← A steady gain is a rate, so keep it a fraction until the end · Multiplicative Comparison and Unit Rate

A steady gain is a rate, so keep it a fraction until the end · 12 practice problems

6.NS.A.16.RP.A.3

Generated variants — 12

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

Variant 1 easy answer: 3:01:45 p.m.

A clock runs fast by a steady 3123\frac{1}{2} minutes every 22 days. It was set exactly right at 33 p.m. one day. What time will it show at 33 p.m. the next day, in hours, minutes and seconds?

Show solution
1 · Understandwhat's really being asked

A clock gains 3 and 1/2 minutes every 2 days. It was correct at 3 p.m., and we want the time it displays at 3 p.m. the next day.

Givens
  • The clock gains 3 and 1/2 minutes in 2 days.
  • It was exactly right at 3 p.m.
  • 1 day pass before we read it.
Unknowns
  • The time the clock displays, to the second.
Constraints
  • The gain is steady, so it is the same amount every day.
2 · Planchoose the strategy

#8 Analyze the Units · also uses: #7 Identify Subproblems

The gain is given per 2 days but wanted per 1. Get one day's gain first, and keep it as a fraction -- turning it into minutes too early throws away the seconds the question asks for.

3 · Execute4 carry out the plan

1Find one day's gain

#7 Identify Subproblems 6.NS.A.1
3 and 1/2 minutes spread evenly over 2 days.
312÷2=134 min3\frac{1}{2} \div 2 = 1\frac{3}{4}\ \text{min}
Every day it runs 1 and 3/4 minutes fast.

2Gain over 1 day

#8 Analyze the Units 6.RP.A.3
Multiply one day's gain by the number of days.
134×1=134 min1\frac{3}{4} \times 1 = 1\frac{3}{4}\ \text{min}
It is 1 and 3/4 minutes ahead.

3Turn the fraction into minutes and seconds

#8 Analyze the Units 6.RP.A.3
A minute is 60 seconds, so the fractional part is 3/4 of 60 seconds.
134 min=1 min 45 s1\frac{3}{4}\ \text{min} = 1\ \text{min}\ 45\ \text{s}
1 minutes and 45 seconds.

4Add it to the true time

#8 Analyze the Units 6.RP.A.3
A fast clock shows a time ahead of the real one, so add the gain to 3 p.m.
3:00:00+1:45=3:01:453\text{:}00\text{:}00 + 1\text{:}45 = 3\text{:}01\text{:}45
The clock reads 3:01:45 p.m.
Answer: 3:01:45 p.m.
4 · Reviewdoes it hold up?

Going the other way: 1 min 45 s is 105 seconds, and 2 times that over 1 day is 210 seconds, which is 3 and 1/2 minutes -- the rate we started from.

Another way: Rounding 3 and 1/2 to the nearest minute first would give a whole number of minutes and lose the seconds the question asks for.

Standardsmin grade 6
  • 6.NS.A.1 Divide fractions by fractions — Dividing the gain by the number of days it covers.
  • 6.RP.A.3 Use ratio and rate reasoning to solve problems — Scaling one day's gain up, then reading it as minutes and seconds.
💡Takeaway. Keep a rate as a fraction while you work with it. Turn it into minutes and seconds only when you are ready to read the clock.
Variant 2 easy answer: 2:01:45 p.m.

A clock runs fast by a steady 5145\frac{1}{4} minutes every 33 days. It was set exactly right at 22 p.m. one day. What time will it show at 22 p.m. the next day, in hours, minutes and seconds?

Show solution
1 · Understandwhat's really being asked

A clock gains 5 and 1/4 minutes every 3 days. It was correct at 2 p.m., and we want the time it displays at 2 p.m. the next day.

Givens
  • The clock gains 5 and 1/4 minutes in 3 days.
  • It was exactly right at 2 p.m.
  • 1 day pass before we read it.
Unknowns
  • The time the clock displays, to the second.
Constraints
  • The gain is steady, so it is the same amount every day.
2 · Planchoose the strategy

#8 Analyze the Units · also uses: #7 Identify Subproblems

The gain is given per 3 days but wanted per 1. Get one day's gain first, and keep it as a fraction -- turning it into minutes too early throws away the seconds the question asks for.

3 · Execute4 carry out the plan

1Find one day's gain

#7 Identify Subproblems 6.NS.A.1
5 and 1/4 minutes spread evenly over 3 days.
514÷3=134 min5\frac{1}{4} \div 3 = 1\frac{3}{4}\ \text{min}
Every day it runs 1 and 3/4 minutes fast.

2Gain over 1 day

#8 Analyze the Units 6.RP.A.3
Multiply one day's gain by the number of days.
134×1=134 min1\frac{3}{4} \times 1 = 1\frac{3}{4}\ \text{min}
It is 1 and 3/4 minutes ahead.

3Turn the fraction into minutes and seconds

#8 Analyze the Units 6.RP.A.3
A minute is 60 seconds, so the fractional part is 3/4 of 60 seconds.
134 min=1 min 45 s1\frac{3}{4}\ \text{min} = 1\ \text{min}\ 45\ \text{s}
1 minutes and 45 seconds.

4Add it to the true time

#8 Analyze the Units 6.RP.A.3
A fast clock shows a time ahead of the real one, so add the gain to 2 p.m.
2:00:00+1:45=2:01:452\text{:}00\text{:}00 + 1\text{:}45 = 2\text{:}01\text{:}45
The clock reads 2:01:45 p.m.
Answer: 2:01:45 p.m.
4 · Reviewdoes it hold up?

Going the other way: 1 min 45 s is 105 seconds, and 3 times that over 1 day is 315 seconds, which is 5 and 1/4 minutes -- the rate we started from.

Another way: Rounding 5 and 1/4 to the nearest minute first would give a whole number of minutes and lose the seconds the question asks for.

Standardsmin grade 6
  • 6.NS.A.1 Divide fractions by fractions — Dividing the gain by the number of days it covers.
  • 6.RP.A.3 Use ratio and rate reasoning to solve problems — Scaling one day's gain up, then reading it as minutes and seconds.
💡Takeaway. Keep a rate as a fraction while you work with it. Turn it into minutes and seconds only when you are ready to read the clock.
Variant 3 easy answer: 4:01:50 p.m.

A clock runs fast by a steady 2342\frac{3}{4} minutes every 33 days. It was set exactly right at 44 p.m. one day. What time will it show at 44 p.m. 22 days later, in hours, minutes and seconds?

Show solution
1 · Understandwhat's really being asked

A clock gains 2 and 3/4 minutes every 3 days. It was correct at 4 p.m., and we want the time it displays at 4 p.m. 2 days later.

Givens
  • The clock gains 2 and 3/4 minutes in 3 days.
  • It was exactly right at 4 p.m.
  • 2 days pass before we read it.
Unknowns
  • The time the clock displays, to the second.
Constraints
  • The gain is steady, so it is the same amount every day.
2 · Planchoose the strategy

#8 Analyze the Units · also uses: #7 Identify Subproblems

The gain is given per 3 days but wanted per 2. Get one day's gain first, and keep it as a fraction -- turning it into minutes too early throws away the seconds the question asks for.

3 · Execute4 carry out the plan

1Find one day's gain

#7 Identify Subproblems 6.NS.A.1
2 and 3/4 minutes spread evenly over 3 days.
234÷3=1112 min2\frac{3}{4} \div 3 = \frac{11}{12}\ \text{min}
Every day it runs 11/12 minutes fast.

2Gain over 2 days

#8 Analyze the Units 6.RP.A.3
Multiply one day's gain by the number of days.
1112×2=156 min\frac{11}{12} \times 2 = 1\frac{5}{6}\ \text{min}
It is 1 and 5/6 minutes ahead.

3Turn the fraction into minutes and seconds

#8 Analyze the Units 6.RP.A.3
A minute is 60 seconds, so the fractional part is 5/6 of 60 seconds.
156 min=1 min 50 s1\frac{5}{6}\ \text{min} = 1\ \text{min}\ 50\ \text{s}
1 minutes and 50 seconds.

4Add it to the true time

#8 Analyze the Units 6.RP.A.3
A fast clock shows a time ahead of the real one, so add the gain to 4 p.m.
4:00:00+1:50=4:01:504\text{:}00\text{:}00 + 1\text{:}50 = 4\text{:}01\text{:}50
The clock reads 4:01:50 p.m.
Answer: 4:01:50 p.m.
4 · Reviewdoes it hold up?

Going the other way: 1 min 50 s is 110 seconds, and 3 times that over 2 days is 165 seconds, which is 2 and 3/4 minutes -- the rate we started from.

Another way: Rounding 2 and 3/4 to the nearest minute first would give a whole number of minutes and lose the seconds the question asks for.

Standardsmin grade 6
  • 6.NS.A.1 Divide fractions by fractions — Dividing the gain by the number of days it covers.
  • 6.RP.A.3 Use ratio and rate reasoning to solve problems — Scaling one day's gain up, then reading it as minutes and seconds.
💡Takeaway. Keep a rate as a fraction while you work with it. Turn it into minutes and seconds only when you are ready to read the clock.
Variant 4 easy answer: 7:02:06 p.m.

A clock runs fast by a steady 8258\frac{2}{5} minutes every 44 days. It was set exactly right at 77 p.m. one day. What time will it show at 77 p.m. the next day, in hours, minutes and seconds?

Show solution
1 · Understandwhat's really being asked

A clock gains 8 and 2/5 minutes every 4 days. It was correct at 7 p.m., and we want the time it displays at 7 p.m. the next day.

Givens
  • The clock gains 8 and 2/5 minutes in 4 days.
  • It was exactly right at 7 p.m.
  • 1 day pass before we read it.
Unknowns
  • The time the clock displays, to the second.
Constraints
  • The gain is steady, so it is the same amount every day.
2 · Planchoose the strategy

#8 Analyze the Units · also uses: #7 Identify Subproblems

The gain is given per 4 days but wanted per 1. Get one day's gain first, and keep it as a fraction -- turning it into minutes too early throws away the seconds the question asks for.

3 · Execute4 carry out the plan

1Find one day's gain

#7 Identify Subproblems 6.NS.A.1
8 and 2/5 minutes spread evenly over 4 days.
825÷4=2110 min8\frac{2}{5} \div 4 = 2\frac{1}{10}\ \text{min}
Every day it runs 2 and 1/10 minutes fast.

2Gain over 1 day

#8 Analyze the Units 6.RP.A.3
Multiply one day's gain by the number of days.
2110×1=2110 min2\frac{1}{10} \times 1 = 2\frac{1}{10}\ \text{min}
It is 2 and 1/10 minutes ahead.

3Turn the fraction into minutes and seconds

#8 Analyze the Units 6.RP.A.3
A minute is 60 seconds, so the fractional part is 1/10 of 60 seconds.
2110 min=2 min 6 s2\frac{1}{10}\ \text{min} = 2\ \text{min}\ 6\ \text{s}
2 minutes and 6 seconds.

4Add it to the true time

#8 Analyze the Units 6.RP.A.3
A fast clock shows a time ahead of the real one, so add the gain to 7 p.m.
7:00:00+2:06=7:02:067\text{:}00\text{:}00 + 2\text{:}06 = 7\text{:}02\text{:}06
The clock reads 7:02:06 p.m.
Answer: 7:02:06 p.m.
4 · Reviewdoes it hold up?

Going the other way: 2 min 6 s is 126 seconds, and 4 times that over 1 day is 504 seconds, which is 8 and 2/5 minutes -- the rate we started from.

Another way: Rounding 8 and 2/5 to the nearest minute first would give a whole number of minutes and lose the seconds the question asks for.

Standardsmin grade 6
  • 6.NS.A.1 Divide fractions by fractions — Dividing the gain by the number of days it covers.
  • 6.RP.A.3 Use ratio and rate reasoning to solve problems — Scaling one day's gain up, then reading it as minutes and seconds.
💡Takeaway. Keep a rate as a fraction while you work with it. Turn it into minutes and seconds only when you are ready to read the clock.
Variant 5 medium answer: 5:02:42 p.m.

A clock runs fast by a steady 4124\frac{1}{2} minutes every 55 days. It was set exactly right at 55 p.m. one day. What time will it show at 55 p.m. 33 days later, in hours, minutes and seconds?

Show solution
1 · Understandwhat's really being asked

A clock gains 4 and 1/2 minutes every 5 days. It was correct at 5 p.m., and we want the time it displays at 5 p.m. 3 days later.

Givens
  • The clock gains 4 and 1/2 minutes in 5 days.
  • It was exactly right at 5 p.m.
  • 3 days pass before we read it.
Unknowns
  • The time the clock displays, to the second.
Constraints
  • The gain is steady, so it is the same amount every day.
2 · Planchoose the strategy

#8 Analyze the Units · also uses: #7 Identify Subproblems

The gain is given per 5 days but wanted per 3. Get one day's gain first, and keep it as a fraction -- turning it into minutes too early throws away the seconds the question asks for.

3 · Execute4 carry out the plan

1Find one day's gain

#7 Identify Subproblems 6.NS.A.1
4 and 1/2 minutes spread evenly over 5 days.
412÷5=910 min4\frac{1}{2} \div 5 = \frac{9}{10}\ \text{min}
Every day it runs 9/10 minutes fast.

2Gain over 3 days

#8 Analyze the Units 6.RP.A.3
Multiply one day's gain by the number of days.
910×3=2710 min\frac{9}{10} \times 3 = 2\frac{7}{10}\ \text{min}
It is 2 and 7/10 minutes ahead.

3Turn the fraction into minutes and seconds

#8 Analyze the Units 6.RP.A.3
A minute is 60 seconds, so the fractional part is 7/10 of 60 seconds.
2710 min=2 min 42 s2\frac{7}{10}\ \text{min} = 2\ \text{min}\ 42\ \text{s}
2 minutes and 42 seconds.

4Add it to the true time

#8 Analyze the Units 6.RP.A.3
A fast clock shows a time ahead of the real one, so add the gain to 5 p.m.
5:00:00+2:42=5:02:425\text{:}00\text{:}00 + 2\text{:}42 = 5\text{:}02\text{:}42
The clock reads 5:02:42 p.m.
Answer: 5:02:42 p.m.
4 · Reviewdoes it hold up?

Going the other way: 2 min 42 s is 162 seconds, and 5 times that over 3 days is 270 seconds, which is 4 and 1/2 minutes -- the rate we started from.

Another way: Rounding 4 and 1/2 to the nearest minute first would give a whole number of minutes and lose the seconds the question asks for.

Standardsmin grade 6
  • 6.NS.A.1 Divide fractions by fractions — Dividing the gain by the number of days it covers.
  • 6.RP.A.3 Use ratio and rate reasoning to solve problems — Scaling one day's gain up, then reading it as minutes and seconds.
💡Takeaway. Keep a rate as a fraction while you work with it. Turn it into minutes and seconds only when you are ready to read the clock.
Variant 6 medium answer: 6:03:30 p.m.

A clock runs fast by a steady 1341\frac{3}{4} minutes every 22 days. It was set exactly right at 66 p.m. one day. What time will it show at 66 p.m. 44 days later, in hours, minutes and seconds?

Show solution
1 · Understandwhat's really being asked

A clock gains 1 and 3/4 minutes every 2 days. It was correct at 6 p.m., and we want the time it displays at 6 p.m. 4 days later.

Givens
  • The clock gains 1 and 3/4 minutes in 2 days.
  • It was exactly right at 6 p.m.
  • 4 days pass before we read it.
Unknowns
  • The time the clock displays, to the second.
Constraints
  • The gain is steady, so it is the same amount every day.
2 · Planchoose the strategy

#8 Analyze the Units · also uses: #7 Identify Subproblems

The gain is given per 2 days but wanted per 4. Get one day's gain first, and keep it as a fraction -- turning it into minutes too early throws away the seconds the question asks for.

3 · Execute4 carry out the plan

1Find one day's gain

#7 Identify Subproblems 6.NS.A.1
1 and 3/4 minutes spread evenly over 2 days.
134÷2=78 min1\frac{3}{4} \div 2 = \frac{7}{8}\ \text{min}
Every day it runs 7/8 minutes fast.

2Gain over 4 days

#8 Analyze the Units 6.RP.A.3
Multiply one day's gain by the number of days.
78×4=312 min\frac{7}{8} \times 4 = 3\frac{1}{2}\ \text{min}
It is 3 and 1/2 minutes ahead.

3Turn the fraction into minutes and seconds

#8 Analyze the Units 6.RP.A.3
A minute is 60 seconds, so the fractional part is 1/2 of 60 seconds.
312 min=3 min 30 s3\frac{1}{2}\ \text{min} = 3\ \text{min}\ 30\ \text{s}
3 minutes and 30 seconds.

4Add it to the true time

#8 Analyze the Units 6.RP.A.3
A fast clock shows a time ahead of the real one, so add the gain to 6 p.m.
6:00:00+3:30=6:03:306\text{:}00\text{:}00 + 3\text{:}30 = 6\text{:}03\text{:}30
The clock reads 6:03:30 p.m.
Answer: 6:03:30 p.m.
4 · Reviewdoes it hold up?

Going the other way: 3 min 30 s is 210 seconds, and 2 times that over 4 days is 105 seconds, which is 1 and 3/4 minutes -- the rate we started from.

Another way: Rounding 1 and 3/4 to the nearest minute first would give a whole number of minutes and lose the seconds the question asks for.

Standardsmin grade 6
  • 6.NS.A.1 Divide fractions by fractions — Dividing the gain by the number of days it covers.
  • 6.RP.A.3 Use ratio and rate reasoning to solve problems — Scaling one day's gain up, then reading it as minutes and seconds.
💡Takeaway. Keep a rate as a fraction while you work with it. Turn it into minutes and seconds only when you are ready to read the clock.
Variant 7 medium answer: 2:04:24 p.m.

A clock runs fast by a steady 6356\frac{3}{5} minutes every 66 days. It was set exactly right at 22 p.m. one day. What time will it show at 22 p.m. 44 days later, in hours, minutes and seconds?

Show solution
1 · Understandwhat's really being asked

A clock gains 6 and 3/5 minutes every 6 days. It was correct at 2 p.m., and we want the time it displays at 2 p.m. 4 days later.

Givens
  • The clock gains 6 and 3/5 minutes in 6 days.
  • It was exactly right at 2 p.m.
  • 4 days pass before we read it.
Unknowns
  • The time the clock displays, to the second.
Constraints
  • The gain is steady, so it is the same amount every day.
2 · Planchoose the strategy

#8 Analyze the Units · also uses: #7 Identify Subproblems

The gain is given per 6 days but wanted per 4. Get one day's gain first, and keep it as a fraction -- turning it into minutes too early throws away the seconds the question asks for.

3 · Execute4 carry out the plan

1Find one day's gain

#7 Identify Subproblems 6.NS.A.1
6 and 3/5 minutes spread evenly over 6 days.
635÷6=1110 min6\frac{3}{5} \div 6 = 1\frac{1}{10}\ \text{min}
Every day it runs 1 and 1/10 minutes fast.

2Gain over 4 days

#8 Analyze the Units 6.RP.A.3
Multiply one day's gain by the number of days.
1110×4=425 min1\frac{1}{10} \times 4 = 4\frac{2}{5}\ \text{min}
It is 4 and 2/5 minutes ahead.

3Turn the fraction into minutes and seconds

#8 Analyze the Units 6.RP.A.3
A minute is 60 seconds, so the fractional part is 2/5 of 60 seconds.
425 min=4 min 24 s4\frac{2}{5}\ \text{min} = 4\ \text{min}\ 24\ \text{s}
4 minutes and 24 seconds.

4Add it to the true time

#8 Analyze the Units 6.RP.A.3
A fast clock shows a time ahead of the real one, so add the gain to 2 p.m.
2:00:00+4:24=2:04:242\text{:}00\text{:}00 + 4\text{:}24 = 2\text{:}04\text{:}24
The clock reads 2:04:24 p.m.
Answer: 2:04:24 p.m.
4 · Reviewdoes it hold up?

Going the other way: 4 min 24 s is 264 seconds, and 6 times that over 4 days is 396 seconds, which is 6 and 3/5 minutes -- the rate we started from.

Another way: Rounding 6 and 3/5 to the nearest minute first would give a whole number of minutes and lose the seconds the question asks for.

Standardsmin grade 6
  • 6.NS.A.1 Divide fractions by fractions — Dividing the gain by the number of days it covers.
  • 6.RP.A.3 Use ratio and rate reasoning to solve problems — Scaling one day's gain up, then reading it as minutes and seconds.
💡Takeaway. Keep a rate as a fraction while you work with it. Turn it into minutes and seconds only when you are ready to read the clock.
Variant 8 medium answer: 1:03:36 p.m.

A clock runs fast by a steady 7157\frac{1}{5} minutes every 44 days. It was set exactly right at 11 p.m. one day. What time will it show at 11 p.m. 22 days later, in hours, minutes and seconds?

Show solution
1 · Understandwhat's really being asked

A clock gains 7 and 1/5 minutes every 4 days. It was correct at 1 p.m., and we want the time it displays at 1 p.m. 2 days later.

Givens
  • The clock gains 7 and 1/5 minutes in 4 days.
  • It was exactly right at 1 p.m.
  • 2 days pass before we read it.
Unknowns
  • The time the clock displays, to the second.
Constraints
  • The gain is steady, so it is the same amount every day.
2 · Planchoose the strategy

#8 Analyze the Units · also uses: #7 Identify Subproblems

The gain is given per 4 days but wanted per 2. Get one day's gain first, and keep it as a fraction -- turning it into minutes too early throws away the seconds the question asks for.

3 · Execute4 carry out the plan

1Find one day's gain

#7 Identify Subproblems 6.NS.A.1
7 and 1/5 minutes spread evenly over 4 days.
715÷4=145 min7\frac{1}{5} \div 4 = 1\frac{4}{5}\ \text{min}
Every day it runs 1 and 4/5 minutes fast.

2Gain over 2 days

#8 Analyze the Units 6.RP.A.3
Multiply one day's gain by the number of days.
145×2=335 min1\frac{4}{5} \times 2 = 3\frac{3}{5}\ \text{min}
It is 3 and 3/5 minutes ahead.

3Turn the fraction into minutes and seconds

#8 Analyze the Units 6.RP.A.3
A minute is 60 seconds, so the fractional part is 3/5 of 60 seconds.
335 min=3 min 36 s3\frac{3}{5}\ \text{min} = 3\ \text{min}\ 36\ \text{s}
3 minutes and 36 seconds.

4Add it to the true time

#8 Analyze the Units 6.RP.A.3
A fast clock shows a time ahead of the real one, so add the gain to 1 p.m.
1:00:00+3:36=1:03:361\text{:}00\text{:}00 + 3\text{:}36 = 1\text{:}03\text{:}36
The clock reads 1:03:36 p.m.
Answer: 1:03:36 p.m.
4 · Reviewdoes it hold up?

Going the other way: 3 min 36 s is 216 seconds, and 4 times that over 2 days is 432 seconds, which is 7 and 1/5 minutes -- the rate we started from.

Another way: Rounding 7 and 1/5 to the nearest minute first would give a whole number of minutes and lose the seconds the question asks for.

Standardsmin grade 6
  • 6.NS.A.1 Divide fractions by fractions — Dividing the gain by the number of days it covers.
  • 6.RP.A.3 Use ratio and rate reasoning to solve problems — Scaling one day's gain up, then reading it as minutes and seconds.
💡Takeaway. Keep a rate as a fraction while you work with it. Turn it into minutes and seconds only when you are ready to read the clock.
Variant 9 hard answer: 8:02:20 p.m.

A clock runs fast by a steady 5565\frac{5}{6} minutes every 55 days. It was set exactly right at 88 p.m. one day. What time will it show at 88 p.m. 22 days later, in hours, minutes and seconds?

Show solution
1 · Understandwhat's really being asked

A clock gains 5 and 5/6 minutes every 5 days. It was correct at 8 p.m., and we want the time it displays at 8 p.m. 2 days later.

Givens
  • The clock gains 5 and 5/6 minutes in 5 days.
  • It was exactly right at 8 p.m.
  • 2 days pass before we read it.
Unknowns
  • The time the clock displays, to the second.
Constraints
  • The gain is steady, so it is the same amount every day.
2 · Planchoose the strategy

#8 Analyze the Units · also uses: #7 Identify Subproblems

The gain is given per 5 days but wanted per 2. Get one day's gain first, and keep it as a fraction -- turning it into minutes too early throws away the seconds the question asks for.

3 · Execute4 carry out the plan

1Find one day's gain

#7 Identify Subproblems 6.NS.A.1
5 and 5/6 minutes spread evenly over 5 days.
556÷5=116 min5\frac{5}{6} \div 5 = 1\frac{1}{6}\ \text{min}
Every day it runs 1 and 1/6 minutes fast.

2Gain over 2 days

#8 Analyze the Units 6.RP.A.3
Multiply one day's gain by the number of days.
116×2=213 min1\frac{1}{6} \times 2 = 2\frac{1}{3}\ \text{min}
It is 2 and 1/3 minutes ahead.

3Turn the fraction into minutes and seconds

#8 Analyze the Units 6.RP.A.3
A minute is 60 seconds, so the fractional part is 1/3 of 60 seconds.
213 min=2 min 20 s2\frac{1}{3}\ \text{min} = 2\ \text{min}\ 20\ \text{s}
2 minutes and 20 seconds.

4Add it to the true time

#8 Analyze the Units 6.RP.A.3
A fast clock shows a time ahead of the real one, so add the gain to 8 p.m.
8:00:00+2:20=8:02:208\text{:}00\text{:}00 + 2\text{:}20 = 8\text{:}02\text{:}20
The clock reads 8:02:20 p.m.
Answer: 8:02:20 p.m.
4 · Reviewdoes it hold up?

Going the other way: 2 min 20 s is 140 seconds, and 5 times that over 2 days is 350 seconds, which is 5 and 5/6 minutes -- the rate we started from.

Another way: Rounding 5 and 5/6 to the nearest minute first would give a whole number of minutes and lose the seconds the question asks for.

Standardsmin grade 6
  • 6.NS.A.1 Divide fractions by fractions — Dividing the gain by the number of days it covers.
  • 6.RP.A.3 Use ratio and rate reasoning to solve problems — Scaling one day's gain up, then reading it as minutes and seconds.
💡Takeaway. Keep a rate as a fraction while you work with it. Turn it into minutes and seconds only when you are ready to read the clock.
Variant 10 hard answer: 9:06:10 p.m.

A clock runs fast by a steady 9149\frac{1}{4} minutes every 66 days. It was set exactly right at 99 p.m. one day. What time will it show at 99 p.m. 44 days later, in hours, minutes and seconds?

Show solution
1 · Understandwhat's really being asked

A clock gains 9 and 1/4 minutes every 6 days. It was correct at 9 p.m., and we want the time it displays at 9 p.m. 4 days later.

Givens
  • The clock gains 9 and 1/4 minutes in 6 days.
  • It was exactly right at 9 p.m.
  • 4 days pass before we read it.
Unknowns
  • The time the clock displays, to the second.
Constraints
  • The gain is steady, so it is the same amount every day.
2 · Planchoose the strategy

#8 Analyze the Units · also uses: #7 Identify Subproblems

The gain is given per 6 days but wanted per 4. Get one day's gain first, and keep it as a fraction -- turning it into minutes too early throws away the seconds the question asks for.

3 · Execute4 carry out the plan

1Find one day's gain

#7 Identify Subproblems 6.NS.A.1
9 and 1/4 minutes spread evenly over 6 days.
914÷6=11324 min9\frac{1}{4} \div 6 = 1\frac{13}{24}\ \text{min}
Every day it runs 1 and 13/24 minutes fast.

2Gain over 4 days

#8 Analyze the Units 6.RP.A.3
Multiply one day's gain by the number of days.
11324×4=616 min1\frac{13}{24} \times 4 = 6\frac{1}{6}\ \text{min}
It is 6 and 1/6 minutes ahead.

3Turn the fraction into minutes and seconds

#8 Analyze the Units 6.RP.A.3
A minute is 60 seconds, so the fractional part is 1/6 of 60 seconds.
616 min=6 min 10 s6\frac{1}{6}\ \text{min} = 6\ \text{min}\ 10\ \text{s}
6 minutes and 10 seconds.

4Add it to the true time

#8 Analyze the Units 6.RP.A.3
A fast clock shows a time ahead of the real one, so add the gain to 9 p.m.
9:00:00+6:10=9:06:109\text{:}00\text{:}00 + 6\text{:}10 = 9\text{:}06\text{:}10
The clock reads 9:06:10 p.m.
Answer: 9:06:10 p.m.
4 · Reviewdoes it hold up?

Going the other way: 6 min 10 s is 370 seconds, and 6 times that over 4 days is 555 seconds, which is 9 and 1/4 minutes -- the rate we started from.

Another way: Rounding 9 and 1/4 to the nearest minute first would give a whole number of minutes and lose the seconds the question asks for.

Standardsmin grade 6
  • 6.NS.A.1 Divide fractions by fractions — Dividing the gain by the number of days it covers.
  • 6.RP.A.3 Use ratio and rate reasoning to solve problems — Scaling one day's gain up, then reading it as minutes and seconds.
💡Takeaway. Keep a rate as a fraction while you work with it. Turn it into minutes and seconds only when you are ready to read the clock.
Variant 11 hard answer: 10:04:30 p.m.

A clock runs fast by a steady 3383\frac{3}{8} minutes every 33 days. It was set exactly right at 1010 p.m. one day. What time will it show at 1010 p.m. 44 days later, in hours, minutes and seconds?

Show solution
1 · Understandwhat's really being asked

A clock gains 3 and 3/8 minutes every 3 days. It was correct at 10 p.m., and we want the time it displays at 10 p.m. 4 days later.

Givens
  • The clock gains 3 and 3/8 minutes in 3 days.
  • It was exactly right at 10 p.m.
  • 4 days pass before we read it.
Unknowns
  • The time the clock displays, to the second.
Constraints
  • The gain is steady, so it is the same amount every day.
2 · Planchoose the strategy

#8 Analyze the Units · also uses: #7 Identify Subproblems

The gain is given per 3 days but wanted per 4. Get one day's gain first, and keep it as a fraction -- turning it into minutes too early throws away the seconds the question asks for.

3 · Execute4 carry out the plan

1Find one day's gain

#7 Identify Subproblems 6.NS.A.1
3 and 3/8 minutes spread evenly over 3 days.
338÷3=118 min3\frac{3}{8} \div 3 = 1\frac{1}{8}\ \text{min}
Every day it runs 1 and 1/8 minutes fast.

2Gain over 4 days

#8 Analyze the Units 6.RP.A.3
Multiply one day's gain by the number of days.
118×4=412 min1\frac{1}{8} \times 4 = 4\frac{1}{2}\ \text{min}
It is 4 and 1/2 minutes ahead.

3Turn the fraction into minutes and seconds

#8 Analyze the Units 6.RP.A.3
A minute is 60 seconds, so the fractional part is 1/2 of 60 seconds.
412 min=4 min 30 s4\frac{1}{2}\ \text{min} = 4\ \text{min}\ 30\ \text{s}
4 minutes and 30 seconds.

4Add it to the true time

#8 Analyze the Units 6.RP.A.3
A fast clock shows a time ahead of the real one, so add the gain to 10 p.m.
10:00:00+4:30=10:04:3010\text{:}00\text{:}00 + 4\text{:}30 = 10\text{:}04\text{:}30
The clock reads 10:04:30 p.m.
Answer: 10:04:30 p.m.
4 · Reviewdoes it hold up?

Going the other way: 4 min 30 s is 270 seconds, and 3 times that over 4 days is 202 seconds, which is 3 and 3/8 minutes -- the rate we started from.

Another way: Rounding 3 and 3/8 to the nearest minute first would give a whole number of minutes and lose the seconds the question asks for.

Standardsmin grade 6
  • 6.NS.A.1 Divide fractions by fractions — Dividing the gain by the number of days it covers.
  • 6.RP.A.3 Use ratio and rate reasoning to solve problems — Scaling one day's gain up, then reading it as minutes and seconds.
💡Takeaway. Keep a rate as a fraction while you work with it. Turn it into minutes and seconds only when you are ready to read the clock.
Variant 12 hard answer: 11:04:30 p.m.

A clock runs fast by a steady 101210\frac{1}{2} minutes every 77 days. It was set exactly right at 1111 p.m. one day. What time will it show at 1111 p.m. 33 days later, in hours, minutes and seconds?

Show solution
1 · Understandwhat's really being asked

A clock gains 10 and 1/2 minutes every 7 days. It was correct at 11 p.m., and we want the time it displays at 11 p.m. 3 days later.

Givens
  • The clock gains 10 and 1/2 minutes in 7 days.
  • It was exactly right at 11 p.m.
  • 3 days pass before we read it.
Unknowns
  • The time the clock displays, to the second.
Constraints
  • The gain is steady, so it is the same amount every day.
2 · Planchoose the strategy

#8 Analyze the Units · also uses: #7 Identify Subproblems

The gain is given per 7 days but wanted per 3. Get one day's gain first, and keep it as a fraction -- turning it into minutes too early throws away the seconds the question asks for.

3 · Execute4 carry out the plan

1Find one day's gain

#7 Identify Subproblems 6.NS.A.1
10 and 1/2 minutes spread evenly over 7 days.
1012÷7=112 min10\frac{1}{2} \div 7 = 1\frac{1}{2}\ \text{min}
Every day it runs 1 and 1/2 minutes fast.

2Gain over 3 days

#8 Analyze the Units 6.RP.A.3
Multiply one day's gain by the number of days.
112×3=412 min1\frac{1}{2} \times 3 = 4\frac{1}{2}\ \text{min}
It is 4 and 1/2 minutes ahead.

3Turn the fraction into minutes and seconds

#8 Analyze the Units 6.RP.A.3
A minute is 60 seconds, so the fractional part is 1/2 of 60 seconds.
412 min=4 min 30 s4\frac{1}{2}\ \text{min} = 4\ \text{min}\ 30\ \text{s}
4 minutes and 30 seconds.

4Add it to the true time

#8 Analyze the Units 6.RP.A.3
A fast clock shows a time ahead of the real one, so add the gain to 11 p.m.
11:00:00+4:30=11:04:3011\text{:}00\text{:}00 + 4\text{:}30 = 11\text{:}04\text{:}30
The clock reads 11:04:30 p.m.
Answer: 11:04:30 p.m.
4 · Reviewdoes it hold up?

Going the other way: 4 min 30 s is 270 seconds, and 7 times that over 3 days is 630 seconds, which is 10 and 1/2 minutes -- the rate we started from.

Another way: Rounding 10 and 1/2 to the nearest minute first would give a whole number of minutes and lose the seconds the question asks for.

Standardsmin grade 6
  • 6.NS.A.1 Divide fractions by fractions — Dividing the gain by the number of days it covers.
  • 6.RP.A.3 Use ratio and rate reasoning to solve problems — Scaling one day's gain up, then reading it as minutes and seconds.
💡Takeaway. Keep a rate as a fraction while you work with it. Turn it into minutes and seconds only when you are ready to read the clock.