Ratio and Proportion Worksheets - Math Monks - Free Printable

To solve the problem of creating equivalent ratios, we need to ensure that the ratios on both sides of the equation are proportional. This means that the cross-products of the ratios must be equal. Let's solve each problem step by step.

Problem (1): \( 8 : 3 = \_\_\_ : 27 \)

We need to find a number \( x \) such that:
\[
\frac{8}{3} = \frac{x}{27}
\]
Cross-multiplying gives:
\[
8 \cdot 27 = 3 \cdot x \implies 216 = 3x \implies x = \frac{216}{3} = 72
\]
So, the answer is:
\[
\boxed{72}
\]

Problem (2): \( 15 : \_\_\_ = 24 : 56 \)

We need to find a number \( y \) such that:
\[
\frac{15}{y} = \frac{24}{56}
\]
First, simplify \( \frac{24}{56} \):
\[
\frac{24}{56} = \frac{3}{7}
\]
So, we have:
\[
\frac{15}{y} = \frac{3}{7}
\]
Cross-multiplying gives:
\[
15 \cdot 7 = 3 \cdot y \implies 105 = 3y \implies y = \frac{105}{3} = 35
\]
So, the answer is:
\[
\boxed{35}
\]

Problem (3): \( 8 : 24 = 1 : \_\_\_ \)

We need to find a number \( z \) such that:
\[
\frac{8}{24} = \frac{1}{z}
\]
Simplify \( \frac{8}{24} \):
\[
\frac{8}{24} = \frac{1}{3}
\]
So, we have:
\[
\frac{1}{3} = \frac{1}{z} \implies z = 3
\]
So, the answer is:
\[
\boxed{3}
\]

Problem (4): \( 12 : \_\_\_ = 16 : 24 \)

We need to find a number \( w \) such that:
\[
\frac{12}{w} = \frac{16}{24}
\]
Simplify \( \frac{16}{24} \):
\[
\frac{16}{24} = \frac{2}{3}
\]
So, we have:
\[
\frac{12}{w} = \frac{2}{3}
\]
Cross-multiplying gives:
\[
12 \cdot 3 = 2 \cdot w \implies 36 = 2w \implies w = \frac{36}{2} = 18
\]
So, the answer is:
\[
\boxed{18}
\]

Problem (5): \( 7 : 56 = \_\_\_ : 72 \)

We need to find a number \( a \) such that:
\[
\frac{7}{56} = \frac{a}{72}
\]
Simplify \( \frac{7}{56} \):
\[
\frac{7}{56} = \frac{1}{8}
\]
So, we have:
\[
\frac{1}{8} = \frac{a}{72}
\]
Cross-multiplying gives:
\[
1 \cdot 72 = 8 \cdot a \implies 72 = 8a \implies a = \frac{72}{8} = 9
\]
So, the answer is:
\[
\boxed{9}
\]

Problem (6): \( \_\_\_ : 20 = 8 : 5 \)

We need to find a number \( b \) such that:
\[
\frac{b}{20} = \frac{8}{5}
\]
Cross-multiplying gives:
\[
b \cdot 5 = 8 \cdot 20 \implies 5b = 160 \implies b = \frac{160}{5} = 32
\]
So, the answer is:
\[
\boxed{32}
\]

Problem (7): \( 40 : 25 = 8 : \_\_\_ \)

We need to find a number \( c \) such that:
\[
\frac{40}{25} = \frac{8}{c}
\]
Simplify \( \frac{40}{25} \):
\[
\frac{40}{25} = \frac{8}{5}
\]
So, we have:
\[
\frac{8}{5} = \frac{8}{c} \implies c = 5
\]
So, the answer is:
\[
\boxed{5}
\]

Problem (8): \( 5 : \_\_\_ = 1 : 7 \)

We need to find a number \( d \) such that:
\[
\frac{5}{d} = \frac{1}{7}
\]
Cross-multiplying gives:
\[
5 \cdot 7 = 1 \cdot d \implies 35 = d \implies d = 35
\]
So, the answer is:
\[
\boxed{35}
\]

Problem (9): \( 12 : 5 = 72 : \_\_\_ \)

We need to find a number \( e \) such that:
\[
\frac{12}{5} = \frac{72}{e}
\]
Cross-multiplying gives:
\[
12 \cdot e = 5 \cdot 72 \implies 12e = 360 \implies e = \frac{360}{12} = 30
\]
So, the answer is:
\[
\boxed{30}
\]

Problem (10): \( 8 : \_\_\_ = 1 : 9 \)

We need to find a number \( f \) such that:
\[
\frac{8}{f} = \frac{1}{9}
\]
Cross-multiplying gives:
\[
8 \cdot 9 = 1 \cdot f \implies 72 = f \implies f = 72
\]
So, the answer is:
\[
\boxed{72}
\]

Problem (11): \( 20 : 28 = 5 : \_\_\_ \)

We need to find a number \( g \) such that:
\[
\frac{20}{28} = \frac{5}{g}
\]
Simplify \( \frac{20}{28} \):
\[
\frac{20}{28} = \frac{5}{7}
\]
So, we have:
\[
\frac{5}{7} = \frac{5}{g} \implies g = 7
\]
So, the answer is:
\[
\boxed{7}
\]

Problem (12): \( 99 : \_\_\_ = 11 : 7 \)

We need to find a number \( h \) such that:
\[
\frac{99}{h} = \frac{11}{7}
\]
Cross-multiplying gives:
\[
99 \cdot 7 = 11 \cdot h \implies 693 = 11h \implies h = \frac{693}{11} = 63
\]
So, the answer is:
\[
\boxed{63}
\]

Problem (13): \( 27 : 72 = 3 : \_\_\_ \)

We need to find a number \( i \) such that:
\[
\frac{27}{72} = \frac{3}{i}
\]
Simplify \( \frac{27}{72} \):
\[
\frac{27}{72} = \frac{3}{8}
\]
So, we have:
\[
\frac{3}{8} = \frac{3}{i} \implies i = 8
\]
So, the answer is:
\[
\boxed{8}
\]

Problem (14): \( 9 : 2 = \_\_\_ : 18 \)

We need to find a number \( j \) such that:
\[
\frac{9}{2} = \frac{j}{18}
\]
Cross-multiplying gives:
\[
9 \cdot 18 = 2 \cdot j \implies 162 = 2j \implies j = \frac{162}{2} = 81
\]
So, the answer is:
\[
\boxed{81}
\]

Problem (15): \( 12 : 7 = \_\_\_ : 56 \)

We need to find a number \( k \) such that:
\[
\frac{12}{7} = \frac{k}{56}
\]
Cross-multiplying gives:
\[
12 \cdot 56 = 7 \cdot k \implies 672 = 7k \implies k = \frac{672}{7} = 96
\]
So, the answer is:
\[
\boxed{96}
\]

Problem (16): \( 5 : 35 = 1 : \_\_\_ \)

We need to find a number \( m \) such that:
\[
\frac{5}{35} = \frac{1}{m}
\]
Simplify \( \frac{5}{35} \):
\[
\frac{5}{35} = \frac{1}{7}
\]
So, we have:
\[
\frac{1}{7} = \frac{1}{m} \implies m = 7
\]
So, the answer is:
\[
\boxed{7}
\]

Problem (17): \( 55 : 25 = \_\_\_ : 5 \)

We need to find a number \( n \) such that:
\[
\frac{55}{25} = \frac{n}{5}
\]
Simplify \( \frac{55}{25} \):
\[
\frac{55}{25} = \frac{11}{5}
\]
So, we have:
\[
\frac{11}{5} = \frac{n}{5} \implies n = 11
\]
So, the answer is:
\[
\boxed{11}
\]

Problem (18): \( 5 : 10 = 25 : \_\_\_ \)

We need to find a number \( p \) such that:
\[
\frac{5}{10} = \frac{25}{p}
\]
Simplify \( \frac{5}{10} \):
\[
\frac{5}{10} = \frac{1}{2}
\]
So, we have:
\[
\frac{1}{2} = \frac{25}{p}
\]
Cross-multiplying gives:
\[
1 \cdot p = 2 \cdot 25 \implies p = 50
\]
So, the answer is:
\[
\boxed{50}
\]

Problem (19): \( 48 : 28 = 12 : \_\_\_ \)

We need to find a number \( q \) such that:
\[
\frac{48}{28} = \frac{12}{q}
\]
Simplify \( \frac{48}{28} \):
\[
\frac{48}{28} = \frac{12}{7}
\]
So, we have:
\[
\frac{12}{7} = \frac{12}{q} \implies q = 7
\]
So, the answer is:
\[
\boxed{7}
\]

Problem (20): \( \_\_\_ : 27 = 8 : 9 \)

We need to find a number \( r \) such that:
\[
\frac{r}{27} = \frac{8}{9}
\]
Cross-multiplying gives:
\[
r \cdot 9 = 8 \cdot 27 \implies 9r = 216 \implies r = \frac{216}{9} = 24
\]
So, the answer is:
\[
\boxed{24}
\]

The final answers are:
\[
\boxed{72, 35, 3, 18, 9, 32, 5, 35, 30, 72, 7, 63, 8, 81, 96, 7, 11, 50, 7, 24}
\]