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Ratios, Percents, and Proportion: Star problems

Ten optional challenges to stretch your reasoning. Work on paper, use hints when you need them, and check the answer or full solution when you are ready. You can skip these problems and continue the course.

1 of 3 stars: Stretch
2 of 3 stars: Challenge
3 of 3 stars: Deep challenge

Stars indicate difficulty within this set.

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Problem 1 of 10

Problem 1 Ratios of overlapping totals

Difficulty: 1 of 3 stars, Stretch

Three positive quantities are $a,b,c$ . Their pairwise totals satisfy $(a+b):(b+c):(c+a)=7:9:8$ , and $a+b+c=144$ . Find $a,b,c$ .

More generally, suppose the ratio of these pairwise totals is $u:v:w$ , where $u,v,w$ are positive real numbers. Give a necessary and sufficient condition on $u,v,w$ for positive $a,b,c$ to exist, and justify it.

Builds on Ratio Problems
1. Hint 1
  Adding the three pairwise totals counts each original quantity twice.
2. Hint 2
  Write the pairwise totals as $ut,vt,wt$ . Subtract one pairwise total from the sum of the other two.
Answer

$(a,b,c)=(36,48,60)$ . In general, each of $u,v,w$ must be smaller than the sum of the other two.

Full solution

Write $a+b=7t$ , $b+c=9t$ , and $c+a=8t$ , where $t>0$ .
Adding gives
$2(a+b+c)=24t.$
Thus $288=24t$ , so $t=12$ .
To isolate $a$ , add the first and third equations and subtract the second:
$2a=(7+8-9)t=6t.$
Similarly, $2b=(7+9-8)t=8t$ and
$2c=(9+8-7)t=10t.$
This gives the stated values.

For the general ratio, the same calculation gives $a=(u+w-v)t/2$ , $b=(u+v-w)t/2$ , and
$c=(v+w-u)t/2.$
Since $t$ is positive, all three quantities are positive exactly when all three parenthesized expressions are positive.

These conditions are necessary by the formulas.
They are also sufficient: if they hold, choose any $t>0$ and use the formulas to construct positive $a,b,c$ with precisely the required pairwise totals.
Equality in one condition would make an original quantity zero, so the inequalities must be strict.

Answer

$(a,b,c)=(36,48,60)$ . In general, each of $u,v,w$ must be smaller than the sum of the other two.

Key idea

Overlapping totals can reveal individual quantities by adding and subtracting; positivity supplies the feasibility test.
Problem 1: Mark complete
Problem 2 Two changes of speed

Difficulty: 1 of 3 stars, Stretch

A courier travels a fixed distance at a constant speed. Increasing the speed by 4 kilometers per hour would shorten the journey by 10 minutes. Decreasing the original speed by 4 kilometers per hour would lengthen it by 15 minutes.

Find the original speed and distance. Explain why the two time changes must be converted to the same units and why there is only one physically possible answer.

Builds on Rate and Work Problems, Conversion Factors, Algebraic Fractions
1. Hint 1
  Let the original speed be $v>4$ and the distance be $D>0$ . Express each time difference using distance divided by speed.
2. Hint 2
  Divide the two time-difference equations. Their common factors cancel, leaving an equation involving $(v+4)/(v-4)$ .
Answer

The original speed is 20 kilometers per hour and the distance is 20 kilometers.

Full solution

Let $v>4$ be the original speed in kilometers per hour and $D>0$ the distance in kilometers.
Since these units make $D/v$ a time in hours, convert 10 minutes to $1/6$ hour and 15 minutes to $1/4$ hour.
The conditions become

$\frac{D}{v}-\frac{D}{v+4}=\frac16,\qquad \frac{D}{v-4}-\frac{D}{v}=\frac14.$

The left sides simplify to $4D/[v(v+4)]$ and $4D/[v(v-4)]$ .
Dividing the second equation by the first cancels the positive common factors and gives
$(v+4)/(v-4)=3/2.$
Hence $2v+8=3v-12$ , so $v=20$ .
Substituting into either time equation gives $D=20$ .

The original journey takes one hour.
At 24 kilometers per hour it takes $5/6$ hour, saving 10 minutes; at 16 it takes $5/4$ hours, adding 15 minutes.
The ratio equation has only one solution, and it satisfies $v>4$ and $D>0$ , proving physical uniqueness.
The distance disappears because both experiments use the same route.

Answer

The original speed is 20 kilometers per hour and the distance is 20 kilometers.

Key idea

When two rate equations share a scale, dividing them can remove that scale before solving.
Problem 2: Mark complete
Problem 3 Recovering an alloy before treatment

Difficulty: 1 of 3 stars, Stretch

An alloy consists only of copper and silver. A treatment removes exactly $20\%$ of its copper and $5\%$ of its silver. It removes no other material and adds nothing. The remaining alloy has equal masses of copper and silver.

Find the original copper-to-silver mass ratio and the exact percentage of the original total mass that was removed. Explain why averaging $20\%$ and $5\%$ would give the wrong total loss.

Builds on Percent Problems, Ratio Problems
1. Hint 1
  The two ingredients retain different fractions of their original masses.
2. Hint 2
  If the original copper and silver masses are $C$ and $S$ , equality after treatment says $0.80C=0.95S$ . Use the resulting ratio as a convenient original batch.
Answer

The original ratio is $19:16$ . The total mass loss is $\frac{92}{7}\%=13\frac17\%$ .

Full solution

Let the original copper and silver masses be $C$ and $S$ .
Their remaining masses are $4C/5$ and $19S/20$ .
Equality therefore gives $16C=19S$ , so $C:S=19:16$ .
Both ingredients are positive because the remaining equal masses are positive.

Use an original batch containing 19 mass units of copper and 16 of silver.
The treatment removes $19/5$ units of copper and $4/5$ unit of silver, for a total loss of $23/5$ units out of 35.
The lost fraction is therefore $23/175$ .
Multiplying by 100 gives
$92/7\%=13\frac17\%.$
As a check, each remaining ingredient has mass $76/5$ units, exactly as required.

The arithmetic mean of the two loss percentages is $12.5\%$ , but that would describe equal starting masses.
Here the ingredient with the larger loss rate also has the larger starting mass.
The correct total loss weights the two percentages by the original masses, not by the number of ingredients.

Answer

The original ratio is $19:16$ . The total mass loss is $\frac{92}{7}\%=13\frac17\%$ .

Key idea

Percentages with different reference amounts must be weighted by those amounts.
Problem 3: Mark complete
Problem 4 Should they train before starting?

Difficulty: 2 of 3 stars, Challenge

Workers A and B have constant rates. Together they complete one standard job in 12 hours. Three hours of A working alone followed by six hours of B working alone would complete $7/20$ of the job.

Before or during a new job, they may hold one 2-hour training session. Neither produces work during training. After training, B works permanently at twice B's original rate; A's rate is unchanged. Whenever they are not training, both work together. They may also choose to skip training.

Find the shortest completion time and determine when training should begin, if it should be used. Prove that delaying the training cannot improve the result.

Builds on Rate and Work Problems
1. Hint 1
  First recover each worker's original rate in jobs per hour.
2. Hint 2
  If training starts after s hours of joint work, $1-s/12$ of the job remains. Express total completion time in terms of s.
Answer

The minimum is $74/7$ hours, achieved only by training immediately for 2 hours and then working together for $60/7$ hours.

Full solution

Let a and b be the original rates.
The observations give $a+b=1/12$ and
$3a+6b=7/20.$
Dividing the second equation by 3 and subtracting the first gives $b=1/30$ ; hence $a=1/20$ .
After training, their combined rate is $1/20+2/30=7/60$ job per hour.

Suppose training begins after s hours of ordinary joint work, where
$0\leq s<12.$
They have completed $s/12$ of the job, so their total time is

$s+2+\frac{1-s/12}{7/60}=\frac{74}{7}+\frac{2s}{7}.$

This expression is smallest at $s=0$ , because every positive delay adds $2s/7$ hours.
Immediate training therefore finishes the job in $2+60/7=74/7$ hours.
Skipping training takes 12 hours, which is longer since $74/7<84/7$ .
Training after completing the job cannot improve its completion time.

Both original rates are positive, the trained rate is correctly doubled only for B, and the proposed schedule completes exactly one job.
The formula covers every permitted training time, proving that the immediate-training schedule is globally optimal.

Answer

The minimum is $74/7$ hours, achieved only by training immediately for 2 hours and then working together for $60/7$ hours.

Key idea

Account for the remaining work when comparing when to make a rate-changing investment.
Problem 4: Mark complete
Problem 5 Two fields and two kinds of worker

Difficulty: 2 of 3 stars, Challenge

A team of 12 people contains skilled workers and trainees. Each skilled worker completes twice as much work per hour as each trainee. Rates are constant, and all working days have the same length.

Two untouched fields require amounts of work in the ratio $7:3$ . The entire team spends the first half-day on the larger field. For the second half-day, six people stay and finish the larger field; the other six start the smaller field. On the next day, one skilled worker works for a full day and finishes the smaller field.

Find every possible number of skilled workers on the original team, and, for each possibility, how many skilled workers stayed on the larger field that afternoon. Prove completeness.

Builds on Rate and Work Problems, Ratio Problems
1. Hint 1
  Use one trainee-day as the unit of work. Let S be the total number of skilled workers and s the number in the group that stays.
2. Hint 2
  The larger field uses $(18+S+s)/2$ work units; the smaller uses $(10+S-s)/2$ . Remember that both afternoon groups contain exactly six people.
Answer

Either there are 6 skilled workers in total and 4 stay, or there are 11 skilled workers in total and 6 stay.

Full solution

Use one trainee-day as a work unit.
A skilled worker produces 2 units per day.
Let S be the total number of skilled workers and s the number staying on the larger field.
The whole team has rate $12+S$ units per day; the staying group has rate $6+s$ ; the other group has rate $6+S-s$ .

The larger field uses $(12+S)/2+(6+s)/2=(18+S+s)/2$ units.
The smaller uses
$(6+S-s)/2+2=(10+S-s)/2.$
The ratio gives $3(18+S+s)=7(10+S-s)$ , or $5s-2S=8$ .

There are six people in each afternoon group, so $0\leq s\leq6$ and $0\leq S-s\leq6$ , with S and s integers.
The equation forces s even.
For $s=0,2,4,6$ , the corresponding S values are $-4,1,6,11$ .
The first is negative, and the second would require more skilled workers in the staying group than in the whole team.
The last two satisfy every count restriction.

For $(S,s)=(6,4)$ the field workloads are 14 and 6; for $(11,6)$ they are $35/2$ and $15/2$ .
Both ratios are $7:3$ , and a skilled worker is available for the last day.
Thus both cases work and the finite count analysis proves completeness.

Answer

Either there are 6 skilled workers in total and 4 stay, or there are 11 skilled workers in total and 6 stay.

Key idea

Equal headcounts need not mean equal work rates; track the composition of every group.
Problem 5: Mark complete
Problem 6 A drift between two powered trips

Difficulty: 2 of 3 stars, Challenge

A boat travels 9 kilometers upstream in 45 minutes. Its engine is then turned off, and it drifts with the current for 10 minutes. The engine is restarted at the same speed relative to the water, and the boat travels downstream to its original starting point in 35 minutes.

Assume the current and the boat's speed relative to the water are constant, and turning takes no time. Find both speeds in kilometers per hour and check that the drift does not carry the boat past its starting point.

Builds on Rate and Work Problems, Conversion Factors
1. Hint 1
  The upstream leg determines the difference between the boat's still-water speed and the current speed.
2. Hint 2
  During the drift the boat moves downstream, so its powered return distance is smaller than 9 kilometers.
Answer

The current is $1.5$ kilometers per hour, and the boat's speed relative to the water is $13.5$ kilometers per hour.

Full solution

Let $v$ be the boat's speed relative to the water and $c$ the current speed, both in kilometers per hour.
The upstream speed is $v-c$ .
Since 45 minutes is $3/4$ hour, the first leg gives $v-c=9/(3/4)=12$ , or $v=c+12$ .

The 10-minute drift lasts $1/6$ hour and carries the boat $c/6$ kilometers downstream.
Hence the powered return distance is $9-c/6$ kilometers.
The return lasts $7/12$ hour at speed $v+c=12+2c$ , so

$9-\frac c6=\frac7{12}(12+2c)=7+\frac{7c}{6}.$

This gives $2=4c/3$ , so $c=3/2$ and $v=27/2$ .
During the drift the boat moves only $1/4$ kilometer, leaving $35/4$ kilometers to return.
Its downstream speed is 15 kilometers per hour, and $(35/4)/15=7/12$ hour, exactly 35 minutes.

The drift therefore stays well short of the starting point.
Also $v>c>0$ , so the assumed upstream travel is possible.
Accounting for the drift distance is essential; using 9 kilometers for both powered legs would model a different trip.

Answer

The current is $1.5$ kilometers per hour, and the boat's speed relative to the water is $13.5$ kilometers per hour.

Key idea

When a trip has several stages, track position as well as elapsed time.
Problem 6: Mark complete
Problem 7 Identifying a proportional relationship

Difficulty: 2 of 3 stars, Challenge

A positive quantity $Q$ is directly proportional to a positive input $a$ and inversely proportional to a positive input $b$ : $Q=ka/b$ , where $k>0$ is fixed. Starting from unknown original values, increasing $a$ by 6 and $b$ by 4 leaves $Q$ unchanged. In a separate experiment from the same original values, increasing $a$ by 6 and decreasing $b$ by 2 doubles $Q$ .

(a) Find the original $a$ and $b$ .

(b) From those original values, by what percentage must $b$ increase if $a$ increases by $25\%$ and $Q$ is to decrease by $10\%$ ?

Builds on Direct and Inverse Proportion, Percent Problems, Solving Linear Equations
1. Hint 1
  In each experiment, compare the new ratio $a/b$ with the original ratio. The constant $k$ cancels.
2. Hint 2
  The first experiment gives $6b=4a$ . For part (b), the multiplier for $Q$ is the multiplier for $a$ divided by the multiplier for $b$ .
Answer

(a) $a=12$ and $b=8$ . (b) $b$ must increase by $\frac{350}{9}\%=38\frac89\%$ .

Full solution

The first experiment gives
$(a+6)/(b+4)=a/b.$
Cross-multiplication yields $6b=4a$ , so $a=3b/2$ .
In particular,
$(a+6)b=a(b+4).$

The second experiment has positive input $b-2$ and gives
$(a+6)/(b-2)=2a/b.$
Thus
$(a+6)b=2a(b-2).$
Replacing its left side using the first experiment and canceling $a>0$ gives $b+4=2b-4$ .
Consequently $b=8$ and $a=12$ .
These values make both modified denominators positive; the ratios are $18/12=12/8$ and $18/6=2(12/8)$ , verifying both observations.

For part (b), let $r$ be the multiplier applied to $b$ .
The multiplier for $a$ is $5/4$ , and the desired multiplier for $Q$ is $9/10$ .
Therefore $(5/4)/r=9/10$ , giving $r=25/18$ .
The fractional increase is $r-1=7/18$ , or $350/9\%$ .

The percent increase is not $25\%+10\%$ : the inverse relationship divides by the new input multiplier, and both changes act multiplicatively.

Answer

(a) $a=12$ and $b=8$ . (b) $b$ must increase by $\frac{350}{9}\%=38\frac89\%$ .

Key idea

Use ratios to remove a proportionality constant; use multipliers to combine percent changes.
Problem 7: Mark complete
Problem 8 One bicycle for two travelers

Difficulty: 3 of 3 stars, Deep challenge

Two travelers A and B start together and must each cover a 24-kilometer straight route. A walks at 4 kilometers per hour and B walks at 6 kilometers per hour. There is one bicycle, which either traveler can ride at 12 kilometers per hour. Only one person can ride it at a time.

Travel is always forward along the route. A traveler may leave the bicycle for the other to pick up later, and either traveler may wait. Switching takes no time. Find the earliest time by which both travelers can finish, give a schedule, and prove that allowing many switches cannot produce a faster result.

Builds on Rate and Work Problems, Conversion Factors
1. Hint 1
  Try a schedule in which A rides first, leaves the bicycle, then walks; B walks to the bicycle and rides the rest.
2. Hint 2
  If A rides r kilometers and B rides s kilometers in any forward-only schedule, then $r+s\leq24$ . Express each traveler's minimum travel time in terms of the ridden distance.
Answer

The minimum is $10/3$ hours. A rides the first 16 kilometers and walks the final 8; B walks the first 16 kilometers and rides the final 8.

Full solution

For the proposed schedule, A takes $16/12+8/4=10/3$ hours.
A leaves the bicycle at kilometer 16 after $4/3$ hours.
B reaches it after $16/6=8/3$ hours, then rides the final 8 kilometers in $2/3$ hour, also finishing at $10/3$ hours.
The bicycle is available when B arrives, so the schedule is feasible.

Now consider any permitted schedule, with any number of switches.
Let r and s be the total distances ridden by A and B.
Because the bicycle only moves forward, $r+s\leq24$ .
Ignoring any waiting, A needs at least $r/12+(24-r)/4=6-r/6$ hours; B needs at least $s/12+(24-s)/6=4-s/12$ hours.

If both finish by time T, these bounds imply $r\geq36-6T$ and
$s\geq48-12T.$
Adding and using $r+s\leq24$ gives $84-18T\leq24$ , hence $T\geq10/3$ .
Waiting cannot weaken these necessary bounds because it adds time.

The schedule attains the universal lower bound, proving it optimal even among schedules with many handoffs.
The forward-only condition is what bounds the total distance traveled by the shared bicycle.

Answer

The minimum is $10/3$ hours. A rides the first 16 kilometers and walks the final 8; B walks the first 16 kilometers and rides the final 8.

Key idea

Bound a shared resource across every possible schedule, then construct a schedule that attains the bound.
Problem 8: Mark complete
Problem 9 Two clocks, one very late reunion

Difficulty: 3 of 3 stars, Deep challenge

Two clocks both show midnight at the correct instant. Each runs at a constant positive rate, and each face repeats after 24 displayed hours. Clock F first completes one full face cycle after 23 real hours. After 20 real hours, the total elapsed time registered by F exceeds the total elapsed time registered by Clock S by exactly one displayed hour; this comparison includes completed cycles, if any.

(a) Find each clock's rate in displayed hours per real hour. Express S's error as seconds lost per real hour.

(b) Find the first positive real time when both faces show midnight together. Give the answer in real days and prove no earlier reunion is possible.

Builds on Conversion Factors, Rate and Work Problems, Ratio Problems
1. Hint 1
  Clock F registers $24/23$ displayed hours per real hour. The difference of the two rates is $1/20$ .
2. Hint 2
  When F shows midnight for the $m$ th time after the start, $23m$ real hours have passed. Determine when S's elapsed displayed hours are a multiple of 24.
Answer

F runs at $24/23$ and S at $457/460$ displayed hours per real hour. S loses $540/23$ seconds per real hour. The first shared midnight is after 460 real days.

Full solution

The rate of F is $24/23$ .
Since its displayed elapsed time leads S by one hour after 20 real hours, its rate exceeds S's by $1/20$ .
Thus S runs at $24/23-1/20=457/460$ displayed hours per real hour.
It loses $1-457/460=3/460$ hour per real hour, or $(3/460)3600=540/23$ seconds per real hour.

Every positive time when F shows midnight has the form $23m$ real hours, where $m$ is a positive integer.
At that time, S has registered $(457/460)(23m)=457m/20$ displayed hours.
Its face also shows midnight exactly when this is a multiple of 24, or when $457m$ is divisible by 480.

The integers 457 and 480 have no common factor: the Euclidean remainders are $480-457=23$ , $457=19(23)+20$ , $23=20+3$ , $20=6(3)+2$ , and $3=2+1$ .
Therefore 480 must divide $m$ .
The smallest possible $m$ is 480.

The first reunion is consequently $23(480)=11040$ real hours, or 460 real days.
At that instant F has completed 480 face cycles and S has completed 457, which directly verifies the reunion.

Answer

F runs at $24/23$ and S at $457/460$ displayed hours per real hour. S loses $540/23$ seconds per real hour. The first shared midnight is after 460 real days.

Key idea

Convert continuous rate information into whole-cycle conditions when an event must repeat exactly.
Problem 9: Mark complete
Problem 10 Two machines, two kinds of ribbon

Difficulty: 3 of 3 stars, Deep challenge

A workshop needs exactly 60 meters of red ribbon and 60 meters of blue ribbon. Machine A makes red ribbon at 6 meters per minute or blue ribbon at 4 meters per minute. Machine B makes either color at 3 meters per minute. Each machine makes only one color at a time, but can switch instantly; both may run simultaneously, and either may be idle. Ribbon can be cut at any real length, so fractional meters are allowed.

Find the shortest possible completion time. Give a schedule achieving it and prove no schedule is faster. Also determine which machine must make the red ribbon in every fastest schedule.

Builds on Rate and Work Problems, Conversion Factors, Direct and Inverse Proportion
1. Hint 1
  Counting total meters alone loses information because the colors take different times on A.
2. Hint 2
  Assign 2 work units to each meter of red and 3 to each meter of blue. Compare the largest work-unit rates of the two machines.
Answer

The minimum is $100/7$ minutes. In every fastest schedule A makes all 60 meters of red; A also makes $120/7$ meters of blue, and B makes $300/7$ meters of blue.

Full solution

Assign an artificial weight of 2 work units to each meter of red and 3 to each meter of blue.
The order requires $60(2)+60(3)=300$ work units.
This weighting makes A equally productive on either color: $6(2)=4(3)=12$ units per minute.
B produces 6 units per minute on red and 9 on blue, so its rate is at most 9.
Together the machines can therefore complete at most 21 work units per minute.
Every schedule needs at least $300/21=100/7$ minutes.

This lower bound is attainable.
Run A on red for 10 minutes, producing all 60 red meters, then on blue for $30/7$ minutes, producing $120/7$ blue meters.
Run B on blue for all $100/7$ minutes, producing $300/7$ blue meters.
The two blue outputs sum to $420/7=60$ meters, and both machines finish by the claimed time.

Equality in the 21-unit rate bound requires B to produce 9 units per minute throughout: any positive time on red or idle would leave a shortfall.
Hence B must make only blue in every fastest schedule, and A must make all the red.
The required remaining blue amounts then follow.
The order of A's color intervals may vary, but its total time on each color is fixed.

Answer

The minimum is $100/7$ minutes. In every fastest schedule A makes all 60 meters of red; A also makes $120/7$ meters of blue, and B makes $300/7$ meters of blue.

Key idea

A carefully chosen common work unit can turn a scheduling problem into a sharp universal bound.
Problem 10: Mark complete