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Ratio Problems: 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 The three teams

A club's $120$ members are split into teams A, B, and C in the ratio $4:7:9$ . Find the number of members on each team, and check that your counts reduce to the given ratio.
1. Hint 1
  The ratio fixes how the teams compare, and the total fixes the size of one part.
2. Hint 2
  Write the team sizes as $4s$ , $7s$ , and $9s$ , and set their sum equal to $120$ .
3. Hint 3
  Solve for $s$ , then multiply it back through each of the three terms.
Answer

Team A has $24$ members, team B has $42$ , and team C has $54$ .

Full solution

Write the team sizes as $4s$ , $7s$ , and $9s$ , where $s$ is the number of members in one part.
Every member is on a team, so the three sizes add to the total.
$4s+7s+9s=120$
Combine the like terms.
$20s=120$
Divide both sides by $20$ .
$s=6$

Multiply back through each term: team A has $4\cdot 6=24$ members, team B has $7\cdot 6=42$ , and team C has $9\cdot 6=54$ members.

Check: the counts add to $24+42+54=120$ as stated, and dividing each count by their common factor $6$ gives $4:7:9$ .

Answer

Team A has $24$ members, team B has $42$ , and team C has $54$ .

Key idea

Dividing a total by the number of parts gives the common multiplier, and each amount is its ratio number times that multiplier.
Problem 1: Mark complete
Problem 2 The two ropes

Two ropes have lengths in the ratio $11:6$ , and the longer rope is $35$ meters longer than the shorter one. Find the length of each rope, and check that your lengths reduce to the given ratio.
1. Hint 1
  A difference between two amounts fixes the size of one part, just as a total does.
2. Hint 2
  Write the lengths as $11s$ and $6s$ meters, and ask how many parts longer the longer rope is.
3. Hint 3
  Set $11s-6s$ equal to $35$ and solve for $s$ .
Answer

The longer rope is $77$ meters long and the shorter rope is $42$ meters long.

Full solution

Write the lengths as $11s$ and $6s$ meters, where $s>0$ is the length of one part.
The longer rope is $35$ meters longer, so the difference of the two lengths is $35$ .
$11s-6s=35$
Combine the like terms.
$5s=35$
Divide both sides by $5$ .
$s=7$

Multiply back: the longer rope is $11\cdot 7=77$ meters and the shorter rope is $6\cdot 7=42$ meters.

Check: the difference is $77-42=35$ meters as stated, and dividing both lengths by their common factor $7$ gives $11:6$ .

Answer

The longer rope is $77$ meters long and the shorter rope is $42$ meters long.

Key idea

A known difference is the difference of the ratio numbers times one part, so it fixes the common multiplier as a total does.
Problem 2: Mark complete
Problem 3 The trail mix batch

A trail mix uses peanuts, raisins, and sunflower seeds in the ratio $5:2:3$ , measured in cups. One batch uses $14$ cups of raisins. Find the number of cups of peanuts and of sunflower seeds in the batch, and check that the three amounts reduce to the given ratio.
1. Hint 1
  One known amount is enough to fix the size of one part, with no total needed.
2. Hint 2
  Raisins are the $2$ term of the ratio, so set $2s$ equal to $14$ .
3. Hint 3
  Multiply the part size by each of the other two ratio numbers.
Answer

$35$ cups of peanuts and $21$ cups of sunflower seeds.

Full solution

Write the amounts as $5s$ cups of peanuts, $2s$ cups of raisins, and $3s$ cups of sunflower seeds.
The batch uses $14$ cups of raisins, and raisins are the $2$ term.
$2s=14$
Divide both sides by $2$ .
$s=7$

Multiply back: the batch uses $5\cdot 7=35$ cups of peanuts and $3\cdot 7=21$ cups of sunflower seeds.

Check: dividing $35$ , $14$ , and $21$ by their common factor $7$ gives $5:2:3$ .
The whole batch is $35+14+21=70$ cups, which also equals
$(5+2+3)s=10\cdot 7.$

Answer

$35$ cups of peanuts and $21$ cups of sunflower seeds.

Key idea

One known amount fixes the common multiplier, so every other amount follows without a total.
Problem 3: Mark complete
Problem 4 The counted trays

Cards are divided among three trays A, B, and C in the ratio $2:3:5$ . A count of A and B gives $30$ cards, and then four cards move from C to A. Find the original counts and the new ratio in simplest whole numbers.
1. Hint 1
  Use the partial total to find the common part size.
2. Hint 2
  The transfer increases A and decreases C by the same count.
Answer

Original: A has $12$ cards, B has $18$ cards, and C has $30$ cards. New ratio: $A:B:C=8:9:13$ .

Full solution

Write the counts as $2s$ , $3s$ , and $5s$ .
The counted trays give
$2s+3s=30$ $s=6$
Thus the original counts are $12$ , $18$ , and $30$ .

After the transfer they are $16$ , $18$ , and $26$ .
Their common factor $2$ gives the new ratio $8:9:13$ .
The total remains $60$ , and the original counts reduce to $2:3:5$ .

Answer

Original: A has $12$ cards, B has $18$ cards, and C has $30$ cards. New ratio: $A:B:C=8:9:13$ .

Key idea

A partial total fixes the common part size, and so every amount, before a transfer changes them.
Problem 4: Mark complete
Problem 5 The shared collection

A collection holds three kinds of folder, A, B, and C, with $A:B=4:3$ and $A:C=2:5$ . There are $21$ more C folders than B folders. Find all three counts and check both given ratios.
1. Hint 1
  The shared term is A, even though it comes first in both ratios.
2. Hint 2
  Match the A terms, then express the difference between C and B in shared parts.
Answer

A: $12$ folders. B: $9$ folders. C: $30$ folders.

Full solution

Scale $A:C=2:5$ by $2$ to get $A:C=4:10$ .
This matches A in the other ratio, so
$A:B:C=4:3:10$

Let the counts be $4s$ , $3s$ , and $10s$ .
The difference gives
$10s-3s=21$ $s=3$
The counts are $12$ , $9$ , and $30$ .
Their difference is $30-9=21$ , while $12:9$ reduces to $4:3$ and $12:30$ reduces to $2:5$ , checking both given ratios.

Answer

A: $12$ folders. B: $9$ folders. C: $30$ folders.

Key idea

Match the quantity shared by two ratios before applying a difference or total.
Problem 5: Mark complete
Problem 6 The exchanged counters

A bag has red and blue counters in the ratio $5:3$ . Four red counters are taken out and replaced by four blue counters. The new ratio is $3:5$ . Find the original counts and verify both ratios.
1. Hint 1
  One color loses exactly what the other gains.
2. Hint 2
  Write the changed amounts as $5s-4$ and $3s+4$ before using the new ratio.
Answer

$10$ red counters and $6$ blue counters.

Full solution

The original counts are $5s$ and $3s$ , with $s>0$ .
Since $3s+4>0$ , the new ratio gives
$\frac{5s-4}{3s+4}=\frac35$
Cross-multiplication yields
$5(5s-4)=3(3s+4)$
Expand both sides.
$25s-20=9s+12$
Collect the $s$ terms on the left and the numbers on the right.
$16s=32$ $s=2$

The original counts are $10$ and $6$ , reducing to $5:3$ .
After the exchange they are $6$ and $10$ , reducing to $3:5$ .
All counts are whole and positive.

Answer

$10$ red counters and $6$ blue counters.

Key idea

An exchange changes both actual amounts, so both terms must be adjusted before the new ratio is imposed.
Problem 6: Mark complete
Problem 7 The two batches

Batch A contains red and white beads in the ratio $2:3$ and has $25$ beads. Batch B has the same color ratio and contains $9$ white beads. The batches are combined. Find the combined red and white counts and the ratio of red to white in the combined batch, in simplest form.
1. Hint 1
  Each batch has its own part size, although the two color comparisons match.
2. Hint 2
  Find each part size before adding matching colors.
Answer

$16$ red beads and $24$ white beads; ratio $2:3$ .

Full solution

For batch A, five parts make $25$ , so each part is $5$ beads.
It has $10$ red and $15$ white.
For batch B, three white parts make $9$ , so each part is $3$ beads.
It has $6$ red and $9$ white.

The combined counts are
$10+6=16$ $15+9=24$
Since $16:24$ reduces to $2:3$ , the combined batch keeps the common ratio.

Answer

$16$ red beads and $24$ white beads; ratio $2:3$ .

Key idea

Batches with the same ratio can have different part sizes and still preserve that ratio when combined.
Problem 7: Mark complete
Problem 8 The doubled amounts

Two positive amounts are in the ratio $4:7$ . Inez doubles both amounts and says the difference between them stays unchanged because the ratio stays unchanged. Is she correct? Explain using a common multiplier.
1. Hint 1
  A ratio and a difference measure different features of a pair.
2. Hint 2
  Write the original amounts as $4s$ and $7s$ , then compare their differences before and after doubling.
Answer

No; the difference doubles.

Full solution

With $s>0$ , the original difference is
$7s-4s=3s$
After doubling, the difference is
$14s-8s=6s$
This is twice the original difference.

The ratio is still $4:7$ , but the actual size of each part has doubled.
Since the original difference is positive, it is not unchanged.

Answer

No; the difference doubles.

Key idea

Scaling preserves a ratio while scaling the difference between the actual amounts.
Problem 8: Mark complete
Problem 9 The reported excess

Three positive amounts have ratio $2:5:6$ . A report says the largest exceeds the sum of the other two by $4$ . Can this report be correct?
1. Hint 1
  Compare the ratio parts before looking for their common size.
2. Hint 2
  Write the three amounts as $2s$ , $5s$ , $6s$ and turn the report into an equation in $s$ .
Answer

No.

Full solution

Write the amounts as $2s$ , $5s$ , and $6s$ , where $s>0$ .
The reported difference would be
$6s-(2s+5s)=4$ $-s=4$

This would require a negative part size.
In fact the largest amount is one positive part smaller than the sum of the other two, so it cannot exceed that sum.

Answer

No.

Key idea

Comparing the ratio parts can reveal an impossible condition before actual amounts are found.
Problem 9: Mark complete
Problem 10 The equal additions

Two positive amounts have ratio $a:b$ , where $a$ and $b$ are positive. Each amount is increased by the same positive amount $h$ . A learner claims the ratio must stay $a:b$ . Decide exactly when the claim is true and justify your answer.
1. Hint 1
  Write the original amounts with one positive multiplier.
2. Hint 2
  Compare the adjusted ratio with $a:b$ by cross-multiplying.
Answer

The ratio stays $a:b$ exactly when $a=b$ .

Full solution

Write the original amounts as $as$ and $bs$ , where $s>0$ .
After the increase the amounts are $as+h$ and $bs+h$ .
For the ratio to stay $a:b$ ,
$\frac{as+h}{bs+h}=\frac{a}{b}$
Both denominators are positive, so cross-multiply.
$b(as+h)=a(bs+h)$
Expand both sides.
$abs+bh=abs+ah$
Subtract $abs$ from both sides.
$bh=ah$
Since $h>0$ , division by $h$ gives $a=b$ .

Conversely, when $a=b$ , the original amounts are equal and adding the same amount keeps them equal.
Both ratios are then $1:1$ .

Answer

The ratio stays $a:b$ exactly when $a=b$ .

Key idea

Adding the same positive amount preserves a positive two-term ratio precisely when its two amounts are equal.
Problem 10: Mark complete

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