CBSE Class 7 Mathematics Working with Fractions Practice Paper Worksheet Set 02 with Answers

This Class 7 Maths Ganita Prakash Chapter 8 Working with Fractions Practice Paper 23 carries 40 marks and has 20 questions spread over Sections A to E, each followed by a worked answer. The paper focuses on multiplying fractions with larger numerators and denominators, cancelling common factors, the reciprocal of a mixed number such as 13 1/9, dividing by a proper fraction, and a fraction multiplied by its reciprocal, with real-life questions on a chef's sugar, cutting rods and ropes, a worker's daily task, packing cocoa powder and the area of a school lawn.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 11. (a) Convert \( 16\frac{2}{7} \) into an improper fraction.
(b) Find its reciprocal.
Answer:
(a) \( 16\frac{2}{7} = \frac{16 \times 7 + 2}{7} = \frac{112 + 2}{7} \) = \( \frac{114}{7} \).
(b) Reciprocal of \( \frac{114}{7} \) = \( \frac{7}{114} \).
In simple words: 16 2/7 is 114 sevenths, and its reciprocal is 7/114.

Teacher's Note:
1. Multiply the whole number by the denominator and add the numerator.
2. Check: 114/7 × 7/114 = 1.

Question 12. (a) Simplify \( \frac{31}{32} \times 32 \)
(b) State the property used.
Answer:
(a) \( \frac{31}{32} \times 32 = \frac{31 \times 32}{32} \) = 31.
(b) \( \frac{31}{32} \times 32 = 31 \times (\frac{1}{32} \times 32) = 31 \times 1 \): a number multiplied by its reciprocal is 1, and then 31 × 1 = 31 by the multiplicative identity property.
So the properties used are “number × reciprocal = 1” and multiplicative identity (a × 1 = a).
In simple words: 32 thirty-seconds make one whole, so 31/32 of 32 is 31.

Teacher's Note:
1. A fraction times its denominator gives back its numerator.
2. The source named only the multiplicative identity; the reciprocal step is what makes 32 and 1/32 disappear.

Question 13. (a) Write expression to find number of equal pieces.
(b) Evaluate: \( \frac{27}{28} \div \frac{1}{28} \)
Answer:
(a) Number of equal pieces = Total quantity ÷ Size of one piece.
(b) \( \frac{27}{28} \div \frac{1}{28} = \frac{27}{28} \times 28 \) = 27.
In simple words: Divide the total by one piece; 27 twenty-eighths contain 27 pieces of 1/28.

Teacher's Note:
1. The reciprocal of a unit fraction 1/n is the whole number n.
2. Check: 27 × 1/28 = 27/28.

Question 14. (a) Is \( \frac{15}{16} \times \frac{17}{15} \) greater than \( \frac{15}{16} \)?
(b) Give reason.
Answer:
(a) \( \frac{15}{16} \times \frac{17}{15} = \frac{17}{16} \) (the 15s cancel). Since \( \frac{17}{16} \gt \frac{15}{16} \), yes, it is greater.
(b) \( \frac{17}{15} \) is greater than 1, and multiplying by a number greater than 1 increases the value.
In simple words: 17/15 is more than 1, so it makes 15/16 bigger, giving 17/16.

Teacher's Note:
1. With the same denominator, the bigger numerator gives the bigger fraction.
2. Always compare the multiplier with 1.

Section C: Short Answer Questions (3 Marks Each)

Question 15. A rope of length \( \frac{32}{33} \) m is cut into pieces of length \( \frac{1}{33} \) m.
(a) Operation used
(b) Number of pieces
(c) Explanation
Answer:
(a) Division (we want how many equal pieces).
(b) \( \frac{32}{33} \div \frac{1}{33} = \frac{32}{33} \times 33 \) = 32 pieces.
(c) The rope is 32 parts of size \( \frac{1}{33} \) m, so the unit fraction \( \frac{1}{33} \) fits into it exactly 32 times.
In simple words: 32 thirty-thirds of a metre cut into 1/33 m pieces gives 32 pieces.

Teacher's Note:
1. Same denominators: divide the numerators, 32 ÷ 1 = 32.
2. Check: 32 × 1/33 = 32/33 m.

Question 16. Find area of rectangle: sides \( 9\frac{1}{5} \) m and \( \frac{5}{46} \) m.
(a) Convert mixed fraction
(b) Multiply
(c) Area
Answer:
(a) \( 9\frac{1}{5} = \frac{9 \times 5 + 1}{5} = \frac{46}{5} \).
(b) \( \frac{46}{5} \times \frac{5}{46} = \frac{230}{230} \) = 1.
(c) Area = 1 m².
In simple words: 46/5 and 5/46 are reciprocals, so the area is exactly 1 square metre.

Teacher's Note:
1. Convert mixed numbers before multiplying.
2. Noticing reciprocals means no long multiplication is needed.

Question 17. Show multiplying a fraction by its reciprocal = 1.
(a) Choose fraction
(b) Multiply
(c) Result
Answer:
(a) Take the fraction \( \frac{14}{17} \); its reciprocal is \( \frac{17}{14} \).
(b) \( \frac{14}{17} \times \frac{17}{14} = \frac{14 \times 17}{17 \times 14} = \frac{238}{238} \).
(c) The result is 1, since the numerator and the denominator are equal.
In simple words: A fraction and its upside-down form always multiply to 1.

Teacher's Note:
1. The numerator of one fraction cancels the denominator of the other.
2. This is why dividing by a fraction is the same as multiplying by its reciprocal.

Section D: Long Answer Question (5 Marks)

Question 18. A worker completes \( \frac{21}{40} \) of a task in one hour.
(a) Work done in 4 hours
(b) Work done in \( \frac{5}{6} \) hour
(c) Time required to complete full work
(d) Operations involved
(e) Conclusion
Answer:
(a) Work in 4 hours = \( 4 \times \frac{21}{40} = \frac{84}{40} \) = \( \frac{21}{10} \) = \( 2\frac{1}{10} \) tasks (that is, if he keeps working on similar tasks, he finishes 2 tasks and \( \frac{1}{10} \) of a third).
(b) Work in \( \frac{5}{6} \) hour = \( \frac{5}{6} \times \frac{21}{40} = \frac{105}{240} \) = \( \frac{7}{16} \) of the task (divide by 15).
(c) Time for the full task = \( 1 \div \frac{21}{40} = \frac{40}{21} \) = \( 1\frac{19}{21} \) hours (a little less than 2 hours).
(d) Multiplication of fractions in (a) and (b), and division of fractions in (c).
(e) The worker finishes one complete task in \( 1\frac{19}{21} \) hours, so in 4 hours he does more than 2 tasks.
In simple words: Multiply the hourly fraction by the time, and divide 1 by 21/40 to get the time for the whole task.

Teacher's Note:
1. Work done = rate × time; time for the whole job = 1 ÷ rate.
2. Always reduce answers like 105/240 to lowest terms (7/16).
3. Check (c): 40/21 × 21/40 = 1 whole task.

Section E: Case Study Based Questions (4 Marks Each)

Question 19. Case Study 1: Packing Cocoa Powder
A cocoa factory produces \( \frac{7}{8} \) kg of cocoa powder daily. It is packed into small sachets, each weighing \( \frac{1}{16} \) kg. The manager calculates how many full sachets can be filled to avoid wastage. This demonstrates division of fractions in real-life application where total quantity is divided into equal fractional units.
Answer the following questions:
(a) Write the expression
(b) Find number of sachets
(c) Reciprocal used
(d) Operation name
Answer:
(a) Number of sachets = \( \frac{7}{8} \div \frac{1}{16} \).
(b) \( \frac{7}{8} \times 16 = \frac{112}{8} \) = 14 sachets.
(c) The reciprocal of \( \frac{1}{16} \), which is 16.
(d) Division of fractions.
In simple words: 7/8 kg filled 1/16 kg at a time fills 14 sachets.

Teacher's Note:
1. Check: 14 × 1/16 = 14/16 = 7/8 kg, so nothing is wasted.
2. Cancel 16 with 8 to get 7 × 2 = 14 quickly.

Question 20. Case Study 2: Calculating School Lawn Area
A rectangular school lawn has a length of \( \frac{25}{26} \) m and a width of \( \frac{13}{25} \) m. To estimate total area, the gardener multiplies length and width. This demonstrates multiplication of fractions for fractional dimensions in real-life applications.
Answer the following questions:
(a) Area formula
(b) Substitute values
(c) Find area
(d) Concept applied
Answer:
(a) Area = length × breadth.
(b) Area = \( \frac{25}{26} \times \frac{13}{25} \).
(c) Cancel 25 with 25 and 13 with 26: \( \frac{1}{2} \times \frac{1}{1} \). Area = \( \frac{1}{2} \) m².
(d) Multiplication of fractions.
In simple words: Multiplying 25/26 by 13/25 gives half a square metre.

Teacher's Note:
1. 26 = 2 × 13, so 13/26 = 1/2.
2. Write the unit m² with every area.