Solutions and Explanations to Grade 7 Fractions and Mixed Numbers Questions
Detailed solutions and full explanations to fractions and mixed numbers grade 7 problems are presented.
Note: Do not use the calculator to solve the questions below.
Solutions
-
Find fraction \(F\) with denominator less than \(8\) such that
\[
\frac{2}{8} + F = 1
\]
Solution
Solve for \(F\):
\[
F = 1 - \frac{2}{8}
\]
\[
= \frac{8}{8} - \frac{2}{8} \quad \text{(common denominator)}
\]
\[
= \frac{6}{8} \quad \text{(subtract numerators)}
\]
\[
= \frac{3}{4} \quad \text{(reduce fraction)}
\]
Find two fractions \(F_1\) and \(F_2\) with same denominator equal to \(6\) such that
\[
F_1 + F_2 = 1 \quad \text{and} \quad F_1 - F_2 = \frac{2}{3}
\]
Solution
Write them with denominator \(6\):
\[
F_1 + F_2 = 1 = \frac{6}{6}
\]
\[
F_1 - F_2 = \frac{2}{3} = \frac{4}{6}
\]
Their numerators must add to \(6\) and differ by \(4\), so:
\[
F_1 = \frac{5}{6}, \quad F_2 = \frac{1}{6}
\]
Which fraction is equivalent to \(16\%\)?
Solution
\[
16\% = \frac{16}{100} = \frac{4}{25}
\]
Which fraction is equivalent to \(\frac{300}{1000}\)?
Solution
\[
\frac{300}{1000} = \frac{3}{10} \quad \text{(divide numerator and denominator by 100)}
\]
\[
\frac{1}{2} + \frac{1}{5} + \frac{1}{6}
\]
Solution
LCM of \(2, 5, 6\):
\[
2 = 2, \quad 5 = 5, \quad 6 = 2 \times 3
\]
\[
\text{LCM} = 2 \times 5 \times 3 = 30
\]
\[
\frac{1}{2} + \frac{1}{5} + \frac{1}{6} =
\frac{15}{30} + \frac{6}{30} + \frac{5}{30}
\]
\[
= \frac{26}{30} = \frac{13}{15}
\]
\[
3\frac{3}{5} + 5\frac{1}{2}
\]
Solution
\[
(3 + 5) + \left( \frac{3}{5} + \frac{1}{2} \right)
= 8 + \left( \frac{6}{10} + \frac{5}{10} \right)
= 8 + \frac{11}{10}
\]
\[
= 8 + 1 + \frac{1}{10} = 9\frac{1}{10}
\]
\[
\frac{1}{7} \times 2\frac{2}{5}
\]
Solution
\[
\frac{1}{7} \times \frac{12}{5} = \frac{12}{35}
\]
\[
\frac{1}{12} \times 0.2
\]
Solution
\[
0.2 = \frac{1}{5}
\]
\[
\frac{1}{12} \times \frac{1}{5} = \frac{1}{60}
\]
\[
\frac{2}{5} \div 6
\]
Solution
\[
\frac{2}{5} \times \frac{1}{6} = \frac{2}{30} = \frac{1}{15}
\]
\[
\frac{9}{7} + 2
\]
Solution
\[
\frac{9}{7} + \frac{14}{7} = \frac{23}{7} = 3\frac{2}{7}
\]
\[
2\frac{1}{3} + \frac{4}{2}
\]
Solution
\[
\frac{4}{2} = 2
\]
\[
2\frac{1}{3} + 2 = 4\frac{1}{3}
\]
\[
3\frac{1}{5} \div 5
\]
Solution
\[
\frac{16}{5} \div \frac{5}{1} = \frac{16}{5} \times \frac{1}{5} = \frac{16}{25}
\]
\[
\frac{1}{2} + 4\frac{1}{3} - 3\frac{2}{5}
\]
Solution
\[
4 - 3 + \frac{1}{2} + \frac{1}{3} - \frac{2}{5}
\]
\[
\text{LCM}(2,3,5) = 30
\]
\[
= 1 + \left( \frac{15}{30} + \frac{10}{30} - \frac{12}{30} \right)
= 1 + \frac{13}{30} = 1\frac{13}{30}
\]
\[
\frac{5}{2} \div \frac{7}{2} - \frac{1}{5}
\]
Solution
\[
\frac{5}{2} \times \frac{2}{7} - \frac{1}{5}
= \frac{5}{7} - \frac{1}{5}
\]
\[
= \frac{25}{35} - \frac{7}{35} = \frac{18}{35}
\]
\[
(0.2 + \frac{1}{5}) \times \frac{2}{7}
\]
Solution
\[
0.2 = \frac{1}{5}
\]
\[
\left( \frac{1}{5} + \frac{1}{5} \right) \times \frac{2}{7}
= \frac{2}{5} \times \frac{2}{7} = \frac{4}{35}
\]
\[
\left( 3\frac{1}{2} + \frac{3}{5} \right) \times \frac{1}{7}
\]
Solution
\[
\frac{7}{2} + \frac{3}{5} = \frac{35}{10} + \frac{6}{10} = \frac{41}{10}
\]
\[
\frac{41}{10} \times \frac{1}{7} = \frac{41}{70}
\]
\[
\frac{40}{4000}
\]
Solution
\[
\frac{40}{4000} = \frac{1}{100} = 1\%
\]
\[
\left(\frac{1}{2} + \frac{2}{3}\right) \div 0.2
\]
Solution
First, add the fractions within the parentheses and convert the decimal $0.2$ to a fraction.
\[ \left(\frac{1}{2} + \frac{2}{3}\right) \div 0.2 = \left(\frac{3}{6} + \frac{4}{6}\right) \div \frac{1}{5}
= \frac{7}{6} \div \frac{1}{5} \]
Next, to divide by a fraction, multiply by its reciprocal.
\[
\frac{7}{6} \times \frac{5}{1} = \frac{35}{6} = 5\frac{5}{6}
\]
Order from least to greatest: \[
3\frac{4}{7}, 3\frac{3}{5}, 3\frac{1}{2}, 3\frac{11}{20}
\]
Solution
To compare the numbers, we find a common denominator for the fractional parts. The Least Common Multiple (LCM) of 7, 5, 2, and 20 is 140. We rewrite each number with this common denominator.
\[ 3\frac{4}{7} = 3\frac{80}{140} \]
\[ 3\frac{3}{5} = 3\frac{84}{140} \]
\[ 3\frac{1}{2} = 3\frac{70}{140} \]
\[ 3\frac{11}{20} = 3\frac{77}{140} \]
By comparing the numerators of the fractional parts (\(70 \lt 77 \lt 80 \lt 84\)), we can order the numbers.
The final order is: \[ 3\frac{1}{2}, 3\frac{11}{20}, 3\frac{4}{7}, 3\frac{3}{5}\]
Order from least to greatest: \[
2\frac{7}{8}, 2.66, 262\%, \frac{25}{8}
\]
Solution
To compare these numbers, we convert them all to decimal form.
\[ 2\frac{7}{8} = 2 + \frac{7}{8} = 2 + 0.875 = 2.875 \]
\[ 2.66 = 2.66 \]
\[ 262\% = \frac{262}{100} = 2.62 \]
\[ \frac{25}{8} = 3.125 \]
Comparing the decimal values (\(2.62 \lt 2.66 \lt 2.875 \lt 3.125\)), we can establish the order.
The final order is: \[ 262\%, 2.66, 2\frac{7}{8}, \frac{25}{8} \]
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