Practice · A steady gain is a rate, so keep it a fraction until the end · Sensim Math

Variant 1 easy

A clock runs fast by a steady $3\frac{1}{2}$ minutes every $2$ days. It was set exactly right at $3$ p.m. one day. What time will it show at $3$ p.m. the next day, in hours, minutes and seconds?

Your answer (check it yourself)

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.

3\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.

1\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.

1\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\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

A clock runs fast by a steady $5\frac{1}{4}$ minutes every $3$ days. It was set exactly right at $2$ p.m. one day. What time will it show at $2$ p.m. the next day, in hours, minutes and seconds?

Your answer (check it yourself)

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.

5\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.

1\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.

1\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\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

A clock runs fast by a steady $2\frac{3}{4}$ minutes every $3$ days. It was set exactly right at $4$ p.m. one day. What time will it show at $4$ p.m. $2$ days later, in hours, minutes and seconds?

Your answer (check it yourself)

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.

2\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.

\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.

1\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\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

A clock runs fast by a steady $8\frac{2}{5}$ minutes every $4$ days. It was set exactly right at $7$ p.m. one day. What time will it show at $7$ p.m. the next day, in hours, minutes and seconds?

Your answer (check it yourself)

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.

8\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.

2\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.

2\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\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

A clock runs fast by a steady $4\frac{1}{2}$ minutes every $5$ days. It was set exactly right at $5$ p.m. one day. What time will it show at $5$ p.m. $3$ days later, in hours, minutes and seconds?

Your answer (check it yourself)

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.

4\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.

\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.

2\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\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

A clock runs fast by a steady $1\frac{3}{4}$ minutes every $2$ days. It was set exactly right at $6$ p.m. one day. What time will it show at $6$ p.m. $4$ days later, in hours, minutes and seconds?

Your answer (check it yourself)

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.

1\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.

\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.

3\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\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

A clock runs fast by a steady $6\frac{3}{5}$ minutes every $6$ days. It was set exactly right at $2$ p.m. one day. What time will it show at $2$ p.m. $4$ days later, in hours, minutes and seconds?

Your answer (check it yourself)

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.

6\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.

1\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.

4\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\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

A clock runs fast by a steady $7\frac{1}{5}$ minutes every $4$ days. It was set exactly right at $1$ p.m. one day. What time will it show at $1$ p.m. $2$ days later, in hours, minutes and seconds?

Your answer (check it yourself)

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.

7\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.

1\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.

3\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\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

A clock runs fast by a steady $5\frac{5}{6}$ minutes every $5$ days. It was set exactly right at $8$ p.m. one day. What time will it show at $8$ p.m. $2$ days later, in hours, minutes and seconds?

Your answer (check it yourself)

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.

5\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.

1\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.

2\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\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

A clock runs fast by a steady $9\frac{1}{4}$ minutes every $6$ days. It was set exactly right at $9$ p.m. one day. What time will it show at $9$ p.m. $4$ days later, in hours, minutes and seconds?

Your answer (check it yourself)

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.

9\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.

1\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.

6\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\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

A clock runs fast by a steady $3\frac{3}{8}$ minutes every $3$ days. It was set exactly right at $10$ p.m. one day. What time will it show at $10$ p.m. $4$ days later, in hours, minutes and seconds?

Your answer (check it yourself)

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.

3\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.

1\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.

4\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\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

A clock runs fast by a steady $10\frac{1}{2}$ minutes every $7$ days. It was set exactly right at $11$ p.m. one day. What time will it show at $11$ p.m. $3$ days later, in hours, minutes and seconds?

Your answer (check it yourself)

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.

10\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.

1\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.

4\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\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.

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

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