← A steady gain is proportional to the time that has passed · Proportion and Proportional Division

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

7.RP.A.27.RP.A.3

Generated variants — 12

Freshly produced from the archetype’s parameters — problem, figure, and solution derived together.

Variant 1 easy answer: 6:10 p.m.

A clock gains a steady 66 minutes a day. If it is set exactly right at 22 a.m. today, what time will it show at 66 p.m. the next day?

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=4024 + 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:□24 : 6 = 40 : \square
One unknown, one equation.

3Solve it

#13 Convert to Algebra 7.RP.A.2
Cross-multiply and divide.
6×40÷24=106 \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=18:1018:00 + 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 answer: 11:12 a.m.

A clock gains a steady 99 minutes a day. If it is set exactly right at 33 a.m. today, what time will it show at 1111 a.m. the next day?

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=3224 + 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:□24 : 9 = 32 : \square
One unknown, one equation.

3Solve it

#13 Convert to Algebra 7.RP.A.2
Cross-multiply and divide.
9×32÷24=129 \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=11:1211:00 + 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 answer: 9:21 p.m.

A clock gains a steady 1212 minutes a day. If it is set exactly right at 33 a.m. today, what time will it show at 99 p.m. the next day?

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=4224 + 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:□24 : 12 = 42 : \square
One unknown, one equation.

3Solve it

#13 Convert to Algebra 7.RP.A.2
Cross-multiply and divide.
12×42÷24=2112 \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=21:2121:00 + 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 answer: 8:17 p.m.

A clock gains a steady 1212 minutes a day. If it is set exactly right at 1010 a.m. today, what time will it show at 88 p.m. the next day?

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=3424 + 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:□24 : 12 = 34 : \square
One unknown, one equation.

3Solve it

#13 Convert to Algebra 7.RP.A.2
Cross-multiply and divide.
12×34÷24=1712 \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=20:1720:00 + 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 answer: 12:04 p.m.

A clock gains a steady 33 minutes a day. If it is set exactly right at 44 a.m. today, what time will it show at 1212 noon the next day?

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=3224 + 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:□24 : 3 = 32 : \square
One unknown, one equation.

3Solve it

#13 Convert to Algebra 7.RP.A.2
Cross-multiply and divide.
3×32÷24=43 \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=12:0412:00 + 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 answer: 12:07 p.m.

A clock gains a steady 66 minutes a day. If it is set exactly right at 88 a.m. today, what time will it show at 1212 noon the next day?

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=2824 + 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:□24 : 6 = 28 : \square
One unknown, one equation.

3Solve it

#13 Convert to Algebra 7.RP.A.2
Cross-multiply and divide.
6×28÷24=76 \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=12:0712:00 + 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 answer: 6:36 p.m.

A clock gains a steady 2424 minutes a day. If it is set exactly right at 66 a.m. today, what time will it show at 66 p.m. the next day?

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=3624 + 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:□24 : 24 = 36 : \square
One unknown, one equation.

3Solve it

#13 Convert to Algebra 7.RP.A.2
Cross-multiply and divide.
24×36÷24=3624 \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=18:3618:00 + 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 answer: 11:25 a.m.

A clock gains a steady 2424 minutes a day. If it is set exactly right at 1010 a.m. today, what time will it show at 1111 a.m. the next day?

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=2524 + 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:□24 : 24 = 25 : \square
One unknown, one equation.

3Solve it

#13 Convert to Algebra 7.RP.A.2
Cross-multiply and divide.
24×25÷24=2524 \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=11:2511:00 + 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 answer: 9:55 p.m.

A clock gains a steady 3333 minutes a day. If it is set exactly right at 55 a.m. today, what time will it show at 99 p.m. the next day?

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=4024 + 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:□24 : 33 = 40 : \square
One unknown, one equation.

3Solve it

#13 Convert to Algebra 7.RP.A.2
Cross-multiply and divide.
33×40÷24=5533 \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=21:5521:00 + 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 answer: 5:55 p.m.

A clock gains a steady 3333 minutes a day. If it is set exactly right at 11 a.m. today, what time will it show at 55 p.m. the next day?

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=4024 + 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:□24 : 33 = 40 : \square
One unknown, one equation.

3Solve it

#13 Convert to Algebra 7.RP.A.2
Cross-multiply and divide.
33×40÷24=5533 \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=17:5517:00 + 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 answer: 6:48 p.m.

A clock gains a steady 3636 minutes a day. If it is set exactly right at 1010 a.m. today, what time will it show at 66 p.m. the next day?

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=3224 + 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:□24 : 36 = 32 : \square
One unknown, one equation.

3Solve it

#13 Convert to Algebra 7.RP.A.2
Cross-multiply and divide.
36×32÷24=4836 \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=18:4818:00 + 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 answer: 11:49 a.m.

A clock gains a steady 4242 minutes a day. If it is set exactly right at 77 a.m. today, what time will it show at 1111 a.m. the next day?

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=2824 + 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:□24 : 42 = 28 : \square
One unknown, one equation.

3Solve it

#13 Convert to Algebra 7.RP.A.2
Cross-multiply and divide.
42×28÷24=4942 \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=11:4911:00 + 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.