CBSE Class 7 Mathematics Fractions and Decimals Assignment Set 06

VII - Mathematics Assignment No. 05 - Fractions and Decimals

Question 1. The sum of two fractions is \(\frac{7}{12}\). If one of the fraction is \(\frac{1}{3}\), find the other fraction.
Answer: Let the other fraction be \(x\).
According to the question:
\(x + \frac{1}{3} = \frac{7}{12}\)
\(x = \frac{7}{12} - \frac{1}{3}\)
LCM of 12 and 3 is 12.
\(x = \frac{7}{12} - \frac{1 \times 4}{3 \times 4}\)
\(x = \frac{7}{12} - \frac{4}{12}\)
\(x = \frac{7 - 4}{12} = \frac{3}{12} = \frac{1}{4}\)
In simple words: To find the missing fraction, subtract the given fraction from the total sum.

Exam Tip: Always reduce the final fraction to its simplest form (lowest terms).

Question 2. What fraction should be added to \(\frac{7}{8}\) so as to get \(1\frac{2}{3}\)?
Answer: Let the required fraction to be added be \(x\).
First, convert the mixed fraction into an improper fraction:
\(1\frac{2}{3} = \frac{1 \times 3 + 2}{3} = \frac{5}{3}\)
According to the question:
\(\frac{7}{8} + x = \frac{5}{3}\)
\(x = \frac{5}{3} - \frac{7}{8}\)
LCM of 3 and 8 is 24.
\(x = \frac{5 \times 8}{3 \times 8} - \frac{7 \times 3}{8 \times 3}\)
\(x = \frac{40}{24} - \frac{21}{24}\)
\(x = \frac{40 - 21}{24} = \frac{19}{24}\)
In simple words: Subtract the starting fraction from the target fraction to find the difference needed.

Exam Tip: Always convert mixed fractions to improper fractions before performing addition or subtraction.

Question 3. What fraction should be subtracted from \(1\frac{1}{3}\) to get \(\frac{4}{5}\)?
Answer: Let the required fraction to be subtracted be \(x\).
Convert the mixed fraction into an improper fraction:
\(1\frac{1}{3} = \frac{1 \times 3 + 1}{3} = \frac{4}{3}\)
According to the question:
\(\frac{4}{3} - x = \frac{4}{5}\)
\(x = \frac{4}{3} - \frac{4}{5}\)
LCM of 3 and 5 is 15.
\(x = \frac{4 \times 5}{3 \times 5} - \frac{4 \times 3}{5 \times 3}\)
\(x = \frac{20}{15} - \frac{15}{15} \rightarrow \frac{20}{15} - \frac{12}{15}\)
\(x = \frac{20 - 12}{15} = \frac{8}{15}\)
In simple words: Rearrange the equation so that you subtract the target fraction from the original fraction.

Exam Tip: Be careful with the order of subtraction when dealing with "subtracted from".

Question 4. Find the sum of \(\frac{2}{3}\) and its reciprocal.
Answer: The given fraction is \(\frac{2}{3}\).
The reciprocal of \(\frac{2}{3}\) is \(\frac{3}{2}\).
Now, find the sum of both fractions:
\(\text{Sum} = \frac{2}{3} + \frac{3}{2}\)
LCM of 3 and 2 is 6.
\(\text{Sum} = \frac{2 \times 2}{3 \times 2} + \frac{3 \times 3}{2 \times 3}\)
\(\text{Sum} = \frac{4}{6} + \frac{9}{6}\)
\(\text{Sum} = \frac{4 + 9}{6} = \frac{13}{6}\)
In simple words: Flip the fraction upside down to get its reciprocal, then add it to the original fraction using a common denominator.

Exam Tip: The reciprocal of a fraction \(\frac{a}{b}\) is simply \(\frac{b}{a}\).

Question 5. Subtract the sum of \(\frac{36}{11}\) and \(\frac{49}{22}\) from the sum of \(\frac{33}{8}\) and \(\frac{19}{4}\).
Answer: First, find the sum of \(\frac{36}{11}\) and \(\frac{49}{22}\):
LCM of 11 and 22 is 22.
\(\text{First Sum} = \frac{36 \times 2}{11 \times 2} + \frac{49}{22}\)
\(\text{First Sum} = \frac{72}{22} + \frac{49}{22} = \frac{72 + 49}{22} = \frac{121}{22} = \frac{11}{2}\)
Next, find the sum of \(\frac{33}{8}\) and \(\frac{19}{4}\):
LCM of 8 and 4 is 8.
\(\text{Second Sum} = \frac{33}{8} + \frac{19 \times 2}{4 \times 2}\)
\(\text{Second Sum} = \frac{33}{8} + \frac{38}{8} = \frac{33 + 38}{8} = \frac{71}{8}\)
Now, subtract the first sum from the second sum:
\(\text{Result} = \frac{71}{8} - \frac{11}{2}\)
LCM of 8 and 2 is 8.
\(\text{Result} = \frac{71}{8} - \frac{11 \times 4}{2 \times 4}\)
\(\text{Result} = \frac{71}{8} - \frac{44}{8} = \frac{71 - 44}{8} = \frac{27}{8} = 3\frac{3}{8}\)
In simple words: Calculate both sums separately, then subtract the first total from the second total.

Exam Tip: Simplify intermediate fractions if possible, or keep a common denominator to make subtraction easier.

Question 6. Sahil weighs \(53\frac{1}{2} \text{ kg}\). What will be his weight after losing \(8\frac{2}{3} \text{ kg}\)?
Answer: Initial weight of Sahil = \(53\frac{1}{2} \text{ kg} = \frac{107}{2} \text{ kg}\)
Weight lost = \(8\frac{2}{3} \text{ kg} = \frac{26}{3} \text{ kg}\)
New weight = Initial weight - Weight lost
\(\text{New Weight} = \frac{107}{2} - \frac{26}{3}\)
LCM of 2 and 3 is 6.
\(\text{New Weight} = \frac{107 \times 3}{2 \times 3} - \frac{26 \times 2}{3 \times 2}\)
\(\text{New Weight} = \frac{321}{6} - \frac{52}{6}\)
\(\text{New Weight} = \frac{321 - 52}{6} = \frac{269}{6} \text{ kg}\)
Converting to mixed fraction:
\(\frac{269}{6} = 44\frac{5}{6} \text{ kg}\)
In simple words: Subtract the weight lost from the original weight after converting mixed fractions into improper fractions.

Exam Tip: For word problems involving measurements, always include the proper unit (\(\text{kg}\)) in your final answer.