CBSE Class 6 Mathematics Fractions Practice Paper Assignment Set 13 with Answers

Practice Paper 21 for CBSE Class 6 Mathematics Ganita Prakash Chapter 7 Fractions has 20 questions worth 40 marks, set out in Sections A to E, with a clear step-by-step solution for each one. The questions cover fractional units of a strip and of a rectangle made of 15 squares, representing 2/7 and 3/5 on a number line, converting improper fractions such as 35/6 and 17/3 into mixed fractions, simplest form of 96/120, checking equivalence of 4/10 and 6/15, and adding and subtracting fractions with denominators like 9, 12 and 18, ending with a chart-work case study and an orange seller comparing 3/4 kg and 5/6 kg.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 11. Represent \( \frac{2}{7} \) on a number line.
Answer:
Draw a number line and mark 0 and 1.
Divide the gap from 0 to 1 into 7 equal parts; each part is \( \frac{1}{7} \).
Count 2 parts from 0. That point (red) is \( \frac{2}{7} \).

In simple words: Cut 0 to 1 into 7 equal pieces and stop after the 2nd piece.

Teacher's Note:
1. Equal parts are essential; measure them with a ruler.
2. Denominator = number of parts, numerator = parts counted.

Question 12. Convert \( \frac{17}{3} \) into a mixed fraction.
Answer:
Divide 17 by 3: 17 = 3 × 5 + 2, so the quotient is 5 and the remainder is 2.
\( \frac{17}{3} \) = \( 5\frac{2}{3} \).
In simple words: 17 thirds make 5 wholes (15 thirds) with 2 thirds left.

Teacher's Note:
1. Check: 5 × 3 + 2 = 17.
2. The denominator stays 3.

Question 13. Add: \( \frac{3}{8} \) + \( \frac{1}{4} \)
Answer:
Make the denominators the same: \( \frac{1}{4} \) = \( \frac{1 \times 2}{4 \times 2} \) = \( \frac{2}{8} \).
\( \frac{3}{8} \) + \( \frac{2}{8} \) = \( \frac{5}{8} \).
In simple words: Change quarters into eighths and add the eighths.

Teacher's Note:
1. 8 is a multiple of 4, so 8 is the common denominator.
2. 5 and 8 have no common factor, so \( \frac{5}{8} \) is final.

Question 14. Reduce \( \frac{96}{120} \) to simplest form.
Answer:
Find the HCF: 96 = 2⁵ × 3 and 120 = 2³ × 3 × 5, so HCF = 2³ × 3 = 24.
\( \frac{96 \div 24}{120 \div 24} \) = \( \frac{4}{5} \).
So the simplest form is \( \frac{4}{5} \).
In simple words: Divide top and bottom by 24, the largest number dividing both.

Teacher's Note:
1. Step by step also works: divide by 2 three times, then by 3.
2. 4 and 5 have no common factor except 1.

Section C: Short Answer Questions (3 Marks Each)

Question 15. Check if \( \frac{4}{10} \) and \( \frac{6}{15} \) are equivalent.
Answer:
Reduce each to lowest terms.
\( \frac{4}{10} \) = \( \frac{4 \div 2}{10 \div 2} \) = \( \frac{2}{5} \).
\( \frac{6}{15} \) = \( \frac{6 \div 3}{15 \div 3} \) = \( \frac{2}{5} \).
Both equal \( \frac{2}{5} \), so yes, \( \frac{4}{10} \) and \( \frac{6}{15} \) are equivalent.
Cross-check: 4 × 15 = 60 and 10 × 6 = 60 are equal.
In simple words: Both fractions simplify to two-fifths.

Teacher's Note:
1. Equivalent fractions have the same lowest form.
2. With denominator 30 both become \( \frac{12}{30} \).

Question 16. Add: \( \frac{4}{9} \) + \( \frac{5}{12} \) + \( \frac{1}{6} \)
Answer:
The LCM of 9, 12 and 6 is 36.
\( \frac{4}{9} \) = \( \frac{16}{36} \), \( \frac{5}{12} \) = \( \frac{15}{36} \), \( \frac{1}{6} \) = \( \frac{6}{36} \).
Sum = \( \frac{16 + 15 + 6}{36} \) = \( \frac{37}{36} \).
37 = 36 × 1 + 1, so the sum is \( 1\frac{1}{36} \) (= \( \frac{37}{36} \)).
In simple words: Change all three into thirty-sixths, add them, then write as a mixed fraction.

Teacher's Note:
1. LCM: 9 = 3², 12 = 2² × 3, 6 = 2 × 3, so LCM = 2² × 3² = 36.
2. The sum is just over 1 whole.

Question 17. Subtract: \( \frac{11}{12} \) − \( \frac{5}{18} \)
Answer:
The LCM of 12 and 18 is 36.
\( \frac{11}{12} \) = \( \frac{33}{36} \) and \( \frac{5}{18} \) = \( \frac{10}{36} \).
\( \frac{33}{36} \) − \( \frac{10}{36} \) = \( \frac{23}{36} \).
23 is prime and does not divide 36, so the answer is \( \frac{23}{36} \).
In simple words: Write both fractions in thirty-sixths and subtract the numerators.

Teacher's Note:
1. 12 = 2² × 3 and 18 = 2 × 3², so LCM = 2² × 3² = 36.
2. Keep the denominator; subtract only numerators.

Section D: Long Answer Question (5 Marks)

Question 18. A rectangle is divided into 15 equal squares.
(a) What is the fractional unit?
(b) Shade 9 squares; write the fraction.
(c) Write it in simplest form.
(d) Mark the simplest form on a number line.
(e) Is \( \frac{9}{15} \) equal to \( \frac{3}{5} \)? Justify.
Answer:
(a) The whole rectangle has 15 equal squares, so the fractional unit is \( \frac{1}{15} \).
(b) 9 squares shaded (the first 3 columns): fraction shaded = \( \frac{9}{15} \).

(c) HCF of 9 and 15 is 3: \( \frac{9 \div 3}{15 \div 3} \) = \( \frac{3}{5} \).
(d) Divide 0 to 1 into 5 equal parts and count 3 parts from 0:

(e) Yes. Dividing the numerator and denominator of \( \frac{9}{15} \) by 3 gives \( \frac{3}{5} \). The picture shows it too: the rectangle has 5 columns and 3 of them are shaded, so \( \frac{3}{5} \) of it is shaded.
In simple words: Nine of fifteen squares is the same as three of five columns, that is three-fifths.

Teacher's Note:
1. Shading whole columns makes the equivalence easy to see.
2. The fractional unit depends on how many equal parts the whole has.

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

Question 19. Case Study 1:
A group of students prepared a poster on “Ancient Indian Fractions”. They divided a large chart into equal parts and allocated work proportionately. One student (student A) completed 5 parts out of 12, while another (student B) completed 7 parts. They compared their contributions, checked for equivalent fractions, and tried converting both to lowest terms. They also checked whether the total work exceeded the full chart or not.
Based on the above, answer the following questions:
(a) Fraction of work by student A.
(b) Fraction of work by student B.
(c) Who worked more?
(d) Write one equivalent fraction of both.
Answer:
(a) Student A did 5 of the 12 equal parts: \( \frac{5}{12} \).
(b) Student B did 7 of the 12 equal parts: \( \frac{7}{12} \).
(c) Same denominator and 7 > 5, so \( \frac{7}{12} \) > \( \frac{5}{12} \). Student B worked more.
(d) Multiply top and bottom by 2: \( \frac{5}{12} \) = \( \frac{10}{24} \) and \( \frac{7}{12} \) = \( \frac{14}{24} \).
Both \( \frac{5}{12} \) and \( \frac{7}{12} \) are already in lowest terms. Together \( \frac{5}{12} \) + \( \frac{7}{12} \) = \( \frac{12}{12} \) = 1, so the total work is exactly the full chart and does not exceed it.
In simple words: A did five-twelfths and B did seven-twelfths, so B did more and together they finished the chart.

Teacher's Note:
1. Like fractions are compared by their numerators.
2. 5 and 12, and 7 and 12, have no common factor, so these fractions cannot be reduced.

Question 20. Case Study 2:
A fruit vendor divides 4 kg of oranges into equal packets of fractional weights. Later, he mixes the packets and sells \( \frac{3}{4} \) kg and \( \frac{5}{6} \) kg to two customers. The customers discuss their purchases using mixed fractions, equivalent fractions, and compare their quantities. They want to identify who purchased more and by how much.
Based on the above, answer the following questions:
(a) Convert \( \frac{5}{6} \) into an equivalent fraction with denominator 12.
(b) Convert \( \frac{3}{4} \) into an equivalent fraction with denominator 12.
(c) Who purchased more?
(d) By how much?
Answer:
(a) \( \frac{5}{6} \) = \( \frac{5 \times 2}{6 \times 2} \) = \( \frac{10}{12} \).
(b) \( \frac{3}{4} \) = \( \frac{3 \times 3}{4 \times 3} \) = \( \frac{9}{12} \).
(c) \( \frac{10}{12} \) > \( \frac{9}{12} \), so the customer who bought \( \frac{5}{6} \) kg purchased more.
(d) Difference = \( \frac{10}{12} \) − \( \frac{9}{12} \) = \( \frac{1}{12} \) kg (about 83 g).
In simple words: Five-sixths kg is ten-twelfths and three-quarters kg is nine-twelfths, so the difference is one-twelfth of a kg.

Teacher's Note:
1. 12 is the LCM of 4 and 6, so it is the best common denominator.
2. \( \frac{1}{12} \) kg = 1000 ÷ 12 g, which is about 83 g.