High School

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(a) [tex]18 \times \frac{5}{9} = [/tex]
(b) [tex]3 \frac{2}{3} \times 1 \frac{2}{5} \times 1 \frac{1}{7} = [/tex]
(c) [tex]\frac{9}{5} \times 3 = [/tex]
(d) [tex]\frac{1}{3} \text{ of } \left(\frac{4}{5} + \frac{2}{3}\right) = [/tex]
(e) [tex]\frac{16}{21} \times \frac{27}{48} = [/tex]
(f) [tex]\left(\frac{1}{3} + \frac{4}{6} - \frac{1}{6}\right) \times \frac{4}{5} = [/tex]
(g) [tex]\frac{100}{25} \times \frac{45}{60} = [/tex]
(h) [tex]4 \frac{1}{3} \times 2 \frac{1}{2} + 6 \times 5 \frac{1}{2} = [/tex]
(i) [tex]3 \frac{4}{5} \times 6 \frac{1}{7} = [/tex]
(j) [tex]5 \frac{1}{3} \times 4 \frac{1}{2} + 3 \frac{1}{4} \times 1 \frac{5}{6} = [/tex]
(k) [tex]\frac{5}{9} \text{ of } \frac{12}{5} = [/tex]
(l) [tex]\frac{2}{5} \text{ of } \frac{9}{16} \text{ of } \frac{12}{5} = [/tex]
(m) [tex]\frac{3}{5} \times \frac{25}{9} = [/tex]
(n) [tex]\frac{2}{5} \text{ of } \left(\frac{5}{4} - \frac{1}{4}\right) = [/tex]

Answer :

(a) To solve $18 \times \frac{5}{9}$, multiply 18 by 5 and then divide by 9.

[tex]= \frac{18 \times 5}{9} = \frac{90}{9} = 10.[/tex]

(b) $3 \frac{2}{3} \times 1 \frac{2}{5} \times 1 \frac{1}{7}$ involves converting mixed numbers to improper fractions.

  • $3 \frac{2}{3} = \frac{11}{3},$
  • $1 \frac{2}{5} = \frac{7}{5},$
  • $1 \frac{1}{7} = \frac{8}{7}$.

Thus,

[tex]= \frac{11}{3} \times \frac{7}{5} \times \frac{8}{7} = \frac{11 \times 7 \times 8}{3 \times 5 \times 7} = \frac{88}{15}.[/tex]

(c) [tex]rac{9}{5} \times 3 = \frac{9 \times 3}{5} = \frac{27}{5}[/tex].

(d) [tex]\frac{1}{3} \text{ of } (\frac{4}{5} + \frac{2}{3})[/tex]

  • First, add the fractions:
    [tex]\frac{4}{5} + \frac{2}{3} = \frac{12}{15} + \frac{10}{15} = \frac{22}{15}.[/tex]

  • Then multiply by [tex]\frac{1}{3}[/tex]:
    [tex]\frac{1}{3} \times \frac{22}{15} = \frac{22}{45}.[/tex]

(e) [tex]\frac{16}{21} \times \frac{27}{48}[/tex]

[tex]= \frac{16 \times 27}{21 \times 48} = \frac{432}{1008},[/tex] which simplifies to [tex]\frac{9}{21},[/tex] giving the final answer [tex]\frac{3}{7}.[/tex]

(f) [tex](\frac{1}{3} + \frac{4}{6} - \frac{1}{6}) \times \frac{4}{5}[/tex]

  • Simplify inside the parentheses:
    [tex]\frac{4}{6} = \frac{2}{3}, \Rightarrow \frac{1}{3} + \frac{2}{3} - \frac{1}{6} = \frac{3}{3} - \frac{1}{6} = \frac{6}{6} - \frac{1}{6} = \frac{5}{6}.[/tex]

  • Multiply by [tex]\frac{4}{5}[/tex]:
    [tex]\frac{5}{6} \times \frac{4}{5} = \frac{20}{30} = \frac{2}{3}.[/tex]

(g) [tex]\frac{100}{25} \times \frac{45}{60}[/tex]

[tex]= 4 \times \frac{45}{60} = 4 \times \frac{3}{4} = 3.[/tex]

(h) $4 \frac{1}{3} \times 2 \frac{1}{2} + 6 \times 5 \frac{1}{2}$

  • First calculation:
    [tex]4 \frac{1}{3} = \frac{13}{3}, \quad 2 \frac{1}{2} = \frac{5}{2}.[/tex]

So,
[tex]= \frac{13}{3} \times \frac{5}{2} = \frac{65}{6} = 10 \frac{5}{6}.[/tex]

  • Second calculation:
    [tex]5 \frac{1}{2} = \frac{11}{2}.[/tex]

So,
[tex]6 \times \frac{11}{2} = \frac{66}{2} = 33.[/tex]

Thus,
[tex]10 \frac{5}{6} + 33 = 43 \frac{5}{6}.[/tex]

(i) $3 \frac{4}{5} \times 6 \frac{1}{7}$

Convert to improper fractions:
[tex]3 \frac{4}{5} = \frac{19}{5}, \quad 6 \frac{1}{7} = \frac{43}{7}.[/tex]

So,
[tex]\frac{19}{5} \times \frac{43}{7} = \frac{817}{35}.[/tex]

(j) $5 \frac{1}{3} \times 4 \frac{1}{2} + 3 \frac{1}{4} \times 1 \frac{5}{6}$

  • First calculation:
    [tex]5 \frac{1}{3} = \frac{16}{3}, \quad 4 \frac{1}{2} = \frac{9}{2}.[/tex]

So,
[tex]\frac{16}{3} \times \frac{9}{2} = \frac{144}{6} = 24.[/tex]

  • Second calculation:
    [tex]3 \frac{1}{4} = \frac{13}{4}, \quad 1 \frac{5}{6} = \frac{11}{6}.[/tex]

So,
[tex]\frac{13}{4} \times \frac{11}{6} = \frac{143}{24}.[/tex]

Adding the two results:
[tex]24 + \frac{143}{24} = \frac{576}{24} + \frac{143}{24} = \frac{719}{24} = 29 \frac{23}{24}.[/tex]

(k) [tex]\frac{5}{9} \text{ of } \frac{12}{5}[/tex]

[tex]= \frac{5}{9} \times \frac{12}{5} = \frac{60}{45} = \frac{4}{3}.[/tex]

(l) [tex]\frac{2}{5} \text{ of } \frac{9}{16} \text{ of } \frac{12}{5}[/tex]

[tex]= \frac{2}{5} \times \frac{9}{16} \times \frac{12}{5} = \frac{216}{400} = \frac{27}{50}.[/tex]

(m) [tex]\frac{3}{5} \times \frac{25}{9}[/tex]

[tex]= \frac{3 \times 25}{5 \times 9} = \frac{75}{45} = \frac{5}{3}.[/tex]

(n) [tex]\frac{2}{5} \text{ of } (\frac{5}{4} - \frac{1}{4})[/tex]

Solve inside the parentheses first:

[tex]\frac{5}{4} - \frac{1}{4} = \frac{4}{4} = 1.[/tex]

Therefore,

[tex]\frac{2}{5} \times 1 = \frac{2}{5}.[/tex]

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