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Word Problems with Systems: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 The filter box

A box holds $x$ packs of three filters and $y$ loose filters. Counting each pack and each loose filter as one item, the box holds $12$ items and $26$ filters altogether. Write the two equations that record these facts.
1. Hint 1
  Count items and filters separately.
2. Hint 2
  Each pack counts as one item but holds three filters.
Answer

$x+y=12$ and $3x+y=26$ .

Full solution

The item count includes each pack and each loose filter once.
$x+y=12$
The filter count includes three filters per pack and one per loose filter.
$3x+y=26$

Here $x$ and $y$ are counts, so they must be whole numbers that are zero or positive.
For example, seven packs and five loose filters check both stated totals.

Answer

$x+y=12$ and $3x+y=26$ .

Key idea

One object can contribute differently to two different totals.
Problem 1: Mark complete
Problem 2 The refund entry

A booth sold $a$ small kits at $5$ dollars each and $b$ large kits at $8$ dollars each. A customer then received a refund of $6$ dollars without returning a kit. The booth kept $102$ dollars. Write the equation that records the money kept.
1. Hint 1
  Attach each price to the count of the matching kit.
2. Hint 2
  The money kept is the sales total less the refund.
Answer

$5a+8b-6=102$ , or equivalently $5a+8b=108$ .

Full solution

Small kits bring in $5a$ dollars and large kits bring in $8b$ dollars.
The refund reduces the amount kept by $6$ dollars, so
$5a+8b-6=102$

Adding the refund back gives sales of $108$ dollars, consistent with the equivalent equation $5a+8b=108$ .

Answer

$5a+8b-6=102$ , or equivalently $5a+8b=108$ .

Key idea

An adjustment after a sale enters the value equation as its own term.
Problem 2: Mark complete
Problem 3 The two class plans

A studio offers two payment plans. Plan A charges a $20$ dollar sign-up fee plus $6$ dollars per class. Plan B charges a $50$ dollar sign-up fee plus $4$ dollars per class. Let $c$ be the number of classes and $y$ the total cost in dollars. Write each plan's cost as an equation, then find the number of classes at which the two plans cost the same, and that cost.
1. Hint 1
  The plans cost the same where one pair $(c, y)$ fits both equations.
2. Hint 2
  Each total is the sign-up fee plus the per-class price times $c$ .
3. Hint 3
  Both equations are solved for $y$ , so set the two expressions equal.
Answer

Plan A: $y=20+6c$ . Plan B: $y=50+4c$ . The two plans cost the same at $15$ classes, where each costs $110$ dollars.

Full solution

Each plan's total is its sign-up fee plus its per-class price times the number of classes.
$y=20+6c$ $y=50+4c$

Both equations are solved for $y$ , so set the two expressions equal.
$20+6c=50+4c$
Subtracting $4c$ and $20$ from both sides gives
$2c=30$ $c=15$
Plan A then costs $20+6(15)=110$ dollars.

Check with Plan B: $50+4(15)=110$ dollars, the same total.
So both plans cost $110$ dollars for $15$ classes.

Answer

Plan A: $y=20+6c$ . Plan B: $y=50+4c$ . The two plans cost the same at $15$ classes, where each costs $110$ dollars.

Key idea

The equal-cost point of two plans is the one pair that satisfies both cost equations.
Problem 3: Mark complete
Problem 4 The prepared batch

A lab tank already has $4$ liters of a $25\%$ dye solution. A technician adds a $5\%$ dye solution and a $35\%$ dye solution to make $10$ liters of a $28\%$ dye solution. Assume the volumes add. Find the liters added from each source. Organize the amount, dye fraction, and dye content for the existing batch and both additions in a table on paper before writing the two equations.
1. Hint 1
  Total the liquid and the dye separately, including what is already present.
2. Hint 2
  If the added amounts are $x$ and $y$ , only six liters remain to be added.
3. Hint 3
  The existing batch supplies one liter of dye.
Answer

$1$ liter of the $5\%$ solution and $5$ liters of the $35\%$ solution.

Full solution

Let $x$ and $y$ be the liters added from the $5\%$ and $35\%$ sources.
The existing-batch row has amount $4$ , fraction $0.25$ , and content $1$ , with amounts and contents in liters.
The two addition rows have amount $x$ , fraction $0.05$ , content $0.05x$ , and amount $y$ , fraction $0.35$ , content $0.35y$ .
The final row has amount $10$ , fraction $0.28$ , and content $2.8$ .

The amount column gives
$4+x+y=10$
and the content column gives
$1+0.05x+0.35y=2.8$
Subtracting the existing batch from each gives
$x+y=6$ $0.05x+0.35y=1.8$
Substitute $x=6-y$ .
$0.05(6-y)+0.35y=1.8$ $0.30+0.30y=1.8$ $0.30y=1.5$ $y=5$
Hence $x=1$ .

The liquid totals $4+1+5=10$ liters.
The dye totals $1+0.05+1.75=2.8$ liters, which is $28\%$ of the finished batch.

Answer

$1$ liter of the $5\%$ solution and $5$ liters of the $35\%$ solution.

Key idea

An existing mixture contributes to both totals before any new ingredients are added.
Problem 4: Mark complete
Problem 5 The patrol distances

A boat maintains the same speed in still water on two river patrols while the current stays constant. It travels downstream for $2$ hours and upstream for $3$ hours. The two distances total $50$ miles, and the downstream distance is $14$ miles longer. Find the boat speed in still water and the current speed, both in miles per hour.
1. Hint 1
  Express each distance using its own travel time and ground speed.
2. Hint 2
  With boat speed $b$ and current speed $c$ , the distances are $2(b+c)$ and $3(b-c)$ .
3. Hint 3
  Write equations for their total and difference.
Answer

Boat: $11$ miles per hour. Current: $5$ miles per hour.

Full solution

Let $b$ and $c$ be the boat and current speeds in miles per hour.
The distance facts give
$2(b+c)+3(b-c)=50$ $2(b+c)-3(b-c)=14$
Adding these equations gives
$4(b+c)=64$ $b+c=16$
The downstream distance is therefore $2(16)=32$ miles, so the upstream distance is $50-32=18$ miles.
Then
$3(b-c)=18$ $b-c=6$

Adding the speed equations gives $2b=22$ , hence $b=11$ .
Then $c=16-11=5$ .
The boat is faster than the current, as required to travel upstream.

The downstream distance checks as $2(11+5)=32$ miles and the upstream distance as $3(11-5)=18$ miles.
They total $50$ miles and differ by $14$ miles.

Answer

Boat: $11$ miles per hour. Current: $5$ miles per hour.

Key idea

Total and difference information about two journeys can be translated through their separate distance equations.
Problem 5: Mark complete
Problem 6 The damaged cards

A fundraiser began with $18$ gift cards, each worth either $4$ dollars or $7$ dollars. Two of the $4$ dollar cards were damaged and discarded. The remaining cards are worth $97$ dollars. Find how many cards of each value the fundraiser began with.
1. Hint 1
  Define the unknowns as the original counts.
2. Hint 2
  Only the count of the lower-value cards changes before the remaining value is totaled.
Answer

$7$ cards worth $4$ dollars each and $11$ cards worth $7$ dollars each.

Full solution

Let $x$ and $y$ be the original counts of $4$ dollar and $7$ dollar cards.
$x+y=18$ $4(x-2)+7y=97$
Rearranging the value equation gives
$4x+7y=105$

Multiply the count equation by $4$ and subtract it from the value equation.
$3y=33$ $y=11$
Thus $x=18-11=7$ .

The original count is $7+11=18$ .
After discarding two lower-value cards, the value is $5(4)+11(7)=97$ dollars.
Both original counts are whole numbers, and there are enough lower-value cards to discard two.

Answer

$7$ cards worth $4$ dollars each and $11$ cards worth $7$ dollars each.

Key idea

Keep original counts separate from changed counts when matching a later value total.
Problem 6: Mark complete
Problem 7 The sample display

A maker produces a batch of pins. Four pins are kept as display samples, and each of the remaining pins is sold for $9$ dollars. The fixed cost is $84$ dollars and materials cost $3$ dollars for every pin produced, including the samples. How many pins must be produced for revenue to equal cost, and how many are sold?
1. Hint 1
  The number produced and the number sold differ by four.
2. Hint 2
  Let $x$ be the number produced and $y$ the common cost and revenue in dollars.
Answer

$20$ pins produced; $16$ pins sold.

Full solution

Let $x$ be the number produced.
At break-even the common amount $y$ , in dollars, satisfies
$y=84+3x$ $y=9(x-4)$
Equating the expressions gives
$84+3x=9x-36$ $120=6x$ $x=20$

There are $20-4=16$ pins sold.
Revenue is $9(16)=144$ dollars, and cost is $84+3(20)=144$ dollars, so the batch breaks even.

Answer

$20$ pins produced; $16$ pins sold.

Key idea

Break-even models must distinguish the items that incur costs from those that produce revenue.
Problem 7: Mark complete
Problem 8 The paired recipes

Recipe A mixes $1$ liter of a $10\%$ solution with $3$ liters of a $30\%$ solution. Recipe B reverses those amounts, using $3$ liters of the $10\%$ solution and $1$ liter of the $30\%$ solution. Assume volumes add. Kim says combining both complete recipes produces a $20\%$ solution. Is Kim correct? Justify the decision using amounts and content.
1. Hint 1
  Track how much comes from each original source after both recipes are combined.
2. Hint 2
  Find the total liters and the total dissolved content of the combined batch, then divide.
Answer

Yes; the combined solution is $20\%$ .

Full solution

Together the recipes use $4$ liters of each source.
The total liquid is $8$ liters, and the total dissolved content is
$0.10(4)+0.30(4)=1.6$
in liters.

The content fraction is
$\frac{1.6}{8}=0.20$
Thus the concentration is $20\%$ , and Kim is correct.
The result follows from equal combined amounts of the two sources, even though neither recipe uses equal amounts by itself.

Answer

Yes; the combined solution is $20\%$ .

Key idea

A combined batch's concentration is its total content divided by its total amount, however its parts were grouped.
Problem 8: Mark complete
Problem 9 The delivery receipts

A seller charges a fixed delivery fee plus a fixed price per bottle. Two bottles delivered cost $18$ dollars, while five bottles delivered cost $30$ dollars. Eli says the delivery fee cannot be found because neither receipt lists it separately. Is Eli correct? Find the bottle price and delivery fee if the records determine them.
1. Hint 1
  The same delivery fee appears once in each total.
2. Hint 2
  Subtract the receipt equations to remove that fee.
Answer

No. Each bottle costs $4$ dollars; delivery costs $10$ dollars.

Full solution

Let $p$ be the bottle price and $f$ the delivery fee, both in dollars.
$2p+f=18$ $5p+f=30$
Subtracting gives
$3p=12$ $p=4$
The first receipt then gives $f=18-8=10$ .

The charges check as $2(4)+10=18$ dollars and $5(4)+10=30$ dollars.
The two different bottle counts provide enough independent information to recover the shared fee, so Eli is incorrect.

Answer

No. Each bottle costs $4$ dollars; delivery costs $10$ dollars.

Key idea

A charge shared by two records can be recovered even when it is not itemized.
Problem 9: Mark complete
Problem 10 The matched increase

A shop makes and sells the same number of items. It has a fixed cost $F>0$ , a material cost of $c$ dollars per item, and a selling price of $p$ dollars per item, with $p>c$ . Sam claims that adding the same positive amount $h$ dollars to both the material cost and selling price leaves the break-even number unchanged. Is Sam correct? Explain algebraically.
1. Hint 1
  Write cost equals revenue before and after the two increases.
2. Hint 2
  The extra cost and extra revenue are each $hx$ for $x$ items.
Answer

Yes; the break-even number remains $\frac{F}{p-c}$ .

Full solution

For $x$ items, the original break-even condition is
$F+cx=px$ $F=(p-c)x$
Since $p>c$ , the difference is nonzero and
$x=\frac{F}{p-c}$

After the increases, the condition is
$F+(c+h)x=(p+h)x$
Subtracting $hx$ from both sides returns the original condition.
Thus exactly the same item count gives break-even, so Sam is correct.

Answer

Yes; the break-even number remains $\frac{F}{p-c}$ .

Key idea

Equal changes to per-item cost and per-item revenue preserve their difference and the break-even condition.
Problem 10: Mark complete

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