Math worksheet featuring four word problems on subtracting fractions, designed for educational practice.

Problem 1:

Sandra had a jar of \(\frac{5}{8}\) litres of water. After she spent some amount of water, she was left with \(\frac{2}{9}\) litres. How much water did she use?

#### Solution:
To find out how much water Sandra used, we need to subtract the amount of water she was left with from the initial amount of water.

1. Initial amount of water: \(\frac{5}{8}\) litres
2. Amount of water left: \(\frac{2}{9}\) litres

We need to subtract \(\frac{2}{9}\) from \(\frac{5}{8}\):

\[
\text{Water used} = \frac{5}{8} - \frac{2}{9}
\]

To perform this subtraction, we need a common denominator. The least common multiple (LCM) of 8 and 9 is 72.

- Convert \(\frac{5}{8}\) to a fraction with a denominator of 72:
\[
\frac{5}{8} = \frac{5 \times 9}{8 \times 9} = \frac{45}{72}
\]

- Convert \(\frac{2}{9}\) to a fraction with a denominator of 72:
\[
\frac{2}{9} = \frac{2 \times 8}{9 \times 8} = \frac{16}{72}
\]

Now, subtract the fractions:
\[
\frac{45}{72} - \frac{16}{72} = \frac{45 - 16}{72} = \frac{29}{72}
\]

Thus, the amount of water Sandra used is:
\[
\boxed{\frac{29}{72}}
\]

---

Problem 2:

A pipe was \(\frac{8}{9}\) m long. A portion was cut from it to resize it to \(\frac{4}{10}\) m. How much pipe was cut?

#### Solution:
To find out how much pipe was cut, we need to subtract the length of the resized pipe from the original length of the pipe.

1. Original length of the pipe: \(\frac{8}{9}\) m
2. Resized length of the pipe: \(\frac{4}{10}\) m

We need to subtract \(\frac{4}{10}\) from \(\frac{8}{9}\):

\[
\text{Length of pipe cut} = \frac{8}{9} - \frac{4}{10}
\]

To perform this subtraction, we need a common denominator. The least common multiple (LCM) of 9 and 10 is 90.

- Convert \(\frac{8}{9}\) to a fraction with a denominator of 90:
\[
\frac{8}{9} = \frac{8 \times 10}{9 \times 10} = \frac{80}{90}
\]

- Convert \(\frac{4}{10}\) to a fraction with a denominator of 90:
\[
\frac{4}{10} = \frac{4 \times 9}{10 \times 9} = \frac{36}{90}
\]

Now, subtract the fractions:
\[
\frac{80}{90} - \frac{36}{90} = \frac{80 - 36}{90} = \frac{44}{90}
\]

Simplify \(\frac{44}{90}\) by dividing the numerator and the denominator by their greatest common divisor (GCD), which is 2:
\[
\frac{44}{90} = \frac{44 \div 2}{90 \div 2} = \frac{22}{45}
\]

Thus, the length of the pipe that was cut is:
\[
\boxed{\frac{22}{45}}
\]

---

Problem 3:

Sam's car had \(\frac{9}{12}\) litres of petrol. He drove to his uncle's place and was left with \(\frac{5}{16}\) litres. How much petrol was consumed in the journey?

#### Solution:
To find out how much petrol was consumed, we need to subtract the amount of petrol left from the initial amount of petrol.

1. Initial amount of petrol: \(\frac{9}{12}\) litres
2. Amount of petrol left: \(\frac{5}{16}\) litres

We need to subtract \(\frac{5}{16}\) from \(\frac{9}{12}\):

\[
\text{Petrol consumed} = \frac{9}{12} - \frac{5}{16}
\]

To perform this subtraction, we need a common denominator. The least common multiple (LCM) of 12 and 16 is 48.

- Convert \(\frac{9}{12}\) to a fraction with a denominator of 48:
\[
\frac{9}{12} = \frac{9 \times 4}{12 \times 4} = \frac{36}{48}
\]

- Convert \(\frac{5}{16}\) to a fraction with a denominator of 48:
\[
\frac{5}{16} = \frac{5 \times 3}{16 \times 3} = \frac{15}{48}
\]

Now, subtract the fractions:
\[
\frac{36}{48} - \frac{15}{48} = \frac{36 - 15}{48} = \frac{21}{48}
\]

Simplify \(\frac{21}{48}\) by dividing the numerator and the denominator by their greatest common divisor (GCD), which is 3:
\[
\frac{21}{48} = \frac{21 \div 3}{48 \div 3} = \frac{7}{16}
\]

Thus, the amount of petrol consumed is:
\[
\boxed{\frac{7}{16}}
\]

---

Problem 4:

Will had \(\frac{11}{15}\) part of a packet of cookies in his kitchen. He ate some cookies, after which \(\frac{5}{24}\) of the packet remained. What fraction of the packet of cookies did he eat?

#### Solution:
To find out what fraction of the packet Will ate, we need to subtract the fraction of the packet that remained from the initial fraction of the packet he had.

1. Initial fraction of the packet: \(\frac{11}{15}\)
2. Fraction of the packet remaining: \(\frac{5}{24}\)

We need to subtract \(\frac{5}{24}\) from \(\frac{11}{15}\):

\[
\text{Fraction of the packet eaten} = \frac{11}{15} - \frac{5}{24}
\]

To perform this subtraction, we need a common denominator. The least common multiple (LCM) of 15 and 24 is 120.

- Convert \(\frac{11}{15}\) to a fraction with a denominator of 120:
\[
\frac{11}{15} = \frac{11 \times 8}{15 \times 8} = \frac{88}{120}
\]

- Convert \(\frac{5}{24}\) to a fraction with a denominator of 120:
\[
\frac{5}{24} = \frac{5 \times 5}{24 \times 5} = \frac{25}{120}
\]

Now, subtract the fractions:
\[
\frac{88}{120} - \frac{25}{120} = \frac{88 - 25}{120} = \frac{63}{120}
\]

Simplify \(\frac{63}{120}\) by dividing the numerator and the denominator by their greatest common divisor (GCD), which is 3:
\[
\frac{63}{120} = \frac{63 \div 3}{120 \div 3} = \frac{21}{40}
\]

Thus, the fraction of the packet of cookies that Will ate is:
\[
\boxed{\frac{21}{40}}
\]

---

Final Answers:

1. \(\boxed{\frac{29}{72}}\)
2. \(\boxed{\frac{22}{45}}\)
3. \(\boxed{\frac{7}{16}}\)
4. \(\boxed{\frac{21}{40}}\)