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Equivalent Fractions and Simplifying: 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 fraction built from products

Write $\frac{3\times14}{8\times21}$ in lowest terms.
1. Hint 1
  The greatest common factor of the top and bottom is built from the primes they share.
2. Hint 2
  Break each of $3$ , $14$ , $8$ and $21$ into prime factors.
3. Hint 3
  Divide out every prime the top and bottom share, then see what is left of each.
Answer

$\frac14$ .

Full solution

Break each number into primes: $14=2\times7$ , $8=2\times2\times2$ and
$21=3\times7.$
So the numerator is $2\times3\times7$ and the denominator is $2\times2\times2\times3\times7$ .

The top has one $2$ and the bottom has three, and each has one $3$ and one $7$ , so taking each shared prime at the smaller of its two counts gives the greatest common factor
$2\times3\times7=42.$
Dividing both by $42$ leaves $1$ on top, because every prime in the numerator is shared, and $2\times2=4$ on the bottom:
$\frac{3\times14}{8\times21}=\frac14.$
The numbers $1$ and $4$ share no factor above $1$ , so the fraction is in lowest terms.

As a check, the numerator is $3\times14=42$ and the denominator is $8\times21=168$ , which is $4\times42$ .

Answer

$\frac14$ .

Key idea

To put a fraction built from products in lowest terms, divide its top and bottom by their greatest common factor, the product of the primes they share, each taken at the smaller of its two counts, read from the factors of each product.
Problem 1: Mark complete
Problem 2 Equal fractions in a range

Which fractions equal to $\frac7{15}$ have whole-number terms and a denominator between $40$ and $80$ ?
1. Hint 1
  Each fraction must name the same amount using a different part size.
2. Hint 2
  Since $7$ and $15$ share no factor above $1$ , any fraction equal to $\frac7{15}$ is $\frac7{15}$ built up by one whole-number multiplier.
3. Hint 3
  Find the multiples of $15$ between $40$ and $80$ , then multiply $7$ by the number that turns $15$ into each one.
Answer

$\frac{21}{45}$ , $\frac{28}{60}$ and $\frac{35}{75}$ .

Full solution

The numbers $7$ and $15$ share no factor above $1$ , so $\frac7{15}$ is in lowest terms.
Any fraction equal to it has the same lowest-terms form.
So dividing the numerator and denominator of such a fraction by their greatest common factor gives $\frac7{15}$ , which means the fraction is $\frac7{15}$ with its numerator and denominator multiplied by that same whole number.

Its denominator is therefore a multiple of $15$ .
The multiples of $15$ between $40$ and $80$ are $45$ , $60$ and $75$ , which are $15$ times $3$ , $4$ and $5$ .

Multiply the numerator by the same number each time:
$\frac{7\times3}{15\times3}=\frac{21}{45}$ $\frac{7\times4}{15\times4}=\frac{28}{60}$ $\frac{7\times5}{15\times5}=\frac{35}{75}$
The next multiple of $15$ is $90$ , which is past $80$ , so these three are all of them.

Answer

$\frac{21}{45}$ , $\frac{28}{60}$ and $\frac{35}{75}$ .

Key idea

Listing the multiples of a fraction's lowest-terms denominator that fall in a range finds every equal fraction with whole-number terms whose denominator lies in that range.
Problem 2: Mark complete
Problem 3 A fraction built from sums

Write $\frac{24+12}{35+19}$ in lowest terms.
1. Hint 1
  The whole expression above the bar is the numerator, and the whole expression below is the denominator.
2. Hint 2
  After finding both numbers, look for their greatest common factor.
Answer

$\frac23$ .

Full solution

Evaluate the grouped parts of the fraction:
$24+12=36$ $35+19=54$

The greatest common factor of $36$ and $54$ is $18$ .
Dividing both by $18$ gives
$\frac{36}{54}=\frac23.$
The remaining numbers $2$ and $3$ have no common factor above $1$ , so the fraction is in lowest terms.

Answer

$\frac23$ .

Key idea

When the top or bottom of a fraction is a sum, evaluate the sum before simplifying the resulting fraction.
Problem 3: Mark complete
Problem 4 A dial setting

A dial stores a setting as $\frac{33}{42}$ . Its screen must show the same value with denominator $84$ , while its printed report must use lowest terms. What should the screen and the report show?
1. Hint 1
  The screen and the report need different names for one value.
2. Hint 2
  The screen denominator is twice the stored denominator.
3. Hint 3
  For the report, find the greatest common factor of $33$ and $42$ .
Answer

Screen: $\frac{66}{84}$ ; report: $\frac{11}{14}$ .

Full solution

The denominator doubles from $42$ to $84$ , so double the numerator as well:
$\frac{33}{42}=\frac{66}{84}.$
This is what the screen shows.

The greatest common factor of $33$ and $42$ is $3$ .
Divide both by $3$ :
$\frac{33}{42}=\frac{11}{14}.$
Since $11$ and $14$ share no factor above $1$ , this is what the report shows.
Both describe the same dial setting.

Answer

Screen: $\frac{66}{84}$ ; report: $\frac{11}{14}$ .

Key idea

A required denominator and lowest terms are different forms that can describe the same value.
Problem 4: Mark complete
Problem 5 An editor’s division

An editor divides the numerator and denominator of a fraction by $6$ and gets $\frac49$ . What was the original fraction, and was $6$ the greatest common factor of its numerator and denominator?
1. Hint 1
  Undoing the editor’s division recovers the original top and bottom, and the greatest common factor is the largest number that divides both.
2. Hint 2
  Write both original numbers as products of primes, then multiply together the primes they share.
Answer

$\frac{24}{54}$ ; yes, $6$ is the greatest common factor of $24$ and $54$ .

Full solution

Before division by $6$ , the numerator was $4\times6=24$ and the denominator was
$9\times6=54.$
Thus the original fraction was
$\frac{24}{54}.$

In primes, $24=2\times2\times2\times3$ and
$54=2\times3\times3\times3.$
The primes they share are one $2$ and one $3$ , so their greatest common factor is
$2\times3=6.$

As a check, $24=6\times4$ and $54=6\times9$ , and the leftovers $4=2\times2$ and $9=3\times3$ share no prime, so $\frac49$ is in lowest terms.
Had $6$ not been the greatest common factor, $4$ and $9$ would still share a factor, so $6$ was the greatest.

Answer

$\frac{24}{54}$ ; yes, $6$ is the greatest common factor of $24$ and $54$ .

Key idea

Reversing a common-factor division restores the original numerator and denominator together, and the divisor was their greatest common factor exactly when the result is in lowest terms.
Problem 5: Mark complete
Problem 6 Two sensor records

Two sensor records show $\frac6{14}$ and $\frac{10}{21}$ . Lower exactly one of the two numerators by $1$ , leaving every other number unchanged, so that the records agree. Which record should change, and what should it become? Justify your decision.
1. Hint 1
  There are two possible edited pairs, and their values determine which change works.
2. Hint 2
  Compare the cross products for each possible pair.
Answer

Change $\frac{10}{21}$ to $\frac9{21}$ ; leave $\frac6{14}$ unchanged.

Full solution

If the first numerator decreases, the records are $\frac5{14}$ and $\frac{10}{21}$ .
Their cross products are
$5\times21=105$ $14\times10=140$
These differ, so that change does not make the records agree.

If the second numerator decreases, the records are $\frac6{14}$ and $\frac9{21}$ .
Their cross products are
$6\times21=126$ $14\times9=126$
These match, so the second record should become $\frac9{21}$ .

As a check, both final records reduce to $\frac37$ .
Both allowed changes have been tested, so this is the one that works.

Answer

Change $\frac{10}{21}$ to $\frac9{21}$ ; leave $\frac6{14}$ unchanged.

Key idea

Cross products can test each allowed change when two fraction records must agree.
Problem 6: Mark complete
Problem 7 Allowed divisions

A machine stores the fraction $\frac{165}{195}$ . It may divide both numbers by any common factor, but its new denominator must be less than $40$ . List every new fraction it can produce in one such division.
1. Hint 1
  Every allowed division uses a number that divides both the numerator and the denominator.
2. Hint 2
  List those common factors, then test which ones bring the denominator below $40$ .
Answer

$\frac{33}{39}$ and $\frac{11}{13}$ .

Full solution

The factors of $165$ are $1$ , $3$ , $5$ , $11$ , $15$ , $33$ , $55$ and $165$ .
Of these, the ones that also divide $195$ are $1$ , $3$ , $5$ and $15$ , so those are the common factors.

Dividing $195$ by them gives denominators $195$ , $65$ , $39$ and $13$ , respectively.

Only division by $5$ or $15$ gives a denominator less than $40$ :
$\frac{165\div5}{195\div5}=\frac{33}{39}$ $\frac{165\div15}{195\div15}=\frac{11}{13}.$
These are all the allowed results because every common factor has been checked.

Answer

$\frac{33}{39}$ and $\frac{11}{13}$ .

Key idea

A restriction on a new denominator can be checked against the complete list of common factors.
Problem 7: Mark complete
Problem 8 Milo’s claim

Milo says two fractions can be equal only if multiplying the numerator and denominator of one by the same whole number gives the other. Is he correct? Support your answer with an example.
1. Hint 1
  Equal fractions share one lowest-terms form.
2. Hint 2
  Test Milo’s claim on two different fractions that simplify to the same lowest-terms form.
3. Hint 3
  Check that no whole number turns one of your denominators into the other.
Answer

No; for example $\frac{27}{45}=\frac{30}{50}$ (any pair of equal fractions in which neither denominator is a whole-number multiple of the other is accepted).

Full solution

One pair of equal fractions in which neither is the other with its numerator and denominator multiplied by one whole number is enough to show that Milo is wrong.
Start from a fraction in lowest terms, such as $\frac35$ .

Since $5\times9=45$ , multiply the numerator and denominator of $\frac35$ by $9$ :
$\frac35=\frac{27}{45}.$
Since $5\times10=50$ , multiply them by $10$ instead:
$\frac35=\frac{30}{50}.$

Both results equal $\frac35$ , so they equal each other.
As a check, their cross products agree:
$27\times50=1350$ $45\times30=1350$

No whole number times $45$ gives $50$ , and no whole number times $50$ gives $45$ .
So neither fraction is the other with its numerator and denominator multiplied by one whole number, and Milo is not correct.
Both are built from the same lowest-terms form, $\frac35$ , by different multipliers.

Answer

No; for example $\frac{27}{45}=\frac{30}{50}$ (any pair of equal fractions in which neither denominator is a whole-number multiple of the other is accepted).

Key idea

Two fractions can be equal even when neither is the other with its numerator and denominator multiplied by one whole number.
Problem 8: Mark complete
Problem 9 A finer pattern

A pattern is divided into nine equal parts, and four of them are colored. Every part is divided into five equal pieces, and then each new piece is divided into two equal pieces. Mara says the colored fraction is now $\frac{40}{90}$ and the colored area has not changed. Is she correct? Explain.
1. Hint 1
  Count what the two stages of cutting do to each original part.
2. Hint 2
  The same two stages apply to every colored part and every part in the whole.
Answer

Yes; $\frac{40}{90}=\frac49$ .

Full solution

Each original ninth becomes $5\times2=10$ equal pieces.
The nine parts of the whole therefore become $9\times10=90$ pieces, and the four colored parts become $4\times10=40$ colored pieces.

The new description is
$\frac{4\times10}{9\times10}=\frac{40}{90}.$
No area was added or removed by drawing the extra dividing lines, so the amount colored is unchanged.
Dividing $40$ and $90$ by $10$ returns $\frac49$ .

Answer

Yes; $\frac{40}{90}=\frac49$ .

Key idea

Repeated equal subdivision changes the counts of all parts together without changing the amount represented.
Problem 9: Mark complete
Problem 10 Kai’s replacement

Kai replaces $\frac{54}{81}$ with $\frac58$ by deleting the last digit of each number. Does the replacement preserve the value? Give the original fraction in lowest terms and justify your decision.
1. Hint 1
  Two fractions are equal exactly when their lowest-terms forms match.
2. Hint 2
  Simplify the original fraction, then compare it with the replacement.
Answer

No; $\frac{54}{81}=\frac23$ , and $\frac23\ne\frac58$ .

Full solution

The greatest common factor of $54$ and $81$ is $27$ , so
$\frac{54}{81}=\frac23.$
The proposed replacement is $\frac58$ .

The cross products for $\frac23$ and $\frac58$ are
$2\times8=16$ $3\times5=15.$
Since they do not match, the replacement changes the value.
Here, deleting the last digits turned $54$ into $5$ and $81$ into $8$ , which is not dividing both numbers by one common factor.

Answer

No; $\frac{54}{81}=\frac23$ , and $\frac23\ne\frac58$ .

Key idea

Simplifying removes a common factor from the numbers, not matching positions from their written digits.
Problem 10: Mark complete

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