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Ratios: 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 Paper shapes

A collection of paper shapes contains only stars and circles. Stars make up $\frac{3}{11}$ of the collection. Write the ratio of circles to stars in simplest form.
1. Hint 1
  Think of the whole as eleven equal parts.
2. Hint 2
  Find how many of those parts belong to circles, then put circles first.
Answer

$8:3$ .

Full solution

Three of the eleven equal parts are stars.
The remaining parts are circles.
$11-3=8$

Circles therefore take eight parts for every three parts of stars, so their ratio is $8:3$ .
The two numbers have no common factor above $1$ .

Answer

$8:3$ .

Key idea

The order of the named groups determines the order of a ratio.
Problem 1: Mark complete
Problem 2 Storage slots

A rack has $32$ slots. Exactly $12$ are empty, and the rest are occupied. Write the ratio of occupied slots to empty slots in simplest form.
1. Hint 1
  Find the occupied count before comparing the two groups.
2. Hint 2
  Divide both counts by their greatest common factor.
Answer

$5:3$ .

Full solution

Subtract the empty slots from the total.
$32-12=20$

The comparison is $20:12$ .
The greatest common factor is $4$ , so dividing both parts by $4$ gives $5:3$ .
$\frac{20}{12}=\frac{5}{3}$

Answer

$5:3$ .

Key idea

A comparison between two parts uses the two part counts, rather than the whole count.
Problem 2: Mark complete
Problem 3 Cutting a strip

A strip consists of $9$ cm of dark material followed by $14$ cm of light material. It is cut into pieces that are each $0.5$ cm long, with no piece crossing the boundary between materials. Write the ratio of the actual number of dark pieces to the actual number of light pieces without simplifying.
1. Hint 1
  Each material length is divided into pieces of the same length.
2. Hint 2
  Find each piece count by dividing its material length by the length of one piece.
Answer

$18:28$ .

Full solution

Each half-centimeter piece uses $0.5$ cm of material.
The dark piece count is
$9\div0.5=18$
and the light piece count is
$14\div0.5=28$

The actual counts give $18:28$ , as requested.
Both counts are twice the corresponding lengths, so this comparison is equivalent to the length ratio $9:14$ .

Answer

$18:28$ .

Key idea

Dividing both quantities by the same positive size produces equivalent comparisons.
Problem 3: Mark complete
Problem 4 Workshop funds

A workshop has 126 dollars. It reserves 22 dollars for delivery and divides the rest between paper and tools in the ratio $3:5$ . How much goes to each of these two purchases?
1. Hint 1
  The ratio applies to the amount left after delivery is reserved.
2. Hint 2
  Count all the ratio parts, find the value of one part, and then find each share.
Answer

Paper: 39 dollars; tools: 65 dollars.

Full solution

The amount to split is the budget after delivery.
$126-22=104$

The ratio has $3+5=8$ parts.
Each part is worth 13 dollars.
$104\div8=13$

Paper receives three parts and tools receive five.
$3\times13=39$ $5\times13=65$
Their shares add to 104 dollars, and delivery brings the total back to 126 dollars.

Answer

Paper: 39 dollars; tools: 65 dollars.

Key idea

Identify the total that a ratio actually splits before valuing its parts.
Problem 4: Mark complete
Problem 5 Audio programs

Program A has $6$ minutes of speech and $11$ minutes of music. Program B has $5$ minutes of speech and $9$ minutes of music. Which program has more speech for the same amount of music? Show why.
1. Hint 1
  Compare speech with music in the same order for both programs.
2. Hint 2
  Write the two comparisons as fractions with a common denominator.
Answer

Program B.

Full solution

The speech-to-music comparisons are $\frac{6}{11}$ and $\frac{5}{9}$ .
Rewrite both over $99$ .
$\frac{6}{11}=\frac{54}{99}$ $\frac{5}{9}=\frac{55}{99}$

Fifty-five ninety-ninths is greater than fifty-four ninety-ninths.
Program B therefore has more speech for the same amount of music.

Answer

Program B.

Key idea

Two ratios can be compared by expressing both as fractions with the same denominator.
Problem 5: Mark complete
Problem 6 Three kinds of fastener

A box holds three kinds of fastener: screws, nails, and $12$ hooks. The ratio of screws to nails is $5:7$ , and there are $42$ nails. What fraction of all the fasteners are screws, in simplest form?
1. Hint 1
  The whole box includes the hooks as well as the two groups in the ratio.
2. Hint 2
  Use the nail count to find the screw count.
Answer

$\frac{5}{14}$ .

Full solution

Seven ratio parts account for $42$ nails, so one part contains six fasteners.
$42\div7=6$
The screw count is five parts.
$5\times6=30$

Include the hooks in the whole.
$30+42+12=84$
The required fraction simplifies by dividing top and bottom by $6$ .
$\frac{30}{84}=\frac{5}{14}$

Answer

$\frac{5}{14}$ .

Key idea

A part-to-whole fraction must include every group that belongs to the whole.
Problem 6: Mark complete
Problem 7 Display pieces

A display uses $28$ small pieces and $48$ large pieces. A worker removes one quarter of the pieces of each size. Find the ratio of small to large pieces that remain, in simplest form, and the total remaining count.
1. Hint 1
  Find the remaining share of each original count.
2. Hint 2
  Compare the remaining counts, then add them for the total.
Answer

$7:12$ ; $57$ pieces remain.

Full solution

Removing one quarter leaves three quarters of each size.
$\frac{3}{4}\times28=21$ $\frac{3}{4}\times48=36$

The remaining ratio is $21:36$ .
Divide both parts by their greatest common factor, $3$ , to get $7:12$ .
The remaining count is
$21+36=57$
Both counts were scaled by the same factor, so the original comparison is preserved.

Answer

$7:12$ ; $57$ pieces remain.

Key idea

Removing the same fractional share from both quantities preserves their ratio.
Problem 7: Mark complete
Problem 8 Tile tray

A tray holds $8$ green tiles and $22$ white tiles. Ana adds $12$ green tiles and $33$ white tiles. She says the tray's green-to-white ratio has not changed. Is she right? Explain.
1. Hint 1
  Two whole-number ratios are equivalent exactly when they reduce to the same simplest form.
2. Hint 2
  Find the new green and white counts, then simplify the ratio before and after the tiles are added.
Answer

Yes; the green-to-white ratio is $4:11$ before and after.

Full solution

Before the change, the ratio is $8:22$ .
The greatest common factor of $8$ and $22$ is $2$ , so it simplifies to $4:11$ .
$\frac{8}{22}=\frac{4}{11}$

After the change, the green count is
$8+12=20$
and the white count is
$22+33=55$
The greatest common factor of $20$ and $55$ is $5$ , so $20:55$ also simplifies to $4:11$ .
$\frac{20}{55}=\frac{4}{11}$

Both ratios reduce to $4:11$ , so Ana is right.
The added tiles explain why: $12:33$ also reduces to $4:11$ , with greatest common factor $3$ , so Ana added three more groups of four green and eleven white tiles to the two groups already there.

Answer

Yes; the green-to-white ratio is $4:11$ before and after.

Key idea

Adding amounts that are in the same ratio as the original keeps the ratio unchanged.
Problem 8: Mark complete
Problem 9 Seat reservation

A room has $63$ seats, all either reserved or available. The ratio of reserved seats to available seats is $2:7$ . Luis calculates the reserved count as $63\times\frac{2}{7}$ . Is that calculation correct? Find the reserved count and explain.
1. Hint 1
  Decide whether the denominator seven represents available seats or every seat.
2. Hint 2
  Count the parts in the entire room before assigning two parts to reserved seats.
Answer

No; $14$ seats are reserved.

Full solution

The $7$ represents available seats, while $63$ counts every seat.
The whole has $2+7=9$ parts, so the reserved fraction is $\frac{2}{9}$ .
$63\times\frac{2}{9}=14$

The other $49$ seats are available.
Reserved to available is then $14:49$ , which reduces to $2:7$ when both counts are divided by $7$ , and the counts add back to $63$ .

Luis's calculation gives $18$ reserved seats, which would leave $45$ available.
Dividing both counts by $9$ reduces $18:45$ to $2:5$ , not $2:7$ , so his result does not fit the room.

Answer

No; $14$ seats are reserved.

Key idea

When the groups in a ratio make up the whole, the denominator for a share of the total is the sum of the ratio parts.
Problem 9: Mark complete
Problem 10 Packing request

A customer requests a box containing exactly $47$ whole pieces, with smooth pieces to ridged pieces in the ratio $4:5$ and no other pieces. Can the request be filled exactly? Justify your answer.
1. Hint 1
  Find the number of pieces in one complete group of the ratio.
2. Hint 2
  Check whether forty-seven pieces can be made up entirely of complete groups of that size.
Answer

No.

Full solution

Any whole-number counts in the ratio $4:5$ reduce to $4:5$ when both are divided by their greatest common factor.
So the counts are $4$ and $5$ times that factor, which means they come in complete groups of four smooth and five ridged pieces.
Each complete group holds nine pieces.
$4+5=9$

Five complete groups give $45$ pieces, and six give $54$ pieces.
The requested $47$ is not a multiple of $9$ , so no whole-piece counts with that total have the required ratio.
As a check, the smooth share would be
$47\times\frac{4}{9}=\frac{188}{9}$
which is not a whole number.

Answer

No.

Key idea

For two whole-piece counts in a simplified ratio, their total must be a multiple of the sum of the ratio parts.
Problem 10: Mark complete

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