CBSE Class 7 Mathematics Working with Fractions Assignment Set 01

Short Answer Questions

Question 1. Aarti’s pet tortoise walks \( \frac{1}{6} \) km in one hour. How far will it walk in 4 hours? Show the working both as a repeated addition and as a multiplication, and give the answer in its lowest form.
Answer:
Repeated addition: \( \frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{4}{6} \) km.
Multiplication: \( 4\times\frac{1}{6}=\frac{4}{6} \) km.
Lowest form: \( \frac{4}{6}=\frac{2}{3} \). So the tortoise walks \( \frac{2}{3} \) km in 4 hours.
In simple words: Four lots of 1/6 km make 4/6 = 2/3 km.

Teacher's Note:
1. Both methods must give the same answer.
2. Divide 4 and 6 by 2 to reduce.

Question 2. Multiply and write each product as a mixed fraction.
(i) \( 8\times\frac{3}{7} \)
(ii) \( \frac{11}{4}\times5 \)
Answer:
Write the whole number over 1, then multiply numerators and denominators.
(i) \( \frac{8}{1}\times\frac{3}{7}=\frac{24}{7} \) = \( 3\frac{3}{7} \).
(ii) \( \frac{11}{4}\times\frac{5}{1}=\frac{55}{4} \) = \( 13\frac{3}{4} \).
In simple words: Multiply the numerator by the whole number, then change to a mixed fraction.

Teacher's Note:
1. 24 ÷ 7 = 3 remainder 3.
2. 55 ÷ 4 = 13 remainder 3.

Question 3. Taking a unit square as the whole, explain how the picture for \( \frac{1}{9}\times\frac{1}{7} \) is made and what fraction is shaded. Then write the general rule for the product of two fractional units \( \frac{1}{b}\times\frac{1}{d} \).
Answer:
Take the unit square as one whole.
Divide it into 9 equal strips one way and 7 equal strips the other way.
The square now has 9 × 7 = 63 equal small parts, and one of them stands for the product.
So \( \frac{1}{9}\times\frac{1}{7}=\frac{1}{63} \); \( \frac{1}{63} \) of the square is shaded.
General rule: \( \frac{1}{b}\times\frac{1}{d}=\frac{1}{b\times d} \), that is, 1 over the product of the denominators.
In simple words: Cut the square 9 ways and 7 ways; one of the 63 pieces is the product.

Teacher's Note:
1. The denominators tell how to cut the square.
2. Multiply the denominators; the numerator stays 1.

Question 4. Use Brahmagupta’s formula \( \frac{a}{b}\times\frac{c}{d}=\frac{a\times c}{b\times d} \) to find the following products, giving each answer in its lowest form.
(i) \( \frac{3}{8}\times\frac{4}{5} \)
(ii) \( \frac{5}{6}\times\frac{9}{10} \)
Answer:
Multiply numerator by numerator and denominator by denominator, then reduce.
(i) \( \frac{3\times4}{8\times5}=\frac{12}{40} \) = \( \frac{3}{10} \).
(ii) \( \frac{5\times9}{6\times10}=\frac{45}{60} \) = \( \frac{3}{4} \).
In simple words: Top times top, bottom times bottom, then reduce.

Teacher's Note:
1. 12/40: divide both by 4.
2. 45/60: divide both by 15.

Question 5. A rectangular tile is \( \frac{2}{3} \) m long and \( \frac{3}{5} \) m wide. Find its area, and state the connection between the area of a rectangle and the multiplication of two fractions.
Answer:
Area = length × breadth = \( \frac{2}{3}\times\frac{3}{5}=\frac{6}{15} \) = \( \frac{2}{5} \) m².
The area of a rectangle whose sides are fractions is exactly the product of those two fractions.
So any product of two fractions can be pictured as the area of a rectangle drawn inside a unit square.
In simple words: The tile's area is 2/5 m², and multiplying two fractions is the same as finding such an area.

Teacher's Note:
1. Area is in square metres.
2. This is the area model of multiplication.

Question 6. Write the reciprocal of each of these numbers: \( \frac{7}{9} \), 5, \( \frac{1}{8} \), \( 2\frac{1}{2} \). What do you always get when a fraction is multiplied by its reciprocal?
Answer:
Swap the numerator and the denominator.
Reciprocal of \( \frac{7}{9} \) is \( \frac{9}{7} \); reciprocal of \( 5=\frac{5}{1} \) is \( \frac{1}{5} \).
Reciprocal of \( \frac{1}{8} \) is 8; \( 2\frac{1}{2}=\frac{5}{2} \), so its reciprocal is \( \frac{2}{5} \).
A fraction multiplied by its reciprocal always gives 1.
In simple words: Flip each number; a number times its flip is 1.

Teacher's Note:
1. Change a mixed number to an improper fraction before flipping.
2. A whole number has denominator 1.

Question 7. Divide, showing the reciprocal step in each case.
(i) \( 6\div\frac{2}{5} \)
(ii) \( \frac{3}{4}\div\frac{9}{8} \)
Answer:
To divide, multiply the dividend by the reciprocal of the divisor.
(i) Reciprocal of \( \frac{2}{5} \) is \( \frac{5}{2} \), so \( 6\times\frac{5}{2}=\frac{30}{2} \) = 15.
(ii) Reciprocal of \( \frac{9}{8} \) is \( \frac{8}{9} \), so \( \frac{3}{4}\times\frac{8}{9}=\frac{24}{36} \) = \( \frac{2}{3} \).
In simple words: Flip the divisor and multiply.

Teacher's Note:
1. Flip only the second number.
2. Reduce the answer to lowest form.

Question 8. Find \( \frac{15}{8}\times\frac{16}{25} \) by cancelling the common factors before multiplying. Why does cancelling not change the value of the product?
Answer:
15 and 25 have the common factor 5, giving 3 and 5.
16 and 8 have the common factor 8, giving 2 and 1.
So the product is \( \frac{3\times2}{1\times5} \) = \( \frac{6}{5}=1\frac{1}{5} \).
Cancelling divides the numerator and the denominator by the same number, and that does not change the value of a fraction. This old Indian idea is called apavartana.
In simple words: Cancel 5 and 8 first to get 6/5; cancelling is just dividing top and bottom by the same number.

Teacher's Note:
1. Cancel a numerator only with a denominator.
2. Smaller numbers mean fewer mistakes.

Question 9. A tap fills \( \frac{4}{5} \) of a tank in 1 hour.
(i) What part of the tank is filled if the tap is open for \( \frac{5}{8} \) hour?
(ii) For how long must the tap run to fill the whole tank?
Answer:
(i) Part filled = \( \frac{5}{8}\times\frac{4}{5}=\frac{20}{40} \) = \( \frac{1}{2} \) of the tank.
(ii) Time = \( 1\div\frac{4}{5}=1\times\frac{5}{4}=\frac{5}{4} \) = \( 1\frac{1}{4} \) hours.
Check: \( \frac{5}{4}\times\frac{4}{5}=1 \), the whole tank.
In simple words: In 5/8 hour it fills half the tank; the full tank takes 1 1/4 hours.

Teacher's Note:
1. Part (i) uses multiplication, part (ii) uses division.
2. 1 ÷ 4/5 is the reciprocal of 4/5.

Question 10. Without carrying out the multiplication, state how the product compares with each of the two numbers being multiplied. Give a reason in each case.
(i) \( \frac{7}{9}\times\frac{5}{6} \)
(ii) \( \frac{9}{4}\times\frac{7}{8} \)
Answer:
The size of the product depends on whether each number is between 0 and 1 or greater than 1.
(i) Both \( \frac{7}{9} \) and \( \frac{5}{6} \) lie between 0 and 1, so the product is less than both of them.
(ii) \( \frac{9}{4} \) is greater than 1, so the product is greater than \( \frac{7}{8} \); \( \frac{7}{8} \) lies between 0 and 1, so the product is less than \( \frac{9}{4} \).
In simple words: A factor below 1 shrinks the other number; a factor above 1 enlarges it.

Teacher's Note:
1. Check each factor against 1.
2. In (ii) the product lies between the two numbers.

Question 11. \( \frac{3}{5} \) litre of milk is poured equally into 6 glasses. Write the operation needed, find how much milk one glass holds, and check your answer by working backwards.
Answer:
The milk is shared equally, so the operation is division.
Milk in one glass = \( \frac{3}{5}\div6=\frac{3}{5}\times\frac{1}{6}=\frac{3}{30} \) = \( \frac{1}{10} \) litre.
Check: \( 6\times\frac{1}{10}=\frac{6}{10}=\frac{3}{5} \) litre, which is the milk we started with.
In simple words: Share 3/5 litre among 6 glasses: each gets 1/10 litre.

Teacher's Note:
1. Dividing by 6 is multiplying by 1/6.
2. Working backwards uses multiplication.

Question 12. Find \( \frac{5}{6}\times\frac{3}{4} \) and \( \frac{3}{4}\times\frac{5}{6} \). What do you notice about the two products? Explain your observation using the area of a rectangle.
Answer:
\( \frac{5}{6}\times\frac{3}{4}=\frac{15}{24}=\frac{5}{8} \) and \( \frac{3}{4}\times\frac{5}{6}=\frac{15}{24}=\frac{5}{8} \).
Both products are the same, so the order of multiplication does not matter.
Reason: both give the area of the same rectangle, once with \( \frac{5}{6} \) as length and once as breadth. Swapping length and breadth does not change the area.
In simple words: Both are 5/8, because turning a rectangle around does not change its area.

Teacher's Note:
1. Area pictures help explain rules.
2. This is true for every pair of fractions.

Question 13. A ribbon of length \( \frac{9}{2} \) m is cut into pieces, each of length \( \frac{3}{8} \) m. How many pieces are obtained? Verify the answer you get.
Answer:
Number of pieces = \( \frac{9}{2}\div\frac{3}{8} \).
Multiply by the reciprocal: \( \frac{9}{2}\times\frac{8}{3}=\frac{72}{6}=12 \). So 12 pieces are obtained.
Verification: \( 12\times\frac{3}{8}=\frac{36}{8}=\frac{9}{2} \) m, the full length of the ribbon.
In simple words: 4 1/2 m holds twelve pieces of 3/8 m each.

Teacher's Note:
1. 'How many pieces' means divide.
2. Verify by multiplying back.

Question 14. A shopkeeper had \( \frac{3}{4} \) of a sack of rice and sold \( \frac{2}{3} \) of it.
(i) What part of a full sack did he sell?
(ii) What part of a full sack is left with him?
Answer:
“Of” means multiplication here.
(i) Part sold = \( \frac{2}{3}\times\frac{3}{4}=\frac{6}{12} \) = \( \frac{1}{2} \) of a full sack.
(ii) Part left = \( \frac{3}{4}-\frac{1}{2}=\frac{3}{4}-\frac{2}{4} \) = \( \frac{1}{4} \) of a full sack.
In simple words: He sold half a sack and has a quarter of a sack left.

Teacher's Note:
1. 2/3 is of the 3/4, not of the full sack.
2. Subtract from 3/4, not from 1.

Question 15. Using the idea “what must the divisor be multiplied by to get the dividend”, explain why \( 1\div\frac{1}{7}=7 \). Hence find the value of \( 1\div\frac{1}{6}+1\div\frac{1}{9} \).
Answer:
\( 1\div\frac{1}{7} \) asks: what must \( \frac{1}{7} \) be multiplied by to give 1?
Since \( \frac{1}{7}\times7=1 \), the answer is 7. The number that gives 1 is the reciprocal of the divisor.
In the same way, \( 1\div\frac{1}{6}=6 \) and \( 1\div\frac{1}{9}=9 \).
Therefore \( 1\div\frac{1}{6}+1\div\frac{1}{9}=6+9 \) = 15.
In simple words: 1 ÷ 1/7 = 7 because 1/7 × 7 = 1; so the sum is 6 + 9 = 15.

Teacher's Note:
1. Do both divisions before adding.
2. 1 ÷ (1/n) = n.

Question 16. Find the missing numbers and show your method.
(i) By what number must \( \frac{5}{8} \) be multiplied to get \( \frac{15}{16} \)?
(ii) By what number must \( \frac{7}{10} \) be multiplied to get 1?
Answer:
The missing number is found by dividing.
(i) Missing number = \( \frac{15}{16}\div\frac{5}{8}=\frac{15}{16}\times\frac{8}{5}=\frac{120}{80} \) = \( \frac{3}{2} \). Check: \( \frac{5}{8}\times\frac{3}{2}=\frac{15}{16} \).
(ii) The number that gives 1 is the reciprocal, so \( \frac{7}{10} \) must be multiplied by \( \frac{10}{7} \).
In simple words: Divide to find a missing factor; to reach 1, use the reciprocal.

Teacher's Note:
1. Missing factor = product ÷ known factor.
2. Always check by multiplying.

Question 17. In a farm, mustard is grown on \( \frac{1}{5} \) of \( \frac{1}{4} \) of \( \frac{2}{3} \) of the whole field. If the whole field measures 480 m², find the fraction of the field under mustard and the area under mustard.
Answer:
Fraction under mustard = \( \frac{1}{5}\times\frac{1}{4}\times\frac{2}{3}=\frac{2}{60} \) = \( \frac{1}{30} \) of the field.
Area under mustard = \( \frac{1}{30}\times480=\frac{480}{30} \) = 16 m².
Check: 30 × 16 = 480 m², the whole field.
In simple words: Mustard is on 1/30 of the field, which is 16 m².

Teacher's Note:
1. Turn the chain of 'of' into one fraction first.
2. Then multiply by the total area.

Question 18. Evaluate \( \left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times\left(1-\frac{1}{5}\right)\times\left(1-\frac{1}{6}\right) \), and state the short rule you notice in the working.
Answer:
First write each bracket as a single fraction: \( \frac{2}{3}\times\frac{3}{4}\times\frac{4}{5}\times\frac{5}{6} \).
Each numerator cancels with the denominator of the bracket before it: 3 with 3, 4 with 4, 5 with 5.
Only the first numerator 2 and the last denominator 6 are left: \( \frac{2}{6} \) = \( \frac{1}{3} \).
Short rule: a chain starting from \( \left(1-\frac{1}{3}\right) \) and ending with \( \left(1-\frac{1}{n}\right) \) equals \( \frac{2}{n} \).
In simple words: Everything cancels except 2 on top and 6 below, giving 1/3.

Teacher's Note:
1. Write the brackets out before multiplying.
2. Look for the cancelling chain.

Question 19. Without actually multiplying, compare the product \( \frac{23}{19}\times\frac{31}{29} \) with (i) \( \frac{23}{19} \), (ii) \( \frac{31}{29} \), and (iii) 1. Justify each comparison.
Answer:
Both \( \frac{23}{19} \) and \( \frac{31}{29} \) are greater than 1, since each numerator is bigger than its denominator.
(i) \( \frac{23}{19} \) is multiplied by a number greater than 1, so the product is greater than \( \frac{23}{19} \).
(ii) In the same way, the product is greater than \( \frac{31}{29} \).
(iii) The product is greater than \( \frac{23}{19} \), which itself is greater than 1, so the product is greater than 1.
In simple words: Both factors are above 1, so the product beats each of them and 1.

Teacher's Note:
1. Compare each factor with 1 first.
2. No multiplication of big numbers is needed.

Question 20. A car covers \( 3\frac{1}{5} \) km using \( \frac{1}{4} \) litre of petrol.
(i) How far will it go on 1 litre of petrol?
(ii) How much petrol is needed for a journey of 64 km?
Answer:
\( 3\frac{1}{5}=\frac{16}{5} \) km on \( \frac{1}{4} \) litre.
(i) Distance on 1 litre = \( \frac{16}{5}\div\frac{1}{4}=\frac{16}{5}\times4=\frac{64}{5} \) = \( 12\frac{4}{5} \) km.
(ii) Petrol for 64 km = \( 64\div\frac{64}{5}=64\times\frac{5}{64} \) = 5 litres.
Check: \( 5\times\frac{64}{5}=64 \) km, as required.
In simple words: The car does 12 4/5 km per litre, so 64 km needs 5 litres.

Teacher's Note:
1. Find the distance per litre first.
2. Keep 64/5 as an improper fraction for part (ii).