CBSE Class 6 Mathematics Fractions Assignment Set 11

Short Answer Questions

Question 1. One litre of milk is poured equally into 8 identical glasses.
(i) Write the fraction of a litre of milk that one glass holds.
(ii) Which is more, a share of \( \frac{1}{8} \) litre or a share of \( \frac{1}{12} \) litre? Give a reason in terms of equal shares.
Answer:
(i) 1 litre is shared equally by 8 glasses, so one glass holds \( \frac{1}{8} \) litre.
(ii) \( \frac{1}{8} \) litre is more. The same 1 litre shared among 8 glasses gives each glass more than when it is shared among 12 glasses.
More sharers means a smaller share, so \( \frac{1}{8} \) > \( \frac{1}{12} \).
In simple words: One litre in 8 glasses gives \( \frac{1}{8} \) litre each, which is more than \( \frac{1}{12} \) litre.

Teacher's Note:
1. Link the fraction to real sharing.
2. Smaller denominator, bigger fractional unit.

Question 2. Three chapatis are shared equally among 5 children in a hostel.
(i) Write the fraction of a chapati that each child gets.
(ii) Write the matching division fact for this sharing.
Answer:
(i) Each child gets \( \frac{3}{5} \) of a chapati.
(ii) Division fact: 3 ÷ 5 = \( \frac{3}{5} \).
Each share is less than one whole chapati, because 3 is smaller than 5.
In simple words: 3 chapatis among 5 children gives \( \frac{3}{5} \) chapati each, and 3 ÷ 5 = \( \frac{3}{5} \).

Teacher's Note:
1. Things shared go on top, sharers go below.
2. A fraction is also a division.

Question 3. A rectangular chikki is cut into 8 equal pieces. Anu picks up 3 of these pieces.
(i) Name the fractional unit of this chikki.
(ii) Write the fraction of the whole chikki that Anu has, and name its numerator and denominator.
Answer:
(i) The chikki is cut into 8 equal pieces, so the fractional unit is \( \frac{1}{8} \).
(ii) Anu has 3 such units, that is \( \frac{3}{8} \) of the chikki.
Numerator = 3 (how many units she has); denominator = 8 (it names the unit \( \frac{1}{8} \)).
In simple words: Each piece is \( \frac{1}{8} \), and 3 pieces make \( \frac{3}{8} \).

Teacher's Note:
1. Say what each number means, not just its name.
2. The denominator names the unit; the numerator counts it.

Question 4. A paper strip of length 1 unit is folded into 6 equal parts.
(i) Write \( \frac{5}{6} \) as a repeated addition of its fractional unit.
(ii) How many such fractional units make one whole strip? Write the result as a fraction equal to 1.
Answer:
(i) The fractional unit is \( \frac{1}{6} \), so
\( \frac{5}{6} \) = \( \frac{1}{6} \) + \( \frac{1}{6} \) + \( \frac{1}{6} \) + \( \frac{1}{6} \) + \( \frac{1}{6} \) = 5 times \( \frac{1}{6} \).
(ii) 6 such units make the whole strip: 6 × \( \frac{1}{6} \) = \( \frac{6}{6} \) = 1.
In simple words: \( \frac{5}{6} \) is five sixths added together, and six sixths make one whole.

Teacher's Note:
1. Count the \( \frac{1}{6} \) terms: there must be 5.
2. Same top and bottom means 1.

Question 5. On a number line, the length from 0 to 1 is divided into 6 equal parts.
(i) Write the fraction for the point that is 5 parts away from 0.
(ii) Write the fraction for the point that is exactly halfway between 0 and 1, using the same fractional unit.
Answer:
(i) Each part is \( \frac{1}{6} \) unit, so 5 parts from 0 gives \( \frac{5}{6} \).
(ii) Halfway is 3 parts out of 6, that is \( \frac{3}{6} \), which in simplest form is \( \frac{1}{2} \).
In simple words: Five sixths is 5 parts along; the middle is 3 sixths, which is one half.

Teacher's Note:
1. Half of 6 parts is 3 parts.
2. The question asks for sixths, so give \( \frac{3}{6} \) first.

Question 6. Convert as asked and show one line of working for each.
(i) Write \( \frac{17}{5} \) as a mixed fraction.
(ii) Write \( 5\frac{3}{8} \) as a fraction with a single numerator and denominator.
Answer:
(i) 17 = 3 × 5 + 2, so \( \frac{17}{5} \) = 3 + \( \frac{2}{5} \) = \( 3\frac{2}{5} \).
(ii) \( 5\frac{3}{8} \) = \( \frac{5\times 8+3}{8} \) = \( \frac{43}{8} \).
The 5 wholes give 5 × 8 = 40 eighths, and 3 more eighths make 43 eighths.
In simple words: Divide to get a mixed fraction; multiply and add to go back.

Teacher's Note:
1. Quotient = whole part, remainder = numerator.
2. Whole × denominator + numerator, over the same denominator.

Question 7. Answer the following about equivalent fractions.
(i) Write three fractions equivalent to \( \frac{3}{7} \).
(ii) Write \( \frac{40}{64} \) in its simplest form and say how you know it is in lowest terms.
Answer:
(i) Multiply the numerator and the denominator by 2, 3 and 4:
\( \frac{3}{7} \) = \( \frac{6}{14} \) = \( \frac{9}{21} \) = \( \frac{12}{28} \). So \( \frac{6}{14} \), \( \frac{9}{21} \), \( \frac{12}{28} \).
(ii) The HCF of 40 and 64 is 8, so \( \frac{40\div 8}{64\div 8} \) = \( \frac{5}{8} \).
5 and 8 have no common factor other than 1, so \( \frac{5}{8} \) is in lowest terms.
In simple words: Multiply top and bottom to get equivalent fractions; divide by the HCF to simplify.

Teacher's Note:
1. Always multiply both numbers by the same number.
2. Lowest terms: only 1 divides both.

Question 8. The chapter tells us how fractions were used in different parts of the ancient world.
(i) Name the two ancient civilisations mentioned in the chapter that mainly used only fractional units, and write the name given today to a number written as a sum of fractional units.
(ii) Show by actual addition that \( \frac{1}{2} \) + \( \frac{1}{5} \) = \( \frac{7}{10} \).
Answer:
(i) The ancient Egyptian and Babylonian civilisations mainly used fractional units (fractions with 1 on top). A number written as a sum of fractional units is called an Egyptian fraction.
(ii) A common multiple of 2 and 5 is 10, so use the unit \( \frac{1}{10} \):
\( \frac{1}{2} \) + \( \frac{1}{5} \) = \( \frac{5}{10} \) + \( \frac{2}{10} \) = \( \frac{7}{10} \).
In simple words: Egyptians and Babylonians used unit fractions, and \( \frac{1}{2} \) + \( \frac{1}{5} \) is indeed \( \frac{7}{10} \).

Teacher's Note:
1. Egyptian fraction = sum of different unit fractions.
2. Use tenths to add halves and fifths.

Question 9. Compare each pair of fractions and give a clear reason for your answer. Do not use a common denominator for part (i).
(i) \( \frac{5}{8} \) and \( \frac{5}{11} \)
(ii) \( \frac{7}{12} \) and \( \frac{11}{12} \)
Answer:
(i) \( \frac{5}{8} \) > \( \frac{5}{11} \). Both have 5 units, but \( \frac{1}{8} \) is a bigger unit than \( \frac{1}{11} \). Think of 5 rotis shared among 8 children and among 11 children: the smaller group gets more each.
(ii) \( \frac{11}{12} \) > \( \frac{7}{12} \). Both use the same unit \( \frac{1}{12} \), so the one with more units is greater.
In simple words: Same top: fewer sharers win. Same bottom: more units win.

Teacher's Note:
1. Part (i) must be done by the sharing idea.
2. Part (ii) needs only the numerators.

Question 10. Find \( \frac{3}{4} \) + \( \frac{2}{5} \) using Brahmagupta’s method. Write the common fractional unit you chose, both equivalent fractions, and the sum as a mixed fraction.
Answer:
Common fractional unit: 4 × 5 = 20, so use \( \frac{1}{20} \).
\( \frac{3}{4} \) = \( \frac{3\times 5}{4\times 5} \) = \( \frac{15}{20} \) and \( \frac{2}{5} \) = \( \frac{2\times 4}{5\times 4} \) = \( \frac{8}{20} \).
\( \frac{15}{20} \) + \( \frac{8}{20} \) = \( \frac{23}{20} \).
As a mixed fraction, \( \frac{23}{20} \) = \( 1\frac{3}{20} \).
In simple words: Change both into twentieths, add, and write 23 twentieths as \( 1\frac{3}{20} \).

Teacher's Note:
1. Show every step the question asks for.
2. Never add the denominators.

Question 11. Find \( \frac{5}{6} \) − \( \frac{3}{8} \). Show the common denominator you used and both equivalent fractions before subtracting.
Answer:
A common multiple of 6 and 8 is 24, so the unit is \( \frac{1}{24} \).
\( \frac{5}{6} \) = \( \frac{5\times 4}{6\times 4} \) = \( \frac{20}{24} \) and \( \frac{3}{8} \) = \( \frac{3\times 3}{8\times 3} \) = \( \frac{9}{24} \).
\( \frac{20}{24} \) − \( \frac{9}{24} \) = \( \frac{11}{24} \).
11 and 24 have no common factor except 1, so this is in lowest terms.
In simple words: Turn both into twenty-fourths and subtract to get \( \frac{11}{24} \).

Teacher's Note:
1. 24 is the smallest common multiple of 6 and 8.
2. 48 would also work but gives bigger numbers.

Question 12. Express \( \frac{84}{126} \) in its lowest terms by dividing step by step, not in one jump. Write each step.
Answer:
Step 1: 84 and 126 are both even, so divide by 2: \( \frac{84}{126} \) = \( \frac{42}{63} \).
Step 2: 42 and 63 are both multiples of 3, so divide by 3: \( \frac{42}{63} \) = \( \frac{14}{21} \).
Step 3: 14 and 21 are both multiples of 7, so divide by 7: \( \frac{14}{21} \) = \( \frac{2}{3} \).
2 and 3 have no common factor other than 1, so \( \frac{84}{126} \) = \( \frac{2}{3} \).
In simple words: Keep dividing top and bottom by common factors until none is left: \( \frac{2}{3} \).

Teacher's Note:
1. 2 × 3 × 7 = 42 is the HCF, used in steps.
2. Stop only when 1 is the only common factor.

Question 13. Ravi walked \( 2\frac{1}{3} \) km in the morning and \( 1\frac{1}{4} \) km in the evening. Find the total distance he walked, and write the answer as a mixed fraction.
Answer:
\( 2\frac{1}{3} \) = \( \frac{7}{3} \) and \( 1\frac{1}{4} \) = \( \frac{5}{4} \).
Common denominator = 3 × 4 = 12: \( \frac{7}{3} \) = \( \frac{28}{12} \) and \( \frac{5}{4} \) = \( \frac{15}{12} \).
\( \frac{28}{12} \) + \( \frac{15}{12} \) = \( \frac{43}{12} \).
43 = 3 × 12 + 7, so the total distance is \( 3\frac{7}{12} \) km.
In simple words: Change to fractions, add as twelfths, then turn \( \frac{43}{12} \) back into \( 3\frac{7}{12} \) km.

Teacher's Note:
1. Check: 2 + 1 = 3 wholes and \( \frac{1}{3} \) + \( \frac{1}{4} \) = \( \frac{7}{12} \).
2. That quick check confirms the answer.

Question 14. Six kilograms of rice are packed equally into 5 packets. Shyam says that each packet would weigh exactly the same if 18 kg of rice were packed equally into 15 packets. Is Shyam correct? Explain using equivalent fractions.
Answer:
Yes, Shyam is correct.
6 kg in 5 packets gives \( \frac{6}{5} \) kg in each packet.
Multiply the numerator and the denominator by 3: \( \frac{6}{5} \) = \( \frac{6\times 3}{5\times 3} \) = \( \frac{18}{15} \).
18 kg in 15 packets gives \( \frac{18}{15} \) kg each, which is equivalent to \( \frac{6}{5} \), so each packet weighs the same.
In simple words: \( \frac{6}{5} \) and \( \frac{18}{15} \) are equivalent, so the packets weigh the same.

Teacher's Note:
1. Both numbers were multiplied by 3.
2. Equivalent fractions mean equal shares.

Question 15. A carpenter measures a wooden strip using a rod that is \( \frac{1}{4} \) m long. The strip is exactly 11 such rods long.
(i) Write the length of the wooden strip as a fraction of a metre.
(ii) Write the same length as a mixed fraction.
Answer:
(i) The strip is 11 times \( \frac{1}{4} \) m: 11 × \( \frac{1}{4} \) = \( \frac{11}{4} \) m.
(ii) 11 = 2 × 4 + 3, so \( \frac{11}{4} \) = \( 2\frac{3}{4} \) m.
In simple words: 11 quarters of a metre is \( \frac{11}{4} \) m, which is \( 2\frac{3}{4} \) m.

Teacher's Note:
1. Measuring with a unit means counting the units.
2. 4 quarters make each whole metre.

Question 16. Explain, step by step, how you would mark \( \frac{7}{4} \) on a number line that already shows 0, 1 and 2. Mention the fractional unit used and the number of units counted from 0.
Answer:
The denominator 4 tells us the fractional unit is \( \frac{1}{4} \).
Divide the gap from 0 to 1 into 4 equal parts, and do the same for the gap from 1 to 2. Each small part is \( \frac{1}{4} \) unit long.
Starting at 0, count 7 such parts.
The 7th mark is between 1 and 2, one small part before 2. That point is \( \frac{7}{4} \), which is also \( 1\frac{3}{4} \).
In simple words: Cut each unit into quarters and count 7 quarters from 0.

Teacher's Note:
1. Every unit must be split into the same number of parts.
2. \( \frac{7}{4} \) is one quarter short of 2.

Question 17. Arrange \( \frac{5}{6} \), \( \frac{7}{9} \) and \( \frac{3}{4} \) in ascending order. Use 36 as the common denominator and show the equivalent fraction for each before ordering.
Answer:
36 is a common multiple of 6, 9 and 4, so use the unit \( \frac{1}{36} \).
\( \frac{5}{6} \) = \( \frac{5\times 6}{6\times 6} \) = \( \frac{30}{36} \); \( \frac{7}{9} \) = \( \frac{7\times 4}{9\times 4} \) = \( \frac{28}{36} \); \( \frac{3}{4} \) = \( \frac{3\times 9}{4\times 9} \) = \( \frac{27}{36} \).
Since 27 < 28 < 30, the ascending order is \( \frac{3}{4} \) < \( \frac{7}{9} \) < \( \frac{5}{6} \).
In simple words: In thirty-sixths they are 27, 28 and 30, so \( \frac{3}{4} \), \( \frac{7}{9} \), \( \frac{5}{6} \).

Teacher's Note:
1. Write the answer with the original fractions.
2. Ascending means smallest first.

Question 18. Rina had a ribbon of length \( 1\frac{1}{2} \) m. She used \( \frac{2}{3} \) m of it to tie a gift.
(i) Find the length of ribbon left, in simplest form.
(ii) Is the piece left longer or shorter than the piece she used? Give a reason.
Answer:
(i) \( 1\frac{1}{2} \) = \( \frac{3}{2} \) m. A common denominator of 2 and 3 is 6.
\( \frac{3}{2} \) = \( \frac{9}{6} \) and \( \frac{2}{3} \) = \( \frac{4}{6} \).
Ribbon left = \( \frac{9}{6} \) − \( \frac{4}{6} \) = \( \frac{5}{6} \) m, which is already in simplest form.
(ii) The piece left is longer. The used piece is \( \frac{2}{3} \) m = \( \frac{4}{6} \) m, and \( \frac{5}{6} \) > \( \frac{4}{6} \) because both use the unit \( \frac{1}{6} \) and 5 units are more than 4.
In simple words: \( \frac{5}{6} \) m is left, which is longer than the \( \frac{4}{6} \) m used.

Teacher's Note:
1. Change the mixed fraction first.
2. Compare using the same unit.

Question 19. The chapter says that different fractional units can be added to give 1.
(i) Show by actual addition that \( \frac{1}{2} \) + \( \frac{1}{3} \) + \( \frac{1}{6} \) = 1.
(ii) Explain why two different fractional units can never add up to 1.
Answer:
(i) A common multiple of 2, 3 and 6 is 6, so use the unit \( \frac{1}{6} \):
\( \frac{1}{2} \) + \( \frac{1}{3} \) + \( \frac{1}{6} \) = \( \frac{3}{6} \) + \( \frac{2}{6} \) + \( \frac{1}{6} \) = \( \frac{6}{6} \) = 1.
(ii) \( \frac{1}{2} \) is the largest fractional unit, and \( \frac{1}{2} \) + \( \frac{1}{2} \) = 1.
If the two units must be different, at least one has to be smaller than \( \frac{1}{2} \), so the sum is less than 1. Hence two different fractional units can never add up to 1.
In simple words: Three different units can make 1, but two different units always fall short.

Teacher's Note:
1. Only two halves make 1 with two units, and they are the same.
2. Any other pair is smaller.

Question 20. Answer both parts.
(i) Find the fraction equivalent to \( \frac{4}{7} \) whose denominator is 63.
(ii) Using your answer, decide which is greater, \( \frac{4}{7} \) or \( \frac{35}{63} \), and by how much.
Answer:
(i) 63 = 7 × 9, so multiply the numerator and the denominator by 9:
\( \frac{4}{7} \) = \( \frac{4\times 9}{7\times 9} \) = \( \frac{36}{63} \).
(ii) Both now use the unit \( \frac{1}{63} \), and 36 > 35, so \( \frac{4}{7} \) > \( \frac{35}{63} \).
Difference = \( \frac{36}{63} \) − \( \frac{35}{63} \) = \( \frac{1}{63} \).
In simple words: \( \frac{4}{7} \) = \( \frac{36}{63} \), which beats \( \frac{35}{63} \) by \( \frac{1}{63} \).

Teacher's Note:
1. Find the multiplier: 63 ÷ 7 = 9.
2. The difference is one fractional unit.