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Adding and Subtracting Fractions: 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 Four contributions

Find $\frac1{16}+\frac5{16}+\frac7{16}+\frac3{16}$ in lowest terms.
1. Hint 1
  All four fractions count the same-sized parts.
2. Hint 2
  Total the numerators and keep the size of each part fixed.
Answer

$1$ .

Full solution

Each piece is a sixteenth, so the denominator stays $16$ .
The total number of pieces is
$1+5+7+3=16.$
Therefore the total is
$\frac{16}{16}=1.$
Sixteen sixteenths make exactly one whole.

Answer

$1$ .

Key idea

Adding several fractions with the same denominator adds their counts of one shared part size.
Problem 1: Mark complete
Problem 2 Twelfths and twentieths

Find $\frac7{12}+\frac3{20}$ in lowest terms.
1. Hint 1
  Twelfths and twentieths are different-sized parts, so they cannot be counted together yet.
2. Hint 2
  Rebuild both fractions over the least common multiple of $12$ and $20$ , then look for a common factor in the result.
Answer

$\frac{11}{15}$ .

Full solution

The multiples of $20$ are $20$ , $40$ , $60$ , and so on, and $60$ is the first that is also a multiple of $12$ .
The least common denominator is therefore $60$ .

Since $12\times5=60$ , multiply the top and bottom of the first fraction by $5$ :
$\frac7{12}=\frac{35}{60}.$
Since $20\times3=60$ , multiply the top and bottom of the second fraction by $3$ :
$\frac3{20}=\frac9{60}.$

Both fractions now count sixtieths, so
$\frac{35}{60}+\frac9{60}=\frac{44}{60}.$
The greatest common factor of $44$ and $60$ is $4$ , so
$\frac{44}{60}=\frac{11}{15}.$
The numbers $11$ and $15$ share no factor above $1$ , so the result is in lowest terms.

Answer

$\frac{11}{15}$ .

Key idea

Unlike fractions are rebuilt over their least common denominator before their numerators are combined, and the result is then simplified.
Problem 2: Mark complete
Problem 3 Thirtieths and tenths

Find $\frac{29}{30}-\frac1{10}$ in lowest terms.
1. Hint 1
  Check whether one denominator is already a multiple of the other.
2. Hint 2
  Rebuild only the tenths as thirtieths, subtract, then look for a common factor.
Answer

$\frac{13}{15}$ .

Full solution

Since $10\times3=30$ , the denominator $30$ is already a multiple of $10$ , so the least common denominator is $30$ .
The fraction $\frac{29}{30}$ stays as it is, and only the tenths need rebuilding: multiply the top and bottom of $\frac1{10}$ by $3$ , so
$\frac1{10}=\frac3{30}.$

Both fractions now count thirtieths, so
$\frac{29}{30}-\frac3{30}=\frac{26}{30}.$
The greatest common factor of $26$ and $30$ is $2$ , so
$\frac{26}{30}=\frac{13}{15}.$
The numbers $13$ and $15$ share no factor above $1$ , so the result is in lowest terms.

Answer

$\frac{13}{15}$ .

Key idea

When one denominator is a multiple of the other, only the fraction with the smaller denominator needs rebuilding.
Problem 3: Mark complete
Problem 4 Paint between cans

Can A contains $\frac34$ of a liter of paint and Can B contains $\frac58$ of a liter. A painter transfers $\frac16$ of a liter from A to B without spilling. How much paint is in each can afterward? Give each amount in lowest terms.
1. Hint 1
  The transferred paint leaves one can and enters the other.
2. Hint 2
  Subtract from A using twelfths, and add to B using twenty-fourths.
Answer

A: $\frac7{12}$ of a liter; B: $\frac{19}{24}$ of a liter.

Full solution

Can A loses $\frac16$ of a liter.
In twelfths,
$\frac34=\frac9{12}$ $\frac16=\frac2{12}$
so A contains
$\frac9{12}-\frac2{12}=\frac7{12}.$

Can B gains the transferred amount.
In twenty-fourths,
$\frac58=\frac{15}{24}$ $\frac16=\frac4{24}$
so B contains
$\frac{15}{24}+\frac4{24}=\frac{19}{24}.$
Both answers are in lowest terms.

As a check, in twenty-fourths $\frac7{12}=\frac{14}{24}$ , so the cans now hold $\frac{14}{24}+\frac{19}{24}=\frac{33}{24}$ of a liter.
At the start, $\frac34=\frac{18}{24}$ and $\frac{18}{24}+\frac{15}{24}=\frac{33}{24}$ , so no paint was lost.

Answer

A: $\frac7{12}$ of a liter; B: $\frac{19}{24}$ of a liter.

Key idea

A transfer is subtracted from its source and added to its destination, preserving the combined amount.
Problem 4: Mark complete
Problem 5 A boat on a canal

A boat on a straight canal is $\frac{11}{20}$ of a kilometer past a lock. It travels $\frac7{10}$ of a kilometer farther from the lock, then turns and travels $\frac35$ of a kilometer back toward it. How far past the lock is the boat now, and how far did it travel on these two trips? Give both as fractions in lowest terms.
1. Hint 1
  The boat’s distance from the lock depends on the direction of each trip, while the distance traveled counts both trips as lengths.
2. Hint 2
  Use twentieths to combine the starting distance from the lock with the trip away and the trip back.
3. Hint 3
  For the distance traveled, rewrite $\frac35$ in tenths and add it to $\frac7{10}$ .
Answer

Distance past the lock: $\frac{13}{20}$ of a kilometer; distance traveled on the two trips: $\frac{13}{10}$ kilometers.

Full solution

Write the trips in twentieths: $\frac7{10}=\frac{14}{20}$ and
$\frac35=\frac{12}{20}.$
The trip away from the lock adds to the boat’s distance from it, and the trip back subtracts, giving
$\frac{11+14-12}{20}=\frac{13}{20}.$

The distance traveled counts both trip lengths, whatever their direction.
Write $\frac35=\frac6{10}$ , so
$\frac7{10}+\frac6{10}=\frac{13}{10}.$
Each resulting fraction is in lowest terms.
The trip away was longer than the trip back, which agrees with the boat finishing farther from the lock than where it started.

Answer

Distance past the lock: $\frac{13}{20}$ of a kilometer; distance traveled on the two trips: $\frac{13}{10}$ kilometers.

Key idea

Total distance traveled adds trip lengths, while a final position also depends on their directions.
Problem 5: Mark complete
Problem 6 A revised walking route

A walking route is $2$ kilometers long. A section of length $\frac78$ of a kilometer is replaced by a new section of length $\frac{11}{12}$ of a kilometer. What is the revised route length as one fraction in lowest terms?
1. Hint 1
  Remove the old section from the total and put the new section in its place.
2. Hint 2
  Write both section lengths and the whole route in twenty-fourths.
Answer

$\frac{49}{24}$ kilometers.

Full solution

The new section replaces, rather than joins, the old one.
The revised length is $2-\frac78+\frac{11}{12}$ .
The least common multiple of $8$ and $12$ is $24$ , and the whole number $2$ can be rebuilt over any denominator, so rewrite the amounts with denominator $24$ :
$2=\frac{48}{24}$ $\frac78=\frac{21}{24}$ $\frac{11}{12}=\frac{22}{24}$

Combine the counts:
$\frac{48-21+22}{24}=\frac{49}{24}.$
The numbers $49$ and $24$ share no factor above $1$ .
The new section is one twenty-fourth of a kilometer longer, so the result being one twenty-fourth above $2$ kilometers is consistent.

Answer

$\frac{49}{24}$ kilometers.

Key idea

Replacing part of a total subtracts the old part and adds its replacement.
Problem 6: Mark complete
Problem 7 Zoe’s shortcut

Zoe says $\frac56+\frac19=\frac6{15}$ , since she added the numerators and added the denominators. Is she correct? Explain, and give the sum in lowest terms.
1. Hint 1
  A denominator names the size of each part, so ask whether sixths and ninths are the same size.
2. Hint 2
  Estimate first: how should the sum compare with $\frac56$ alone, and how does Zoe’s result compare with it?
3. Hint 3
  Rebuild both fractions over the least common multiple of $6$ and $9$ , then combine the numerators.
Answer

No; the sum is $\frac{17}{18}$ .

Full solution

Zoe’s result simplifies to $\frac25$ , which is less than $\frac12$ .
But $\frac56$ alone is more than $\frac12$ , and adding $\frac19$ can only make the total larger.
So $\frac6{15}$ cannot be the sum.

The denominators name part sizes, and sixths and ninths are different-sized parts.
Their counts cannot be added until both are measured in one part size.
Adding $6$ and $9$ instead names fifteenths, a part size that neither fraction uses.

The least common multiple of $6$ and $9$ is $18$ , so rebuild both fractions in eighteenths:
$\frac56=\frac{15}{18}$ $\frac19=\frac2{18}$
Now both count eighteenths, so
$\frac{15}{18}+\frac2{18}=\frac{17}{18}.$
The number $17$ is prime and is not a factor of $18$ , so the sum is in lowest terms, and Zoe is not correct.

Answer

No; the sum is $\frac{17}{18}$ .

Key idea

Fractions add only as counts of one part size, so their denominators are never added.
Problem 7: Mark complete
Problem 8 Two submitted totals

For $\frac14+\frac5{14}$ , Lina submits $\frac{17}{28}$ and Max submits $\frac{34}{56}$ . Lina says both answers have the correct value, although only hers is in lowest terms. Is her statement correct? Explain.
1. Hint 1
  Check the sum independently before comparing the two submitted forms.
2. Hint 2
  Twenty-eighths can describe both starting fractions, and a common factor can relate the two submitted answers.
Answer

Yes; both values are $\frac{17}{28}$ , and only Lina’s form is in lowest terms.

Full solution

The least common denominator is $28$ , the least common multiple of $4$ and $14$ .
Rewrite $\frac14=\frac7{28}$ and
$\frac5{14}=\frac{10}{28}.$
Then
$\frac7{28}+\frac{10}{28}=\frac{17}{28}.$

Max’s form is what the larger common denominator $56$ , the product of $4$ and $14$ , produces: $\frac14=\frac{14}{56}$ and $\frac5{14}=\frac{20}{56}$ , which total $\frac{34}{56}$ .
Dividing Max’s numerator and denominator by $2$ gives
$\frac{34}{56}=\frac{17}{28}.$
Both answers therefore have the correct value.
The numbers $17$ and $28$ share no factor above $1$ , while $34$ and $56$ share $2$ , so Lina’s full statement is correct.

Answer

Yes; both values are $\frac{17}{28}$ , and only Lina’s form is in lowest terms.

Key idea

Any common denominator gives the correct sum, but one larger than the least, as the product of the denominators often is, leaves an extra common factor to divide out.
Problem 8: Mark complete
Problem 9 A grouped calculation

Tao rewrites $\frac89-\frac5{12}-\frac16$ as $\frac89-\left(\frac5{12}-\frac16\right)$ . Does this preserve the value? Find the value of each expression and explain.
1. Hint 1
  In the original expression, both later fractions are taken away.
2. Hint 2
  In the new expression, the parentheses form one difference that is taken away.
3. Hint 3
  Thirty-sixths allow every term to be compared directly.
Answer

No; original: $\frac{11}{36}$ ; rewritten: $\frac{23}{36}$ .

Full solution

Write $\frac89=\frac{32}{36}$ , $\frac5{12}=\frac{15}{36}$ , and
$\frac16=\frac6{36}.$
The original expression gives
$\frac{32-15-6}{36}=\frac{11}{36}.$

The parentheses in the rewrite give
$\frac{15-6}{36}=\frac9{36}.$
Taking that difference away gives
$\frac{32-9}{36}=\frac{23}{36}.$

The rewrite takes away a smaller amount.
Its inner subtraction reduces what is removed, whereas the original removes both fractions.
The unequal results show that the value is not preserved.

Answer

No; original: $\frac{11}{36}$ ; rewritten: $\frac{23}{36}$ .

Key idea

Grouping a string of subtractions can change which amounts are taken away.
Problem 9: Mark complete
Problem 10 Matching box entries

The same whole number must fill both boxes in $\frac{\square}{10}+\frac{\square}{15}=1$ . Find that number and explain why no other whole number works.
1. Hint 1
  Think of the two fractions as counts of one shared part size.
2. Hint 2
  In thirtieths, each copy of the box contributes three parts from the first fraction and two from the second.
3. Hint 3
  Find how many groups of that combined size make thirty thirtieths.
Answer

$6$ .

Full solution

In thirtieths, the first numerator becomes three times the box and the second becomes twice the box.
Together they count five times the box in thirtieths.
One whole needs thirty thirtieths, so the box is
$30\div5=6.$

The check is
$\frac6{10}=\frac{18}{30}$ $\frac6{15}=\frac{12}{30}$
and their sum is
$\frac{30}{30}=1.$
A smaller whole number supplies fewer than thirty parts and a larger one supplies more, so no other whole number works.

Answer

$6$ .

Key idea

A repeated missing count can be found by counting how many common-sized parts each copy contributes.
Problem 10: Mark complete

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