CBSE Class 7 Mathematics Working with Fractions Practice Paper Assignment Set 04 with Answers

Here is Class 7 Maths Ganita Prakash Chapter 8 Working with Fractions Practice Paper 26, a 40-mark assessment of 20 questions in Sections A to E, solved step by step for self-study and revision. Its questions deal with fertiliser for part of a plant, cutting ribbon strips, a product that is smaller than both its factors, simplifying 20/27 × 9/10, the reciprocal of 8 3/7, dividing by 3/4 and multiplying a fraction by its reciprocal, along with case studies on filling honey bottles and the floor area of a school hall.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 11. (a) Convert \( 12\frac{4}{9} \) into an improper fraction.
(b) Find its reciprocal.
Answer:
(a) \( 12\frac{4}{9} = \frac{12 \times 9 + 4}{9} = \frac{108 + 4}{9} \) = \( \frac{112}{9} \).
(b) Reciprocal = \( \frac{9}{112} \).
In simple words: 12 4/9 is 112 ninths; its reciprocal is 9/112.

Teacher's Note:
1. 12 × 9 = 108, plus 4 is 112.
2. Check: 112/9 × 9/112 = 1.

Question 12. (a) Simplify \( \frac{21}{22} \times 22 \)
(b) State the property used.
Answer:
(a) \( \frac{21}{22} \times 22 = \frac{21 \times 22}{22} \) = 21.
(b) \( \frac{21}{22} \times 22 = 21 \times (\frac{1}{22} \times 22) = 21 \times 1 = 21 \). A number times its reciprocal is 1, and 21 × 1 = 21 by the multiplicative identity.
Properties used: “number × reciprocal = 1” and multiplicative identity.
In simple words: 22 twenty-seconds make a whole, so 21/22 of 22 is 21.

Teacher's Note:
1. A fraction multiplied by its denominator gives its numerator.
2. The source gives only “multiplicative identity”; the reciprocal step is also needed.

Question 13. (a) Write expression to find number of equal pieces.
(b) Evaluate: \( \frac{19}{20} \div \frac{1}{20} \)
Answer:
(a) Number of equal pieces = Total quantity ÷ Size of one piece.
(b) \( \frac{19}{20} \div \frac{1}{20} = \frac{19}{20} \times 20 \) = 19.
In simple words: 19/20 holds 19 pieces of 1/20.

Teacher's Note:
1. Dividing by 1/20 is the same as multiplying by 20.
2. Check: 19 × 1/20 = 19/20.

Question 14. (a) Is \( \frac{10}{11} \times \frac{13}{10} \) greater than \( \frac{10}{11} \)?
(b) Give reason.
Answer:
(a) \( \frac{10}{11} \times \frac{13}{10} = \frac{13}{11} \) (the 10s cancel). \( \frac{13}{11} \gt \frac{10}{11} \), so yes.
(b) \( \frac{13}{10} \) is greater than 1, and multiplying by a number greater than 1 increases the value.
In simple words: 13/10 is more than 1, so it makes 10/11 bigger.

Teacher's Note:
1. Multiplier greater than 1 → bigger result; less than 1 → smaller result.
2. Cancelling the 10s shows the answer immediately.

Section C: Short Answer Questions (3 Marks Each)

Question 15. A rope of length \( \frac{24}{25} \) m is cut into pieces of length \( \frac{1}{25} \) m.
(a) Operation used
(b) Number of pieces
(c) Explanation
Answer:
(a) Division.
(b) \( \frac{24}{25} \div \frac{1}{25} = \frac{24}{25} \times 25 \) = 24 pieces.
(c) \( \frac{24}{25} \) m is 24 parts of \( \frac{1}{25} \) m, so each \( \frac{1}{25} \) m piece fits 24 times and nothing is left over.
In simple words: A 24/25 m rope makes 24 pieces of 1/25 m.

Teacher's Note:
1. Total ÷ one piece gives the number of pieces.
2. Check: 24 × 1/25 = 24/25 m.

Question 16. Find area of rectangle: sides \( 3\frac{2}{7} \) m and \( \frac{7}{23} \) m.
(a) Convert mixed fraction
(b) Multiply
(c) Area
Answer:
(a) \( 3\frac{2}{7} = \frac{3 \times 7 + 2}{7} = \frac{23}{7} \).
(b) \( \frac{23}{7} \times \frac{7}{23} = \frac{161}{161} \) = 1.
(c) Area = 1 m².
In simple words: 23/7 and 7/23 are reciprocals, so the area is 1 square metre.

Teacher's Note:
1. Convert the mixed number first, then multiply.
2. A number times its reciprocal is always 1.

Question 17. Show that multiplying by reciprocal = 1.
(a) Choose fraction
(b) Multiply
(c) Result
Answer:
(a) Take \( \frac{9}{17} \); its reciprocal is \( \frac{17}{9} \).
(b) \( \frac{9}{17} \times \frac{17}{9} = \frac{9 \times 17}{17 \times 9} = \frac{153}{153} \).
(c) The product is 1.
In simple words: A fraction times its reciprocal always equals 1.

Teacher's Note:
1. The reciprocal is found by swapping the numerator and the denominator.
2. Every number except 0 has a reciprocal.

Section D: Long Answer Question (5 Marks)

Question 18. A worker completes \( \frac{13}{20} \) of a task in one hour.
(a) Work done in 4 hours
(b) Work done in \( \frac{3}{5} \) hour
(c) Time to complete full work
(d) Operations involved
(e) Conclusion
Answer:
(a) Work in 4 hours = \( 4 \times \frac{13}{20} = \frac{52}{20} \) = \( \frac{13}{5} \) = \( 2\frac{3}{5} \) tasks (if he keeps doing similar tasks).
(b) Work in \( \frac{3}{5} \) hour = \( \frac{3}{5} \times \frac{13}{20} \) = \( \frac{39}{100} \) of the task.
(c) Time for the full task = \( 1 \div \frac{13}{20} = \frac{20}{13} \) = \( 1\frac{7}{13} \) hours.
(d) Multiplication of fractions in (a) and (b); division of fractions in (c).
(e) The worker completes one task in \( 1\frac{7}{13} \) hours (less than 2 hours), so in 4 hours he can do the work of \( 2\frac{3}{5} \) such tasks.
In simple words: Multiply 13/20 by the hours to get work done; divide 1 by 13/20 to get the time for one task.

Teacher's Note:
1. Time for a whole job = 1 ÷ (part done in 1 hour).
2. 39/100 is already in lowest terms because 39 and 100 have no common factor.
3. Check (c): 20/13 × 13/20 = 1.

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

Question 19. Case Study 1: Packing Honey Bottles
A honey factory produces \( \frac{3}{5} \) kg of honey daily. The honey is poured into small bottles of \( \frac{1}{15} \) kg each. The supervisor calculates how many full bottles can be prepared so that no honey is wasted. This demonstrates division of fractions in a real-life scenario where a total quantity is divided into equal fractional units.
Answer the following questions:
(a) Write the expression
(b) Find number of bottles
(c) Reciprocal used
(d) Operation name
Answer:
(a) Number of bottles = \( \frac{3}{5} \div \frac{1}{15} \).
(b) \( \frac{3}{5} \times 15 = \frac{45}{5} \) = 9 bottles.
(c) Reciprocal of \( \frac{1}{15} \) = 15.
(d) Division of fractions.
In simple words: 3/5 kg of honey poured 1/15 kg at a time fills 9 bottles.

Teacher's Note:
1. Cancel 15 with 5: 3 × 3 = 9.
2. Check: 9 × 1/15 = 9/15 = 3/5 kg.

Question 20. Case Study 2: Measuring a School Hall
A rectangular school hall has a length of \( \frac{17}{18} \) m and a width of \( \frac{6}{17} \) m. To plan for painting the floor, the principal calculates the area. This demonstrates multiplication of fractions with fractional dimensions to find the area in real-life application.
Answer the following questions:
(a) Area formula
(b) Substitute values
(c) Find area
(d) Concept applied
Answer:
(a) Area = length × breadth.
(b) Area = \( \frac{17}{18} \times \frac{6}{17} \).
(c) Cancel 17 with 17: Area = \( \frac{6}{18} \) = \( \frac{1}{3} \) m².
(d) Multiplication of fractions.
In simple words: 17/18 × 6/17 = 1/3, so the floor area is one-third of a square metre.

Teacher's Note:
1. 6/18 reduces to 1/3 on dividing by 6.
2. Give area in square units (m²).