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Percent: 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 Share label

A label describes a share as $32\%$ . Write the share as a fraction in lowest terms.
1. Hint 1
  A percent counts hundredths of a whole.
2. Hint 2
  Write the hundredths as a fraction, then divide top and bottom by their greatest common factor.
Answer

$\frac{8}{25}$ .

Full solution

A percent counts hundredths, so thirty-two percent is thirty-two hundredths.
$32\%=\frac{32}{100}$

The greatest common factor of $32$ and $100$ is $4$ .
Dividing top and bottom by four gives
$\frac{32}{100}=\frac{8}{25}$

Answer

$\frac{8}{25}$ .

Key idea

Writing a whole-number percent over one hundred and dividing out the greatest common factor gives its fraction in lowest terms.
Problem 1: Mark complete
Problem 2 Control settings

A volume slider starts at $0\%$ and moves in equal steps of $0.5\%$ of its full setting. Write, as a decimal, the share of the full setting that the slider reaches at the seventh step above zero.
1. Hint 1
  First find the percent represented by seven steps.
2. Hint 2
  The percent sign divides that number by one hundred.
Answer

$0.035$ .

Full solution

Seven equal steps give a percent of
$7\times0.5=3.5$

Convert that percent to a decimal by dividing by one hundred.
$3.5\%=0.035$
Thus the seventh step represents $0.035$ of the full setting.

Answer

$0.035$ .

Key idea

The percent sign divides by one hundred even when the number before it is a decimal.
Problem 2: Mark complete
Problem 3 Inspection record

An inspector checks exactly $100$ pieces and finds $31$ scratched pieces. All the remaining pieces are unscratched. What percent of the pieces are unscratched?
1. Hint 1
  Find the count that belongs to the requested group.
2. Hint 2
  A count out of one hundred is already on the percent scale.
Answer

$69\%$ .

Full solution

The unscratched count is
$100-31=69$

Sixty-nine of the hundred checked pieces are unscratched.
$\frac{69}{100}=69\%$
The scratched and unscratched counts together account for every piece.

Answer

$69\%$ .

Key idea

A percent states how many parts belong to a group out of one hundred equal parts.
Problem 3: Mark complete
Problem 4 Card assignments

A box has $24$ cards marked for an activity, representing $30\%$ of all its cards. The organizer needs $65\%$ of the box, using only unmarked cards, for another activity. How many cards does the organizer need for that activity?
1. Hint 1
  Find the whole box count from the known marked share.
2. Hint 2
  Apply the requested share to that whole, not to the twenty-four marked cards.
Answer

$52$ cards.

Full solution

Let $w$ be the positive whole box count.
The known share gives
$\frac{24}{w}=\frac{30}{100}$
Multiply the known diagonal, $24$ and $100$ , and divide by the $30$ opposite the unknown.
$w=\frac{24\times100}{30}$ $w=80$

The new activity needs sixty-five percent of that whole.
$0.65\times80=52$
As a check, thirty percent of eighty is the stated twenty-four cards.
The marked and new cards total $24+52=76$ , which fits in the box of $80$ .

Answer

$52$ cards.

Key idea

Recover the whole from a known percent share before finding another share of that whole.
Problem 4: Mark complete
Problem 5 Editing progress

A manuscript has $160$ pages. An editor finishes $52$ pages on Monday and $44$ different pages on Tuesday. What percent of the manuscript has the editor finished across these two days?
1. Hint 1
  The two completed groups do not overlap, so their counts combine.
2. Hint 2
  Compare their total with the fixed manuscript length.
Answer

$60\%$ .

Full solution

The completed count is
$52+44=96$
pages.
The whole remains $160$ pages.

Write completed pages over all pages.
Since $160\div16=10$ , dividing top and bottom by $16$ gives tenths.
$\frac{96}{160}=\frac{6}{10}$
Multiplying top and bottom by $10$ gives
$\frac{6}{10}=\frac{60}{100}$
Therefore the editor has finished $60\%$ of the manuscript.

Answer

$60\%$ .

Key idea

Add separate part counts before comparing their combined amount with the fixed whole.
Problem 5: Mark complete
Problem 6 Materials reports

Report A says $\frac{7}{16}$ of its material is recycled. Report B says $0.44$ of its material is recycled. Give each share as a percent and say which report describes the larger recycled share.
1. Hint 1
  Put both shares on a single scale before comparing them.
2. Hint 2
  Rewrite each share as a number of hundredths.
3. Hint 3
  Sixteen does not divide evenly into one hundred, so set up a proportion with an unknown number of hundredths.
Answer

A: $43.75\%$ ; B: $44\%$ ; Report B has the larger share.

Full solution

Sixteen does not divide evenly into one hundred, so let $p$ be the unknown number of hundredths.
$\frac{7}{16}=\frac{p}{100}$
Multiply the known diagonal, $7$ and $100$ , and divide by the $16$ opposite $p$ .
$p=\frac{7\times100}{16}$ $p=43.75$
So Report A describes $43.75\%$ .

The decimal already names forty-four hundredths.
$0.44=44\%$
Since $44$ is greater than $43.75$ , Report B describes the larger share.
This compares shares rather than total recycled amounts.

Answer

A: $43.75\%$ ; B: $44\%$ ; Report B has the larger share.

Key idea

A common percent scale makes shares given in different forms directly comparable.
Problem 6: Mark complete
Problem 7 Fundraising target

A group collects $140\%$ of its fundraising target. The collected amount exceeds the target by 96 dollars. What was the target, and how much did the group collect?
1. Hint 1
  The target itself represents one hundred percent.
2. Hint 2
  The excess is the part above one hundred percent, so use it to recover the target.
Answer

Target: 240 dollars; collected: 336 dollars.

Full solution

A percent counts hundredths, so $140\%$ of the target is $140$ hundredths of it, and the target itself is $100$ of those hundredths.
The excess is the other $140-100=40$ hundredths, which is $40\%$ of the target.
So 96 dollars is $40\%$ of the target, and the target is
$96\div0.40=240$
dollars.

The collected amount is
$1.40\times240=336$
dollars.
Checking the difference gives $336-240=96$ , as stated.

Answer

Target: 240 dollars; collected: 336 dollars.

Key idea

When an amount is more than $100\%$ of a target, subtract $100\%$ from its percent; what remains, taken of the target, is the amount above the target.
Problem 7: Mark complete
Problem 8 Two calculations

Teo claims that $72\%$ of $25$ equals $25\%$ of $72$ . Is this correct? Explain why the two results agree or differ.
1. Hint 1
  Write each percent as a fraction over one hundred.
2. Hint 2
  Compare the products in the two numerators.
Answer

Yes; both equal $18$ .

Full solution

The first amount is
$\frac{72\times25}{100}=18$
The second is
$\frac{25\times72}{100}=18$

The numerator factors are the same in reversed order, and reversing multiplication does not change a product.
Both amounts therefore equal $18$ .

Answer

Yes; both equal $18$ .

Key idea

Interchanging a percent number and its whole gives the same numerical amount because the resulting products have the same factors.
Problem 8: Mark complete
Problem 9 Completion reports

Team A completes $12$ of its $25$ assigned jobs, and Team B completes $19$ of its $40$ assigned jobs. A report ranks B ahead because it completed more jobs. Does that ranking also hold for the completed share of each team's assignment? Explain using percents.
1. Hint 1
  A count and a share answer different questions when the assignments differ in size.
2. Hint 2
  Express each completed share out of one hundred before comparing.
Answer

No; A completed $48\%$ , while B completed $47.5\%$ .

Full solution

Team A completed
$\frac{12}{25}=\frac{48}{100}$
or $48\%$ of its assignment.
Team B completed
$19\div40=0.475$
or $47.5\%$ of its assignment.

Team B completed more jobs, but Team A completed the larger share.
Once both shares are expressed out of one hundred, their percent numbers can be compared directly.

Answer

No; A completed $48\%$ , while B completed $47.5\%$ .

Key idea

A larger completed count can represent a smaller share when the total assignment is larger.
Problem 9: Mark complete
Problem 10 Unmarked objects

A record says that $0\%$ of a collection is marked, and $0$ objects are marked. The collection has at least one object. Noor says this information identifies how many objects the collection has. Is Noor right? Justify your answer.
1. Hint 1
  Consider what zero percent does to different positive totals.
2. Hint 2
  Test two distinct whole-number collection sizes with no marked objects.
Answer

No. For example, total counts of $8$ and $23$ both work.

Full solution

Zero percent is zero of any whole.
For example,
$0\times8=0$ $0\times23=0$
Both collections have the recorded marked count and percent.

The total is therefore not determined.
Recovering a whole by dividing the part by its percent would require division by zero here, which is undefined.

Answer

No. For example, total counts of $8$ and $23$ both work.

Key idea

A zero count representing zero percent does not determine the positive whole it came from.
Problem 10: Mark complete

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