Printable primary math worksheet for math grades 1 to 6 based on ... - Free Printable

Let's solve each of these fractions by simplifying them to their lowest terms. To do this, we find the greatest common divisor (GCD) of the numerator and denominator, then divide both by that number.

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🔹 Step-by-step solutions:

1. $\frac{27}{81}$
GCD(27, 81) = 27
$\frac{27 \div 27}{81 \div 27} = \frac{1}{3}$

2. $\frac{49}{98}$
GCD(49, 98) = 49
$\frac{49 \div 49}{98 \div 49} = \frac{1}{2}$

3. $\frac{72}{96}$
GCD(72, 96) = 24
$\frac{72 \div 24}{96 \div 24} = \frac{3}{4}$

4. $\frac{5}{60}$
GCD(5, 60) = 5
$\frac{5 \div 5}{60 \div 5} = \frac{1}{12}$

5. $\frac{12}{40}$
GCD(12, 40) = 4
$\frac{12 \div 4}{40 \div 4} = \frac{3}{10}$

6. $\frac{55}{75}$
GCD(55, 75) = 5
$\frac{55 \div 5}{75 \div 5} = \frac{11}{15}$

7. $\frac{13}{78}$
GCD(13, 78) = 13
$\frac{13 \div 13}{78 \div 13} = \frac{1}{6}$

8. $\frac{35}{45}$
GCD(35, 45) = 5
$\frac{35 \div 5}{45 \div 5} = \frac{7}{9}$

9. $\frac{3}{30}$
GCD(3, 30) = 3
$\frac{3 \div 3}{30 \div 3} = \frac{1}{10}$

10. $\frac{3}{45}$
GCD(3, 45) = 3
$\frac{3 \div 3}{45 \div 3} = \frac{1}{15}$

11. $\frac{21}{56}$
GCD(21, 56) = 7
$\frac{21 \div 7}{56 \div 7} = \frac{3}{8}$

12. $\frac{16}{80}$
GCD(16, 80) = 16
$\frac{16 \div 16}{80 \div 16} = \frac{1}{5}$

13. $\frac{6}{84}$
GCD(6, 84) = 6
$\frac{6 \div 6}{84 \div 6} = \frac{1}{14}$

14. $\frac{36}{80}$
GCD(36, 80) = 4
$\frac{36 \div 4}{80 \div 4} = \frac{9}{20}$

15. $\frac{3}{66}$
GCD(3, 66) = 3
$\frac{3 \div 3}{66 \div 3} = \frac{1}{22}$

16. $\frac{27}{42}$
GCD(27, 42) = 3
$\frac{27 \div 3}{42 \div 3} = \frac{9}{14}$

17. $\frac{12}{90}$
GCD(12, 90) = 6
$\frac{12 \div 6}{90 \div 6} = \frac{2}{15}$

18. $\frac{8}{21}$
GCD(8, 21) = 1 → already in lowest terms
$\frac{8}{21}$

19. $\frac{20}{75}$
GCD(20, 75) = 5
$\frac{20 \div 5}{75 \div 5} = \frac{4}{15}$

20. $\frac{28}{76}$
GCD(28, 76) = 4
$\frac{28 \div 4}{76 \div 4} = \frac{7}{19}$

21. $\frac{12}{44}$
GCD(12, 44) = 4
$\frac{12 \div 4}{44 \div 4} = \frac{3}{11}$

22. $\frac{16}{72}$
GCD(16, 72) = 8
$\frac{16 \div 8}{72 \div 8} = \frac{2}{9}$

23. $\frac{51}{68}$
GCD(51, 68) = 17
$\frac{51 \div 17}{68 \div 17} = \frac{3}{4}$

24. $\frac{66}{99}$
GCD(66, 99) = 33
$\frac{66 \div 33}{99 \div 33} = \frac{2}{3}$

25. $\frac{42}{70}$
GCD(42, 70) = 14
$\frac{42 \div 14}{70 \div 14} = \frac{3}{5}$

26. $\frac{15}{90}$
GCD(15, 90) = 15
$\frac{15 \div 15}{90 \div 15} = \frac{1}{6}$

27. $\frac{12}{51}$
GCD(12, 51) = 3
$\frac{12 \div 3}{51 \div 3} = \frac{4}{17}$

28. $\frac{12}{96}$
GCD(12, 96) = 12
$\frac{12 \div 12}{96 \div 12} = \frac{1}{8}$

29. $\frac{20}{70}$
GCD(20, 70) = 10
$\frac{20 \div 10}{70 \div 10} = \frac{2}{7}$

30. $\frac{14}{91}$
GCD(14, 91) = 7
$\frac{14 \div 7}{91 \div 7} = \frac{2}{13}$

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✔ Final Answers:

$$
\begin{aligned}
&\frac{27}{81} = \frac{1}{3} & \frac{49}{98} = \frac{1}{2} & \frac{72}{96} = \frac{3}{4} \\
&\frac{5}{60} = \frac{1}{12} & \frac{12}{40} = \frac{3}{10} & \frac{55}{75} = \frac{11}{15} \\
&\frac{13}{78} = \frac{1}{6} & \frac{35}{45} = \frac{7}{9} & \frac{3}{30} = \frac{1}{10} & \frac{3}{45} = \frac{1}{15} \\
&\frac{21}{56} = \frac{3}{8} & \frac{16}{80} = \frac{1}{5} & \frac{6}{84} = \frac{1}{14} & \frac{36}{80} = \frac{9}{20} \\
&\frac{3}{66} = \frac{1}{22} & \frac{27}{42} = \frac{9}{14} & \frac{12}{90} = \frac{2}{15} & \frac{8}{21} = \frac{8}{21} \\
&\frac{20}{75} = \frac{4}{15} & \frac{28}{76} = \frac{7}{19} & \frac{12}{44} = \frac{3}{11} & \frac{16}{72} = \frac{2}{9} \\
&\frac{51}{68} = \frac{3}{4} & \frac{66}{99} = \frac{2}{3} & \frac{42}{70} = \frac{3}{5} & \frac{15}{90} = \frac{1}{6} \\
&\frac{12}{51} = \frac{4}{17} & \frac{12}{96} = \frac{1}{8} & \frac{20}{70} = \frac{2}{7} & \frac{14}{91} = \frac{2}{13} \\
\end{aligned}
$$

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📝 Summary:

All fractions have been simplified using the GCD method. The key idea is to divide both numerator and denominator by their greatest common divisor until they can no longer be reduced.

Let me know if you'd like a printable version or explanation for any specific one!