Practice Ratio and proportion practice: 6 questions with answers Six ratio and proportion practice questions, from easy to hard, with worked explanations. Pick an answer to check it instantly.
These questions come from our SHSAT, TACHS and SSAT practice apps. For parents
A ratio compares two amounts, like 3 cups of flour for every 2 cups of sugar. A proportion says two ratios are equal. Most ratio questions can be solved with the same three steps.
A three-step method
Write the ratio as a fraction, with labels. “3 cups of flour for every 2 cups of sugar” becomes flour over sugar, 3/2.Set up a proportion with the unknown. If the recipe uses 12 cups of flour, then 3/2 = 12/s, where s is the cups of sugar.Solve and check that the answer makes sense. Here 3s = 24, so s = 8. Since there’s more flour than sugar in the recipe, 8 cups of sugar with 12 cups of flour fits.
For a ratio with three parts, like 3 : 4 : 7, add the parts first (3 + 4 + 7 = 14). Each part is then that fraction of the total.
The questions start easy and get harder. Try each one before you look at the explanation.
Practice questions Each question and explanation was reviewed by a person. Pick an answer to see if you're right.
Question 1 What value of
n n n makes the proportion
n 12 = 3 4 \frac{n}{12} = \frac{3}{4} 12 n = 4 3 true?
A 3 B 9 C 12 D 16 E 36
Cross-multiply:
4 n = 3 × 12 = 36 4n = 3 \times 12 = 36 4 n = 3 × 12 = 36 , so
n = 36 ÷ 4 = 9 n = 36 \div 4 = 9 n = 36 ÷ 4 = 9 . Sanity check:
9 12 \frac{9}{12} 12 9 reduces by dividing top and bottom by 3 to
3 4 \frac{3}{4} 4 3 , matching the right side. The value of
n n n is 9.
Question 2 To make a shade of paint, blue and white are mixed in a ratio of
2 2 2 cups of blue to
5 5 5 cups of white. If a painter uses
8 8 8 cups of blue, how many cups of white are needed to keep the same ratio?
A 4 B 11 C 20 D 28 E 40
Set up the proportion
2 5 = 8 w \frac{2}{5} = \frac{8}{w} 5 2 = w 8 . Since
8 = 2 × 4 8 = 2 \times 4 8 = 2 × 4 , scale the white part by the same factor:
5 × 4 = 20 5 \times 4 = 20 5 × 4 = 20 cups. Check:
8 20 = 2 5 \frac{8}{20} = \frac{2}{5} 20 8 = 5 2 .
Question 3 A garden drip line delivers 3 gallons of water every 12 minutes at a constant rate. The relationship between
y y y , the total gallons delivered, and
x x x , the time in minutes, can be written as
y = k x y = kx y = k x . What is the value of
k k k ?
A 1 4 \frac{1}{4} 4 1 B 3 C 4 D 12 E 36
In
y = k x y = kx y = k x , the constant
k k k is the unit rate in gallons per minute:
k = y x = 3 gallons 12 minutes = 1 4 k = \frac{y}{x} = \frac{3 \text{ gallons}}{12 \text{ minutes}} = \frac{1}{4} k = x y = 12 minutes 3 gallons = 4 1 gallon per minute. Second check: with
k = 1 4 k = \frac{1}{4} k = 4 1 , the equation gives
y = 1 4 × 12 = 3 y = \frac{1}{4} \times 12 = 3 y = 4 1 × 12 = 3 gallons in 12 minutes, matching the given information. The reciprocal, 4, is minutes per gallon, which would be the constant only if the equation were
x = k y x = ky x = k y . Sanity check: in 60 minutes the line should deliver
5 × 3 = 15 5 \times 3 = 15 5 × 3 = 15 gallons, and
1 4 × 60 = 15 \frac{1}{4} \times 60 = 15 4 1 × 60 = 15 agrees, so
k = 1 4 k = \frac{1}{4} k = 4 1 .
Question 4 Out of every $8 that Maya earns, she puts $3 into her savings account. Last month Maya earned $96. How much of that $96 did she put into her savings account?
A $12 B $36 C $48 D $60 E $72
Savings are
3 8 \frac{3}{8} 8 3 of earnings. Compute
3 8 × 96 \frac{3}{8} \times 96 8 3 × 96 : first
96 ÷ 8 = 12 96 \div 8 = 12 96 ÷ 8 = 12 , then
12 × 3 = 36 12 \times 3 = 36 12 × 3 = 36 . She saved $36. Sanity check: $96 is twelve groups of $8, and each group contributes $3 to savings, so
12 × 3 = 36 12 \times 3 = 36 12 × 3 = 36 . The amount saved is $36.
Question 5 A \$2{,}100 grant is divided among three clubs in the ratio
3 : 4 : 7 3:4:7 3 : 4 : 7 . How much more money does the club with the largest share receive than the club with the smallest share?
A \$150 B \$450 C \$525 D \$600 E \$1{,}050
The ratio
3 : 4 : 7 3:4:7 3 : 4 : 7 has
3 + 4 + 7 = 14 3+4+7 = 14 3 + 4 + 7 = 14 parts, so one part is
2100 ÷ 14 = $ 150 2100\div 14 = \$150 2100 ÷ 14 = $150 . The largest share is
7 × 150 = $ 1,050 7\times 150 = \$1{,}050 7 × 150 = $1 , 050 and the smallest is
3 × 150 = $ 450 3\times 150 = \$450 3 × 150 = $450 . The difference is
1050 − 450 = $ 600 1050 - 450 = \$600 1050 − 450 = $600 . Independent check: the gap is
( 7 − 3 ) = 4 (7-3)=4 ( 7 − 3 ) = 4 parts, and
4 × 150 = $ 600 4\times 150 = \$600 4 × 150 = $600 .
Question 6 A tour boat is carrying adults and children in the ratio
4 : 7 4:7 4 : 7 . At the next dock, a whole number of additional adults board; no children board and no one gets off. After the adults board, the ratio of adults to children on the boat is
3 : 4 3:4 3 : 4 . The boat is legally allowed to carry at most
200 200 200 passengers. What is the greatest possible number of children on the boat?
A 84 B 112 C 114 D 126 E 196
Let the original counts be
4 m 4m 4 m adults and
7 m 7m 7 m children (the children never change). If
a a a adults board, then
4 m + a 7 m = 3 4 ⇒ 16 m + 4 a = 21 m ⇒ a = 5 m 4 \frac{4m+a}{7m}=\frac{3}{4}\Rightarrow 16m+4a=21m\Rightarrow a=\frac{5m}{4} 7 m 4 m + a = 4 3 ⇒ 16 m + 4 a = 21 m ⇒ a = 4 5 m . Because
a a a must be a whole number of people,
m m m must be a multiple of
4 4 4 ; write
m = 4 s m=4s m = 4 s . Then the number of children is
7 m = 28 s 7m=28s 7 m = 28 s and the total aboard afterward is
( 4 m + a ) + 7 m = ( 16 s + 5 s ) + 28 s = 49 s (4m+a)+7m=(16s+5s)+28s=49s ( 4 m + a ) + 7 m = ( 16 s + 5 s ) + 28 s = 49 s . The legal cap gives
49 s ≤ 200 ⇒ s ≤ 4 49s\le 200\Rightarrow s\le 4 49 s ≤ 200 ⇒ s ≤ 4 (since
49 ⋅ 4 = 196 49\cdot4=196 49 ⋅ 4 = 196 but
49 ⋅ 5 = 245 > 200 49\cdot5=245>200 49 ⋅ 5 = 245 > 200 ). Children are greatest at
s = 4 s=4 s = 4 :
28 ⋅ 4 = 112 28\cdot4=112 28 ⋅ 4 = 112 children (with
196 196 196 aboard). Verify by matching the unchanged part: scale
4 : 7 4:7 4 : 7 and
3 : 4 3:4 3 : 4 to a common children value, LCM
( 7 , 4 ) = 28 (7,4)=28 ( 7 , 4 ) = 28 , giving before
16 : 28 16:28 16 : 28 and after
21 : 28 21:28 21 : 28 . So each valid boat is a whole multiple
g g g : children
= 28 g =28g = 28 g , total
= 49 g =49g = 49 g ;
49 ⋅ 4 = 196 ≤ 200 49\cdot4=196\le200 49 ⋅ 4 = 196 ≤ 200 is the largest that fits, so
28 ⋅ 4 = 112 28\cdot4=112 28 ⋅ 4 = 112 .
For parents
These practice questions come from our SHSAT, TACHS and SSAT practice apps. Each app has more than 1,200 practice questions, every one with a worked explanation.
Checked against official sources. How we make this site Wednesday, September 30, 2026