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Word Problems with Linear Equations: 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 Five integers in a row

Five consecutive integers have a sum of $185$ . What is the largest of the five?
1. Hint 1
  Consecutive integers are tied together: each one is $1$ more than the one before it. Name just one of them with a letter, and every other one follows from it.
2. Hint 2
  Let $n$ be the smallest of the five. Write the other four in terms of $n$ , add all five, and set the total equal to $185$ .
3. Hint 3
  Solving gives the smallest integer, which is not the one asked for. Count up from it to the largest, and check the sum of all five against the words.
Answer

$39$ (the five integers are $35$ , $36$ , $37$ , $38$ and $39$ ).

Full solution

Let $n$ be the smallest of the five integers.
Each integer is $1$ more than the one before it, so the five integers are $n$ , $n + 1$ , $n + 2$ , $n + 3$ and $n + 4$ .

Adding them gives five $n$ terms and the constants $1 + 2 + 3 + 4 = 10$ , so the sum is $5n + 10$ .
The sum is $185$ , so subtract $10$ from both sides and divide both sides by $5$ :
$\begin{aligned} 5n + 10 &= 185 \\ 5n &= 175 \\ n &= 35. \end{aligned}$

The five integers are $35$ , $36$ , $37$ , $38$ and $39$ .
Check against the words: they are consecutive, and
$35 + 36 + 37 + 38 + 39 = 185.$

The question asks for the largest, not the smallest, so the answer is
$n + 4 = 39.$

Answer

$39$ (the five integers are $35$ , $36$ , $37$ , $38$ and $39$ ).

Key idea

Name the smallest of a run of consecutive integers, write each of the others by adding $1$ at a time, and report the one the question asks for.
Problem 1: Mark complete
Problem 2 A square and a triangle with one perimeter

A square and an equilateral triangle (a triangle whose three sides are equal) have the same perimeter. Each side of the triangle is $5$ cm longer than each side of the square. Find the side length of each shape.
1. Hint 1
  The two shapes are linked by one fact about their sides, so a single letter can describe both. The sentence about the perimeters is the one that states an equality.
2. Hint 2
  Let $s$ be the side of the square in centimeters, so each side of the triangle is $s + 5$ . The square's perimeter adds four equal sides, and the triangle's adds three.
3. Hint 3
  Set the two perimeters equal and collect the $s$ terms on one side. Remember that $s$ is only the square's side, and the question asks about both shapes.
Answer

Each side of the square is $15$ cm and each side of the triangle is $20$ cm (both perimeters are $60$ cm).

Full solution

Let $s$ be the side length of the square, in centimeters.
Each side of the triangle is $5$ cm longer, so it is $s + 5$ centimeters.

The square has four sides of length $s$ , so its perimeter is $4s$ .
The triangle has three sides of length $s + 5$ , so its perimeter is $3(s + 5)$ .
The two perimeters are equal:
$4s = 3(s + 5).$

Distribute the $3$ , then subtract $3s$ from both sides:
$\begin{aligned} 4s &= 3s + 15 \\ s &= 15. \end{aligned}$

So the square's side is $15$ cm, and the triangle's side is $15 + 5 = 20$ cm.
Check against the words: the square's perimeter is $4 \times 15 = 60$ cm, the triangle's is $3 \times 20 = 60$ cm, and $20$ is $5$ more than $15$ .

Answer

Each side of the square is $15$ cm and each side of the triangle is $20$ cm (both perimeters are $60$ cm).

Key idea

When two shapes are linked through their sides, name one side, write the other from it, and let the shared perimeter supply the equation.
Problem 2: Mark complete
Problem 3 Walkers and bus riders

In one class, $40\%$ of the students walk to school, and all the other students in the class take the bus. There are $9$ more bus riders than walkers. How many students are in the class?
1. Hint 1
  Both groups are percentages of the same whole class, so name the size of the class with a letter and describe both groups from it.
2. Hint 2
  If $40\%$ of the class walk, the rest of the class, $60\%$ of it, take the bus. Write each group as a decimal times your letter.
3. Hint 3
  The equality in the words compares the two groups: the bus riders minus the walkers is $9$ . Once you have the class size, work out both groups and check that comparison.
Answer

$45$ students ( $18$ walk and $27$ take the bus).

Full solution

Let $n$ be the number of students in the class.
The walkers are $40\%$ of the class, which is $0.4n$ students.
Everyone else takes the bus, and removing $40\%$ of the class leaves $60\%$ of it, so there are $0.6n$ bus riders.

There are $9$ more bus riders than walkers, so the bus riders minus the walkers is $9$ :
$0.6n - 0.4n = 9.$

Combine like terms, then divide both sides by $0.2$ :
$\begin{aligned} 0.2n &= 9 \\ n &= 45. \end{aligned}$

Check against the words: $40\%$ of $45$ is $0.4 \times 45 = 18$ walkers, and the other $45 - 18 = 27$ students take the bus.
That is $27 - 18 = 9$ more bus riders than walkers, as stated.
So the class has $45$ students.

Answer

$45$ students ( $18$ walk and $27$ take the bus).

Key idea

When a whole splits into a percentage and the rest, name the whole, write both parts as percentages of it, and translate the comparison between them.
Problem 3: Mark complete
Problem 4 Three kinds of coin in one jar

A jar holds only nickels ( $5$ cents each), dimes ( $10$ cents each) and quarters ( $25$ cents each). There are twice as many dimes as nickels, and $3$ more quarters than nickels. Altogether the coins are worth $4$ dollars and $25$ cents. How many coins of each kind are in the jar?
1. Hint 1
  Two of the three counts are described by comparing them with the nickels, so a single letter for the number of nickels can carry all three counts.
2. Hint 2
  Write the number of dimes and the number of quarters in terms of the number of nickels. The value of each kind, in cents, is its count times the value of one coin.
3. Hint 3
  Change the total to cents before writing the equation, so every value in it has the same unit. After solving, work out all three counts, not only the nickels.
Answer

$7$ nickels, $14$ dimes and $10$ quarters.

Full solution

Let $n$ be the number of nickels.
There are twice as many dimes, so there are $2n$ dimes.
There are $3$ more quarters than nickels, so there are $n + 3$ quarters.

Work in cents, so that every value is a whole number: $4$ dollars and $25$ cents is $425$ cents.
Each kind of coin is worth its count times the value of one coin, so the nickels are worth $5n$ cents, the dimes $10(2n)$ cents and the quarters $25(n + 3)$ cents.
Together they are worth $425$ cents:
$5n + 10(2n) + 25(n + 3) = 425.$

Multiply out the parentheses, collect like terms, subtract $75$ from both sides and divide both sides by $50$ :
$\begin{aligned} 5n + 20n + 25n + 75 &= 425 \\ 50n + 75 &= 425 \\ 50n &= 350 \\ n &= 7. \end{aligned}$

So there are $7$ nickels, $2 \times 7 = 14$ dimes and $7 + 3 = 10$ quarters.
Check against the words: $14$ is twice $7$ , $10$ is $3$ more than $7$ , and the coins are worth $35 + 140 + 250 = 425$ cents, which is $4$ dollars and $25$ cents.

Answer

$7$ nickels, $14$ dimes and $10$ quarters.

Key idea

Count every kind of coin from one letter, change every amount to one unit, and total the values as count times value per coin.
Problem 4: Mark complete
Problem 5 An equation with its words missing

A $96$ cm board is cut into three pieces. The second piece is twice as long as the first, and the third piece is $8$ cm longer than the first. Someone translated this situation and wrote down only the equation
$x + 2x + (x + 8) = 96.$
Say what $x$ stands for, units included, and which piece each of the other two terms on the left describes. Then find the length of the longest piece.
1. Hint 1
  Only one of the pieces is described on its own; the other two are described by comparing them with it. That piece is the natural one for the letter to name.
2. Hint 2
  Doubling a length and adding $8$ to it are two different instructions; find the phrase of the situation that gives each one, and the piece that phrase describes.
3. Hint 3
  The solved value is the length of one particular piece. Work out all three lengths before deciding which piece is the longest.
Answer

$x$ is the first piece's length in centimeters, $2x$ is the second piece and $x + 8$ the third. The longest is the second piece, $44$ cm (the pieces are $22$ , $44$ and $30$ cm).

Full solution

Only the first piece is described on its own terms; the other two are described by comparing them with it.
So $x$ is the length of the first piece, in centimeters.

Read the left side one term at a time.
The term $x$ is the first piece.
The phrase "twice as long as the first" doubles it, giving the second piece, $2x$ .
The phrase " $8$ cm longer than the first" adds $8$ , giving the third piece, $x + 8$ .
The right side is the whole board: cutting loses no length, so the three pieces together make up the $96$ cm.

Combine like terms, subtract $8$ from both sides and divide both sides by $4$ :
$\begin{aligned} 4x + 8 &= 96 \\ 4x &= 88 \\ x &= 22. \end{aligned}$

The letter names the first piece, so $22$ cm is the first piece and not yet the answer.
The second piece is $2 \times 22 = 44$ cm and the third is $22 + 8 = 30$ cm.
Check against the words: $22 + 44 + 30 = 96$ , the second is twice the first, and the third is $8$ cm longer than the first.
The longest piece is the second, at $44$ cm.

Answer

$x$ is the first piece's length in centimeters, $2x$ is the second piece and $x + 8$ the third. The longest is the second piece, $44$ cm (the pieces are $22$ , $44$ and $30$ cm).

Key idea

In a sound translation each term traces back to a phrase of the words, and the solved value belongs to whichever quantity the letter names.
Problem 5: Mark complete
Problem 6 Two names for the same pair

Two numbers have a sum of $60$ , and the larger is $14$ more than the smaller.

Write and solve an equation for this situation twice: first naming the smaller number as your unknown, then naming the larger. Report the two numbers each time. Then explain why your two equations have different solutions even though they describe the same pair of numbers.
1. Hint 1
  The letter is a name you choose for one quantity, not something the problem hands you. Before any equation, write a sentence saying which number your letter stands for.
2. Hint 2
  Naming the smaller number, the larger is $14$ more than it. Naming the larger number, read the relationship from the other end: the smaller is $14$ less than the larger.
3. Hint 3
  For the explanation, ask what each solved value is a value of. Two values that measure different quantities are not in competition, however different they look.
Answer

Naming the smaller $n$ : $n + (n + 14) = 60$ gives $n = 23$ . Naming the larger $m$ : $m + (m - 14) = 60$ gives $m = 37$ . Each route gives $23$ and $37$ . The solutions differ: $n$ is the smaller number and $m$ the larger.

Full solution

First route: let $n$ be the smaller number.
The larger is $14$ more than the smaller, so it is $n + 14$ .
The sum of the two is $60$ , so combine like terms, subtract $14$ from both sides and divide both sides by $2$ :
$\begin{aligned} n + (n + 14) &= 60 \\ 2n + 14 &= 60 \\ 2n &= 46 \\ n &= 23. \end{aligned}$

So the smaller number is $23$ , and the larger is
$23 + 14 = 37.$

Second route: let $m$ be the larger number.
Now the relationship is read from the other end: the smaller is $14$ less than the larger, so it is $m - 14$ .
The sum is still $60$ , so combine like terms, add $14$ to both sides and divide both sides by $2$ :
$\begin{aligned} m + (m - 14) &= 60 \\ 2m - 14 &= 60 \\ 2m &= 74 \\ m &= 37. \end{aligned}$

So the larger number is $37$ , and the smaller is $37 - 14 = 23$ , the same pair as before.
Check against the words: $23 + 37 = 60$ , and $37$ is $14$ more than $23$ .

The two solutions differ because the two letters name different quantities.
The value $n = 23$ belongs to the smaller number and the value $m = 37$ to the larger, so they answer two different questions, and both routes report the same pair.
A contradiction appears only if the meaning of each letter is dropped, which is why the method writes down what the letter stands for before any equation is written.

Answer

Naming the smaller $n$ : $n + (n + 14) = 60$ gives $n = 23$ . Naming the larger $m$ : $m + (m - 14) = 60$ gives $m = 37$ . Each route gives $23$ and $37$ . The solutions differ: $n$ is the smaller number and $m$ the larger.

Key idea

A solved value belongs to the quantity its letter was defined to name, so write that definition down before writing the equation.
Problem 6: Mark complete
Problem 7 A ride to the lake and back

A cyclist leaves home at 8:00 in the morning and rides to a lake without stopping, at a steady $15$ kilometers per hour. She then rides home along the same road at a steady $10$ kilometers per hour. Her riding time for the whole trip, there and back, is $5$ hours. At what time does she reach the lake, and how far is the lake from her home?
1. Hint 1
  The ride out and the ride home cover the same stretch of road, so they cover the same distance, though not in the same time.
2. Hint 2
  Let $t$ be the number of hours the ride out takes. The two rides take $5$ hours together, so the ride home takes $5 - t$ hours. Write each distance as rate times time.
3. Hint 3
  Set the two distances equal and solve. The question asks for a time of day and a distance, so turn your value of $t$ into a clock time and work out how far either ride goes.
Answer

She reaches the lake at 10:00 in the morning, and the lake is $30$ kilometers from her home.

Full solution

Let $t$ be the number of hours the ride out to the lake takes.
She rides for $5$ hours in all, so the ride home takes the rest of that time, $5 - t$ hours.

Distance is rate times time, $d = rt$ .
So the ride out covers $15t$ kilometers and the ride home covers $10(5 - t)$ kilometers.
Both rides run along the same road between her home and the lake, so the two distances are equal:
$15t = 10(5 - t).$

Distribute the $10$ , add $10t$ to both sides, and divide both sides by $25$ :
$\begin{aligned} 15t &= 50 - 10t \\ 25t &= 50 \\ t &= 2. \end{aligned}$

The value $t = 2$ counts hours of riding out, so it answers neither question yet.
Two hours after 8:00 is 10:00 in the morning, and in those $2$ hours she rides $15 \times 2 = 30$ kilometers.

Check against the words: the ride home takes $5 - 2 = 3$ hours at $10$ kilometers per hour, which covers $10 \times 3 = 30$ kilometers, the same distance as the ride out, and the two rides take $2 + 3 = 5$ hours in all.
So she reaches the lake at 10:00 in the morning, and the lake is $30$ kilometers from her home.

Answer

She reaches the lake at 10:00 in the morning, and the lake is $30$ kilometers from her home.

Key idea

On a trip there and back along the same road, the two legs cover equal distances, and a known total time lets one letter give both travel times.
Problem 7: Mark complete
Problem 8 Three times as old, and six times

A father is $45$ years old today, and his son is $9$ .

In how many years will the father be three times as old as his son? Then write and solve an equation for the father being six times as old as his son, and use its solution to decide whether that will happen at some future date.
1. Hint 1
  Everyone ages at the same rate, so one letter can count the years from today for both of them.
2. Hint 2
  In $x$ years the father is $45 + x$ and the son is $9 + x$ . "Three times as old" multiplies the son's age, because he is the younger.
3. Hint 3
  Solve the six-times equation the same way, then read the sign of $x$ : a positive $x$ is a date in the future, and a negative $x$ is a date in the past.
Answer

In $9$ years, when the father is $54$ and the son $18$ . Six times: $45 + x = 6(9 + x)$ gives $x = -1.8$ , which is $1.8$ years ago; it happens at no future date.

Full solution

Let $x$ be the number of years from today.
Both ages grow by the same $x$ , so in $x$ years the father is $45 + x$ and the son is $9 + x$ .

Three times as old multiplies the son's age, since he is the younger.
Distribute the $3$ , subtract $x$ and $27$ from both sides, and divide both sides by $2$ :
$\begin{aligned} 45 + x &= 3(9 + x) \\ 45 + x &= 27 + 3x \\ 18 &= 2x \\ x &= 9. \end{aligned}$

Check against the words: in $9$ years the father is $54$ and the son is $18$ , and
$54 = 3 \times 18.$

Six times as old sets up the same way.
Distribute the $6$ , subtract $x$ and $54$ from both sides, and divide both sides by $5$ :
$\begin{aligned} 45 + x &= 6(9 + x) \\ 45 + x &= 54 + 6x \\ -9 &= 5x \\ x &= -1.8. \end{aligned}$

This equation has exactly one solution, and it is negative.
A negative number of years from today is a date in the past: $1.8$ years ago the father was $43.2$ and the son was $7.2$ , and
$43.2 = 6 \times 7.2.$
So the father was six times as old as his son $1.8$ years ago, and there is no future date on which he is.

Reasoning from the ages directly gives the same verdict.
The father is always $36$ years older than his son, so being six times as old needs those $36$ years to be five times the son's age, which makes the son $36 \div 5 = 7.2$ years old.
The son is already $9$ and only gets older, so that moment has passed.

Answer

In $9$ years, when the father is $54$ and the son $18$ . Six times: $45 + x = 6(9 + x)$ gives $x = -1.8$ , which is $1.8$ years ago; it happens at no future date.

Key idea

Read a solved value back into the situation: a negative number of years from today is a date in the past, not a future one.
Problem 8: Mark complete
Problem 9 Putting the discount back on

A shop advertises $30\%$ off every item. Rosa pays $91$ dollars for a coat in the sale and wants to know its price before the sale. She reasons: "The sale took $30\%$ off, so I put $30\%$ back on. $30\%$ of $91$ is $27.30$ , and $91 + 27.30 = 118.30$ , so the coat was $118.30$ dollars."

Check Rosa's figure against the advertisement, then find the coat's price before the sale. Finally, explain which amount each $30\%$ was taken of, hers and the shop's, and what that does to her answer.
1. Hint 1
  A percentage is always a percentage of some amount. Decide which amount of money the shop's $30\%$ was taken of: the price before the sale, or the price after it.
2. Hint 2
  Let $p$ be the price before the sale, in dollars. The sale removes $30\%$ of $p$ , and what is left is the $91$ dollars Rosa paid.
3. Hint 3
  Removing $30\%$ of a price leaves $70\%$ of it. To test any candidate price, take $30\%$ off it and compare the result with $91$ dollars.
Answer

Rosa's figure fails: $30\%$ off $118.30$ dollars leaves $82.81$ , not $91$ . The pre-sale price was $130$ dollars. The shop's $30\%$ was of $130$ , hers of $91$ ; she put back $27.30$ of the $39$ dollars removed, and her figure is too low.

Full solution

First, test Rosa's figure against the advertisement.
Had the coat cost $118.30$ dollars, the sale would have taken $0.3 \times 118.30 = 35.49$ dollars off, leaving $118.30 - 35.49 = 82.81$ dollars.
Rosa paid $91$ dollars, so $118.30$ dollars was not the price before the sale.

The advertisement means that the amount removed is $30\%$ of the price the coat had before the sale.
That price is the unknown, so let $p$ be the price before the sale, in dollars.

The sale removes $0.3p$ dollars, and what remains is the $91$ dollars Rosa paid:
$p - 0.3p = 91.$

One whole $p$ less three tenths of it leaves seven tenths of it, so combine like terms, then divide both sides by $0.7$ :
$\begin{aligned} 0.7p &= 91 \\ p &= 130. \end{aligned}$

Check against the advertisement: $30\%$ of $130$ dollars is $39$ dollars, and $130 - 39 = 91$ dollars, the amount Rosa paid.

The shop's $30\%$ was taken of the price before the sale, $130$ dollars, so the sale removed $39$ dollars.
Rosa took her $30\%$ of $91$ dollars, the price after the sale, and $30\%$ of that smaller amount is only $27.30$ dollars.
Adding back less money than the sale removed cannot restore the original price, which is why her figure falls short of $130$ dollars.

Answer

Rosa's figure fails: $30\%$ off $118.30$ dollars leaves $82.81$ , not $91$ . The pre-sale price was $130$ dollars. The shop's $30\%$ was of $130$ , hers of $91$ ; she put back $27.30$ of the $39$ dollars removed, and her figure is too low.

Key idea

A percentage is a fraction of a particular amount, so name the amount it is taken of before writing the equation.
Problem 9: Mark complete
Problem 10 One sum, two kinds of integer

Decide whether there are three consecutive even integers with a sum of $96$ , and whether there are three consecutive odd integers with a sum of $96$ . Where such integers exist, find them; where they do not, explain why not.
1. Hint 1
  Consecutive even integers and consecutive odd integers both step by $2$ , so both kinds can be written as $n$ , $n + 2$ and $n + 4$ . What differs is the kind of number $n$ has to be.
2. Hint 2
  Add $n$ , $n + 2$ and $n + 4$ , set the total equal to $96$ , and solve. The equation does not know which kind of integer you had in mind.
3. Hint 3
  Check your value of $n$ against the words for each kind in turn.
Answer

Even: yes, $30$ , $32$ and $34$ . Odd: no three consecutive odd integers have a sum of $96$ ; the equation's only solution, $n = 30$ , is not odd.

Full solution

Consecutive even integers step by $2$ , and so do consecutive odd integers.
So for either kind, name the smallest $n$ , and the three integers are $n$ , $n + 2$ and $n + 4$ .
The kind of integer appears only as a condition on $n$ : it must be even for the first question and odd for the second.

The three integers have a sum of $96$ , so combine like terms, subtract $6$ from both sides and divide both sides by $3$ :
$\begin{aligned} n + (n + 2) + (n + 4) &= 96 \\ 3n + 6 &= 96 \\ 3n &= 90 \\ n &= 30. \end{aligned}$

The equation has exactly one solution, $n = 30$ , and it serves both questions.

For the even question, $30$ is even, so the integers are $30$ , $32$ and $34$ .
Check against the words: they are consecutive even integers, and
$30 + 32 + 34 = 96.$

For the odd question, the smallest integer would have to satisfy the same equation, and its only solution, $30$ , is not odd.
So no three consecutive odd integers have a sum of $96$ .

A second argument confirms it.
An odd number plus an odd number is even, and adding a third odd number makes the total odd, so the sum of three odd integers is odd, and $96$ is even.

Answer

Even: yes, $30$ , $32$ and $34$ . Odd: no three consecutive odd integers have a sum of $96$ ; the equation's only solution, $n = 30$ , is not odd.

Key idea

The form $n$ , $n + 2$ , $n + 4$ fits consecutive even and consecutive odd integers alike, so check that the solved $n$ is the kind of integer the words ask for.
Problem 10: Mark complete

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