Practice · A steady gain is proportional to the time that has passed · Sensim Math

Variant 1 easy

A clock gains a steady $6$ minutes a day. If it is set exactly right at $2$ a.m. today, what time will it show at $6$ p.m. the next day?

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A clock gains 6 minutes every day. It was right at 2 a.m., and we want what it shows at 18 the next day.

Givens

The clock gains 6 minutes a day.
It was exactly right at 2 a.m.
It is read the next day.

Unknowns

The time the clock displays.

Constraints

The gain is steady, so it is proportional to the time elapsed.

2 · Planchoose the strategy

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

A steady gain is proportional to time, so this is one proportion once the elapsed hours are known. Counting those hours is the part worth care -- crossing midnight is more than a day, not less.

3 · Execute4 carry out the plan

1Count the hours that pass

#7 Identify Subproblems 7.RP.A.3

From 2 a.m. to 2 a.m. the next day is a full day, and then 16 hours more.

24 + 16 = 40

40 hours in all.

2Set up the proportion

#13 Convert to Algebra 7.RP.A.2

6 minutes belong to 24 hours, so the gain over 40 hours is in the same ratio.

24 : 6 = 40 : \square

One unknown, one equation.

3Solve it

#13 Convert to Algebra 7.RP.A.2

Cross-multiply and divide.

6 \times 40 \div 24 = 10

It is 10 minutes ahead.

4Add the gain to the true time

#8 Analyze the Units 7.RP.A.3

A clock that runs fast shows a time ahead of the real one.

18:00 + 10 = 18\text{:}10

The clock reads 6:10 p.m..

Answer: 6:10 p.m.

4 · Reviewdoes it hold up?

40 hours is a little more than a day, and the gain, 10 minutes, is a little more than the 6 minutes a full day would give.

Another way: Working per hour first -- 6 over 24 minutes an hour -- and multiplying by 40 gives the same 10, through a rate rather than a proportion.

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Treating the steady gain as a proportional relationship.
7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Counting the elapsed hours and applying the gain.

💡Takeaway. A clock that gains steadily gains in proportion to the time. Count the hours carefully and the rest is one proportion.

Variant 2 easy

A clock gains a steady $9$ minutes a day. If it is set exactly right at $3$ a.m. today, what time will it show at $11$ a.m. the next day?

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A clock gains 9 minutes every day. It was right at 3 a.m., and we want what it shows at 11 the next day.

Givens

The clock gains 9 minutes a day.
It was exactly right at 3 a.m.
It is read the next day.

Unknowns

The time the clock displays.

Constraints

The gain is steady, so it is proportional to the time elapsed.

2 · Planchoose the strategy

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

A steady gain is proportional to time, so this is one proportion once the elapsed hours are known. Counting those hours is the part worth care -- crossing midnight is more than a day, not less.

3 · Execute4 carry out the plan

1Count the hours that pass

#7 Identify Subproblems 7.RP.A.3

From 3 a.m. to 3 a.m. the next day is a full day, and then 8 hours more.

24 + 8 = 32

32 hours in all.

2Set up the proportion

#13 Convert to Algebra 7.RP.A.2

9 minutes belong to 24 hours, so the gain over 32 hours is in the same ratio.

24 : 9 = 32 : \square

One unknown, one equation.

3Solve it

#13 Convert to Algebra 7.RP.A.2

Cross-multiply and divide.

9 \times 32 \div 24 = 12

It is 12 minutes ahead.

4Add the gain to the true time

#8 Analyze the Units 7.RP.A.3

A clock that runs fast shows a time ahead of the real one.

11:00 + 12 = 11\text{:}12

The clock reads 11:12 a.m..

Answer: 11:12 a.m.

4 · Reviewdoes it hold up?

32 hours is a little more than a day, and the gain, 12 minutes, is a little more than the 9 minutes a full day would give.

Another way: Working per hour first -- 9 over 24 minutes an hour -- and multiplying by 32 gives the same 12, through a rate rather than a proportion.

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Treating the steady gain as a proportional relationship.
7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Counting the elapsed hours and applying the gain.

💡Takeaway. A clock that gains steadily gains in proportion to the time. Count the hours carefully and the rest is one proportion.

Variant 3 easy

A clock gains a steady $12$ minutes a day. If it is set exactly right at $3$ a.m. today, what time will it show at $9$ p.m. the next day?

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A clock gains 12 minutes every day. It was right at 3 a.m., and we want what it shows at 21 the next day.

Givens

The clock gains 12 minutes a day.
It was exactly right at 3 a.m.
It is read the next day.

Unknowns

The time the clock displays.

Constraints

The gain is steady, so it is proportional to the time elapsed.

2 · Planchoose the strategy

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

A steady gain is proportional to time, so this is one proportion once the elapsed hours are known. Counting those hours is the part worth care -- crossing midnight is more than a day, not less.

3 · Execute4 carry out the plan

1Count the hours that pass

#7 Identify Subproblems 7.RP.A.3

From 3 a.m. to 3 a.m. the next day is a full day, and then 18 hours more.

24 + 18 = 42

42 hours in all.

2Set up the proportion

#13 Convert to Algebra 7.RP.A.2

12 minutes belong to 24 hours, so the gain over 42 hours is in the same ratio.

24 : 12 = 42 : \square

One unknown, one equation.

3Solve it

#13 Convert to Algebra 7.RP.A.2

Cross-multiply and divide.

12 \times 42 \div 24 = 21

It is 21 minutes ahead.

4Add the gain to the true time

#8 Analyze the Units 7.RP.A.3

A clock that runs fast shows a time ahead of the real one.

21:00 + 21 = 21\text{:}21

The clock reads 9:21 p.m..

Answer: 9:21 p.m.

4 · Reviewdoes it hold up?

42 hours is a little more than a day, and the gain, 21 minutes, is a little more than the 12 minutes a full day would give.

Another way: Working per hour first -- 12 over 24 minutes an hour -- and multiplying by 42 gives the same 21, through a rate rather than a proportion.

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Treating the steady gain as a proportional relationship.
7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Counting the elapsed hours and applying the gain.

💡Takeaway. A clock that gains steadily gains in proportion to the time. Count the hours carefully and the rest is one proportion.

Variant 4 medium

A clock gains a steady $12$ minutes a day. If it is set exactly right at $10$ a.m. today, what time will it show at $8$ p.m. the next day?

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A clock gains 12 minutes every day. It was right at 10 a.m., and we want what it shows at 20 the next day.

Givens

The clock gains 12 minutes a day.
It was exactly right at 10 a.m.
It is read the next day.

Unknowns

The time the clock displays.

Constraints

The gain is steady, so it is proportional to the time elapsed.

2 · Planchoose the strategy

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

A steady gain is proportional to time, so this is one proportion once the elapsed hours are known. Counting those hours is the part worth care -- crossing midnight is more than a day, not less.

3 · Execute4 carry out the plan

1Count the hours that pass

#7 Identify Subproblems 7.RP.A.3

From 10 a.m. to 10 a.m. the next day is a full day, and then 10 hours more.

24 + 10 = 34

34 hours in all.

2Set up the proportion

#13 Convert to Algebra 7.RP.A.2

12 minutes belong to 24 hours, so the gain over 34 hours is in the same ratio.

24 : 12 = 34 : \square

One unknown, one equation.

3Solve it

#13 Convert to Algebra 7.RP.A.2

Cross-multiply and divide.

12 \times 34 \div 24 = 17

It is 17 minutes ahead.

4Add the gain to the true time

#8 Analyze the Units 7.RP.A.3

A clock that runs fast shows a time ahead of the real one.

20:00 + 17 = 20\text{:}17

The clock reads 8:17 p.m..

Answer: 8:17 p.m.

4 · Reviewdoes it hold up?

34 hours is a little more than a day, and the gain, 17 minutes, is a little more than the 12 minutes a full day would give.

Another way: Working per hour first -- 12 over 24 minutes an hour -- and multiplying by 34 gives the same 17, through a rate rather than a proportion.

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Treating the steady gain as a proportional relationship.
7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Counting the elapsed hours and applying the gain.

💡Takeaway. A clock that gains steadily gains in proportion to the time. Count the hours carefully and the rest is one proportion.

Variant 5 medium

A clock gains a steady $3$ minutes a day. If it is set exactly right at $4$ a.m. today, what time will it show at $12$ noon the next day?

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A clock gains 3 minutes every day. It was right at 4 a.m., and we want what it shows at noon the next day.

Givens

The clock gains 3 minutes a day.
It was exactly right at 4 a.m.
It is read the next day.

Unknowns

The time the clock displays.

Constraints

The gain is steady, so it is proportional to the time elapsed.

2 · Planchoose the strategy

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

A steady gain is proportional to time, so this is one proportion once the elapsed hours are known. Counting those hours is the part worth care -- crossing midnight is more than a day, not less.

3 · Execute4 carry out the plan

1Count the hours that pass

#7 Identify Subproblems 7.RP.A.3

From 4 a.m. to 4 a.m. the next day is a full day, and then 8 hours more.

24 + 8 = 32

32 hours in all.

2Set up the proportion

#13 Convert to Algebra 7.RP.A.2

3 minutes belong to 24 hours, so the gain over 32 hours is in the same ratio.

24 : 3 = 32 : \square

One unknown, one equation.

3Solve it

#13 Convert to Algebra 7.RP.A.2

Cross-multiply and divide.

3 \times 32 \div 24 = 4

It is 4 minutes ahead.

4Add the gain to the true time

#8 Analyze the Units 7.RP.A.3

A clock that runs fast shows a time ahead of the real one.

12:00 + 4 = 12\text{:}04

The clock reads 12:04 p.m..

Answer: 12:04 p.m.

4 · Reviewdoes it hold up?

32 hours is a little more than a day, and the gain, 4 minutes, is a little more than the 3 minutes a full day would give.

Another way: Working per hour first -- 3 over 24 minutes an hour -- and multiplying by 32 gives the same 4, through a rate rather than a proportion.

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Treating the steady gain as a proportional relationship.
7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Counting the elapsed hours and applying the gain.

💡Takeaway. A clock that gains steadily gains in proportion to the time. Count the hours carefully and the rest is one proportion.

Variant 6 easy

A clock gains a steady $6$ minutes a day. If it is set exactly right at $8$ a.m. today, what time will it show at $12$ noon the next day?

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A clock gains 6 minutes every day. It was right at 8 a.m., and we want what it shows at noon the next day.

Givens

The clock gains 6 minutes a day.
It was exactly right at 8 a.m.
It is read the next day.

Unknowns

The time the clock displays.

Constraints

The gain is steady, so it is proportional to the time elapsed.

2 · Planchoose the strategy

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

A steady gain is proportional to time, so this is one proportion once the elapsed hours are known. Counting those hours is the part worth care -- crossing midnight is more than a day, not less.

3 · Execute4 carry out the plan

1Count the hours that pass

#7 Identify Subproblems 7.RP.A.3

From 8 a.m. to 8 a.m. the next day is a full day, and then 4 hours more.

24 + 4 = 28

28 hours in all.

2Set up the proportion

#13 Convert to Algebra 7.RP.A.2

6 minutes belong to 24 hours, so the gain over 28 hours is in the same ratio.

24 : 6 = 28 : \square

One unknown, one equation.

3Solve it

#13 Convert to Algebra 7.RP.A.2

Cross-multiply and divide.

6 \times 28 \div 24 = 7

It is 7 minutes ahead.

4Add the gain to the true time

#8 Analyze the Units 7.RP.A.3

A clock that runs fast shows a time ahead of the real one.

12:00 + 7 = 12\text{:}07

The clock reads 12:07 p.m..

Answer: 12:07 p.m.

4 · Reviewdoes it hold up?

28 hours is a little more than a day, and the gain, 7 minutes, is a little more than the 6 minutes a full day would give.

Another way: Working per hour first -- 6 over 24 minutes an hour -- and multiplying by 28 gives the same 7, through a rate rather than a proportion.

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Treating the steady gain as a proportional relationship.
7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Counting the elapsed hours and applying the gain.

💡Takeaway. A clock that gains steadily gains in proportion to the time. Count the hours carefully and the rest is one proportion.

Variant 7 medium

A clock gains a steady $24$ minutes a day. If it is set exactly right at $6$ a.m. today, what time will it show at $6$ p.m. the next day?

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A clock gains 24 minutes every day. It was right at 6 a.m., and we want what it shows at 18 the next day.

Givens

The clock gains 24 minutes a day.
It was exactly right at 6 a.m.
It is read the next day.

Unknowns

The time the clock displays.

Constraints

The gain is steady, so it is proportional to the time elapsed.

2 · Planchoose the strategy

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

A steady gain is proportional to time, so this is one proportion once the elapsed hours are known. Counting those hours is the part worth care -- crossing midnight is more than a day, not less.

3 · Execute4 carry out the plan

1Count the hours that pass

#7 Identify Subproblems 7.RP.A.3

From 6 a.m. to 6 a.m. the next day is a full day, and then 12 hours more.

24 + 12 = 36

36 hours in all.

2Set up the proportion

#13 Convert to Algebra 7.RP.A.2

24 minutes belong to 24 hours, so the gain over 36 hours is in the same ratio.

24 : 24 = 36 : \square

One unknown, one equation.

3Solve it

#13 Convert to Algebra 7.RP.A.2

Cross-multiply and divide.

24 \times 36 \div 24 = 36

It is 36 minutes ahead.

4Add the gain to the true time

#8 Analyze the Units 7.RP.A.3

A clock that runs fast shows a time ahead of the real one.

18:00 + 36 = 18\text{:}36

The clock reads 6:36 p.m..

Answer: 6:36 p.m.

4 · Reviewdoes it hold up?

36 hours is a little more than a day, and the gain, 36 minutes, is a little more than the 24 minutes a full day would give.

Another way: Working per hour first -- 24 over 24 minutes an hour -- and multiplying by 36 gives the same 36, through a rate rather than a proportion.

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Treating the steady gain as a proportional relationship.
7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Counting the elapsed hours and applying the gain.

💡Takeaway. A clock that gains steadily gains in proportion to the time. Count the hours carefully and the rest is one proportion.

Variant 8 medium

A clock gains a steady $24$ minutes a day. If it is set exactly right at $10$ a.m. today, what time will it show at $11$ a.m. the next day?

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A clock gains 24 minutes every day. It was right at 10 a.m., and we want what it shows at 11 the next day.

Givens

The clock gains 24 minutes a day.
It was exactly right at 10 a.m.
It is read the next day.

Unknowns

The time the clock displays.

Constraints

The gain is steady, so it is proportional to the time elapsed.

2 · Planchoose the strategy

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

A steady gain is proportional to time, so this is one proportion once the elapsed hours are known. Counting those hours is the part worth care -- crossing midnight is more than a day, not less.

3 · Execute4 carry out the plan

1Count the hours that pass

#7 Identify Subproblems 7.RP.A.3

From 10 a.m. to 10 a.m. the next day is a full day, and then 1 hours more.

24 + 1 = 25

25 hours in all.

2Set up the proportion

#13 Convert to Algebra 7.RP.A.2

24 minutes belong to 24 hours, so the gain over 25 hours is in the same ratio.

24 : 24 = 25 : \square

One unknown, one equation.

3Solve it

#13 Convert to Algebra 7.RP.A.2

Cross-multiply and divide.

24 \times 25 \div 24 = 25

It is 25 minutes ahead.

4Add the gain to the true time

#8 Analyze the Units 7.RP.A.3

A clock that runs fast shows a time ahead of the real one.

11:00 + 25 = 11\text{:}25

The clock reads 11:25 a.m..

Answer: 11:25 a.m.

4 · Reviewdoes it hold up?

25 hours is a little more than a day, and the gain, 25 minutes, is a little more than the 24 minutes a full day would give.

Another way: Working per hour first -- 24 over 24 minutes an hour -- and multiplying by 25 gives the same 25, through a rate rather than a proportion.

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Treating the steady gain as a proportional relationship.
7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Counting the elapsed hours and applying the gain.

💡Takeaway. A clock that gains steadily gains in proportion to the time. Count the hours carefully and the rest is one proportion.

Variant 9 hard

A clock gains a steady $33$ minutes a day. If it is set exactly right at $5$ a.m. today, what time will it show at $9$ p.m. the next day?

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A clock gains 33 minutes every day. It was right at 5 a.m., and we want what it shows at 21 the next day.

Givens

The clock gains 33 minutes a day.
It was exactly right at 5 a.m.
It is read the next day.

Unknowns

The time the clock displays.

Constraints

The gain is steady, so it is proportional to the time elapsed.

2 · Planchoose the strategy

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

A steady gain is proportional to time, so this is one proportion once the elapsed hours are known. Counting those hours is the part worth care -- crossing midnight is more than a day, not less.

3 · Execute4 carry out the plan

1Count the hours that pass

#7 Identify Subproblems 7.RP.A.3

From 5 a.m. to 5 a.m. the next day is a full day, and then 16 hours more.

24 + 16 = 40

40 hours in all.

2Set up the proportion

#13 Convert to Algebra 7.RP.A.2

33 minutes belong to 24 hours, so the gain over 40 hours is in the same ratio.

24 : 33 = 40 : \square

One unknown, one equation.

3Solve it

#13 Convert to Algebra 7.RP.A.2

Cross-multiply and divide.

33 \times 40 \div 24 = 55

It is 55 minutes ahead.

4Add the gain to the true time

#8 Analyze the Units 7.RP.A.3

A clock that runs fast shows a time ahead of the real one.

21:00 + 55 = 21\text{:}55

The clock reads 9:55 p.m..

Answer: 9:55 p.m.

4 · Reviewdoes it hold up?

40 hours is a little more than a day, and the gain, 55 minutes, is a little more than the 33 minutes a full day would give.

Another way: Working per hour first -- 33 over 24 minutes an hour -- and multiplying by 40 gives the same 55, through a rate rather than a proportion.

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Treating the steady gain as a proportional relationship.
7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Counting the elapsed hours and applying the gain.

💡Takeaway. A clock that gains steadily gains in proportion to the time. Count the hours carefully and the rest is one proportion.

Variant 10 hard

A clock gains a steady $33$ minutes a day. If it is set exactly right at $1$ a.m. today, what time will it show at $5$ p.m. the next day?

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A clock gains 33 minutes every day. It was right at 1 a.m., and we want what it shows at 17 the next day.

Givens

The clock gains 33 minutes a day.
It was exactly right at 1 a.m.
It is read the next day.

Unknowns

The time the clock displays.

Constraints

The gain is steady, so it is proportional to the time elapsed.

2 · Planchoose the strategy

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

A steady gain is proportional to time, so this is one proportion once the elapsed hours are known. Counting those hours is the part worth care -- crossing midnight is more than a day, not less.

3 · Execute4 carry out the plan

1Count the hours that pass

#7 Identify Subproblems 7.RP.A.3

From 1 a.m. to 1 a.m. the next day is a full day, and then 16 hours more.

24 + 16 = 40

40 hours in all.

2Set up the proportion

#13 Convert to Algebra 7.RP.A.2

33 minutes belong to 24 hours, so the gain over 40 hours is in the same ratio.

24 : 33 = 40 : \square

One unknown, one equation.

3Solve it

#13 Convert to Algebra 7.RP.A.2

Cross-multiply and divide.

33 \times 40 \div 24 = 55

It is 55 minutes ahead.

4Add the gain to the true time

#8 Analyze the Units 7.RP.A.3

A clock that runs fast shows a time ahead of the real one.

17:00 + 55 = 17\text{:}55

The clock reads 5:55 p.m..

Answer: 5:55 p.m.

4 · Reviewdoes it hold up?

40 hours is a little more than a day, and the gain, 55 minutes, is a little more than the 33 minutes a full day would give.

Another way: Working per hour first -- 33 over 24 minutes an hour -- and multiplying by 40 gives the same 55, through a rate rather than a proportion.

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Treating the steady gain as a proportional relationship.
7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Counting the elapsed hours and applying the gain.

💡Takeaway. A clock that gains steadily gains in proportion to the time. Count the hours carefully and the rest is one proportion.

Variant 11 hard

A clock gains a steady $36$ minutes a day. If it is set exactly right at $10$ a.m. today, what time will it show at $6$ p.m. the next day?

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A clock gains 36 minutes every day. It was right at 10 a.m., and we want what it shows at 18 the next day.

Givens

The clock gains 36 minutes a day.
It was exactly right at 10 a.m.
It is read the next day.

Unknowns

The time the clock displays.

Constraints

The gain is steady, so it is proportional to the time elapsed.

2 · Planchoose the strategy

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

A steady gain is proportional to time, so this is one proportion once the elapsed hours are known. Counting those hours is the part worth care -- crossing midnight is more than a day, not less.

3 · Execute4 carry out the plan

1Count the hours that pass

#7 Identify Subproblems 7.RP.A.3

From 10 a.m. to 10 a.m. the next day is a full day, and then 8 hours more.

24 + 8 = 32

32 hours in all.

2Set up the proportion

#13 Convert to Algebra 7.RP.A.2

36 minutes belong to 24 hours, so the gain over 32 hours is in the same ratio.

24 : 36 = 32 : \square

One unknown, one equation.

3Solve it

#13 Convert to Algebra 7.RP.A.2

Cross-multiply and divide.

36 \times 32 \div 24 = 48

It is 48 minutes ahead.

4Add the gain to the true time

#8 Analyze the Units 7.RP.A.3

A clock that runs fast shows a time ahead of the real one.

18:00 + 48 = 18\text{:}48

The clock reads 6:48 p.m..

Answer: 6:48 p.m.

4 · Reviewdoes it hold up?

32 hours is a little more than a day, and the gain, 48 minutes, is a little more than the 36 minutes a full day would give.

Another way: Working per hour first -- 36 over 24 minutes an hour -- and multiplying by 32 gives the same 48, through a rate rather than a proportion.

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Treating the steady gain as a proportional relationship.
7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Counting the elapsed hours and applying the gain.

💡Takeaway. A clock that gains steadily gains in proportion to the time. Count the hours carefully and the rest is one proportion.

Variant 12 hard

A clock gains a steady $42$ minutes a day. If it is set exactly right at $7$ a.m. today, what time will it show at $11$ a.m. the next day?

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A clock gains 42 minutes every day. It was right at 7 a.m., and we want what it shows at 11 the next day.

Givens

The clock gains 42 minutes a day.
It was exactly right at 7 a.m.
It is read the next day.

Unknowns

The time the clock displays.

Constraints

The gain is steady, so it is proportional to the time elapsed.

2 · Planchoose the strategy

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

A steady gain is proportional to time, so this is one proportion once the elapsed hours are known. Counting those hours is the part worth care -- crossing midnight is more than a day, not less.

3 · Execute4 carry out the plan

1Count the hours that pass

#7 Identify Subproblems 7.RP.A.3

From 7 a.m. to 7 a.m. the next day is a full day, and then 4 hours more.

24 + 4 = 28

28 hours in all.

2Set up the proportion

#13 Convert to Algebra 7.RP.A.2

42 minutes belong to 24 hours, so the gain over 28 hours is in the same ratio.

24 : 42 = 28 : \square

One unknown, one equation.

3Solve it

#13 Convert to Algebra 7.RP.A.2

Cross-multiply and divide.

42 \times 28 \div 24 = 49

It is 49 minutes ahead.

4Add the gain to the true time

#8 Analyze the Units 7.RP.A.3

A clock that runs fast shows a time ahead of the real one.

11:00 + 49 = 11\text{:}49

The clock reads 11:49 a.m..

Answer: 11:49 a.m.

4 · Reviewdoes it hold up?

28 hours is a little more than a day, and the gain, 49 minutes, is a little more than the 42 minutes a full day would give.

Another way: Working per hour first -- 42 over 24 minutes an hour -- and multiplying by 28 gives the same 49, through a rate rather than a proportion.

Standardsmin grade 7

7.RP.A.2 Recognize and represent proportional relationships between quantities — Treating the steady gain as a proportional relationship.
7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems — Counting the elapsed hours and applying the gain.

💡Takeaway. A clock that gains steadily gains in proportion to the time. Count the hours carefully and the rest is one proportion.

A steady gain is proportional to the time that has passed · 12 practice problems

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