CBSE Class 7 Mathematics Fractions Assignment Set 05

FRACTIONS

Q1. Shalu studies for \(5 \frac{2}{3}\) daily. She devotes \(2 \frac{4}{5}\) hours of her time for science and mathematics. How much time does she devote for other subjects?
Answer: Total study time = \(5 \frac{2}{3} \text{ hours} = \frac{17}{3} \text{ hours}\)
Time spent on science and mathematics = \(2 \frac{4}{5} \text{ hours} = \frac{14}{5} \text{ hours}\)
Time devoted for other subjects = Total time - (Time for science and mathematics)
\(= \frac{17}{3} - \frac{14}{5}\)
LCM of 3 and 5 is 15
\(= \frac{17 \times 5}{3 \times 5} - \frac{14 \times 3}{5 \times 3}\)
\(= \frac{85}{15} - \frac{42}{15}\)
\(= \frac{43}{15} \text{ hours} = 2 \frac{13}{15} \text{ hours}\)
In simple words: Subtract the time spent on science and math from Shalu's total daily study time to find the remaining hours for other subjects.

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

Q2. What should be added to \(5 \frac{3}{7}\) to get 12?
Answer: Let the required number to be added be \(x\).
\(5 \frac{3}{7} + x = 12\)
Convert mixed fraction into improper fraction:
\(5 \frac{3}{7} = \frac{38}{7}\)
\(\frac{38}{7} + x = 12\)
\(x = 12 - \frac{38}{7}\)
\(x = \frac{12 \times 7}{7} - \frac{38}{7}\)
\(x = \frac{84}{7} - \frac{38}{7}\)
\(x = \frac{46}{7} = 6 \frac{4}{7}\)
In simple words: To find what number must be added to a given number to reach a target sum, subtract the given number from the target sum.

Exam Tip: Treat whole numbers like 12 as \(\frac{12}{1}\) when finding a common denominator for fraction subtraction.

Q3. A piece of wire is of length \(12 \frac{3}{4} \text{ m}\). If it cuts into two pieces in such a way that the length of one piece is \(5 \frac{1}{4} \text{ m}\), find out the length of the other piece?
Answer: Total length of the wire = \(12 \frac{3}{4} \text{ m} = \frac{51}{4} \text{ m}\)
Length of the first piece = \(5 \frac{1}{4} \text{ m} = \frac{21}{4} \text{ m}\)
Length of the other piece = Total length - Length of the first piece
\(= \frac{51}{4} - \frac{21}{4}\)
\(= \frac{51 - 21}{4}\)
\(= \frac{30}{4}\)
Simplify the fraction by dividing numerator and denominator by 2:
\(= \frac{15}{2} \text{ m} = 7 \frac{1}{2} \text{ m}\)
In simple words: Subtract the length of the known piece from the total length of the wire to find the remaining length.

Exam Tip: When fractions already share a common denominator, you can subtract their numerators directly without finding a new LCM.

Q4. A rectangular sheet of paper is \(12 \frac{1}{2} \text{ cm}\) long and \(10 \frac{2}{3} \text{ m}\) wide. Find its Perimeter.
Answer: Length (\(l\)) = \(12 \frac{1}{2} \text{ cm} = \frac{25}{2} \text{ cm}\)
Width (\(b\)) = \(10 \frac{2}{3} \text{ cm} = \frac{32}{3} \text{ cm}\)
Perimeter of a rectangle = \(2 \times (l + b)\)
\(= 2 \times \left( \frac{25}{2} + \frac{32}{3} \right)\)
Find the LCM of 2 and 3, which is 6:
\(= 2 \times \left( \frac{25 \times 3}{2 \times 3} + \frac{32 \times 2}{3 \times 2} \right)\)
\(= 2 \times \left( \frac{75}{6} + \frac{64}{6} \right)\)
\(= 2 \times \left( \frac{139}{6} \right)\)
\(= \frac{139}{3} \text{ cm} = 46 \frac{1}{3} \text{ cm}\)
In simple words: Add the length and width together, multiply the sum by 2, and simplify to get the perimeter.

Exam Tip: Remember that perimeter formula requires multiplying the sum of length and breadth by 2. Do not forget to simplify fractions at the end.

Q5. The cost of Add gel pen is Rs. \(20 \frac{1}{2}\) and that of uni ball is Rs. \(25 \frac{3}{4}\). Which is costlier and by how much?
Answer: Cost of Add gel pen = Rs. \(20 \frac{1}{2} = \frac{41}{2}\) = Rs. \(20.50\)
Cost of uni ball = Rs. \(25 \frac{3}{4} = \frac{103}{4}\) = Rs. \(25.75\)
Comparing the two costs, \(25.75 > 20.50\), so uni ball is costlier.
Difference in cost = Cost of uni ball - Cost of Add gel pen
\(= \frac{103}{4} - \frac{41}{2}\)
Make denominators equal (LCM of 4 and 2 is 4):
\(= \frac{103}{4} - \frac{41 \times 2}{2 \times 2}\)
\(= \frac{103}{4} - \frac{82}{4}\)
\(= \frac{103 - 82}{4}\)
\(= \frac{21}{4} = 5 \frac{1}{4}\)
Therefore, uni ball is costlier by Rs. \(5 \frac{1}{4}\) (or Rs. 5.25).
In simple words: Compare the fractional values by converting them to like fractions or decimals, then subtract the smaller price from the larger price to find the difference.

Exam Tip: To compare mixed numbers easily, convert them to improper fractions with a common denominator or convert them to decimals.