CBSE Class 7 Mathematics Working with Fractions Worksheet Set 01

Long Answer Questions

Question 1. A snail crawls \( \frac{2}{7} \) m in one minute.
(i) Find the distance it crawls in 5 minutes, writing the calculation as a repeated addition and then as a multiplication.
(ii) In the product \( \frac{3}{4}\times8 \), name the multiplier and the multiplicand, and explain clearly the two steps used to find this product.
(iii) Find \( \frac{9}{5}\times7 \) and write the answer as a mixed fraction.
Answer:
(i) \( \frac{2}{7}+\frac{2}{7}+\frac{2}{7}+\frac{2}{7}+\frac{2}{7}=5\times\frac{2}{7}=\frac{10}{7} \) = \( 1\frac{3}{7} \) m.
(ii) \( \frac{3}{4} \) is the multiplier and 8 is the multiplicand.
Step 1: divide the multiplicand by the denominator of the multiplier: 8 ÷ 4 = 2.
Step 2: multiply this result by the numerator of the multiplier: 3 × 2 = 6. So \( \frac{3}{4}\times8 \) = 6. (Also \( \frac{3\times8}{4}=\frac{24}{4}=6 \).)
(iii) Writing 7 as \( \frac{7}{1} \): \( \frac{9}{5}\times\frac{7}{1}=\frac{63}{5} \) = \( 12\frac{3}{5} \).
Since \( \frac{9}{5} \) is greater than 1, the product \( 12\frac{3}{5} \) is greater than 7, as expected.
In simple words: Multiplying by a fraction means divide by its denominator and multiply by its numerator.

Teacher's Note:
1. Name the parts correctly: the fraction in front is the multiplier.
2. Check the size of the answer against the numbers used.

Question 2. A unit square is used as the “whole” to show the product \( \frac{2}{5}\times\frac{3}{4} \).
(i) Into how many equal parts one way and how many the other way is the unit square divided, and how many equal parts does the whole square now have?
(ii) How many of these parts stand for the product, and what is the product in its lowest form?
(iii) State Brahmagupta’s formula for multiplying two fractions and use it to check the product you found.
(iv) Hence write the area of a rectangle whose sides are \( \frac{2}{5} \) cm and \( \frac{3}{4} \) cm.
Answer:
(i) The square is cut into 5 equal parts one way and 4 equal parts the other way, giving 5 × 4 = 20 equal parts.
(ii) The block for the product is 2 parts wide and 3 parts tall, so it has 2 × 3 = 6 parts. Product = \( \frac{6}{20} \) = \( \frac{3}{10} \).
(iii) Brahmagupta (628 CE): \( \frac{a}{b}\times\frac{c}{d}=\frac{a\times c}{b\times d} \). Using it, \( \frac{2\times3}{5\times4}=\frac{6}{20}=\frac{3}{10} \), the same as the picture gives.
(iv) The area of a rectangle with fractional sides is the product of its sides, so the area is \( \frac{3}{10} \) cm².
In simple words: The square is cut into 20 parts, 6 of them make the product, and 6/20 = 3/10.

Teacher's Note:
1. Denominators decide the cuts; numerators decide the block.
2. The picture and the formula must agree.

Question 3. Division can always be turned into a multiplication with a missing number.
(i) Rewrite 20 ÷ 5 as a multiplication question with a missing number, and answer it.
(ii) Using the same idea, find \( 1\div\frac{3}{8} \). What special name is given to the number you get?
(iii) Hence find \( 4\div\frac{3}{8} \) and write the answer as a mixed fraction.
(iv) State the general rule for dividing one fraction by another.
Answer:
(i) 20 ÷ 5 asks: 5 × ? = 20. Since 5 × 4 = 20, the quotient is 4.
(ii) \( 1\div\frac{3}{8} \) asks: \( \frac{3}{8}\times ? = 1 \). Since \( \frac{3}{8}\times\frac{8}{3}=\frac{24}{24}=1 \), the answer is \( \frac{8}{3} \). It is called the reciprocal of \( \frac{3}{8} \).
(iii) \( \frac{8}{3} \) gives 1, so four times it gives 4: \( 4\div\frac{3}{8}=4\times\frac{8}{3}=\frac{32}{3} \) = \( 10\frac{2}{3} \).
(iv) To divide by a fraction, multiply by its reciprocal: \( \frac{a}{b}\div\frac{c}{d}=\frac{a}{b}\times\frac{d}{c}=\frac{a\times d}{b\times c} \).
This rule was given by Brahmagupta, and Bhāskara II explained it using the reciprocal in the Līlāvatī.
In simple words: Dividing by a fraction is the same as multiplying by its reciprocal.

Teacher's Note:
1. Every division has a matching multiplication.
2. Check (iii): 32/3 × 3/8 = 4.

Question 4. A shopkeeper has 12 m of lace on a roll.
(i) She cuts the whole roll into pieces, each of length \( \frac{3}{4} \) m. Write the division needed and find the number of pieces.
(ii) She sells each piece for ₹15. Find the money she gets from the whole roll.
(iii) The roll is only 12 m long, yet she gets more than 12 pieces. Explain why this happens.
Answer:
(i) We need how many lengths of \( \frac{3}{4} \) m fit into 12 m: \( 12\div\frac{3}{4}=12\times\frac{4}{3}=\frac{48}{3}=16 \). She gets 16 pieces.
Check: \( 16\times\frac{3}{4}=12 \) m, the whole roll.
(ii) Money = 16 × 15 = ₹240.
(iii) Each piece is shorter than 1 m, so \( \frac{3}{4} \) m fits into one metre more than once. When the divisor lies between 0 and 1, the quotient is greater than the dividend, so 16 > 12.
In simple words: Pieces shorter than a metre give more pieces than metres: 16 pieces earning ₹240.

Teacher's Note:
1. 'How many pieces' always means divide.
2. A small divisor gives a big quotient.

Question 5. The product of two numbers is not always greater than the numbers multiplied.
(i) Take two numbers that are both greater than 1, find their product, and compare the product with each number.
(ii) Take two numbers that both lie between 0 and 1, find their product, and compare it with each number.
(iii) Take one number between 0 and 1 and one number greater than 1, find their product, and compare it with each number.
(iv) Complete and explain the two general statements: “When one of the numbers being multiplied is between 0 and 1, the product is ______ than the other number” and “When one of the numbers being multiplied is greater than 1, the product is ______ than the other number”.
Answer:
(i) Take \( \frac{5}{3} \) and 3: \( \frac{5}{3}\times3=5 \). Since 5 > \( \frac{5}{3} \) and 5 > 3, the product is greater than both.
(ii) Take \( \frac{2}{5} \) and \( \frac{5}{6} \): \( \frac{2}{5}\times\frac{5}{6}=\frac{10}{30}=\frac{1}{3} \). As \( \frac{1}{3}<\frac{2}{5} \) and \( \frac{1}{3}<\frac{5}{6} \), the product is less than both.
(iii) Take \( \frac{4}{7} \) and 3: \( \frac{4}{7}\times3=\frac{12}{7} \). Here \( \frac{12}{7}<3 \) but \( \frac{12}{7}>\frac{4}{7} \), so the product lies between the two numbers.
(iv) When one number is between 0 and 1, the product is less than the other number, because we take only a part of it.
When one number is greater than 1, the product is greater than the other number, because we take it more than once.
In simple words: Multiplying by less than 1 shrinks a number; multiplying by more than 1 enlarges it.

Teacher's Note:
1. Any correct examples of each kind are fine.
2. Compare fractions using a common denominator if needed.

Question 6. Look at how the quotient behaves in a division.
(i) Find 8 ÷ 4, \( 8\div\frac{1}{3} \), \( \frac{1}{6}\div\frac{1}{2} \) and \( \frac{1}{6}\div2 \).
(ii) In which of these is the quotient greater than the dividend?
(iii) Write the general rule connecting the size of the divisor with the size of the quotient.
(iv) Explain, using the question “how many \( \frac{1}{3} \) fit into 8?”, why dividing by a number between 0 and 1 makes the quotient grow.
Answer:
(i) 8 ÷ 4 = 2; \( 8\div\frac{1}{3}=8\times3 \) = 24; \( \frac{1}{6}\div\frac{1}{2}=\frac{1}{6}\times2=\frac{2}{6} \) = \( \frac{1}{3} \); \( \frac{1}{6}\div2=\frac{1}{6}\times\frac{1}{2} \) = \( \frac{1}{12} \).
(ii) In \( 8\div\frac{1}{3} \) (24 > 8) and in \( \frac{1}{6}\div\frac{1}{2} \) (\( \frac{1}{3}>\frac{1}{6} \)). In the other two, the quotient is smaller than the dividend.
(iii) When the divisor lies between 0 and 1, the quotient is greater than the dividend. When the divisor is greater than 1, the quotient is less than the dividend.
(iv) 8 ÷ \( \frac{1}{3} \) asks how many pieces of size \( \frac{1}{3} \) fit into 8. Three such pieces fit into each 1, so 8 × 3 = 24 fit into 8. The smaller the piece, the more of them fit in, so the quotient grows.
In simple words: Dividing by a number smaller than 1 gives a bigger answer; dividing by a number bigger than 1 gives a smaller one.

Teacher's Note:
1. Look at the divisor, not the dividend.
2. Think of how many small pieces fit in.

Question 7. The panchayat takes \( \frac{1}{5} \) of a farmer’s field to build a canal. Of the land that is left, the farmer gives one half to his son and one fourth to his daughter, and keeps the rest for himself.
(i) What part of the original field is left after the canal is built?
(ii) What part of the original field does the son get?
(iii) What part of the original field does the daughter get?
(iv) What part does the farmer keep? Check that all the four parts together make one whole field.
Answer:
(i) Land left = \( 1-\frac{1}{5} \) = \( \frac{4}{5} \) of the field.
(ii) Son = \( \frac{1}{2}\times\frac{4}{5}=\frac{4}{10} \) = \( \frac{2}{5} \) of the field.
(iii) Daughter = \( \frac{1}{4}\times\frac{4}{5}=\frac{4}{20} \) = \( \frac{1}{5} \) of the field.
(iv) Given away from the \( \frac{4}{5} \): \( \frac{2}{5}+\frac{1}{5}=\frac{3}{5} \), so the farmer keeps \( \frac{4}{5}-\frac{3}{5} \) = \( \frac{1}{5} \) of the field.
Check: canal + son + daughter + farmer = \( \frac{1}{5}+\frac{2}{5}+\frac{1}{5}+\frac{1}{5}=\frac{5}{5}=1 \), one whole field.
In simple words: Canal 1/5, son 2/5, daughter 1/5 and farmer 1/5 make the whole field.

Teacher's Note:
1. The son's and daughter's shares are of the remaining land.
2. The final check must give exactly 1.

Question 8. Baudhāyana’s Śhulbasūtra has problems about covering a surface with square bricks. A floor patch of area \( 4\frac{1}{2} \) square units is to be covered completely with square bricks whose side is \( \frac{1}{4} \) unit.
(i) Find the area of one brick.
(ii) Find the number of bricks needed, writing the division you use.
(iii) If one brick costs ₹25, find the total cost of the bricks.
(iv) How many bricks would be needed if each brick had a side of \( \frac{1}{2} \) unit instead?
Answer:
(i) Area of one brick = \( \frac{1}{4}\times\frac{1}{4} \) = \( \frac{1}{16} \) square unit.
(ii) Total area = \( 4\frac{1}{2}=\frac{9}{2} \) square units. Number of bricks = \( \frac{9}{2}\div\frac{1}{16}=\frac{9}{2}\times16=\frac{144}{2} \) = 72 bricks.
Check: \( 72\times\frac{1}{16}=\frac{72}{16}=\frac{9}{2} \) square units.
(iii) Cost = 72 × 25 = ₹1800.
(iv) Area of a bigger brick = \( \frac{1}{2}\times\frac{1}{2}=\frac{1}{4} \) square unit, so bricks = \( \frac{9}{2}\div\frac{1}{4}=\frac{9}{2}\times4 \) = 18 bricks. Bigger bricks, fewer bricks.
In simple words: Each small brick covers 1/16 square unit, so 72 are needed; bricks of 1/4 square unit need only 18.

Teacher's Note:
1. Divide by the brick's area, not by its side.
2. Change the mixed number before dividing.

Question 9. Three pipes can fill a tank. Pipe A alone fills it in \( \frac{1}{2} \) hour, pipe B alone fills it in \( \frac{1}{3} \) hour and pipe C alone fills it in \( \frac{1}{6} \) hour.
(i) How many times can each pipe fill the tank in one hour?
(ii) If all three pipes are opened together, how many tanks will they fill in one hour, and how long will they take to fill one tank?
(iii) If only pipes A and C are opened, how long will they take to fill the tank?
(iv) Explain why the time taken by all three pipes together is less than the time taken by the fastest pipe alone.
Answer:
(i) Filling in \( \frac{1}{2} \) hour means the job can be done \( 1\div\frac{1}{2}=2 \) times in an hour. So A: 2, B: \( 1\div\frac{1}{3} \) = 3, C: \( 1\div\frac{1}{6} \) = 6 tanks in an hour.
(ii) Together: 2 + 3 + 6 = 11 tanks in an hour. One tank takes 1 ÷ 11 = \( \frac{1}{11} \) hour.
(iii) A and C: 2 + 6 = 8 tanks in an hour, so one tank takes \( \frac{1}{8} \) hour.
(iv) Working together, all the pipes pour water in at once, so the tank fills faster. The fastest pipe C needs \( \frac{1}{6} \) hour, while all three need only \( \frac{1}{11} \) hour, and \( \frac{1}{11}<\frac{1}{6} \). More water per hour means less time for each tank.
In simple words: Count how many tanks each pipe fills in an hour, add them, and take the reciprocal.

Teacher's Note:
1. Add the rates, not the times.
2. Time for one tank = 1 ÷ (tanks per hour).

Question 10. A square park is divided into four equal square plots. The top-left plot is cut by a diagonal into two equal triangles. One of these triangles is then divided into 3 equal parts and 2 of those parts are covered with grass.
(i) What fraction of the whole park is one small plot?
(ii) What fraction of the whole park is one of the two triangles?
(iii) What fraction of the whole park is covered with grass?
(iv) If the area of the park is 144 m², find the area covered with grass, and describe in one or two sentences the general method you used.
Answer:
(i) The park has 4 equal plots, so one plot is \( \frac{1}{4} \) of the park.
(ii) Each triangle is half of the plot: \( \frac{1}{2}\times\frac{1}{4} \) = \( \frac{1}{8} \) of the park.
(iii) Grass covers \( \frac{2}{3} \) of the triangle: \( \frac{2}{3}\times\frac{1}{8}=\frac{2}{24} \) = \( \frac{1}{12} \) of the park.
(iv) Area of grass = \( \frac{1}{12}\times144 \) = 12 m². Check: 12 × 12 = 144 m², the whole park.
Method: take the whole figure as 1, write each new piece as a fraction of the piece just before it, and multiply all these fractions. Multiplying the result by the total area gives the actual area.
In simple words: 1/4, then 1/8, then 2/3 of 1/8 = 1/12 of the park, which is 12 m².

Teacher's Note:
1. Each step is a fraction of the previous piece.
2. Multiply the chain of fractions.

Case Study Based Questions

Question 11. Meera halves a recipe
Meera’s grandmother has written down a halwa recipe that serves 8 people. For 8 people it needs \( \frac{3}{4} \) kg of suji, \( \frac{1}{2} \) kg of sugar, \( \frac{2}{5} \) kg of ghee and \( 1\frac{1}{2} \) litres of milk. On Sunday only four people are at home, so Meera decides to make exactly half of this recipe. Before going to the kitchen she writes “half of” as multiplication by \( \frac{1}{2} \), and works out every quantity on a sheet of paper. Her younger brother is surprised that she needs less of each ingredient than the recipe card shows.
(i) How much suji does Meera need?
(ii) How much sugar and how much ghee does she need?
(iii) How much milk does she need?
(iv) Why is each quantity in Meera’s list smaller than the quantity on the recipe card?
Answer:
(i) Suji = \( \frac{1}{2}\times\frac{3}{4} \) = \( \frac{3}{8} \) kg.
(ii) Sugar = \( \frac{1}{2}\times\frac{1}{2} \) = \( \frac{1}{4} \) kg; ghee = \( \frac{1}{2}\times\frac{2}{5}=\frac{2}{10} \) = \( \frac{1}{5} \) kg.
(iii) Milk = \( \frac{1}{2}\times1\frac{1}{2}=\frac{1}{2}\times\frac{3}{2} \) = \( \frac{3}{4} \) litre.
(iv) Every quantity is multiplied by \( \frac{1}{2} \), which lies between 0 and 1. Multiplying by such a number gives a product less than the number, so each new quantity is smaller.
In simple words: Half of each amount: 3/8 kg suji, 1/4 kg sugar, 1/5 kg ghee and 3/4 litre milk.

Teacher's Note:
1. 'Half of' means multiply by 1/2.
2. Change 1 1/2 into 3/2 first.

Question 12. Anil’s ribbon bows
Anil has bought a roll of 6 m of red ribbon to tie bows on the gift packets for his sister’s birthday. Each bow uses exactly \( \frac{2}{5} \) m of ribbon, and he cuts the pieces one after another without wasting any ribbon. His friend guesses that he cannot make more than 6 bows because the roll is only 6 m long. Anil says that the number of bows will be much more than 6, and he shows his friend the division he has written in his notebook.
(i) Write the expression that gives the number of bows Anil can make.
(ii) How many bows can he make from the whole roll?
(iii) If he makes only 12 bows, how much ribbon is used and how much is left?
(iv) Explain why the number of bows is greater than 6.
Answer:
(i) Number of bows = \( 6\div\frac{2}{5} \).
(ii) \( 6\times\frac{5}{2}=\frac{30}{2} \) = 15 bows. Check: \( 15\times\frac{2}{5}=6 \) m.
(iii) Used = \( 12\times\frac{2}{5}=\frac{24}{5} \) = \( 4\frac{4}{5} \) m; left = \( 6-\frac{24}{5}=\frac{6}{5} \) = \( 1\frac{1}{5} \) m.
(iv) Each bow needs less than 1 m, so more than one bow comes out of every metre. When the divisor is between 0 and 1, the quotient is greater than the dividend, so 15 > 6.
In simple words: 6 ÷ 2/5 = 15 bows, because each bow takes less than a metre.

Teacher's Note:
1. 'How many bows' means divide.
2. Write 6 as 30/5 to subtract.

Question 13. The overhead tank
A school has an empty overhead tank that is filled by a single tap. The caretaker has noticed that when the tap is kept open for exactly 1 hour, \( \frac{3}{8} \) of the tank gets filled. The tap gives water at the same steady rate all through the morning, so the part of the tank that gets filled is found by multiplying \( \frac{3}{8} \) by the number of hours for which the tap is open. On Monday the caretaker opened the tap at 7 a.m. with the tank completely empty.
(i) What part of the tank is filled by 9 a.m.?
(ii) What part of the tank would be filled if the tap were open for only \( \frac{1}{2} \) hour?
(iii) For how long must the tap run to fill the empty tank completely?
(iv) What part of the tank is still empty at 9 a.m.?
Answer:
(i) 7 a.m. to 9 a.m. is 2 hours: \( 2\times\frac{3}{8}=\frac{6}{8} \) = \( \frac{3}{4} \) of the tank.
(ii) \( \frac{1}{2}\times\frac{3}{8} \) = \( \frac{3}{16} \) of the tank.
(iii) Time = \( 1\div\frac{3}{8}=\frac{8}{3} \) = \( 2\frac{2}{3} \) hours (2 hours 40 minutes).
(iv) Empty part = \( 1-\frac{3}{4} \) = \( \frac{1}{4} \) of the tank.
In simple words: The tank is 3/4 full at 9 a.m. and needs 2 2/3 hours to fill completely.

Teacher's Note:
1. Count the hours carefully from the times given.
2. The full tank is 1.

Question 14. Hamid’s field
Hamid owns a field of area \( \frac{3}{4} \) hectare. In the rabi season he sows wheat on \( \frac{2}{3} \) of the field. On \( \frac{3}{4} \) of the land that is still left he sows mustard, and the small piece that remains after this is kept fallow so that the soil can rest. Hamid records every area in hectares in his notebook, and his son checks the work by adding all the pieces to see whether they give back the area of the whole field.
(i) What area is under wheat?
(ii) What area of the field is left after sowing wheat?
(iii) What area is under mustard?
(iv) What area is kept fallow? Check that the three areas add up to \( \frac{3}{4} \) hectare.
Answer:
(i) Wheat = \( \frac{2}{3}\times\frac{3}{4}=\frac{6}{12} \) = \( \frac{1}{2} \) hectare.
(ii) Left = \( \frac{3}{4}-\frac{1}{2}=\frac{3}{4}-\frac{2}{4} \) = \( \frac{1}{4} \) hectare.
(iii) Mustard = \( \frac{3}{4}\times\frac{1}{4} \) = \( \frac{3}{16} \) hectare.
(iv) Fallow = \( \frac{1}{4}-\frac{3}{16}=\frac{4}{16}-\frac{3}{16} \) = \( \frac{1}{16} \) hectare.
Check: \( \frac{1}{2}+\frac{3}{16}+\frac{1}{16}=\frac{8+3+1}{16}=\frac{12}{16}=\frac{3}{4} \) hectare, the whole field.
In simple words: Wheat 1/2 ha, mustard 3/16 ha and fallow 1/16 ha make 3/4 ha.

Teacher's Note:
1. Mustard is 3/4 of what is left, not of the field.
2. Use sixteenths to add at the end.

Question 15. The tailor’s cloth roll
A tailor receives a fresh roll of 24 m of shirting cloth. For one school shirt of the senior size he needs \( 1\frac{3}{4} \) m of cloth, and he cuts the cloth carefully so that nothing is wasted between two shirts. The school has ordered as many shirts as can be made from this single roll, and the tailor must also report how much cloth will be left over at the end, because the leftover piece is returned to the school store. He first changes the mixed fraction into an improper fraction before dividing.
(i) Write \( 1\frac{3}{4} \) as an improper fraction.
(ii) How much cloth is needed for 8 such shirts?
(iii) What is the greatest number of complete shirts that can be made from the roll?
(iv) How much cloth is left over after making that many shirts?
Answer:
(i) \( 1\frac{3}{4}=\frac{4+3}{4} \) = \( \frac{7}{4} \) m.
(ii) \( 8\times\frac{7}{4}=\frac{56}{4} \) = 14 m.
(iii) \( 24\div\frac{7}{4}=24\times\frac{4}{7}=\frac{96}{7}=13\frac{5}{7} \). Part of a shirt is no use, so 13 complete shirts can be made.
(iv) Cloth used = \( 13\times\frac{7}{4}=\frac{91}{4}=22\frac{3}{4} \) m, so left over = \( 24-\frac{91}{4}=\frac{96-91}{4}=\frac{5}{4} \) = \( 1\frac{1}{4} \) m.
In simple words: The roll makes 13 full shirts and leaves 1 1/4 m of cloth.

Teacher's Note:
1. Round the number of shirts down, not up.
2. Leftover = roll − cloth used.

Question 16. Ravi’s gift money
On his birthday Ravi received ₹960 from his relatives. He first spent \( \frac{1}{4} \) of this money on story books. Out of the money that was then left with him, he spent \( \frac{1}{3} \) on stationery for the new school year. The money still left after these two purchases was put into his savings box. His father asked him to write the whole calculation step by step, and also to say what fraction of the original gift money finally reached the savings box.
(i) How much did Ravi spend on story books?
(ii) How much money was left with him after buying the books?
(iii) How much did he spend on stationery?
(iv) How much money went into the savings box, and what fraction is this of the original ₹960?
Answer:
(i) Books = \( \frac{1}{4}\times960 \) = ₹240.
(ii) Left = 960 − 240 = ₹720.
(iii) Stationery = \( \frac{1}{3}\times720 \) = ₹240.
(iv) Saved = 720 − 240 = ₹480. Fraction = \( \frac{480}{960} \) = \( \frac{1}{2} \), so exactly half the gift money was saved.
In simple words: ₹240 on books, ₹240 on stationery, and ₹480 (half) saved.

Teacher's Note:
1. The 1/3 is of the money left, not of ₹960.
2. Check: 240 + 240 + 480 = 960.

Question 17. Packing milk at the dairy
A small dairy collects \( 7\frac{1}{2} \) litres of milk every morning and packs all of it into sealed packets before sending it to the shops. On Monday the dairy used packets that hold \( \frac{3}{4} \) litre each, and every packet was filled completely. Each such packet is sold for ₹32. On Tuesday the same quantity of milk was collected, but the dairy had only smaller packets of \( \frac{1}{2} \) litre each. The owner wanted to know in advance how the number of packets would change when a smaller packet is used.
(i) How many packets were filled on Monday?
(ii) How much money does the dairy get from Monday’s packets?
(iii) How many packets were filled on Tuesday?
(iv) Explain why using a smaller packet gives a larger number of packets.
Answer:
(i) \( 7\frac{1}{2}\div\frac{3}{4}=\frac{15}{2}\times\frac{4}{3}=\frac{60}{6} \) = 10 packets. Check: \( 10\times\frac{3}{4}=\frac{15}{2} \) litres.
(ii) Money = 10 × 32 = ₹320.
(iii) \( \frac{15}{2}\div\frac{1}{2}=\frac{15}{2}\times2 \) = 15 packets.
(iv) The same milk is measured out in smaller amounts, so more packets are needed. In a division, a smaller divisor gives a larger quotient.
In simple words: 7 1/2 litres fill 10 packets of 3/4 litre but 15 packets of 1/2 litre.

Teacher's Note:
1. Change 7 1/2 to 15/2 first.
2. Smaller divisor, bigger quotient.

Question 18. Riya’s grid picture
In her notebook Riya draws a unit square and takes it as one whole. She divides it into equal parts by drawing straight lines across and down, and then colours one rectangular block of these parts, as shown in the figure. She tells her group that the coloured block is really a rectangle whose length and breadth are fractions of the side of the unit square, so its area must be the product of those two fractions. The group uses her picture to read off the product without doing any written multiplication.
(i) Into how many equal parts is the whole square divided?
(ii) What fraction of the whole square is coloured?
(iii) What are the length and the breadth of the coloured rectangle, and which multiplication of two fractions does the picture show?
(iv) Using the same picture, what is the value of \( \frac{1}{5}\times\frac{1}{3} \)?
Answer:
(i) The square is cut into 5 equal parts one way and 3 equal parts the other way, so there are 5 × 3 = 15 equal parts.
(ii) 4 of the 15 parts are coloured, so \( \frac{4}{15} \) of the square is coloured.
(iii) The coloured block is 2 parts out of 5 along one side and 2 parts out of 3 along the other, so its sides are \( \frac{2}{5} \) and \( \frac{2}{3} \) of the side. The picture shows \( \frac{2}{5}\times\frac{2}{3}=\frac{4}{15} \).
(iv) One small cell is \( \frac{1}{5} \) of the side one way and \( \frac{1}{3} \) the other way, and 15 such cells fill the square, so \( \frac{1}{5}\times\frac{1}{3} \) = \( \frac{1}{15} \).
In simple words: The square has 15 cells; the 2 by 2 coloured block shows 2/5 × 2/3 = 4/15.

Teacher's Note:
1. Count columns and rows to find the denominators.
2. Count the coloured block's columns and rows for the numerators.

Question 19. A coin from the Līlāvatī
In the Līlāvatī, written in 1150 CE, Bhāskarāchārya poses playful problems about money. A dramma was a silver coin of those times, and the tale says that 1 dramma was worth 1280 cowrie shells, the smallest money of all. In one such problem a miser gives a beggar \( \frac{1}{2} \) of \( \frac{2}{3} \) of \( \frac{3}{5} \) of \( \frac{1}{8} \) of a dramma, and the poet asks the reader to tell, in cowrie shells, exactly how much the beggar received. The joke lies in how little the miser actually parts with.
(i) Write the gift as a single fraction of a dramma in its lowest form.
(ii) How many cowrie shells did the beggar receive?
(iii) What fraction of a dramma is one cowrie shell?
(iv) Had the miser given \( \frac{1}{5} \) of a dramma instead, how many cowrie shells would that be, and how many times the earlier gift is it?
Answer:
(i) \( \frac{1}{2}\times\frac{2}{3}=\frac{1}{3} \); then \( \frac{1}{3}\times\frac{3}{5}=\frac{1}{5} \); then \( \frac{1}{5}\times\frac{1}{8}=\frac{1}{40} \). The gift is \( \frac{1}{40} \) of a dramma.
(ii) \( \frac{1}{40}\times1280=\frac{1280}{40} \) = 32 cowrie shells.
(iii) 1280 cowrie shells make one dramma, so one cowrie shell is \( \frac{1}{1280} \) of a dramma.
(iv) \( \frac{1}{5}\times1280 \) = 256 cowrie shells. Since 256 ÷ 32 = 8, it would be 8 times the earlier gift.
In simple words: The miser gave just 1/40 of a dramma, only 32 cowrie shells.

Teacher's Note:
1. Multiply the chain step by step and cancel.
2. This is a real puzzle from Bhāskara II.

Question 20. The sherbet stall
At a summer fair, a stall prepares \( 7\frac{1}{2} \) litres of lemon sherbet in the morning. By noon, \( \frac{2}{5} \) of the whole quantity has been sold. Of the sherbet still left in the pot, \( \frac{1}{3} \) is sold during the afternoon, and whatever remains after that is sold in the evening. The sherbet is served only in glasses that hold \( \frac{1}{4} \) litre each, and every glass is filled to the brim. At closing time the stall owner writes down how much was sold in each part of the day and checks that the three amounts add up to the whole.
(i) How many litres were sold by noon?
(ii) How many litres were sold during the afternoon?
(iii) How many litres were sold in the evening, and how many glasses does this make?
(iv) What fraction of the whole quantity was sold in the evening?
Answer:
(i) Total = \( 7\frac{1}{2}=\frac{15}{2} \) litres. Sold by noon = \( \frac{2}{5}\times\frac{15}{2} \) = 3 litres.
(ii) Left in the pot = \( \frac{15}{2}-3=\frac{9}{2} \) litres. Afternoon = \( \frac{1}{3}\times\frac{9}{2}=\frac{3}{2} \) = \( 1\frac{1}{2} \) litres.
(iii) Evening = \( \frac{9}{2}-\frac{3}{2} \) = 3 litres. Glasses = \( 3\div\frac{1}{4}=3\times4 \) = 12 glasses.
(iv) Fraction = \( 3\div\frac{15}{2}=3\times\frac{2}{15}=\frac{6}{15} \) = \( \frac{2}{5} \) of the whole. Check: \( 3+1\frac{1}{2}+3=7\frac{1}{2} \) litres, the whole lot.
In simple words: 3 L by noon, 1 1/2 L in the afternoon and 3 L (12 glasses) in the evening.

Teacher's Note:
1. The afternoon fraction is of what is left after noon.
2. Check that the three parts add to 7 1/2 litres.