $70$ km apart.
Going downstream, the boat's own speed is $18$ km/h.
The round trip must take $5$ hours or less.

Unknowns

The smallest possible own speed of the boat on the way upstream, in km/h.

Constraints

Downstream, the current adds to the boat's speed; upstream, it subtracts.
The boat's speed is a whole number of km/h.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#8 Analyze the Units#6 Guess and Check

The downstream half is completely known, so finish it first: its time comes off the $5$ -hour budget. What is left is a time limit for the upstream half, which becomes one inequality in the unknown speed. Last, round up to a whole number and check the neighbors.

3 · Execute6 carry out the plan

1Speed over the ground going downstream

#7 Identify Subproblems 6.RP.A.3

Going with the current, the river carries the boat along, so its speed over the ground is its own speed plus the current.

18 + 2 = 20

Downstream the boat really covers

20

km every hour.

2Time for the downstream trip

#8 Analyze the Units 6.RP.A.3

Time is distance divided by speed: hours

=

km

\div

(km per hour).

\frac{70}{20} = \frac{7}{2}

Going downstream takes

\frac{7}{2}

hours, between

3

and

4

hours.

3Time left for the upstream trip

#7 Identify Subproblems 7.EE.B.4

Subtract the downstream time from the

5

-hour limit.

5 - \frac{7}{2} = \frac{3}{2}

The upstream trip may take at most

\frac{3}{2}

hours.

4Write the inequality

#13 Convert to Algebra 7.EE.B.4

Let the boat's own speed upstream be

x

km/h. The current pushes against it, so its speed over the ground is

x - 2

km/h, and the upstream time must fit in what is left.

\frac{70}{x - 2} \le \frac{3}{2}

A faster boat makes the left side smaller, so the inequality sets a lower limit on

x

.

5Clear the fractions

#13 Convert to Algebra 7.EE.B.4

The boat must make progress upstream, so

x - 2 > 0

. Multiplying both sides by the positive number

2(x - 2)

keeps the direction of the inequality.

140 \le 3(x - 2), \quad 146 \le 3x, \quad x \ge \frac{146}{3} = 48\frac{2}{3}

Multiplying by a positive quantity never flips an inequality.

6Round up and check

#6 Guess and Check 7.EE.B.4

The speed is a whole number at least

48\frac{2}{3}

, so the smallest choice is

49

. Check it and the number just below.

\frac{70}{47} \approx 1.489 \le 1.5, \qquad \frac{70}{46} \approx 1.522 > 1.5

At

49

km/h the trip fits within

5

hours; at

48

km/h it does not.

Answer: at least

49

km/h

4 · Reviewdoes it hold up?

At $49$ km/h the upstream ground speed is $47$ km/h, so the round trip takes $\frac{7}{2} + \frac{70}{47} \approx 3.500 + 1.489 = 4.989$ hours, just under $5$ . The upstream speed has to be much larger than $18$ because the downstream leg already used more than $3$ of the $5$ hours.

Another way: Work backwards from the time left: to cover $70$ km in $\frac{3}{2}$ hours, the ground speed must be at least $70 \div \frac{3}{2} = \frac{140}{3} = 46\frac{2}{3}$ km/h. Add back the $2$ km/h current to get an own speed of at least $48\frac{2}{3}$ , so $49$ .

Standardsmin grade 7

6.RP.A.3 Use ratio and rate reasoning to solve problems — Finding ground speeds with the current and times as distance divided by speed.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the upstream time limit as an inequality in the boat's speed and solving $140 \le 3(x - 2)$.

💡Takeaway. When a trip has a known half, spend its time first -- the leftover time turns into a clean inequality for the unknown half.

Variant 2 easy

A river flows at a steady $1$ km per hour. One point lies upstream and another lies downstream, $80$ km apart. A boat that moves at $34$ km per hour (its own speed, in still water) makes a round trip between the two points: it goes downstream at that speed, and for the trip back upstream it raises its own speed. To finish the round trip within $4$ hours, at least how many km per hour must the boat's own speed be on the way upstream? (The boat's speed is a whole number of km per hour.)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A boat goes $80$ km downstream at its own speed of $34$ km/h on a river flowing at $1$ km/h, then comes back upstream at a higher own speed. The whole round trip must take at most $4$ hours. Find the smallest whole-number own speed for the upstream trip.

Givens

The river flows at $1$ km/h.
The two points are $80$ km apart.
Going downstream, the boat's own speed is $34$ km/h.
The round trip must take $4$ hours or less.

Unknowns

The smallest possible own speed of the boat on the way upstream, in km/h.

Constraints

Downstream, the current adds to the boat's speed; upstream, it subtracts.
The boat's speed is a whole number of km/h.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#8 Analyze the Units#6 Guess and Check

The downstream half is completely known, so finish it first: its time comes off the $4$ -hour budget. What is left is a time limit for the upstream half, which becomes one inequality in the unknown speed. Last, round up to a whole number and check the neighbors.

3 · Execute6 carry out the plan

1Speed over the ground going downstream

#7 Identify Subproblems 6.RP.A.3

Going with the current, the river carries the boat along, so its speed over the ground is its own speed plus the current.

34 + 1 = 35

Downstream the boat really covers

35

km every hour.

2Time for the downstream trip

#8 Analyze the Units 6.RP.A.3

Time is distance divided by speed: hours

=

km

\div

(km per hour).

\frac{80}{35} = \frac{16}{7}

Going downstream takes

\frac{16}{7}

hours, a little over

2

hours.

3Time left for the upstream trip

#7 Identify Subproblems 7.EE.B.4

Subtract the downstream time from the

4

-hour limit.

4 - \frac{16}{7} = \frac{12}{7}

The upstream trip may take at most

\frac{12}{7}

hours.

4Write the inequality

#13 Convert to Algebra 7.EE.B.4

Let the boat's own speed upstream be

x

km/h. The current pushes against it, so its speed over the ground is

x - 1

km/h, and the upstream time must fit in what is left.

\frac{80}{x - 1} \le \frac{12}{7}

A faster boat makes the left side smaller, so the inequality sets a lower limit on

x

.

5Clear the fractions

#13 Convert to Algebra 7.EE.B.4

The boat must make progress upstream, so

x - 1 > 0

. Multiplying both sides by the positive number

7(x - 1)

keeps the direction of the inequality.

560 \le 12(x - 1), \quad 572 \le 12x, \quad x \ge \frac{143}{3} = 47\frac{2}{3}

Multiplying by a positive quantity never flips an inequality.

6Round up and check

#6 Guess and Check 7.EE.B.4

The speed is a whole number at least

47\frac{2}{3}

, so the smallest choice is

48

. Check it and the number just below.

\frac{80}{47} \approx 1.702 \le \frac{12}{7} \approx 1.714, \qquad \frac{80}{46} \approx 1.739 > \frac{12}{7} \approx 1.714

At

48

km/h the trip fits within

4

hours; at

47

km/h it does not.

Answer: at least

48

km/h

4 · Reviewdoes it hold up?

At $48$ km/h the upstream ground speed is $47$ km/h, so the round trip takes $\frac{16}{7} + \frac{80}{47} \approx 2.286 + 1.702 = 3.988$ hours, just under $4$ . The upstream speed has to be much larger than $34$ because the downstream leg already used more than $2$ of the $4$ hours.

Another way: Work backwards from the time left: to cover $80$ km in $\frac{12}{7}$ hours, the ground speed must be at least $80 \div \frac{12}{7} = \frac{140}{3} = 46\frac{2}{3}$ km/h. Add back the $1$ km/h current to get an own speed of at least $47\frac{2}{3}$ , so $48$ .

Standardsmin grade 7

6.RP.A.3 Use ratio and rate reasoning to solve problems — Finding ground speeds with the current and times as distance divided by speed.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the upstream time limit as an inequality in the boat's speed and solving $560 \le 12(x - 1)$.

💡Takeaway. When a trip has a known half, spend its time first -- the leftover time turns into a clean inequality for the unknown half.

Variant 3 medium

A river flows at a steady $2$ km per hour. One point lies upstream and another lies downstream, $90$ km apart. A boat that moves at $23$ km per hour (its own speed, in still water) makes a round trip between the two points: it goes downstream at that speed, and for the trip back upstream it raises its own speed. To finish the round trip within $6$ hours, at least how many km per hour must the boat's own speed be on the way upstream? (The boat's speed is a whole number of km per hour.)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A boat goes $90$ km downstream at its own speed of $23$ km/h on a river flowing at $2$ km/h, then comes back upstream at a higher own speed. The whole round trip must take at most $6$ hours. Find the smallest whole-number own speed for the upstream trip.

Givens

The river flows at $2$ km/h.
The two points are $90$ km apart.
Going downstream, the boat's own speed is $23$ km/h.
The round trip must take $6$ hours or less.

Unknowns

The smallest possible own speed of the boat on the way upstream, in km/h.

Constraints

Downstream, the current adds to the boat's speed; upstream, it subtracts.
The boat's speed is a whole number of km/h.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#8 Analyze the Units#6 Guess and Check

The downstream half is completely known, so finish it first: its time comes off the $6$ -hour budget. What is left is a time limit for the upstream half, which becomes one inequality in the unknown speed. Last, round up to a whole number and check the neighbors.

3 · Execute6 carry out the plan

1Speed over the ground going downstream

#7 Identify Subproblems 6.RP.A.3

Going with the current, the river carries the boat along, so its speed over the ground is its own speed plus the current.

23 + 2 = 25

Downstream the boat really covers

25

km every hour.

2Time for the downstream trip

#8 Analyze the Units 6.RP.A.3

Time is distance divided by speed: hours

=

km

\div

(km per hour).

\frac{90}{25} = \frac{18}{5}

Going downstream takes

\frac{18}{5}

hours, between

3

and

4

hours.

3Time left for the upstream trip

#7 Identify Subproblems 7.EE.B.4

Subtract the downstream time from the

6

-hour limit.

6 - \frac{18}{5} = \frac{12}{5}

The upstream trip may take at most

\frac{12}{5}

hours.

4Write the inequality

#13 Convert to Algebra 7.EE.B.4

Let the boat's own speed upstream be

x

km/h. The current pushes against it, so its speed over the ground is

x - 2

km/h, and the upstream time must fit in what is left.

\frac{90}{x - 2} \le \frac{12}{5}

A faster boat makes the left side smaller, so the inequality sets a lower limit on

x

.

5Clear the fractions

#13 Convert to Algebra 7.EE.B.4

The boat must make progress upstream, so

x - 2 > 0

. Multiplying both sides by the positive number

5(x - 2)

keeps the direction of the inequality.

450 \le 12(x - 2), \quad 474 \le 12x, \quad x \ge \frac{79}{2} = 39\frac{1}{2}

Multiplying by a positive quantity never flips an inequality.

6Round up and check

#6 Guess and Check 7.EE.B.4

The speed is a whole number at least

39\frac{1}{2}

, so the smallest choice is

40

. Check it and the number just below.

\frac{90}{38} \approx 2.368 \le 2.4, \qquad \frac{90}{37} \approx 2.432 > 2.4

At

40

km/h the trip fits within

6

hours; at

39

km/h it does not.

Answer: at least

40

km/h

4 · Reviewdoes it hold up?

At $40$ km/h the upstream ground speed is $38$ km/h, so the round trip takes $\frac{18}{5} + \frac{90}{38} \approx 3.600 + 2.368 = 5.968$ hours, just under $6$ . The upstream speed has to be much larger than $23$ because the downstream leg already used more than $3$ of the $6$ hours.

Another way: Work backwards from the time left: to cover $90$ km in $\frac{12}{5}$ hours, the ground speed must be at least $90 \div \frac{12}{5} = \frac{75}{2} = 37\frac{1}{2}$ km/h. Add back the $2$ km/h current to get an own speed of at least $39\frac{1}{2}$ , so $40$ .

Standardsmin grade 7

6.RP.A.3 Use ratio and rate reasoning to solve problems — Finding ground speeds with the current and times as distance divided by speed.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the upstream time limit as an inequality in the boat's speed and solving $450 \le 12(x - 2)$.

💡Takeaway. When a trip has a known half, spend its time first -- the leftover time turns into a clean inequality for the unknown half.

Variant 4 easy

A river flows at a steady $5$ km per hour. One point lies upstream and another lies downstream, $90$ km apart. A boat that moves at $20$ km per hour (its own speed, in still water) makes a round trip between the two points: it goes downstream at that speed, and for the trip back upstream it raises its own speed. To finish the round trip within $5$ hours, at least how many km per hour must the boat's own speed be on the way upstream? (The boat's speed is a whole number of km per hour.)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A boat goes $90$ km downstream at its own speed of $20$ km/h on a river flowing at $5$ km/h, then comes back upstream at a higher own speed. The whole round trip must take at most $5$ hours. Find the smallest whole-number own speed for the upstream trip.

Givens

The river flows at $5$ km/h.
The two points are $90$ km apart.
Going downstream, the boat's own speed is $20$ km/h.
The round trip must take $5$ hours or less.

Unknowns

The smallest possible own speed of the boat on the way upstream, in km/h.

Constraints

Downstream, the current adds to the boat's speed; upstream, it subtracts.
The boat's speed is a whole number of km/h.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#8 Analyze the Units#6 Guess and Check

The downstream half is completely known, so finish it first: its time comes off the $5$ -hour budget. What is left is a time limit for the upstream half, which becomes one inequality in the unknown speed. Last, round up to a whole number and check the neighbors.

3 · Execute6 carry out the plan

1Speed over the ground going downstream

#7 Identify Subproblems 6.RP.A.3

Going with the current, the river carries the boat along, so its speed over the ground is its own speed plus the current.

20 + 5 = 25

Downstream the boat really covers

25

km every hour.

2Time for the downstream trip

#8 Analyze the Units 6.RP.A.3

Time is distance divided by speed: hours

=

km

\div

(km per hour).

\frac{90}{25} = \frac{18}{5}

Going downstream takes

\frac{18}{5}

hours, between

3

and

4

hours.

3Time left for the upstream trip

#7 Identify Subproblems 7.EE.B.4

Subtract the downstream time from the

5

-hour limit.

5 - \frac{18}{5} = \frac{7}{5}

The upstream trip may take at most

\frac{7}{5}

hours.

4Write the inequality

#13 Convert to Algebra 7.EE.B.4

Let the boat's own speed upstream be

x

km/h. The current pushes against it, so its speed over the ground is

x - 5

km/h, and the upstream time must fit in what is left.

\frac{90}{x - 5} \le \frac{7}{5}

A faster boat makes the left side smaller, so the inequality sets a lower limit on

x

.

5Clear the fractions

#13 Convert to Algebra 7.EE.B.4

The boat must make progress upstream, so

x - 5 > 0

. Multiplying both sides by the positive number

5(x - 5)

keeps the direction of the inequality.

450 \le 7(x - 5), \quad 485 \le 7x, \quad x \ge \frac{485}{7} = 69\frac{2}{7}

Multiplying by a positive quantity never flips an inequality.

6Round up and check

#6 Guess and Check 7.EE.B.4

The speed is a whole number at least

69\frac{2}{7}

, so the smallest choice is

70

. Check it and the number just below.

\frac{90}{65} \approx 1.385 \le 1.4, \qquad \frac{90}{64} \approx 1.406 > 1.4

At

70

km/h the trip fits within

5

hours; at

69

km/h it does not.

Answer: at least

70

km/h

4 · Reviewdoes it hold up?

At $70$ km/h the upstream ground speed is $65$ km/h, so the round trip takes $\frac{18}{5} + \frac{90}{65} \approx 3.600 + 1.385 = 4.985$ hours, just under $5$ . The upstream speed has to be much larger than $20$ because the downstream leg already used more than $3$ of the $5$ hours.

Another way: Work backwards from the time left: to cover $90$ km in $\frac{7}{5}$ hours, the ground speed must be at least $90 \div \frac{7}{5} = \frac{450}{7} = 64\frac{2}{7}$ km/h. Add back the $5$ km/h current to get an own speed of at least $69\frac{2}{7}$ , so $70$ .

Standardsmin grade 7

6.RP.A.3 Use ratio and rate reasoning to solve problems — Finding ground speeds with the current and times as distance divided by speed.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the upstream time limit as an inequality in the boat's speed and solving $450 \le 7(x - 5)$.

💡Takeaway. When a trip has a known half, spend its time first -- the leftover time turns into a clean inequality for the unknown half.

Variant 5 easy

A river flows at a steady $4$ km per hour. One point lies upstream and another lies downstream, $90$ km apart. A boat that moves at $36$ km per hour (its own speed, in still water) makes a round trip between the two points: it goes downstream at that speed, and for the trip back upstream it raises its own speed. To finish the round trip within $4$ hours, at least how many km per hour must the boat's own speed be on the way upstream? (The boat's speed is a whole number of km per hour.)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A boat goes $90$ km downstream at its own speed of $36$ km/h on a river flowing at $4$ km/h, then comes back upstream at a higher own speed. The whole round trip must take at most $4$ hours. Find the smallest whole-number own speed for the upstream trip.

Givens

The river flows at $4$ km/h.
The two points are $90$ km apart.
Going downstream, the boat's own speed is $36$ km/h.
The round trip must take $4$ hours or less.

Unknowns

The smallest possible own speed of the boat on the way upstream, in km/h.

Constraints

Downstream, the current adds to the boat's speed; upstream, it subtracts.
The boat's speed is a whole number of km/h.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#8 Analyze the Units#6 Guess and Check

The downstream half is completely known, so finish it first: its time comes off the $4$ -hour budget. What is left is a time limit for the upstream half, which becomes one inequality in the unknown speed. Last, round up to a whole number and check the neighbors.

3 · Execute6 carry out the plan

1Speed over the ground going downstream

#7 Identify Subproblems 6.RP.A.3

Going with the current, the river carries the boat along, so its speed over the ground is its own speed plus the current.

36 + 4 = 40

Downstream the boat really covers

40

km every hour.

2Time for the downstream trip

#8 Analyze the Units 6.RP.A.3

Time is distance divided by speed: hours

=

km

\div

(km per hour).

\frac{90}{40} = \frac{9}{4}

Going downstream takes

\frac{9}{4}

hours, a little over

2

hours.

3Time left for the upstream trip

#7 Identify Subproblems 7.EE.B.4

Subtract the downstream time from the

4

-hour limit.

4 - \frac{9}{4} = \frac{7}{4}

The upstream trip may take at most

\frac{7}{4}

hours.

4Write the inequality

#13 Convert to Algebra 7.EE.B.4

Let the boat's own speed upstream be

x

km/h. The current pushes against it, so its speed over the ground is

x - 4

km/h, and the upstream time must fit in what is left.

\frac{90}{x - 4} \le \frac{7}{4}

A faster boat makes the left side smaller, so the inequality sets a lower limit on

x

.

5Clear the fractions

#13 Convert to Algebra 7.EE.B.4

The boat must make progress upstream, so

x - 4 > 0

. Multiplying both sides by the positive number

4(x - 4)

keeps the direction of the inequality.

360 \le 7(x - 4), \quad 388 \le 7x, \quad x \ge \frac{388}{7} = 55\frac{3}{7}

Multiplying by a positive quantity never flips an inequality.

6Round up and check

#6 Guess and Check 7.EE.B.4

The speed is a whole number at least

55\frac{3}{7}

, so the smallest choice is

56

. Check it and the number just below.

\frac{90}{52} \approx 1.731 \le 1.75, \qquad \frac{90}{51} \approx 1.765 > 1.75

At

56

km/h the trip fits within

4

hours; at

55

km/h it does not.

Answer: at least

56

km/h

4 · Reviewdoes it hold up?

At $56$ km/h the upstream ground speed is $52$ km/h, so the round trip takes $\frac{9}{4} + \frac{90}{52} \approx 2.250 + 1.731 = 3.981$ hours, just under $4$ . The upstream speed has to be much larger than $36$ because the downstream leg already used more than $2$ of the $4$ hours.

Another way: Work backwards from the time left: to cover $90$ km in $\frac{7}{4}$ hours, the ground speed must be at least $90 \div \frac{7}{4} = \frac{360}{7} = 51\frac{3}{7}$ km/h. Add back the $4$ km/h current to get an own speed of at least $55\frac{3}{7}$ , so $56$ .

Standardsmin grade 7

6.RP.A.3 Use ratio and rate reasoning to solve problems — Finding ground speeds with the current and times as distance divided by speed.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the upstream time limit as an inequality in the boat's speed and solving $360 \le 7(x - 4)$.

💡Takeaway. When a trip has a known half, spend its time first -- the leftover time turns into a clean inequality for the unknown half.

Variant 6 medium

A river flows at a steady $2$ km per hour. One point lies upstream and another lies downstream, $100$ km apart. A boat that moves at $30$ km per hour (its own speed, in still water) makes a round trip between the two points: it goes downstream at that speed, and for the trip back upstream it raises its own speed. To finish the round trip within $5$ hours, at least how many km per hour must the boat's own speed be on the way upstream? (The boat's speed is a whole number of km per hour.)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A boat goes $100$ km downstream at its own speed of $30$ km/h on a river flowing at $2$ km/h, then comes back upstream at a higher own speed. The whole round trip must take at most $5$ hours. Find the smallest whole-number own speed for the upstream trip.

Givens

The river flows at $2$ km/h.
The two points are $100$ km apart.
Going downstream, the boat's own speed is $30$ km/h.
The round trip must take $5$ hours or less.

Unknowns

The smallest possible own speed of the boat on the way upstream, in km/h.

Constraints

Downstream, the current adds to the boat's speed; upstream, it subtracts.
The boat's speed is a whole number of km/h.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#8 Analyze the Units#6 Guess and Check

The downstream half is completely known, so finish it first: its time comes off the $5$ -hour budget. What is left is a time limit for the upstream half, which becomes one inequality in the unknown speed. Last, round up to a whole number and check the neighbors.

3 · Execute6 carry out the plan

1Speed over the ground going downstream

#7 Identify Subproblems 6.RP.A.3

Going with the current, the river carries the boat along, so its speed over the ground is its own speed plus the current.

30 + 2 = 32

Downstream the boat really covers

32

km every hour.

2Time for the downstream trip

#8 Analyze the Units 6.RP.A.3

Time is distance divided by speed: hours

=

km

\div

(km per hour).

\frac{100}{32} = \frac{25}{8}

Going downstream takes

\frac{25}{8}

hours, a little over

3

hours.

3Time left for the upstream trip

#7 Identify Subproblems 7.EE.B.4

Subtract the downstream time from the

5

-hour limit.

5 - \frac{25}{8} = \frac{15}{8}

The upstream trip may take at most

\frac{15}{8}

hours.

4Write the inequality

#13 Convert to Algebra 7.EE.B.4

Let the boat's own speed upstream be

x

km/h. The current pushes against it, so its speed over the ground is

x - 2

km/h, and the upstream time must fit in what is left.

\frac{100}{x - 2} \le \frac{15}{8}

A faster boat makes the left side smaller, so the inequality sets a lower limit on

x

.

5Clear the fractions

#13 Convert to Algebra 7.EE.B.4

The boat must make progress upstream, so

x - 2 > 0

. Multiplying both sides by the positive number

8(x - 2)

keeps the direction of the inequality.

800 \le 15(x - 2), \quad 830 \le 15x, \quad x \ge \frac{166}{3} = 55\frac{1}{3}

Multiplying by a positive quantity never flips an inequality.

6Round up and check

#6 Guess and Check 7.EE.B.4

The speed is a whole number at least

55\frac{1}{3}

, so the smallest choice is

56

. Check it and the number just below.

\frac{100}{54} \approx 1.852 \le 1.875, \qquad \frac{100}{53} \approx 1.887 > 1.875

At

56

km/h the trip fits within

5

hours; at

55

km/h it does not.

Answer: at least

56

km/h

4 · Reviewdoes it hold up?

At $56$ km/h the upstream ground speed is $54$ km/h, so the round trip takes $\frac{25}{8} + \frac{100}{54} \approx 3.125 + 1.852 = 4.977$ hours, just under $5$ . The upstream speed has to be much larger than $30$ because the downstream leg already used more than $3$ of the $5$ hours.

Another way: Work backwards from the time left: to cover $100$ km in $\frac{15}{8}$ hours, the ground speed must be at least $100 \div \frac{15}{8} = \frac{160}{3} = 53\frac{1}{3}$ km/h. Add back the $2$ km/h current to get an own speed of at least $55\frac{1}{3}$ , so $56$ .

Standardsmin grade 7

6.RP.A.3 Use ratio and rate reasoning to solve problems — Finding ground speeds with the current and times as distance divided by speed.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the upstream time limit as an inequality in the boat's speed and solving $800 \le 15(x - 2)$.

💡Takeaway. When a trip has a known half, spend its time first -- the leftover time turns into a clean inequality for the unknown half.

Variant 7 medium

A river flows at a steady $1$ km per hour. One point lies upstream and another lies downstream, $100$ km apart. A boat that moves at $35$ km per hour (its own speed, in still water) makes a round trip between the two points: it goes downstream at that speed, and for the trip back upstream it raises its own speed. To finish the round trip within $4$ hours, at least how many km per hour must the boat's own speed be on the way upstream? (The boat's speed is a whole number of km per hour.)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A boat goes $100$ km downstream at its own speed of $35$ km/h on a river flowing at $1$ km/h, then comes back upstream at a higher own speed. The whole round trip must take at most $4$ hours. Find the smallest whole-number own speed for the upstream trip.

Givens

The river flows at $1$ km/h.
The two points are $100$ km apart.
Going downstream, the boat's own speed is $35$ km/h.
The round trip must take $4$ hours or less.

Unknowns

The smallest possible own speed of the boat on the way upstream, in km/h.

Constraints

Downstream, the current adds to the boat's speed; upstream, it subtracts.
The boat's speed is a whole number of km/h.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#8 Analyze the Units#6 Guess and Check

The downstream half is completely known, so finish it first: its time comes off the $4$ -hour budget. What is left is a time limit for the upstream half, which becomes one inequality in the unknown speed. Last, round up to a whole number and check the neighbors.

3 · Execute6 carry out the plan

1Speed over the ground going downstream

#7 Identify Subproblems 6.RP.A.3

Going with the current, the river carries the boat along, so its speed over the ground is its own speed plus the current.

35 + 1 = 36

Downstream the boat really covers

36

km every hour.

2Time for the downstream trip

#8 Analyze the Units 6.RP.A.3

Time is distance divided by speed: hours

=

km

\div

(km per hour).

\frac{100}{36} = \frac{25}{9}

Going downstream takes

\frac{25}{9}

hours, between

2

and

3

hours.

3Time left for the upstream trip

#7 Identify Subproblems 7.EE.B.4

Subtract the downstream time from the

4

-hour limit.

4 - \frac{25}{9} = \frac{11}{9}

The upstream trip may take at most

\frac{11}{9}

hours.

4Write the inequality

#13 Convert to Algebra 7.EE.B.4

Let the boat's own speed upstream be

x

km/h. The current pushes against it, so its speed over the ground is

x - 1

km/h, and the upstream time must fit in what is left.

\frac{100}{x - 1} \le \frac{11}{9}

A faster boat makes the left side smaller, so the inequality sets a lower limit on

x

.

5Clear the fractions

#13 Convert to Algebra 7.EE.B.4

The boat must make progress upstream, so

x - 1 > 0

. Multiplying both sides by the positive number

9(x - 1)

keeps the direction of the inequality.

900 \le 11(x - 1), \quad 911 \le 11x, \quad x \ge \frac{911}{11} = 82\frac{9}{11}

Multiplying by a positive quantity never flips an inequality.

6Round up and check

#6 Guess and Check 7.EE.B.4

The speed is a whole number at least

82\frac{9}{11}

, so the smallest choice is

83

. Check it and the number just below.

\frac{100}{82} \approx 1.220 \le \frac{11}{9} \approx 1.222, \qquad \frac{100}{81} \approx 1.235 > \frac{11}{9} \approx 1.222

At

83

km/h the trip fits within

4

hours; at

82

km/h it does not.

Answer: at least

83

km/h

4 · Reviewdoes it hold up?

At $83$ km/h the upstream ground speed is $82$ km/h, so the round trip takes $\frac{25}{9} + \frac{100}{82} \approx 2.778 + 1.220 = 3.998$ hours, just under $4$ . The upstream speed has to be much larger than $35$ because the downstream leg already used more than $2$ of the $4$ hours.

Another way: Work backwards from the time left: to cover $100$ km in $\frac{11}{9}$ hours, the ground speed must be at least $100 \div \frac{11}{9} = \frac{900}{11} = 81\frac{9}{11}$ km/h. Add back the $1$ km/h current to get an own speed of at least $82\frac{9}{11}$ , so $83$ .

Standardsmin grade 7

6.RP.A.3 Use ratio and rate reasoning to solve problems — Finding ground speeds with the current and times as distance divided by speed.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the upstream time limit as an inequality in the boat's speed and solving $900 \le 11(x - 1)$.

💡Takeaway. When a trip has a known half, spend its time first -- the leftover time turns into a clean inequality for the unknown half.

Variant 8 medium

A river flows at a steady $2$ km per hour. One point lies upstream and another lies downstream, $120$ km apart. A boat that moves at $31$ km per hour (its own speed, in still water) makes a round trip between the two points: it goes downstream at that speed, and for the trip back upstream it raises its own speed. To finish the round trip within $6$ hours, at least how many km per hour must the boat's own speed be on the way upstream? (The boat's speed is a whole number of km per hour.)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A boat goes $120$ km downstream at its own speed of $31$ km/h on a river flowing at $2$ km/h, then comes back upstream at a higher own speed. The whole round trip must take at most $6$ hours. Find the smallest whole-number own speed for the upstream trip.

Givens

The river flows at $2$ km/h.
The two points are $120$ km apart.
Going downstream, the boat's own speed is $31$ km/h.
The round trip must take $6$ hours or less.

Unknowns

The smallest possible own speed of the boat on the way upstream, in km/h.

Constraints

Downstream, the current adds to the boat's speed; upstream, it subtracts.
The boat's speed is a whole number of km/h.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#8 Analyze the Units#6 Guess and Check

The downstream half is completely known, so finish it first: its time comes off the $6$ -hour budget. What is left is a time limit for the upstream half, which becomes one inequality in the unknown speed. Last, round up to a whole number and check the neighbors.

3 · Execute6 carry out the plan

1Speed over the ground going downstream

#7 Identify Subproblems 6.RP.A.3

Going with the current, the river carries the boat along, so its speed over the ground is its own speed plus the current.

31 + 2 = 33

Downstream the boat really covers

33

km every hour.

2Time for the downstream trip

#8 Analyze the Units 6.RP.A.3

Time is distance divided by speed: hours

=

km

\div

(km per hour).

\frac{120}{33} = \frac{40}{11}

Going downstream takes

\frac{40}{11}

hours, between

3

and

4

hours.

3Time left for the upstream trip

#7 Identify Subproblems 7.EE.B.4

Subtract the downstream time from the

6

-hour limit.

6 - \frac{40}{11} = \frac{26}{11}

The upstream trip may take at most

\frac{26}{11}

hours.

4Write the inequality

#13 Convert to Algebra 7.EE.B.4

Let the boat's own speed upstream be

x

km/h. The current pushes against it, so its speed over the ground is

x - 2

km/h, and the upstream time must fit in what is left.

\frac{120}{x - 2} \le \frac{26}{11}

A faster boat makes the left side smaller, so the inequality sets a lower limit on

x

.

5Clear the fractions

#13 Convert to Algebra 7.EE.B.4

The boat must make progress upstream, so

x - 2 > 0

. Multiplying both sides by the positive number

11(x - 2)

keeps the direction of the inequality.

1320 \le 26(x - 2), \quad 1372 \le 26x, \quad x \ge \frac{686}{13} = 52\frac{10}{13}

Multiplying by a positive quantity never flips an inequality.

6Round up and check

#6 Guess and Check 7.EE.B.4

The speed is a whole number at least

52\frac{10}{13}

, so the smallest choice is

53

. Check it and the number just below.

\frac{120}{51} \approx 2.353 \le \frac{26}{11} \approx 2.364, \qquad \frac{120}{50} = 2.4 > \frac{26}{11} \approx 2.364

At

53

km/h the trip fits within

6

hours; at

52

km/h it does not.

Answer: at least

53

km/h

4 · Reviewdoes it hold up?

At $53$ km/h the upstream ground speed is $51$ km/h, so the round trip takes $\frac{40}{11} + \frac{120}{51} \approx 3.636 + 2.353 = 5.989$ hours, just under $6$ . The upstream speed has to be much larger than $31$ because the downstream leg already used more than $3$ of the $6$ hours.

Another way: Work backwards from the time left: to cover $120$ km in $\frac{26}{11}$ hours, the ground speed must be at least $120 \div \frac{26}{11} = \frac{660}{13} = 50\frac{10}{13}$ km/h. Add back the $2$ km/h current to get an own speed of at least $52\frac{10}{13}$ , so $53$ .

Standardsmin grade 7

6.RP.A.3 Use ratio and rate reasoning to solve problems — Finding ground speeds with the current and times as distance divided by speed.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the upstream time limit as an inequality in the boat's speed and solving $1320 \le 26(x - 2)$.

💡Takeaway. When a trip has a known half, spend its time first -- the leftover time turns into a clean inequality for the unknown half.

Variant 9 hard

A river flows at a steady $3$ km per hour. One point lies upstream and another lies downstream, $130$ km apart. A boat that moves at $27$ km per hour (its own speed, in still water) makes a round trip between the two points: it goes downstream at that speed, and for the trip back upstream it raises its own speed. To finish the round trip within $7$ hours, at least how many km per hour must the boat's own speed be on the way upstream? (The boat's speed is a whole number of km per hour.)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A boat goes $130$ km downstream at its own speed of $27$ km/h on a river flowing at $3$ km/h, then comes back upstream at a higher own speed. The whole round trip must take at most $7$ hours. Find the smallest whole-number own speed for the upstream trip.

Givens

The river flows at $3$ km/h.
The two points are $130$ km apart.
Going downstream, the boat's own speed is $27$ km/h.
The round trip must take $7$ hours or less.

Unknowns

The smallest possible own speed of the boat on the way upstream, in km/h.

Constraints

Downstream, the current adds to the boat's speed; upstream, it subtracts.
The boat's speed is a whole number of km/h.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#8 Analyze the Units#6 Guess and Check

The downstream half is completely known, so finish it first: its time comes off the $7$ -hour budget. What is left is a time limit for the upstream half, which becomes one inequality in the unknown speed. Last, round up to a whole number and check the neighbors.

3 · Execute6 carry out the plan

1Speed over the ground going downstream

#7 Identify Subproblems 6.RP.A.3

Going with the current, the river carries the boat along, so its speed over the ground is its own speed plus the current.

27 + 3 = 30

Downstream the boat really covers

30

km every hour.

2Time for the downstream trip

#8 Analyze the Units 6.RP.A.3

Time is distance divided by speed: hours

=

km

\div

(km per hour).

\frac{130}{30} = \frac{13}{3}

Going downstream takes

\frac{13}{3}

hours, a little over

4

hours.

3Time left for the upstream trip

#7 Identify Subproblems 7.EE.B.4

Subtract the downstream time from the

7

-hour limit.

7 - \frac{13}{3} = \frac{8}{3}

The upstream trip may take at most

\frac{8}{3}

hours.

4Write the inequality

#13 Convert to Algebra 7.EE.B.4

Let the boat's own speed upstream be

x

km/h. The current pushes against it, so its speed over the ground is

x - 3

km/h, and the upstream time must fit in what is left.

\frac{130}{x - 3} \le \frac{8}{3}

A faster boat makes the left side smaller, so the inequality sets a lower limit on

x

.

5Clear the fractions

#13 Convert to Algebra 7.EE.B.4

The boat must make progress upstream, so

x - 3 > 0

. Multiplying both sides by the positive number

3(x - 3)

keeps the direction of the inequality.

390 \le 8(x - 3), \quad 414 \le 8x, \quad x \ge \frac{207}{4} = 51\frac{3}{4}

Multiplying by a positive quantity never flips an inequality.

6Round up and check

#6 Guess and Check 7.EE.B.4

The speed is a whole number at least

51\frac{3}{4}

, so the smallest choice is

52

. Check it and the number just below.

\frac{130}{49} \approx 2.653 \le \frac{8}{3} \approx 2.667, \qquad \frac{130}{48} \approx 2.708 > \frac{8}{3} \approx 2.667

At

52

km/h the trip fits within

7

hours; at

51

km/h it does not.

Answer: at least

52

km/h

4 · Reviewdoes it hold up?

At $52$ km/h the upstream ground speed is $49$ km/h, so the round trip takes $\frac{13}{3} + \frac{130}{49} \approx 4.333 + 2.653 = 6.986$ hours, just under $7$ . The upstream speed has to be much larger than $27$ because the downstream leg already used more than $4$ of the $7$ hours.

Another way: Work backwards from the time left: to cover $130$ km in $\frac{8}{3}$ hours, the ground speed must be at least $130 \div \frac{8}{3} = \frac{195}{4} = 48\frac{3}{4}$ km/h. Add back the $3$ km/h current to get an own speed of at least $51\frac{3}{4}$ , so $52$ .

Standardsmin grade 7

6.RP.A.3 Use ratio and rate reasoning to solve problems — Finding ground speeds with the current and times as distance divided by speed.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the upstream time limit as an inequality in the boat's speed and solving $390 \le 8(x - 3)$.

💡Takeaway. When a trip has a known half, spend its time first -- the leftover time turns into a clean inequality for the unknown half.

Variant 10 hard

A river flows at a steady $3$ km per hour. One point lies upstream and another lies downstream, $140$ km apart. A boat that moves at $46$ km per hour (its own speed, in still water) makes a round trip between the two points: it goes downstream at that speed, and for the trip back upstream it raises its own speed. To finish the round trip within $5$ hours, at least how many km per hour must the boat's own speed be on the way upstream? (The boat's speed is a whole number of km per hour.)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A boat goes $140$ km downstream at its own speed of $46$ km/h on a river flowing at $3$ km/h, then comes back upstream at a higher own speed. The whole round trip must take at most $5$ hours. Find the smallest whole-number own speed for the upstream trip.

Givens

The river flows at $3$ km/h.
The two points are $140$ km apart.
Going downstream, the boat's own speed is $46$ km/h.
The round trip must take $5$ hours or less.

Unknowns

The smallest possible own speed of the boat on the way upstream, in km/h.

Constraints

Downstream, the current adds to the boat's speed; upstream, it subtracts.
The boat's speed is a whole number of km/h.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#8 Analyze the Units#6 Guess and Check

The downstream half is completely known, so finish it first: its time comes off the $5$ -hour budget. What is left is a time limit for the upstream half, which becomes one inequality in the unknown speed. Last, round up to a whole number and check the neighbors.

3 · Execute6 carry out the plan

1Speed over the ground going downstream

#7 Identify Subproblems 6.RP.A.3

Going with the current, the river carries the boat along, so its speed over the ground is its own speed plus the current.

46 + 3 = 49

Downstream the boat really covers

49

km every hour.

2Time for the downstream trip

#8 Analyze the Units 6.RP.A.3

Time is distance divided by speed: hours

=

km

\div

(km per hour).

\frac{140}{49} = \frac{20}{7}

Going downstream takes

\frac{20}{7}

hours, between

2

and

3

hours.

3Time left for the upstream trip

#7 Identify Subproblems 7.EE.B.4

Subtract the downstream time from the

5

-hour limit.

5 - \frac{20}{7} = \frac{15}{7}

The upstream trip may take at most

\frac{15}{7}

hours.

4Write the inequality

#13 Convert to Algebra 7.EE.B.4

Let the boat's own speed upstream be

x

km/h. The current pushes against it, so its speed over the ground is

x - 3

km/h, and the upstream time must fit in what is left.

\frac{140}{x - 3} \le \frac{15}{7}

A faster boat makes the left side smaller, so the inequality sets a lower limit on

x

.

5Clear the fractions

#13 Convert to Algebra 7.EE.B.4

The boat must make progress upstream, so

x - 3 > 0

. Multiplying both sides by the positive number

7(x - 3)

keeps the direction of the inequality.

980 \le 15(x - 3), \quad 1025 \le 15x, \quad x \ge \frac{205}{3} = 68\frac{1}{3}

Multiplying by a positive quantity never flips an inequality.

6Round up and check

#6 Guess and Check 7.EE.B.4

The speed is a whole number at least

68\frac{1}{3}

, so the smallest choice is

69

. Check it and the number just below.

\frac{140}{66} \approx 2.121 \le \frac{15}{7} \approx 2.143, \qquad \frac{140}{65} \approx 2.154 > \frac{15}{7} \approx 2.143

At

69

km/h the trip fits within

5

hours; at

68

km/h it does not.

Answer: at least

69

km/h

4 · Reviewdoes it hold up?

At $69$ km/h the upstream ground speed is $66$ km/h, so the round trip takes $\frac{20}{7} + \frac{140}{66} \approx 2.857 + 2.121 = 4.978$ hours, just under $5$ . The upstream speed has to be much larger than $46$ because the downstream leg already used more than $2$ of the $5$ hours.

Another way: Work backwards from the time left: to cover $140$ km in $\frac{15}{7}$ hours, the ground speed must be at least $140 \div \frac{15}{7} = \frac{196}{3} = 65\frac{1}{3}$ km/h. Add back the $3$ km/h current to get an own speed of at least $68\frac{1}{3}$ , so $69$ .

Standardsmin grade 7

6.RP.A.3 Use ratio and rate reasoning to solve problems — Finding ground speeds with the current and times as distance divided by speed.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the upstream time limit as an inequality in the boat's speed and solving $980 \le 15(x - 3)$.

💡Takeaway. When a trip has a known half, spend its time first -- the leftover time turns into a clean inequality for the unknown half.

Variant 11 hard

A river flows at a steady $5$ km per hour. One point lies upstream and another lies downstream, $140$ km apart. A boat that moves at $44$ km per hour (its own speed, in still water) makes a round trip between the two points: it goes downstream at that speed, and for the trip back upstream it raises its own speed. To finish the round trip within $5$ hours, at least how many km per hour must the boat's own speed be on the way upstream? (The boat's speed is a whole number of km per hour.)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A boat goes $140$ km downstream at its own speed of $44$ km/h on a river flowing at $5$ km/h, then comes back upstream at a higher own speed. The whole round trip must take at most $5$ hours. Find the smallest whole-number own speed for the upstream trip.

Givens

The river flows at $5$ km/h.
The two points are $140$ km apart.
Going downstream, the boat's own speed is $44$ km/h.
The round trip must take $5$ hours or less.

Unknowns

The smallest possible own speed of the boat on the way upstream, in km/h.

Constraints

Downstream, the current adds to the boat's speed; upstream, it subtracts.
The boat's speed is a whole number of km/h.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#8 Analyze the Units#6 Guess and Check

The downstream half is completely known, so finish it first: its time comes off the $5$ -hour budget. What is left is a time limit for the upstream half, which becomes one inequality in the unknown speed. Last, round up to a whole number and check the neighbors.

3 · Execute6 carry out the plan

1Speed over the ground going downstream

#7 Identify Subproblems 6.RP.A.3

Going with the current, the river carries the boat along, so its speed over the ground is its own speed plus the current.

44 + 5 = 49

Downstream the boat really covers

49

km every hour.

2Time for the downstream trip

#8 Analyze the Units 6.RP.A.3

Time is distance divided by speed: hours

=

km

\div

(km per hour).

\frac{140}{49} = \frac{20}{7}

Going downstream takes

\frac{20}{7}

hours, between

2

and

3

hours.

3Time left for the upstream trip

#7 Identify Subproblems 7.EE.B.4

Subtract the downstream time from the

5

-hour limit.

5 - \frac{20}{7} = \frac{15}{7}

The upstream trip may take at most

\frac{15}{7}

hours.

4Write the inequality

#13 Convert to Algebra 7.EE.B.4

Let the boat's own speed upstream be

x

km/h. The current pushes against it, so its speed over the ground is

x - 5

km/h, and the upstream time must fit in what is left.

\frac{140}{x - 5} \le \frac{15}{7}

A faster boat makes the left side smaller, so the inequality sets a lower limit on

x

.

5Clear the fractions

#13 Convert to Algebra 7.EE.B.4

The boat must make progress upstream, so

x - 5 > 0

. Multiplying both sides by the positive number

7(x - 5)

keeps the direction of the inequality.

980 \le 15(x - 5), \quad 1055 \le 15x, \quad x \ge \frac{211}{3} = 70\frac{1}{3}

Multiplying by a positive quantity never flips an inequality.

6Round up and check

#6 Guess and Check 7.EE.B.4

The speed is a whole number at least

70\frac{1}{3}

, so the smallest choice is

71

. Check it and the number just below.

\frac{140}{66} \approx 2.121 \le \frac{15}{7} \approx 2.143, \qquad \frac{140}{65} \approx 2.154 > \frac{15}{7} \approx 2.143

At

71

km/h the trip fits within

5

hours; at

70

km/h it does not.

Answer: at least

71

km/h

4 · Reviewdoes it hold up?

At $71$ km/h the upstream ground speed is $66$ km/h, so the round trip takes $\frac{20}{7} + \frac{140}{66} \approx 2.857 + 2.121 = 4.978$ hours, just under $5$ . The upstream speed has to be much larger than $44$ because the downstream leg already used more than $2$ of the $5$ hours.

Another way: Work backwards from the time left: to cover $140$ km in $\frac{15}{7}$ hours, the ground speed must be at least $140 \div \frac{15}{7} = \frac{196}{3} = 65\frac{1}{3}$ km/h. Add back the $5$ km/h current to get an own speed of at least $70\frac{1}{3}$ , so $71$ .

Standardsmin grade 7

6.RP.A.3 Use ratio and rate reasoning to solve problems — Finding ground speeds with the current and times as distance divided by speed.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the upstream time limit as an inequality in the boat's speed and solving $980 \le 15(x - 5)$.

💡Takeaway. When a trip has a known half, spend its time first -- the leftover time turns into a clean inequality for the unknown half.

Variant 12 hard

A river flows at a steady $4$ km per hour. One point lies upstream and another lies downstream, $150$ km apart. A boat that moves at $32$ km per hour (its own speed, in still water) makes a round trip between the two points: it goes downstream at that speed, and for the trip back upstream it raises its own speed. To finish the round trip within $6$ hours, at least how many km per hour must the boat's own speed be on the way upstream? (The boat's speed is a whole number of km per hour.)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A boat goes $150$ km downstream at its own speed of $32$ km/h on a river flowing at $4$ km/h, then comes back upstream at a higher own speed. The whole round trip must take at most $6$ hours. Find the smallest whole-number own speed for the upstream trip.

Givens

The river flows at $4$ km/h.
The two points are $150$ km apart.
Going downstream, the boat's own speed is $32$ km/h.
The round trip must take $6$ hours or less.

Unknowns

The smallest possible own speed of the boat on the way upstream, in km/h.

Constraints

Downstream, the current adds to the boat's speed; upstream, it subtracts.
The boat's speed is a whole number of km/h.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#8 Analyze the Units#6 Guess and Check

The downstream half is completely known, so finish it first: its time comes off the $6$ -hour budget. What is left is a time limit for the upstream half, which becomes one inequality in the unknown speed. Last, round up to a whole number and check the neighbors.

3 · Execute6 carry out the plan

1Speed over the ground going downstream

#7 Identify Subproblems 6.RP.A.3

Going with the current, the river carries the boat along, so its speed over the ground is its own speed plus the current.

32 + 4 = 36

Downstream the boat really covers

36

km every hour.

2Time for the downstream trip

#8 Analyze the Units 6.RP.A.3

Time is distance divided by speed: hours

=

km

\div

(km per hour).

\frac{150}{36} = \frac{25}{6}

Going downstream takes

\frac{25}{6}

hours, a little over

4

hours.

3Time left for the upstream trip

#7 Identify Subproblems 7.EE.B.4

Subtract the downstream time from the

6

-hour limit.

6 - \frac{25}{6} = \frac{11}{6}

The upstream trip may take at most

\frac{11}{6}

hours.

4Write the inequality

#13 Convert to Algebra 7.EE.B.4

Let the boat's own speed upstream be

x

km/h. The current pushes against it, so its speed over the ground is

x - 4

km/h, and the upstream time must fit in what is left.

\frac{150}{x - 4} \le \frac{11}{6}

A faster boat makes the left side smaller, so the inequality sets a lower limit on

x

.

5Clear the fractions

#13 Convert to Algebra 7.EE.B.4

The boat must make progress upstream, so

x - 4 > 0

. Multiplying both sides by the positive number

6(x - 4)

keeps the direction of the inequality.

900 \le 11(x - 4), \quad 944 \le 11x, \quad x \ge \frac{944}{11} = 85\frac{9}{11}

Multiplying by a positive quantity never flips an inequality.

6Round up and check

#6 Guess and Check 7.EE.B.4

The speed is a whole number at least

85\frac{9}{11}

, so the smallest choice is

86

. Check it and the number just below.

\frac{150}{82} \approx 1.829 \le \frac{11}{6} \approx 1.833, \qquad \frac{150}{81} \approx 1.852 > \frac{11}{6} \approx 1.833

At

86

km/h the trip fits within

6

hours; at

85

km/h it does not.

Answer: at least

86

km/h

4 · Reviewdoes it hold up?

At $86$ km/h the upstream ground speed is $82$ km/h, so the round trip takes $\frac{25}{6} + \frac{150}{82} \approx 4.167 + 1.829 = 5.996$ hours, just under $6$ . The upstream speed has to be much larger than $32$ because the downstream leg already used more than $4$ of the $6$ hours.

Another way: Work backwards from the time left: to cover $150$ km in $\frac{11}{6}$ hours, the ground speed must be at least $150 \div \frac{11}{6} = \frac{900}{11} = 81\frac{9}{11}$ km/h. Add back the $4$ km/h current to get an own speed of at least $85\frac{9}{11}$ , so $86$ .

Standardsmin grade 7

6.RP.A.3 Use ratio and rate reasoning to solve problems — Finding ground speeds with the current and times as distance divided by speed.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the upstream time limit as an inequality in the boat's speed and solving $900 \le 11(x - 4)$.

💡Takeaway. When a trip has a known half, spend its time first -- the leftover time turns into a clean inequality for the unknown half.

Spend downstream time first, bound upstream speed · 12 practice problems

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