Star problems Advanced ← Back to chapter

Ratios, Rates, and Percents: 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.

0 of 10 completed

Progress saved in this browser.

Problem 1 of 10

Problem 1 A fourth color

Difficulty: 1 of 3 stars, Stretch

A collection contains positive whole numbers of red, blue, green, and yellow tiles. The ratio of red to blue tiles is $2:3$ , and the ratio of blue to green tiles is $4:5$ . The number of red tiles plus the number of yellow tiles equals the number of green tiles.

Find the smallest possible total number of tiles. Then describe every possible total and prove that your description is complete.
1. Hint 1
  Make the blue entries in the two ratios agree before combining them.
2. Hint 2
  Once the four-color ratio is known, check whether its scale factor must be a whole number.
Answer

The smallest total is 42. Every possible total is a positive multiple of 42; the four counts are $8t,12t,15t,7t$ for a positive integer $t$ .

Full solution

The ratios $2:3$ and $4:5$ share blue, so rewrite them as $8:12$ and $12:15$ .
Thus red, blue, and green have ratio $8:12:15$ .
Since red plus yellow equals green, yellow contributes the difference $15-8=7$ parts.
All four colors therefore have ratio $8:12:15:7$ , containing 42 parts altogether.

We must still justify that a fractional scale cannot create a smaller collection.
Let one part represent $t$ tiles, so red and green contain $8t$ and $15t$ tiles.
Both are whole numbers.
Their combination $2(8t)-15t=t$ is consequently a whole number too.
Positivity makes $t$ a positive integer.

Taking $t=1$ gives counts $8,12,15,7$ , which satisfy every condition and total 42.
Every positive integer $t$ gives a valid collection, and the argument above shows that every collection has this form.
Hence the possible totals are exactly $42,84,126,\ldots$ .

Answer

The smallest total is 42. Every possible total is a positive multiple of 42; the four counts are $8t,12t,15t,7t$ for a positive integer $t$ .

Key idea

Matching ratio parts determines proportions; whole-number conditions determine the allowed scale.
Problem 1: Mark complete
Problem 2 Half the time or half the distance?

Difficulty: 1 of 3 stars, Stretch

Two carts start together and travel for the same positive amount of time. Cart A travels at 6 kilometers per hour for the first half of its travel time and at 12 kilometers per hour for the second half.

Cart B travels at 6 kilometers per hour for the first half of its total distance and at 12 kilometers per hour for the second half. Neither cart stops.

(a) Which cart travels farther, and by what percentage of the shorter distance?

(b) Keep Cart B's first-half speed at 6 kilometers per hour. What second-half speed would make it travel the same distance as Cart A in the same time?
1. Hint 1
  For Cart A, use one hour as a convenient travel time. For Cart B, use two equal distance pieces.
2. Hint 2
  For part (b), suppose the common time is one hour. Cart B must cover 4.5 kilometers in each distance half.
Answer

Cart A travels $12.5\%$ farther. Cart B needs a second-half speed of 18 kilometers per hour to tie Cart A.

Full solution

The comparison does not depend on the common travel time, so use one hour.
Cart A covers $6\times\frac12+12\times\frac12=9$ kilometers.
Its equal-time average speed is 9 kilometers per hour.

For Cart B, consider a journey consisting of 6 kilometers at each speed.
The slow half takes one hour; the fast half takes half an hour.
Its average speed is therefore $12\div\frac32=8$ kilometers per hour.
In the common one-hour period, Cart B travels 8 kilometers.
Cart A travels 1 kilometer farther, which is $\frac18=12.5\%$ of Cart B's distance.

To tie Cart A, Cart B must travel 9 kilometers in one hour.
Its first 4.5 kilometers at 6 kilometers per hour take $\frac34$ hour.
Only $\frac14$ hour remains for the other 4.5 kilometers, requiring $4.5\div\frac14=18$ kilometers per hour.

Averaging 6 and 12 gives the right speed only when the two speeds are used for equal times.
Equal distances give more time to the slower speed.

Answer

Cart A travels $12.5\%$ farther. Cart B needs a second-half speed of 18 kilometers per hour to tie Cart A.

Key idea

An average speed must weight speeds by time, even when the journey is described by distance.
Problem 2: Mark complete
Problem 3 Reordering four price changes

Difficulty: 1 of 3 stars, Stretch

A positive price undergoes four changes: an increase of $25\%$ , an increase of $20\%$ , a decrease of $20\%$ , and a decrease of $25\%$ . Each change is applied once to the price then in effect, in any order.

(a) Does the order affect the final price? Find the final percentage change.

(b) Across all orders, what are the greatest and least prices that can appear immediately after a change, as percentages of the original price? Prove both bounds are attainable and cannot be exceeded.
1. Hint 1
  Represent each change by the fraction by which it multiplies the current price.
2. Hint 2
  To bound an intermediate price, ask which completed changes could increase its product and which could decrease it.
Answer

The final price is $90\%$ of the original, a $10\%$ decrease, in every order. Intermediate prices can be as high as $150\%$ or as low as $60\%$ of the original.

Full solution

The four multipliers are $\frac54,\frac65,\frac45,\frac34$ .
The final price is the original price multiplied by all four.
Multiplication can be reordered, and the product is

$\frac54\times\frac65\times\frac45\times\frac34=\frac9{10}.$

Thus every order finishes at $90\%$ of the original price.
Adding the signed percentages would incorrectly predict no change, because the percentages use changing prices as their wholes.

At any intermediate stage, the price multiplier is the product of some of these four factors.
A largest possible product includes both factors greater than 1 and neither factor less than 1.
It is $\frac54\times\frac65=\frac32$ , attained by applying both increases first.

Similarly, a smallest possible product includes both decrease factors and neither increase factor.
It is $\frac45\times\frac34=\frac35$ , attained by applying both decreases first.
Adding any increase raises this product; omitting any decrease also raises it.
These arguments prove the $150\%$ and $60\%$ bounds for every order.

Answer

The final price is $90\%$ of the original, a $10\%$ decrease, in every order. Intermediate prices can be as high as $150\%$ or as low as $60\%$ of the original.

Key idea

Percent changes multiply, and intermediate extremes can be bounded by choosing which factors are present.
Problem 3: Mark complete
Problem 4 A simultaneous exchange

Difficulty: 2 of 3 stars, Challenge

Jug A contains 9 liters of a well-mixed drink that is $20\%$ concentrate. Jug B contains 6 liters of a well-mixed drink that is $70\%$ concentrate. Volumes add normally when the drinks are mixed.

The same quantity is removed from each jug into separate clean containers before either quantity is poured into the other jug. After the exchange, each jug is mixed thoroughly. The two final drinks have the same concentration.

How many liters were exchanged in each direction, and what is the final concentration? Explain why no other exchange quantity works.
1. Hint 1
  The exchange changes neither the total drink volume nor the total amount of concentrate.
2. Hint 2
  Each liter exchanged increases the concentrate in Jug A by the difference between the original concentrations.
Answer

$\frac{18}{5}=3.6$ liters in each direction; both final concentrations are $40\%$ .

Full solution

The original concentrate amounts are $9\times0.20=1.8$ liters and $6\times0.70=4.2$ liters.
Altogether, 15 liters of drink contain 6 liters of concentrate.
If both jugs finish at the same concentration, that concentration must therefore be $6/15=40\%$ .

Jug A keeps its volume of 9 liters.
To reach $40\%$ , its concentrate amount must become 3.6 liters, a gain of 1.8 liters.
Each liter removed from A takes away 0.2 liter of concentrate; each liter received from the original mixture in B brings 0.7 liter.
The net gain is 0.5 liter per exchanged liter.

Consequently, the required exchange is $1.8\div0.5=3.6$ liters.
This is feasible because it is less than either jug's initial volume.
Jug B then loses 1.8 liters of concentrate, leaving $4.2-1.8=2.4$ liters in 6 liters of drink, also $40\%$ .

Any equal final concentration must be $40\%$ , and only this exchange quantity produces A's required gain.
The simultaneous-removal condition matters: both transferred portions have their original concentrations.

Answer

$\frac{18}{5}=3.6$ liters in each direction; both final concentrations are $40\%$ .

Key idea

Conservation fixes the common target before any transfer calculation is needed.
Problem 4: Mark complete
Problem 5 Reconstructing a rounded survey

Difficulty: 2 of 3 stars, Challenge

A survey of at most 100 students asks each student to choose exactly one of hiking or cycling. The reported percentage choosing hiking is $57\%$ , rounded to the nearest whole percent. A percentage exactly halfway between two whole percentages rounds upward.

Later, exactly four hiking voters switch to cycling, and the two choices then have equal numbers of voters. No one else changes a vote.

Find every possible number of surveyed students and the original counts for each choice. Prove that the list is complete.
1. Hint 1
  Four voters changing sides reduces the difference between the groups by eight.
2. Hint 2
  The original hiking share is one-half of the total, plus four students. A reported $57\%$ means the true percentage is at least $56.5\%$ and less than $57.5\%$ .
Answer

Possible totals are 54, 56, 58, and 60. The corresponding original hiking/cycling counts are $(31,23)$ , $(32,24)$ , $(33,25)$ , and $(34,26)$ .

Full solution

Let $N$ be the total.
After the switch, each group has $N/2$ students, so $N$ is even.
Originally, hiking had $N/2+4$ voters and cycling had $N/2-4$ .
Thus the true hiking share exceeds $50\%$ by the fraction $4/N$ of the full survey.

To round to $57\%$ , that extra fraction must be at least $0.065$ and less than $0.075$ .
Therefore

$0.065\leq\frac4N<0.075.$

Since $N$ is positive, the upper inequality gives
$N>4/0.075=53\frac13.$
The lower inequality gives
$N\leq4/0.065=61\frac7{13}.$
The only even integers in this interval are $54,56,58,60$ .

Using $N/2+4$ and $N/2-4$ gives the four stated pairs.
Every pair has positive whole-number counts, the initial difference is eight, and its hiking share lies in the required rounding interval.
Switching four voters therefore gives equality in every case.
All possible surveys had to satisfy the interval and evenness conditions, so there are no others.

Answer

Possible totals are 54, 56, 58, and 60. The corresponding original hiking/cycling counts are $(31,23)$ , $(32,24)$ , $(33,25)$ , and $(34,26)$ .

Key idea

A rounded percentage gives an interval; exact count information can reduce that interval to a finite list.
Problem 5: Mark complete
Problem 6 The final partial cycle

Difficulty: 2 of 3 stars, Challenge

An empty tank has two pumps and a leak. With the leak sealed, Pump A alone would fill the tank in 12 minutes and Pump B alone would fill it in 20 minutes. The leak removes water at a constant rate of one full tank per 30 minutes whenever water is available.

The leak stays open. Starting at time zero, both pumps run together for 2 minutes, then both are off for 1 minute. This 3-minute pattern repeats until the tank first becomes full.

Exactly how long does filling take? Justify why the tank has not already filled during an earlier pump-on interval.
1. Hint 1
  Find the net filling rate while both pumps run, then the net gain over one complete 3-minute cycle.
2. Hint 2
  The tank may become full before the final cycle is complete. Check the highest water level in the previous cycle.
Answer

$16\frac23$ minutes, or 16 minutes 40 seconds.

Full solution

Measure water in tankfuls.
While both pumps run, the net rate is

$\frac1{12}+\frac1{20}-\frac1{30}=\frac1{10}$

tank per minute.
Each 2-minute pumping interval adds $1/5$ of a tank.
The following off minute removes $1/30$ , so a complete cycle gains $1/5-1/30=1/6$ of a tank.
Even in the first off minute there is enough water for this full leak amount.

After five complete cycles, 15 minutes have elapsed and the tank contains $5/6$ of a tank.
The remaining $1/6$ takes $(1/6)\div(1/10)=5/3$ minutes of pumping.
This is less than the available 2 minutes, so filling finishes at $15+5/3=16\frac23$ minutes.

To verify that no earlier filling was overlooked, examine the fifth pumping interval, the highest of the first five.
It starts with $4/6$ and ends with $4/6+1/5=13/15$ , below full.
Earlier pumping peaks are still lower.
Thus 16 minutes 40 seconds is the first filling time.

Answer

$16\frac23$ minutes, or 16 minutes 40 seconds.

Key idea

A net gain per cycle locates the final cycle; the within-cycle rate determines the exact finish.
Problem 6: Mark complete
Problem 7 A discount with a cap

Difficulty: 2 of 3 stars, Challenge

Three items cost 20, 30, and 50 dollars. You have three coupons: one for $20\%$ off one item, one for $30\%$ off one item, and one for $50\%$ off one item. The $50\%$ coupon has a maximum saving of 12 dollars; the other coupons have no cap. Every coupon must be used, exactly one per item. There are no other charges.

Find the greatest and least possible savings as percentages of the original total price. Find every coupon assignment attaining either extreme, and give a systematic argument proving completeness.
1. Hint 1
  The capped coupon does not necessarily belong on the most expensive item. First calculate its saving on each of the three items.
2. Hint 2
  Split into three cases according to which item receives the capped coupon. Within each case, decide how to assign the $20\%$ and $30\%$ coupons for greatest or least saving.
Answer

Greatest saving: $31\%$ , with coupon percentages $(50,20,30)$ or $(20,50,30)$ on the items costing 20, 30, and 50 dollars, respectively. Least saving: $24\%$ , with coupon percentages $(30,20,50)$ in that order.

Full solution

The original total is 100 dollars, so the number of dollars saved is also the percentage saved.
The capped $50\%$ coupon saves 10 dollars on the 20-dollar item and 12 dollars on either of the other items.

Once that coupon is placed, only the uncapped $20\%$ and $30\%$ coupons remain.
Giving the larger percentage to the dearer remaining item gives the larger saving.
Swapping the two changes the saving by $10\%$ of the difference between those two prices, so the comparison is strict.

If the capped coupon is on the 20-dollar item, the largest total saving is $10+6+15=31$ dollars; the smallest is $10+9+10=29$ dollars.

If it is on the 30-dollar item, the largest saving is $12+4+15=31$ dollars; the smallest is $12+6+10=28$ dollars.

If it is on the 50-dollar item, the largest saving is $12+4+9=25$ dollars; the smallest is $12+6+6=24$ dollars.

The three cases include every location of the capped coupon, and within each case both remaining orders have been resolved.
Thus the greatest saving is 31 dollars, attained by exactly the two assignments stated, and the least is 24 dollars, attained by exactly one assignment.
The percentages are therefore $31\%$ and $24\%$ .

Answer

Greatest saving: $31\%$ , with coupon percentages $(50,20,30)$ or $(20,50,30)$ on the items costing 20, 30, and 50 dollars, respectively. Least saving: $24\%$ , with coupon percentages $(30,20,50)$ in that order.

Key idea

A cap can change which assignment is best. Separate the exceptional rule into cases before using a familiar comparison.
Problem 7: Mark complete
Problem 8 Exactly half the concentration

Difficulty: 3 of 3 stars, Deep challenge

A full jug contains a well-mixed drink with a positive concentrate percentage. Choose integers $a$ and $b$ , each at least 2. Remove exactly $1/a$ of the jug's contents, replace it with the same volume of pure water, and mix thoroughly. Then remove exactly $1/b$ of the jug's contents, replace it with pure water, and mix again.

The final concentrate percentage is exactly half the original percentage. Find all ordered pairs $(a,b)$ that work and prove completeness.
1. Hint 1
  Each step multiplies the amount of concentrate by the fraction of the drink left behind.
2. Hint 2
  The two multipliers commute. Temporarily assume $a\leq b$ , and consider $a=2$ , $a=3$ , and $a\geq4$ .
Answer

Exactly $(a,b)=(3,4)$ and $(4,3)$ .

Full solution

The jug finishes each step with its original volume, so concentrate percentage and concentrate amount change by the same factor.
The first step retains the fraction $(a-1)/a$ of the concentrate; the second retains $(b-1)/b$ of what remains.
Their product must equal $1/2$ .

These factors can be multiplied in either order, so first classify pairs with $a\leq b$ .
If $a=2$ , the first factor is already $1/2$ .
The second factor is strictly less than 1 for every finite integer $b$ , making the product less than $1/2$ .
Thus $a=2$ is impossible.

If $a\geq4$ , then $b\geq4$ as well.
Each retained fraction is at least $3/4$ , so their product is at least $9/16>1/2$ .
This case is also impossible.

Only $a=3$ remains.
After retaining $2/3$ , the next step must retain $(1/2)\div(2/3)=3/4$ , so $b=4$ .
Indeed,
$(2/3)(3/4)=1/2.$
Restoring both possible orders gives $(3,4)$ and $(4,3)$ , and the bounds excluded every other pair.

Answer

Exactly $(a,b)=(3,4)$ and $(4,3)$ .

Key idea

Symmetry and rough bounds can reduce an infinite integer search to one case.
Problem 8: Mark complete
Problem 9 Both groups improve, but the total falls

Difficulty: 3 of 3 stars, Deep challenge

Two problem-solving events each have the same total number $N$ of participants. Every participant belongs to one of two categories, experienced or newer; both categories attend each event. The people and the category proportions may differ between events.

At the first event, $90\%$ of experienced participants and $40\%$ of newer participants solve the final problem. Overall, $60\%$ solve it. At the second event, the corresponding percentages are $100\%$ , $50\%$ , and $55\%$ . All these percentages are exact.

Find the smallest possible positive integer $N$ . For that $N$ , give the category sizes and numbers of solvers at both events. Explain how the overall success rate can fall even though both category rates rise.
1. Hint 1
  For each event, imagine everyone had the newer group's success rate. How much does the experienced group add to the overall percentage?
2. Hint 2
  The first event forces one denominator on $N$ ; the second event forces another. All category sizes and solver counts must be whole numbers.
Answer

The smallest total is $N=100$ . Event 1: experienced 40 with 36 solvers; newer 60 with 24 solvers. Event 2: experienced 10 with 10 solvers; newer 90 with 45 solvers.

Full solution

At Event 1, the experienced rate is 50 percentage points above the newer rate.
The overall rate is 20 points above the newer rate.
Thus experienced participants form $20/50=2/5$ of all participants.
Newer participants form $3/5$ .
The experienced solver count is
$(9/10)(2N/5)=9N/25.$
Since 9 and 25 share no factor, this count is a whole number only if $N$ is a multiple of 25.
That condition also makes the other Event 1 counts whole.

At Event 2, the overall rate is 5 points above the newer rate, while the experienced rate is 50 points above it.
Experienced participants therefore form $5/50=1/10$ of the total; newer participants form $9/10$ .
The newer solver count is $9N/20$ , forcing $N$ to be a multiple of 20.

The smallest common positive multiple of 25 and 20 is 100.
The listed counts then give 60 solvers at Event 1 and 55 at Event 2, verifying every exact percentage.

The newer group, which has the lower success rate at either event, grows from $60\%$ to $90\%$ of attendance.
This change in weighting more than offsets the improvement within both groups.

Answer

The smallest total is $N=100$ . Event 1: experienced 40 with 36 solvers; newer 60 with 24 solvers. Event 2: experienced 10 with 10 solvers; newer 90 with 45 solvers.

Key idea

An overall percentage depends on both group rates and group sizes; integer counts add divisibility constraints.
Problem 9: Mark complete
Problem 10 Whole-number percentages that undo each other

Difficulty: 3 of 3 stars, Deep challenge

A positive price is increased by $p\%$ and then decreased by $q\%$ of the increased price. Both $p$ and $q$ are whole numbers from 1 through 99, inclusive. The final price equals the original price.

Find every possible ordered pair $(p,q)$ and prove there are no others. Would reversing the two changes affect which pairs work?
1. Hint 1
  Write restoration as an equation involving the two price multipliers.
2. Hint 2
  The integer $100+p$ must be a divisor of 10000 strictly between 100 and 200. Use $10000=2^4\times5^4$ .
Answer

Only $(p,q)=(25,20)$ works. Reversing the order also restores the price for this pair and creates no additional pairs.

Full solution

Since the original price is positive, restoring it requires the multiplier equation

$\frac{100+p}{100}\times\frac{100-q}{100}=1.$

Equivalently,
$(100+p)(100-q)=10000.$
Thus $100+p$ is an integer divisor of 10000 lying from 101 through 199.
We now find all divisors in that interval, rather than testing 99 percentages.

Every divisor of $10000=2^4\times5^4$ has the form $2^u5^v$ , where $u$ and $v$ are integers from 0 through 4.
For $v=0$ or 1, the largest such divisor is 80, too small.
For $v=2$ , the possibilities are $25,50,100,200,400$ , none in the interval.
For $v=3$ , they begin $125,250,500,\ldots$ ; only 125 fits.
For $v=4$ , even the smallest is 625, too large.

Therefore $100+p=125$ , so $p=25$ .
Its partner factor is $10000/125=80$ , giving $q=20$ .
The product $1.25\times0.80=1$ verifies restoration.
Reversing the changes merely reverses two multiplication factors, so the same pair works and the uniqueness proof still applies.

Answer

Only $(p,q)=(25,20)$ works. Reversing the order also restores the price for this pair and creates no additional pairs.

Key idea

An integer percentage condition can turn a percent problem into a short divisor search.
Problem 10: Mark complete