CBSE Class 7 Mathematics Working with Fractions Practice Paper Assignment Set 03 with Answers

Practice Paper 24 for Class 7 Maths Ganita Prakash Chapter 8 Working with Fractions is a 40-mark paper with 20 questions in five sections (A to E), and the solutions below explain each step. Here students practise the flour needed for part of a batch, how many short ropes fit in a long rope, deciding when a product is greater than 1, simplifying products such as 32/45 × 15/16, the reciprocal of 11 2/5 and why dividing by 8/9 makes a number bigger. Section E applies division and multiplication of fractions to packing almonds and finding the area of a garden.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 11. (a) Convert \( 15\frac{1}{6} \) into an improper fraction.
(b) Find its reciprocal.
Answer:
(a) \( 15\frac{1}{6} = \frac{15 \times 6 + 1}{6} = \frac{90 + 1}{6} \) = \( \frac{91}{6} \).
(b) Reciprocal = \( \frac{6}{91} \).
In simple words: 15 1/6 is 91 sixths, and its reciprocal is 6/91.

Teacher's Note:
1. Keep the same denominator when converting a mixed number.
2. Check: 91/6 × 6/91 = 1.

Question 12. (a) Simplify \( \frac{27}{28} \times 28 \)
(b) State the property used.
Answer:
(a) \( \frac{27}{28} \times 28 = \frac{27 \times 28}{28} \) = 27.
(b) \( \frac{27}{28} \times 28 = 27 \times (\frac{1}{28} \times 28) = 27 \times 1 = 27 \). A number times its reciprocal is 1, and 1 is the multiplicative identity (a × 1 = a).
So the properties used are “number × reciprocal = 1” and multiplicative identity.
In simple words: 28 twenty-eighths make one whole, so 27/28 of 28 is 27.

Teacher's Note:
1. Multiplying a fraction by its denominator leaves just the numerator.
2. The source mentions only the multiplicative identity; the reciprocal step should also be named.

Question 13. (a) Write expression to find number of equal parts.
(b) Evaluate: \( \frac{25}{26} \div \frac{1}{26} \)
Answer:
(a) Number of equal parts = Total quantity ÷ Size of one part.
(b) \( \frac{25}{26} \div \frac{1}{26} = \frac{25}{26} \times 26 \) = 25.
In simple words: 25 twenty-sixths contain 25 parts of size 1/26.

Teacher's Note:
1. Dividing by 1/26 is the same as multiplying by 26.
2. Check: 25 × 1/26 = 25/26.

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

Teacher's Note:
1. Look at the multiplier: more than 1 means bigger, less than 1 means smaller.
2. 16/15 is even greater than 1.

Section C: Short Answer Questions (3 Marks Each)

Question 15. A rope of length \( \frac{30}{31} \) m is cut into pieces of length \( \frac{1}{31} \) m.
(a) Operation used
(b) Number of pieces
(c) Explanation
Answer:
(a) Division.
(b) \( \frac{30}{31} \div \frac{1}{31} = \frac{30}{31} \times 31 \) = 30 pieces.
(c) \( \frac{30}{31} \) m is made of 30 parts of \( \frac{1}{31} \) m each, so exactly 30 pieces can be cut with nothing left over.
In simple words: Thirty 1/31 m pieces make up a 30/31 m rope.

Teacher's Note:
1. “Cut into pieces of length …” tells us to divide.
2. Check: 30 × 1/31 = 30/31 m.

Question 16. Find area of rectangle: sides \( 8\frac{1}{7} \) m and \( \frac{7}{57} \) m.
(a) Convert mixed fraction
(b) Multiply
(c) Area
Answer:
(a) \( 8\frac{1}{7} = \frac{8 \times 7 + 1}{7} = \frac{57}{7} \).
(b) \( \frac{57}{7} \times \frac{7}{57} = \frac{399}{399} \) = 1.
(c) Area = 1 m².
In simple words: 57/7 and 7/57 are reciprocals, so the area is 1 square metre.

Teacher's Note:
1. Area of a rectangle = length × breadth.
2. When two sides are reciprocals of each other, the area is always 1 square unit.

Question 17. Show multiplying a fraction by its reciprocal = 1.
(a) Choose fraction
(b) Multiply
(c) Result
Answer:
(a) Take \( \frac{13}{18} \); its reciprocal is \( \frac{18}{13} \).
(b) \( \frac{13}{18} \times \frac{18}{13} = \frac{13 \times 18}{18 \times 13} = \frac{234}{234} \).
(c) The product is 1.
In simple words: Any fraction times its flipped form equals 1.

Teacher's Note:
1. The numerator and denominator of the product are the same numbers multiplied in a different order.
2. Zero has no reciprocal, because nothing times 0 gives 1.

Section D: Long Answer Question (5 Marks)

Question 18. A worker completes \( \frac{19}{40} \) of a task in one hour.
(a) Work done in 5 hours
(b) Work done in \( \frac{7}{8} \) hour
(c) Time required to complete full work
(d) Operations involved
(e) Conclusion
Answer:
(a) Work in 5 hours = \( 5 \times \frac{19}{40} = \frac{95}{40} \) = \( \frac{19}{8} \) = \( 2\frac{3}{8} \) tasks (if he goes on doing similar tasks).
(b) Work in \( \frac{7}{8} \) hour = \( \frac{7}{8} \times \frac{19}{40} \) = \( \frac{133}{320} \) of the task.
(c) Time for the whole task = \( 1 \div \frac{19}{40} = \frac{40}{19} \) = \( 2\frac{2}{19} \) hours (just over 2 hours).
(d) Multiplication of fractions for (a) and (b); division of fractions for (c).
(e) The worker needs \( 2\frac{2}{19} \) hours to finish one task; in 5 hours he can do the work of \( 2\frac{3}{8} \) such tasks.
In simple words: Multiply 19/40 by the time worked, and divide 1 by 19/40 to find the time for the whole task.

Teacher's Note:
1. 133/320 cannot be simplified: 133 = 7 × 19 and 320 = 2⁶ × 5 share no factor.
2. Time for one complete job = reciprocal of the part done in one hour.
3. Check (c): 40/19 × 19/40 = 1.

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

Question 19. Case Study 1: Packing Dry Fruits
A dry fruit company produces \( \frac{6}{7} \) kg of almonds daily. The almonds are packed into small packets, each weighing \( \frac{1}{21} \) kg. The supervisor calculates how many full packets can be prepared to avoid wastage. This scenario demonstrates division of fractions in real-life application where a total quantity is divided into equal fractional units.
Answer the following questions:
(a) Write the expression
(b) Find number of packets
(c) Reciprocal used
(d) Operation name
Answer:
(a) Number of packets = \( \frac{6}{7} \div \frac{1}{21} \).
(b) \( \frac{6}{7} \times 21 = \frac{126}{7} \) = 18 packets.
(c) Reciprocal of \( \frac{1}{21} \) = 21.
(d) Division of fractions.
In simple words: 6/7 kg of almonds packed 1/21 kg at a time makes 18 packets.

Teacher's Note:
1. Cancel 21 with 7 to get 6 × 3 = 18.
2. Check: 18 × 1/21 = 18/21 = 6/7 kg.

Question 20. Case Study 2: Calculating Garden Area
A rectangular garden has a length of \( \frac{23}{24} \) m and a width of \( \frac{12}{23} \) m. To plan planting, the gardener calculates the area. This demonstrates multiplication of fractions to find area with fractional dimensions in real-life scenarios.
Answer the following questions:
(a) Area formula
(b) Substitute values
(c) Find area
(d) Concept applied
Answer:
(a) Area = length × breadth.
(b) Area = \( \frac{23}{24} \times \frac{12}{23} \).
(c) Cancel 23 with 23 and 12 with 24: Area = \( \frac{12}{24} \) = \( \frac{1}{2} \) m².
(d) Multiplication of fractions.
In simple words: 23/24 × 12/23 = 1/2, so the garden is half a square metre.

Teacher's Note:
1. Cancel common factors across numerators and denominators before multiplying.
2. Area answers need square units.