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Multiplying and Dividing 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 A whole number in a product

Find $\frac23\times9\times\frac5{12}$ in lowest terms.
1. Hint 1
  A whole number can be written as a fraction over $1$ , so the whole product becomes one fraction.
2. Hint 2
  Look for a factor that a numerator shares with a denominator, and cancel it before multiplying.
Answer

$\frac52$ .

Full solution

Write $9$ as $\frac91$ .
The product is then one fraction, with numerator $2\times9\times5$ and denominator $3\times1\times12$ .

The numerator $9$ and the denominator $3$ share the factor $3$ , leaving $3$ and $1$ .
The numerator $2$ and the denominator $12$ share the factor $2$ , leaving $1$ and $6$ .

The remaining $3$ on top and $6$ below still share the factor $3$ , leaving $1$ and $2$ .
What is left is
$\frac{1\times1\times5}{1\times1\times2}=\frac52.$
Multiplying first instead gives $\frac{90}{36}$ , which also simplifies to $\frac52$ by the greatest common factor $18$ .

Answer

$\frac52$ .

Key idea

A whole number in a product is a fraction over $1$ , and a factor shared by a numerator and a denominator may be removed anywhere across the product.
Problem 1: Mark complete
Problem 2 A sequence of operations

Find $\left(\frac38\div\frac9{10}\right)\div\frac{25}6$ in lowest terms.
1. Hint 1
  Division by a nonzero fraction asks for multiplication by its reciprocal.
2. Hint 2
  First find the value in parentheses, then divide it by the last fraction.
Answer

$\frac1{10}$ .

Full solution

Divide by $\frac9{10}$ by multiplying by $\frac{10}9$ :
$\frac38\div\frac9{10}=\frac38\times\frac{10}9.$
Remove a factor of $3$ from $3$ and $9$ , and a factor of $2$ from $10$ and $8$ , giving
$\frac38\times\frac{10}9=\frac5{12}.$

Now divide by $\frac{25}6$ by multiplying by its reciprocal $\frac6{25}$ .
Cancel $5$ from the $5$ and the $25$ , leaving $1$ and $5$ , and cancel $6$ from the $6$ and the $12$ , leaving $1$ and $2$ .
The leftovers give
$\frac5{12}\times\frac6{25}=\frac1{10}.$
The result is in lowest terms.

Written as one product from the start, $\frac38\times\frac{10}9\times\frac6{25}$ has numerator $180$ and denominator $1800$ , which also gives $\frac1{10}$ .

Answer

$\frac1{10}$ .

Key idea

In a chain of divisions by nonzero fractions, rewrite each as multiplication by the divisor’s reciprocal; the chain then becomes one product to simplify.
Problem 2: Mark complete
Problem 3 A reciprocal in lowest terms

Find the reciprocal of $\frac{30}{42}$ , giving your answer in lowest terms, and check that the two numbers multiply to $1$ .
1. Hint 1
  A reciprocal swaps the numerator and the denominator, and simplifying a fraction does not change its value.
2. Hint 2
  Divide $30$ and $42$ by their greatest common factor, then turn the simpler fraction over.
Answer

$\frac75$ ; $\frac{30}{42}\times\frac75=1$ .

Full solution

The greatest common factor of $30$ and $42$ is $6$ , so
$\frac{30}{42}=\frac57.$
Swapping the numerator and denominator of $\frac57$ gives the reciprocal $\frac75$ , which is in lowest terms since $7$ and $5$ share no factor above $1$ .

Flipping first gives $\frac{42}{30}$ , which also simplifies to $\frac75$ by the same factor $6$ .

For the check, the product of $\frac{30}{42}$ and $\frac75$ has numerator $30\times7=210$ and denominator
$42\times5=210.$
A fraction whose numerator equals its denominator is $1$ , so the two numbers multiply to $1$ .

Answer

$\frac75$ ; $\frac{30}{42}\times\frac75=1$ .

Key idea

The reciprocal of a nonzero fraction swaps its numerator and denominator, and the two multiply to $1$ .
Problem 3: Mark complete
Problem 4 A patterned panel

A rectangular panel is divided into six equal rows and eight equal columns. Four rows are selected. In each selected row, three spaces are left plain and the other spaces are colored. Every unselected row is left plain. What fraction of the whole panel is colored? Give the fraction in lowest terms, and write it as a product of two fractions, one for the rows and one for the spaces in a row.
1. Hint 1
  Count colored spaces in one selected row, then count across all selected rows.
2. Hint 2
  The denominator counts every space in the panel, including those outside the selected rows.
3. Hint 3
  The selected rows are a fraction of all the rows, and the colored spaces in one selected row are a fraction of that row’s spaces.
Answer

$\frac5{12}$ , which is $\frac46\times\frac58$ (or $\frac23\times\frac58$ ).

Full solution

The cuts create $6\times8=48$ equal spaces.
Each selected row has $8-3=5$ colored spaces, so the four selected rows contain $4\times5=20$ colored spaces.

Dividing the top and bottom by their greatest common factor $4$ , the fraction colored is
$\frac{20}{48}=\frac5{12}.$

The same count is the product $\frac46\times\frac58$ .
Its first factor is the four selected rows out of all six rows, and its second is the five colored spaces out of the eight spaces in each selected row.
The product’s denominator $6\times8$ counts all $48$ spaces and its numerator $4\times5$ counts the $20$ colored ones, with no common denominator needed.

Answer

$\frac5{12}$ , which is $\frac46\times\frac58$ (or $\frac23\times\frac58$ ).

Key idea

In an area model, the product of the numerators counts the colored spaces and the product of the denominators counts all the equal spaces.
Problem 4: Mark complete
Problem 5 Clay for small models

A workshop has $\frac79$ of a kilogram of clay. It sets aside $\frac13$ of a kilogram and uses all the rest for identical models requiring $\frac2{27}$ of a kilogram each. How many models can it make?
1. Hint 1
  Find the clay available after the amount set aside is removed.
2. Hint 2
  The model count asks how many portions of size $\frac2{27}$ fit in that remainder.
Answer

$6$ models.

Full solution

Subtract the amount set aside:
$\frac79-\frac13=\frac49.$
Divide the remaining clay by the amount per model:
$\frac49\div\frac2{27}=\frac49\times\frac{27}2.$

The factors $27$ and $9$ reduce to $3$ and $1$ , while $4$ and $2$ reduce to $2$ and $1$ .
The product is $2\times3=6$ , so six models can be made.

Checking the clay used, $6\times\frac2{27}=\frac49$ of a kilogram, exactly the remainder.

Answer

$6$ models.

Key idea

Find the available amount before dividing it into portions of a specified size.
Problem 5: Mark complete
Problem 6 Sam’s quotient

Sam computes $\frac{10}{39}\div\frac6{13}$ as $\frac{39}{10}\times\frac6{13}=\frac95$ . Find the quotient in lowest terms, and use multiplication by the divisor to show which of the two answers is right.
1. Hint 1
  A correct quotient, multiplied by the divisor, gives back the number that was divided.
2. Hint 2
  Rewrite the division as a multiplication yourself, then compare it with Sam’s first step.
3. Hint 3
  Cancel factors shared by a numerator and a denominator before multiplying when you compute the quotient yourself.
Answer

The quotient is $\frac59$ , not $\frac95$ .

Full solution

Dividing by $\frac6{13}$ asks for the number that, multiplied by $\frac6{13}$ , gives back $\frac{10}{39}$ .
So a correct quotient must pass the test of multiplying it by the divisor.

In Sam’s test product no numerator shares a factor with a denominator, so
$\frac95\times\frac6{13}=\frac{54}{65}.$
This is not $\frac{10}{39}$ , since $54$ is more than half of $65$ while $10$ is less than half of $39$ .
His answer fails the test.

Sam turned over the first fraction, $\frac{10}{39}$ , and left the divisor as it was.
The rule flips only the divisor, because multiplying by $\frac{13}6$ is what undoes a multiplication by $\frac6{13}$ .

The correct rewrite is
$\frac{10}{39}\div\frac6{13}=\frac{10}{39}\times\frac{13}6.$
Cancel $13$ from the $13$ and the $39$ , leaving $1$ and $3$ , and cancel $2$ from the $10$ and the $6$ , leaving $5$ and $3$ .
That gives
$\frac53\times\frac13=\frac59.$

The test now works.
Canceling $3$ from the $6$ and the $9$ leaves $\frac53\times\frac2{13}$ , so
$\frac59\times\frac6{13}=\frac{10}{39}.$

Answer

The quotient is $\frac59$ , not $\frac95$ .

Key idea

Only the divisor is flipped, and a correct quotient multiplied by the divisor gives back the number that was divided.
Problem 6: Mark complete
Problem 7 Water through a filter

Four identical pump cycles use $\frac{21}{20}$ of a liter of water altogether. In each cycle, $\frac57$ of its water passes through a filter. How much water passes through the filter in one cycle? Give your answer in lowest terms.
1. Hint 1
  Find the amount in one cycle before taking the fraction that is filtered.
2. Hint 2
  Divide the total by $4$ , then multiply the result by $\frac57$ .
Answer

$\frac3{16}$ of a liter.

Full solution

One cycle uses $\frac{21}{20}\div4$ of a liter, which is $\frac{21}{20}\times\frac14$ .
No numerator shares a factor with a denominator here, so one cycle uses $\frac{21}{80}$ of a liter.

Five sevenths of that amount passes through the filter.
Canceling $7$ from the $21$ and the $7$ , and $5$ from the $5$ and the $80$ , gives
$\frac{21}{80}\times\frac57=\frac3{16}.$
So $\frac3{16}$ of a liter is filtered in one cycle.

As a check, all four cycles filter $4\times\frac3{16}=\frac34$ of a liter.
Five sevenths of the full $\frac{21}{20}$ of a liter is $\frac{105}{140}$ , which is also $\frac34$ of a liter.

Answer

$\frac3{16}$ of a liter.

Key idea

An equal share of a total can be found before selecting a fraction of that share.
Problem 7: Mark complete
Problem 8 Two ways to shrink a product

Noor rewrites $\frac9{22}\times\frac{11}{12}$ as $\frac3{22}\times\frac{11}4$ , and Eli rewrites it as $\frac9{11}\times\frac{11}6$ . Which rewrite keeps the product, and what is the product in lowest terms?
1. Hint 1
  Write the original product as one fraction to see what changed on its top and bottom.
2. Hint 2
  For each rewrite, check whether one common factor left the top and the bottom of the single fraction.
Answer

Noor’s rewrite; the product is $\frac38$ .

Full solution

The original product is $\frac{9\times11}{22\times12}$ .
Noor divided its numerator by $3$ through the first factor, turning $9$ into $3$ , and its denominator by $3$ through the second factor, turning $12$ into $4$ .
Dividing the top and bottom by the same nonzero number preserves the value.

Eli divided the denominator by $2$ through each factor, turning $22$ into $11$ and $12$ into $6$ , and divided the numerator by nothing.
Taking a factor from the bottom alone changes the value.
His product gives
$\frac9{11}\times\frac{11}6=\frac{99}{66}.$
That simplifies to $\frac32$ , which is four times the original product, so his rewrite does not keep it.

Noor’s product gives
$\frac3{22}\times\frac{11}4=\frac{33}{88}.$
Dividing $33$ and $88$ by their common factor $11$ gives
$\frac{33}{88}=\frac38.$
The original product $\frac{99}{264}$ also simplifies to $\frac38$ , by the common factor $33$ .

Answer

Noor’s rewrite; the product is $\frac38$ .

Key idea

Canceling is valid when the full product’s numerator and denominator are divided by the same nonzero factor.
Problem 8: Mark complete
Problem 9 A machine’s return setting

A machine multiplies a positive fraction by $\frac8{15}$ . A second setting must multiply the result by one fixed fraction to restore every starting value. What fraction should that setting use, and why does it work? Check your choice with a starting value of $\frac56$ .
1. Hint 1
  The two multipliers together must have the effect of multiplying by $1$ .
2. Hint 2
  Find the reciprocal of the first multiplier, then follow the two stages on the given starting value.
Answer

$\frac{15}8$ ; the check returns $\frac56$ .

Full solution

The required multiplier is $\frac{15}8$ , since
$\frac8{15}\times\frac{15}8=1.$
Multiplying any starting value by these two factors therefore leaves that value unchanged.

For the requested check, the first stage gives
$\frac56\times\frac8{15}=\frac49.$
The second gives
$\frac49\times\frac{15}8=\frac56.$
It restores the original value as required.

Answer

$\frac{15}8$ ; the check returns $\frac56$ .

Key idea

Multiplication by a nonzero fraction is undone by multiplication by its reciprocal.
Problem 9: Mark complete
Problem 10 Pia’s prediction

Pia says that a positive proper fraction multiplied by an improper fraction must give a value greater than $1$ . Is the claim true? Justify your answer.
1. Hint 1
  Think about how much multiplying by an improper fraction, which is at least $1$ , can enlarge a positive number.
2. Hint 2
  Try a small positive unit fraction and an improper fraction only a little above $1$ .
Answer

False; for example, $\frac15\times\frac32=\frac3{10}<1$ .

Full solution

Choose the proper fraction $\frac15$ and the improper fraction $\frac32$ .
Their product is
$\frac15\times\frac32=\frac3{10}.$

The product is less than $1$ because $3<10$ , although the second factor is improper.
The multiplier enlarged $\frac15$ but did not enlarge it enough to reach a whole, so this example disproves the claim.

Answer

False; for example, $\frac15\times\frac32=\frac3{10}<1$ .

Key idea

A factor greater than one increases a positive value without necessarily making the product exceed one.
Problem 10: Mark complete

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