CBSE Class 6 Mathematics Fractions Worksheet Set 08

Long Answer Questions

Question 1. Answer all three parts about fractional units and equal shares.
(i) What is a fractional unit? Write any three fractional units.
(ii) Find the share of each child in these three situations: 1 roti shared equally between 2 children; 3 rotis shared equally among 4 children; 1 roti shared equally among 8 children.
(iii) Arvin says that \( \frac{1}{9} \) is greater than \( \frac{1}{5} \) because 9 is greater than 5. Explain, using the idea of equal shares, why Arvin is wrong.
Answer:
(i) When one whole is divided into equal parts, each part is called a fractional unit. Three examples: \( \frac{1}{2} \), \( \frac{1}{4} \) and \( \frac{1}{10} \). They are also called unit fractions because the numerator is 1.
(ii) 1 roti between 2 children: each gets \( \frac{1}{2} \) roti (1 ÷ 2 = \( \frac{1}{2} \)).
3 rotis among 4 children: each gets \( \frac{3}{4} \) roti (3 ÷ 4 = \( \frac{3}{4} \)).
1 roti among 8 children: each gets \( \frac{1}{8} \) roti.
(iii) Arvin is looking only at the denominators. The denominator tells how many equal parts the whole is cut into.
One roti shared among 9 children gives each \( \frac{1}{9} \) roti; the same roti shared among 5 children gives each \( \frac{1}{5} \) roti.
More children share the roti in the first case, so each share is smaller. Hence \( \frac{1}{9} \) < \( \frac{1}{5} \): for fractional units, the bigger the denominator, the smaller the fraction.
In simple words: A fractional unit is one equal part; more sharers make each share smaller, so \( \frac{1}{9} \) is less than \( \frac{1}{5} \).

Teacher's Note:
1. Write the division fact with each share.
2. Use a real sharing picture to correct Arvin.

Question 2. A paper strip is taken to be 1 unit long.
(i) The strip is folded into two equal parts, then folded again, and then folded once more. Write the fractional unit obtained after each fold.
(ii) Using the fractional unit of the third fold, write \( \frac{7}{8} \) as a repeated addition and also as ‘so many times the fractional unit’. Then show that the whole strip is \( \frac{8}{8} \).
(iii) Describe how you would mark \( \frac{7}{8} \) on a number line between 0 and 1.
Answer:
(i) First fold: 2 equal parts, unit \( \frac{1}{2} \). Second fold: 4 equal parts, unit \( \frac{1}{4} \). Third fold: 8 equal parts, unit \( \frac{1}{8} \).
(ii) \( \frac{7}{8} \) = \( \frac{1}{8} \) + \( \frac{1}{8} \) + \( \frac{1}{8} \) + \( \frac{1}{8} \) + \( \frac{1}{8} \) + \( \frac{1}{8} \) + \( \frac{1}{8} \) = 7 times \( \frac{1}{8} \).
This tells us the fractional unit (\( \frac{1}{8} \)) and how many of them there are (7).
The whole strip has 8 such units: 8 × \( \frac{1}{8} \) = \( \frac{8}{8} \) = 1.
(iii) Take the length from 0 to 1 as one unit and divide it into 8 equal parts; each small part is \( \frac{1}{8} \).
Starting from 0, count 7 parts to the right. The point reached is \( \frac{7}{8} \). It is one small part before 1, because 1 = \( \frac{8}{8} \) and \( \frac{7}{8} \) is one unit less.
In simple words: Each fold doubles the parts: halves, quarters, eighths; \( \frac{7}{8} \) is 7 eighths, one short of the whole.

Teacher's Note:
1. Each fold doubles the number of parts.
2. \( \frac{7}{8} \) sits just one small part before 1.

Question 3. Mixed fractions and fractions greater than one.
(i) Write \( \frac{31}{4} \) and \( \frac{29}{6} \) as mixed fractions. Show the whole part and the fractional part clearly.
(ii) Write \( 3\frac{5}{8} \) and \( 5\frac{2}{3} \) as fractions with a single numerator and denominator.
(iii) Explain how you can tell, just by looking at a fraction, whether it is less than 1, equal to 1, or greater than 1. Give one example of each.
Answer:
(i) 31 = 7 × 4 + 3, so \( \frac{31}{4} \) = 7 + \( \frac{3}{4} \) = \( 7\frac{3}{4} \). Whole part 7, fractional part \( \frac{3}{4} \).
29 = 4 × 6 + 5, so \( \frac{29}{6} \) = 4 + \( \frac{5}{6} \) = \( 4\frac{5}{6} \). Whole part 4, fractional part \( \frac{5}{6} \).
(ii) \( 3\frac{5}{8} \) = \( \frac{3\times 8+5}{8} \) = \( \frac{29}{8} \), because each whole makes 8 eighths.
\( 5\frac{2}{3} \) = \( \frac{5\times 3+2}{3} \) = \( \frac{17}{3} \).
(iii) Compare the numerator with the denominator:
Numerator smaller than denominator: less than 1, e.g. \( \frac{3}{7} \).
Numerator equal to denominator: exactly 1, e.g. \( \frac{5}{5} \).
Numerator bigger than denominator: more than 1, e.g. \( \frac{9}{4} \). Only such fractions can be written as mixed fractions.
In simple words: Divide to get mixed fractions, multiply and add to go back, and compare top with bottom to compare with 1.

Teacher's Note:
1. The fractional part of a mixed fraction must be less than 1.
2. Check by converting back.

Question 4. Three paper strips of the same length are shaded as shown.
(i) Write the fraction of each strip that is shaded.
(ii) What do you notice about the three shaded lengths? What name is given to such fractions?
(iii) Show the same fact by equal sharing: 2 rotis among 3 children, 4 rotis among 6 children, and 6 rotis among 9 children. Also write the simplest form of these fractions.
Answer:
(i) Strip X: 3 equal parts, 2 shaded, so \( \frac{2}{3} \). Strip Y: 6 equal parts, 4 shaded, so \( \frac{4}{6} \). Strip Z: 9 equal parts, 6 shaded, so \( \frac{6}{9} \).
(ii) The strips are of the same length and the three shaded lengths are exactly equal, even though the fractional units differ.
Fractions that stand for the same length or the same share are called equivalent fractions: \( \frac{2}{3} \) = \( \frac{4}{6} \) = \( \frac{6}{9} \).
(iii) 2 rotis among 3 children: \( \frac{2}{3} \) roti each. 4 rotis among 6 children: \( \frac{4}{6} \) roti each. 6 rotis among 9 children: \( \frac{6}{9} \) roti each.
In each case there are 2 rotis for every 3 children, so every share is the same.
\( \frac{4\div 2}{6\div 2} \) = \( \frac{2}{3} \) and \( \frac{6\div 3}{9\div 3} \) = \( \frac{2}{3} \). Since 2 and 3 have no common factor other than 1, the simplest form is \( \frac{2}{3} \).
In simple words: All three strips have the same shaded length, so \( \frac{2}{3} \), \( \frac{4}{6} \) and \( \frac{6}{9} \) are equivalent fractions.

Teacher's Note:
1. Line up the strips to see the equal shaded lengths.
2. Equal sharing gives the same idea without a picture.

Question 5. Four fractions are given: \( \frac{3}{4} \), \( \frac{5}{6} \), \( \frac{7}{12} \) and \( \frac{2}{3} \).
(i) Choose a suitable common denominator and explain why you chose it.
(ii) Write the equivalent fraction of each with that denominator.
(iii) Arrange the four given fractions in descending order and state which one is the smallest.
Answer:
(i) The denominators are 4, 6, 12 and 3. 12 is a multiple of all four, so it is a common denominator and the common unit is \( \frac{1}{12} \). It is the smallest such number, which keeps the working simple.
(ii) \( \frac{3}{4} \) = \( \frac{3\times 3}{4\times 3} \) = \( \frac{9}{12} \); \( \frac{5}{6} \) = \( \frac{5\times 2}{6\times 2} \) = \( \frac{10}{12} \); \( \frac{7}{12} \) stays \( \frac{7}{12} \); \( \frac{2}{3} \) = \( \frac{2\times 4}{3\times 4} \) = \( \frac{8}{12} \).
(iii) All four use the unit \( \frac{1}{12} \), so compare numerators: 10 > 9 > 8 > 7.
Descending order: \( \frac{5}{6} \) > \( \frac{3}{4} \) > \( \frac{2}{3} \) > \( \frac{7}{12} \).
The smallest is \( \frac{7}{12} \).
In simple words: In twelfths they are 9, 10, 7 and 8, so the order is \( \frac{5}{6} \), \( \frac{3}{4} \), \( \frac{2}{3} \), \( \frac{7}{12} \).

Teacher's Note:
1. Descending means largest first.
2. Give the answer with the original fractions.

Question 6. Gauri spends \( \frac{1}{3} \) of a day sleeping, \( \frac{1}{4} \) of a day in school and \( \frac{1}{8} \) of a day playing.
(i) Convert the three fractions into equivalent fractions with a common fractional unit.
(ii) Find the total part of the day spent on these three activities.
(iii) Find the part of the day left for everything else, and say which of the three activities takes the largest part of her day, with a reason.
Answer:
(i) A common multiple of 3, 4 and 8 is 24, so use the unit \( \frac{1}{24} \).
\( \frac{1}{3} \) = \( \frac{8}{24} \) (sleep), \( \frac{1}{4} \) = \( \frac{6}{24} \) (school), \( \frac{1}{8} \) = \( \frac{3}{24} \) (play).
(ii) \( \frac{8}{24} \) + \( \frac{6}{24} \) + \( \frac{3}{24} \) = \( \frac{17}{24} \) of the day. 17 and 24 have no common factor other than 1, so this is in lowest terms.
(iii) The whole day is \( \frac{24}{24} \), so the part left is \( \frac{24}{24} \) − \( \frac{17}{24} \) = \( \frac{7}{24} \).
Sleeping takes the largest part. All three are fractional units, so the one with the smallest denominator is the biggest: \( \frac{1}{3} \) > \( \frac{1}{4} \) > \( \frac{1}{8} \). In twenty-fourths too, 8 > 6 > 3.
In simple words: In twenty-fourths she spends 8 + 6 + 3 = 17 parts, leaving \( \frac{7}{24} \), and sleep is the biggest part.

Teacher's Note:
1. 24 is the smallest common multiple of 3, 4 and 8.
2. One whole day = \( \frac{24}{24} \).

Question 7. In Group A, 3 chikkis are shared equally among 4 children. In Group B, 5 chikkis are shared equally among 6 children.
(i) Write the share of one child in each group as a fraction.
(ii) Convert both shares into equivalent fractions having the same fractional unit.
(iii) Decide in which group a child gets more chikki, and find exactly how much more.
Answer:
(i) Group A: \( \frac{3}{4} \) chikki each. Group B: \( \frac{5}{6} \) chikki each.
(ii) A common multiple of 4 and 6 is 12, so the unit is \( \frac{1}{12} \).
\( \frac{3}{4} \) = \( \frac{3\times 3}{4\times 3} \) = \( \frac{9}{12} \) and \( \frac{5}{6} \) = \( \frac{5\times 2}{6\times 2} \) = \( \frac{10}{12} \).
(iii) Both are now in twelfths, and 10 > 9, so \( \frac{5}{6} \) > \( \frac{3}{4} \). A child in Group B gets more.
Extra amount = \( \frac{10}{12} \) − \( \frac{9}{12} \) = \( \frac{1}{12} \) chikki.
In simple words: Group B gets \( \frac{10}{12} \) against Group A's \( \frac{9}{12} \), so \( \frac{1}{12} \) chikki more.

Teacher's Note:
1. Shares: things over sharers.
2. Same unit first, then compare and subtract.

Question 8. Ramesh travels 5 km from his home to his uncle’s house. He cycles \( 1\frac{1}{4} \) km, then goes \( 2\frac{1}{2} \) km by rickshaw, and walks the rest of the way.
(i) Find the total distance he covers by cycle and rickshaw together. Write the answer as a mixed fraction.
(ii) Find the distance he walks.
(iii) Compare the walking part with the cycling part and state what you find.
Answer:
(i) \( 1\frac{1}{4} \) = \( \frac{5}{4} \) km and \( 2\frac{1}{2} \) = \( \frac{5}{2} \) km. A common denominator of 4 and 2 is 4.
\( \frac{5}{4} \) + \( \frac{5}{2} \) = \( \frac{5}{4} \) + \( \frac{10}{4} \) = \( \frac{15}{4} \) km.
15 = 3 × 4 + 3, so this is \( 3\frac{3}{4} \) km.
(ii) The whole journey is 5 km = \( \frac{20}{4} \) km.
Walking distance = \( \frac{20}{4} \) − \( \frac{15}{4} \) = \( \frac{5}{4} \) km = \( 1\frac{1}{4} \) km.
(iii) Walking is \( \frac{5}{4} \) km and cycling is also \( \frac{5}{4} \) km. Same unit, same number of units, so the two distances are exactly equal: Ramesh walks as much as he cycles.
In simple words: Cycle and rickshaw cover \( 3\frac{3}{4} \) km, he walks \( 1\frac{1}{4} \) km, which equals the cycling distance.

Teacher's Note:
1. Write 5 km as quarters before subtracting.
2. Compare using the same fractional unit.

Question 9. Three points P, Q and R are marked on the number line shown below. Each unit has been divided into equal parts.
(i) Write the fractional unit used on this number line, and write the fraction for each of P, Q and R.
(ii) Write Q and R as mixed fractions.
(iii) Find the distance between P and R, and arrange P, Q and R in descending order of their values.
Answer:
(i) Each unit length is divided into 4 equal parts, so the fractional unit is \( \frac{1}{4} \).
P is 3 small parts from 0, so P = \( \frac{3}{4} \). Q is 5 parts from 0, so Q = \( \frac{5}{4} \). R is 7 parts from 0, so R = \( \frac{7}{4} \).
P lies between 0 and 1; Q and R lie between 1 and 2, because their numerators are greater than their denominators.
(ii) 5 = 1 × 4 + 1, so Q = \( \frac{5}{4} \) = \( 1\frac{1}{4} \). 7 = 1 × 4 + 3, so R = \( \frac{7}{4} \) = \( 1\frac{3}{4} \).
(iii) Distance between P and R = \( \frac{7}{4} \) − \( \frac{3}{4} \) = \( \frac{4}{4} \) = 1 unit. This fits the picture: P is one small part before 1 and R is one small part before 2.
All three use the unit \( \frac{1}{4} \), so compare numerators: 7 > 5 > 3. Descending order: \( \frac{7}{4} \) > \( \frac{5}{4} \) > \( \frac{3}{4} \), that is R, Q, P.
In simple words: The line is in quarters: P = \( \frac{3}{4} \), Q = \( \frac{5}{4} \), R = \( \frac{7}{4} \), and R − P = 1.

Teacher's Note:
1. Count the small parts from 0, not from the nearest whole number.
2. Same unit: just compare numerators.

Question 10. Fractions, as we write them today, grew out of Indian mathematics.
(i) The chapter shows that some fractions can be written as a sum of different fractional units. Show by actual addition that \( \frac{1}{2} \) + \( \frac{1}{6} \) + \( \frac{1}{8} \) gives \( \frac{19}{24} \).
(ii) What was a fraction called in ancient India, and what do those words mean? Name the old manuscript that wrote \( \frac{1}{2} \) almost the way we write it today, and give its approximate date.
(iii) State Brahmagupta’s method of adding fractions in your own words, and use it to find \( \frac{5}{6} \) + \( \frac{3}{8} \).
Answer:
(i) A common multiple of 2, 6 and 8 is 24, so the unit is \( \frac{1}{24} \).
\( \frac{1}{2} \) = \( \frac{12}{24} \), \( \frac{1}{6} \) = \( \frac{4}{24} \), \( \frac{1}{8} \) = \( \frac{3}{24} \).
\( \frac{12}{24} \) + \( \frac{4}{24} \) + \( \frac{3}{24} \) = \( \frac{19}{24} \). So \( \frac{19}{24} \) is a sum of three different fractional units; such a sum is called an Egyptian fraction.
(ii) A fraction was called bhinna, a Sanskrit word meaning “broken”. It was also called bhaga or ansha, meaning “part” or “piece”.
The Bakshali manuscript, from around 300 CE, wrote the numerator above the denominator, very close to how we write \( \frac{1}{2} \) today. The line between the two numbers was added much later.
(iii) Brahmagupta’s method: first change all the fractions into equivalent fractions with the same fractional unit, using a common multiple of the denominators. Then add the numerators and keep the common denominator. Finally, reduce to lowest terms if needed.
For \( \frac{5}{6} \) + \( \frac{3}{8} \), a common multiple of 6 and 8 is 24: \( \frac{5}{6} \) = \( \frac{20}{24} \) and \( \frac{3}{8} \) = \( \frac{9}{24} \).
\( \frac{20}{24} \) + \( \frac{9}{24} \) = \( \frac{29}{24} \) = \( 1\frac{5}{24} \).
In simple words: Make the units the same, then add the numerators: \( \frac{5}{6} \) + \( \frac{3}{8} \) = \( \frac{29}{24} \) = \( 1\frac{5}{24} \).

Teacher's Note:
1. Bhinna, Bakshali, Brahmagupta: three key history words.
2. Show the common unit before adding.

Case Study Based Questions

Question 11. Lunch at Kamla’s house.
Kamla has made rotis for lunch. Her two youngest nephews are not very hungry, so she gives them one roti to share equally between themselves. Four older cousins are sitting at the other end of the mat. For them Kamla puts 5 rotis on a plate and asks them to share the rotis equally. A little later another roti is shared equally among a bigger group of children sitting near the door. Kamla tells the children that every share must be exactly equal, because only then can the share be written as a fraction.
(i) What fraction of a roti does each of the two youngest nephews get?
(ii) How much roti does each of the four older cousins get? Write the answer as a fraction.
(iii) Write the division fact for the sharing in part (ii), and write the same share as a mixed fraction.
(iv) If one roti is shared equally among 4 children and another roti is shared equally among 6 children, in which group does a child get more? Give a reason.
Answer:
(i) One roti between 2 children: each nephew gets \( \frac{1}{2} \) roti.
(ii) 5 rotis among 4 children: each cousin gets \( \frac{5}{4} \) roti.
(iii) Division fact: 5 ÷ 4 = \( \frac{5}{4} \). Since 5 = 1 × 4 + 1, the share is \( 1\frac{1}{4} \) roti, that is one whole roti and a quarter.
(iv) A child in the group of 4 gets more. One roti among 4 gives \( \frac{1}{4} \) each, while among 6 it gives only \( \frac{1}{6} \) each. More sharers means smaller shares, so \( \frac{1}{4} \) > \( \frac{1}{6} \).
In simple words: Shares are \( \frac{1}{2} \) and \( 1\frac{1}{4} \) roti, and fewer sharers get bigger pieces.

Teacher's Note:
1. Things shared on top, sharers below.
2. Write the division fact alongside the fraction.

Question 12. A birthday cake.
It is Hasan’s birthday. His mother brings a round cake and cuts it into 8 equal pieces, so that every piece is exactly the same size. Hasan takes 3 pieces for himself and his two sisters. His father then takes 4 pieces for the guests who are waiting outside. Hasan’s mother says that the cake is now easy to talk about with fractions, because the whole cake has been divided into equal parts and every piece is one such part of the whole.
(i) Name the fractional unit of this cake.
(ii) What fraction of the cake did Hasan take?
(iii) Write the fraction of the cake taken by Hasan’s father, in its simplest form.
(iv) After Hasan and his father took their pieces, what fraction of the cake was left? Is this more or less than what the father took?
Answer:
(i) The cake is cut into 8 equal pieces, so the fractional unit is \( \frac{1}{8} \).
(ii) Hasan takes 3 units: \( \frac{3}{8} \) of the cake.
(iii) The father takes 4 pieces, that is \( \frac{4}{8} \). Dividing both numbers by 4: \( \frac{4\div 4}{8\div 4} \) = \( \frac{1}{2} \). He took half the cake.
(iv) Pieces taken = 3 + 4 = 7, so 1 piece is left: \( \frac{1}{8} \) of the cake.
\( \frac{1}{8} \) is less than \( \frac{4}{8} \) (same unit, fewer units), so less is left than the father took.
In simple words: Each piece is \( \frac{1}{8} \); Hasan took \( \frac{3}{8} \), his father \( \frac{1}{2} \), and only \( \frac{1}{8} \) was left.

Teacher's Note:
1. Count pieces, then write them over 8.
2. Simplify \( \frac{4}{8} \) by dividing by 4.

Question 13. Ribbon for rakhis.
Rehana buys a roll of ribbon that is 2 m long to make rakhis for her brothers. For each rakhi she needs a piece of ribbon that is exactly \( \frac{1}{5} \) m long, so she uses a small paper measure of that length and marks the ribbon carefully before cutting. While working, she joins seven of these small pieces end to end to check the total length of ribbon used for seven rakhis. Her mother tells her that measuring a length with a fractional unit is just like counting how many such units fit into it.
(i) How many pieces of length \( \frac{1}{5} \) m fit into 1 m of ribbon?
(ii) Write, as a fraction of a metre, the total length of the seven pieces joined end to end.
(iii) Write the answer of part (ii) as a mixed fraction.
(iv) How many rakhis in all can Rehana make from the full 2 m roll?
Answer:
(i) 5 × \( \frac{1}{5} \) = \( \frac{5}{5} \) = 1, so 5 pieces fit into 1 m.
(ii) 7 × \( \frac{1}{5} \) = \( \frac{7}{5} \) m.
(iii) 7 = 1 × 5 + 2, so \( \frac{7}{5} \) = \( 1\frac{2}{5} \) m.
(iv) 2 m holds 2 × 5 = 10 pieces of \( \frac{1}{5} \) m, so she can make 10 rakhis.
In simple words: Five fifths make a metre, so 2 m gives 10 pieces; seven pieces are \( 1\frac{2}{5} \) m.

Teacher's Note:
1. Measuring = counting fractional units.
2. Each metre holds 5 fifths.

Question 14. A water bottle for the day.
Nita’s water bottle has marks on it that divide the water inside into 5 equal parts. On Monday she drinks 2 parts of the water before the morning assembly and 1 more part during the lunch break. She keeps the rest for the games period. Her friend Salma has a bottle of exactly the same size and drinks 4 of its parts by lunch time. The two girls compare their bottles and try to write what they have drunk as fractions of a full bottle.
(i) Name the fractional unit marked on Nita’s bottle.
(ii) What fraction of the bottle has Nita drunk by lunch time?
(iii) What fraction of the bottle is left for the games period?
(iv) Who has drunk more by lunch time, Nita or Salma? Give a reason.
Answer:
(i) The water is divided into 5 equal parts, so the fractional unit is \( \frac{1}{5} \) of a bottle.
(ii) \( \frac{2}{5} \) + \( \frac{1}{5} \) = \( \frac{3}{5} \) of the bottle (same unit, so just add the numbers of units).
(iii) The whole bottle is \( \frac{5}{5} \), so the water left is \( \frac{5}{5} \) − \( \frac{3}{5} \) = \( \frac{2}{5} \) of the bottle.
(iv) Salma has drunk more. The bottles are the same size and both use the unit \( \frac{1}{5} \), and \( \frac{4}{5} \) > \( \frac{3}{5} \) because 4 units are more than 3.
In simple words: Nita drank \( \frac{3}{5} \) and has \( \frac{2}{5} \) left; Salma drank \( \frac{4}{5} \), which is more.

Teacher's Note:
1. Same bottle size makes the comparison fair.
2. Full bottle = \( \frac{5}{5} \).

Question 15. A farmer’s field.
Mohan’s field has been divided into 10 strips of exactly the same size. He has sown wheat in some strips, mustard in some and gram in some, while one strip has been kept empty for the time being. The picture below shows how the field looks from above.
Mohan’s daughter says that since all the strips are equal, each strip is one fractional unit of the whole field, and so every crop can be described by a fraction of the field.
(i) Name the fractional unit of this field.
(ii) What fraction of the field is under wheat? Write it in its simplest form.
(iii) What fraction of the field has been sown in all, and what fraction is not sown?
(iv) How much more of the field is under wheat than under gram? Write the answer in simplest form.
Answer:
(i) The field has 10 equal strips, so the fractional unit is \( \frac{1}{10} \) of the field.
(ii) Wheat is on 4 strips: \( \frac{4}{10} \). Dividing both numbers by 2: \( \frac{4\div 2}{10\div 2} \) = \( \frac{2}{5} \).
(iii) Sown strips = 4 + 3 + 2 = 9, so \( \frac{9}{10} \) of the field is sown, and \( \frac{10}{10} \) − \( \frac{9}{10} \) = \( \frac{1}{10} \) is not sown.
(iv) Wheat: 4 strips; gram: 2 strips. Difference = \( \frac{4}{10} \) − \( \frac{2}{10} \) = \( \frac{2}{10} \) = \( \frac{1}{5} \) of the field.
In simple words: Each strip is \( \frac{1}{10} \); wheat is \( \frac{2}{5} \), \( \frac{9}{10} \) is sown, and wheat beats gram by \( \frac{1}{5} \).

Teacher's Note:
1. Count the strips of each crop from the picture.
2. Simplify at the end.

Question 16. Sharing a chocolate bar.
Ravi brings a chocolate bar to school. The bar is marked into 12 equal small squares. During the lunch break he breaks it along the marks and shares it with his friends. Ravi takes 3 squares, Sara takes 4 squares and Mina takes 2 squares. The remaining squares are wrapped up and kept for Ravi’s little brother. The three friends decide to write each person’s share as a fraction of the whole bar, and then to compare these shares.
(i) Name the fractional unit of this chocolate bar.
(ii) Write Sara’s share as a fraction of the bar and also in its simplest form.
(iii) What fraction of the bar did the three friends eat in all? Write the answer in simplest form.
(iv) Arrange the three shares in ascending order and state what fraction of the bar was kept for Ravi’s brother.
Answer:
(i) The bar has 12 equal squares, so the fractional unit is \( \frac{1}{12} \).
(ii) Sara’s share = \( \frac{4}{12} \). Dividing both numbers by 4 gives \( \frac{1}{3} \).
(iii) 3 + 4 + 2 = 9 squares, that is \( \frac{9}{12} \). Dividing both numbers by 3 gives \( \frac{3}{4} \) of the bar.
(iv) Mina \( \frac{2}{12} \), Ravi \( \frac{3}{12} \), Sara \( \frac{4}{12} \), so the ascending order is \( \frac{1}{6} \) < \( \frac{1}{4} \) < \( \frac{1}{3} \) (Mina, Ravi, Sara).
Squares left = 12 − 9 = 3, so \( \frac{3}{12} \) = \( \frac{1}{4} \) of the bar was kept for Ravi’s brother.
In simple words: In twelfths the shares are 2, 3 and 4; together \( \frac{3}{4} \) is eaten and \( \frac{1}{4} \) is kept.

Teacher's Note:
1. Compare in twelfths before simplifying.
2. Ascending means smallest first.

Question 17. At the sweet shop.
Before Diwali, Lata goes to a sweet shop with her father. She asks the shopkeeper for \( \frac{2}{3} \) kg of jalebi and \( \frac{3}{5} \) kg of barfi. The shopkeeper weighs the two sweets in separate boxes and puts both boxes in one bag. On the way home Lata’s father asks her to work out how much sweet they are carrying altogether, and whether they have bought more jalebi or more barfi. Lata says she must first change both fractions so that they have the same fractional unit.
(i) Which common denominator should Lata use for \( \frac{2}{3} \) and \( \frac{3}{5} \)? Write both equivalent fractions.
(ii) Which sweet did she buy more of? Give a reason.
(iii) Find the total weight of the sweets bought.
(iv) Write the total weight as a mixed fraction, and find how much more jalebi than barfi was bought.
Answer:
(i) A common multiple of 3 and 5 is 15, so use the unit \( \frac{1}{15} \): \( \frac{2}{3} \) = \( \frac{10}{15} \) and \( \frac{3}{5} \) = \( \frac{9}{15} \).
(ii) She bought more jalebi, because both weights now have the same unit and \( \frac{10}{15} \) > \( \frac{9}{15} \).
(iii) Total weight = \( \frac{10}{15} \) + \( \frac{9}{15} \) = \( \frac{19}{15} \) kg.
(iv) 19 = 1 × 15 + 4, so the total is \( 1\frac{4}{15} \) kg. Extra jalebi = \( \frac{10}{15} \) − \( \frac{9}{15} \) = \( \frac{1}{15} \) kg.
In simple words: In fifteenths: jalebi 10, barfi 9, total 19, so \( 1\frac{4}{15} \) kg with \( \frac{1}{15} \) kg more jalebi.

Teacher's Note:
1. 3 × 5 = 15 is the common denominator.
2. One set of equivalent fractions answers every part.

Question 18. A jug of lemonade.
On a hot afternoon Dadi makes lemonade and fills a jug that holds \( 1\frac{1}{2} \) litres. The glasses on the tray are small: each glass holds exactly \( \frac{1}{6} \) litre when it is filled to the brim. Dadi fills five such glasses for the children playing in the courtyard and keeps the rest of the lemonade in the jug for later. The children want to know how many full glasses the jug can give in all, and how much lemonade is still left after five glasses have been poured.
(i) Write the capacity of the jug as a fraction of a litre.
(ii) How many full glasses of \( \frac{1}{6} \) litre can be poured from the full jug?
(iii) How much lemonade is left in the jug after five glasses have been poured? Give the answer in simplest form.
(iv) Another set of glasses holds \( \frac{1}{8} \) litre each. Which glass holds more lemonade, and why?
Answer:
(i) \( 1\frac{1}{2} \) = \( \frac{1\times 2+1}{2} \) = \( \frac{3}{2} \) litres.
(ii) \( \frac{3}{2} \) = \( \frac{9}{6} \), which is 9 times \( \frac{1}{6} \). So 9 full glasses can be poured.
(iii) Five glasses hold 5 × \( \frac{1}{6} \) = \( \frac{5}{6} \) litre. Lemonade left = \( \frac{9}{6} \) − \( \frac{5}{6} \) = \( \frac{4}{6} \) = \( \frac{2}{3} \) litre.
(iv) The \( \frac{1}{6} \) litre glass holds more. Both are fractional units, and the one with the smaller denominator is the larger share, so \( \frac{1}{6} \) > \( \frac{1}{8} \).
In simple words: The jug holds \( \frac{9}{6} \) litre, so 9 glasses; after 5 glasses \( \frac{2}{3} \) litre is left.

Teacher's Note:
1. Write the jug's capacity in sixths.
2. Simplify \( \frac{4}{6} \) to \( \frac{2}{3} \).

Question 19. Reading a story book.
Aarav borrows a story book from the school library and promises to return it on Monday. On Saturday he reads \( \frac{2}{5} \) of the book. On Sunday he reads another \( \frac{3}{8} \) of the same book, and then falls asleep. The rest of the book is still unread on Monday morning. His sister asks him how much of the book he has read in all, how much is still left, and on which day he read more. Aarav remembers that two fractions can be compared or added only after they are written with the same fractional unit.
(i) Write a common denominator for \( \frac{2}{5} \) and \( \frac{3}{8} \), and write both equivalent fractions.
(ii) What fraction of the book did Aarav read in the two days together?
(iii) What fraction of the book was still unread on Monday morning?
(iv) On which day did he read more, and by what fraction of the book?
Answer:
(i) A common multiple of 5 and 8 is 40, so the unit is \( \frac{1}{40} \): \( \frac{2}{5} \) = \( \frac{2\times 8}{5\times 8} \) = \( \frac{16}{40} \) and \( \frac{3}{8} \) = \( \frac{3\times 5}{8\times 5} \) = \( \frac{15}{40} \).
(ii) Read in two days = \( \frac{16}{40} \) + \( \frac{15}{40} \) = \( \frac{31}{40} \) of the book.
(iii) The whole book is \( \frac{40}{40} \), so the unread part is \( \frac{40}{40} \) − \( \frac{31}{40} \) = \( \frac{9}{40} \) of the book.
(iv) Since \( \frac{16}{40} \) > \( \frac{15}{40} \), he read more on Saturday, by \( \frac{16}{40} \) − \( \frac{15}{40} \) = \( \frac{1}{40} \) of the book.
In simple words: In fortieths he read 16 + 15 = 31, left 9 unread, and read 1 more on Saturday.

Teacher's Note:
1. 5 × 8 = 40 is the common denominator.
2. The whole book is \( \frac{40}{40} \).

Question 20. The tailor’s cloth.
Shamim is a tailor. She receives a roll of cloth 10 m long and an order for school frocks. For one frock of the plain design she needs exactly \( 1\frac{1}{2} \) m of cloth. She first stitches 4 frocks of this plain design and keeps the remaining cloth aside. The next day a customer asks for a frock of a fancy design with extra pleats, which needs \( 1\frac{2}{3} \) m of cloth. Shamim must now work out how much cloth is left, how many more plain frocks she can still cut, and how much extra cloth the fancy design needs.
(i) Write \( 1\frac{1}{2} \) m as a fraction of a metre.
(ii) How much cloth is used for the 4 plain frocks?
(iii) How much cloth is left after the 4 plain frocks, and how many more plain frocks can be cut from it?
(iv) How much more cloth does the fancy design need than the plain design? Give the answer in simplest form.
Answer:
(i) \( 1\frac{1}{2} \) = \( \frac{1\times 2+1}{2} \) = \( \frac{3}{2} \) m.
(ii) 4 × \( \frac{3}{2} \) = \( \frac{12}{2} \) = 6 m.
(iii) Cloth left = 10 − 6 = 4 m = \( \frac{8}{2} \) m. Each frock needs \( \frac{3}{2} \) m, so two frocks need \( \frac{6}{2} \) = 3 m.
So 2 more plain frocks can be cut, and 4 − 3 = 1 m will still be left (not enough for a third frock).
(iv) \( 1\frac{2}{3} \) = \( \frac{5}{3} \) m and \( 1\frac{1}{2} \) = \( \frac{3}{2} \) m. With the unit \( \frac{1}{6} \): \( \frac{5}{3} \) = \( \frac{10}{6} \) and \( \frac{3}{2} \) = \( \frac{9}{6} \).
The fancy design needs \( \frac{10}{6} \) − \( \frac{9}{6} \) = \( \frac{1}{6} \) m more cloth.
In simple words: 4 frocks use 6 m, the 4 m left gives 2 more frocks, and the fancy frock needs \( \frac{1}{6} \) m extra.

Teacher's Note:
1. Halves make the counting of frocks easy.
2. Use sixths to compare halves and thirds.