CBSE Class 7 Mathematics Working with Fractions Worksheet Set 01

Very Short Answer Questions

Question 1. Write the reciprocal of \( \frac{9}{11} \).
Answer:
Swapping the numerator and the denominator, the reciprocal of \( \frac{9}{11} \) is \( \frac{11}{9} \).

Teacher's Note:
1. Check: 9/11 × 11/9 = 1.
2. Swap top and bottom.

Question 2. Find the product \( 4\times\frac{3}{7} \).
Answer:
\( 4\times\frac{3}{7}=\frac{4\times3}{7}=\frac{12}{7} \) = \( 1\frac{5}{7} \).

Teacher's Note:
1. Only the numerator is multiplied by 4.
2. 12 ÷ 7 = 1 remainder 5.

Question 3. Find \( \frac{1}{3}\times\frac{1}{8} \) and \( \frac{1}{8}\times\frac{1}{3} \). What do you notice about the two products?
Answer:
Both products are \( \frac{1}{3\times8}=\frac{1}{24} \).
So the order of multiplication does not change the product.

Teacher's Note:
1. For fractional units, multiply the denominators.
2. Order never matters in multiplication.

Question 4. What do we get when a fraction is multiplied by its reciprocal?
Answer:
We always get 1. For example, \( \frac{4}{7}\times\frac{7}{4}=\frac{28}{28}=1 \).

Teacher's Note:
1. This is the meaning of 'reciprocal'.
2. Give an example to support the answer.

Question 5. Write the mixed number \( 3\frac{2}{5} \) as an improper fraction.
Answer:
\( 3\frac{2}{5}=\frac{3\times5+2}{5} \) = \( \frac{17}{5} \).

Teacher's Note:
1. Whole × denominator + numerator.
2. The denominator stays the same.

Question 6. State in words Brahmagupta’s rule for multiplying two fractions.
Answer:
The product of two fractions is the product of the numerators divided by the product of the denominators, that is, \( \frac{a}{b}\times\frac{c}{d}=\frac{a\times c}{b\times d} \).

Teacher's Note:
1. Top times top, bottom times bottom.
2. Brahmagupta gave it in 628 CE.

Question 7. Find the value of \( \frac{5}{6}\times12 \).
Answer:
\( \frac{5}{6}\times12=\frac{60}{6} \) = 10.

Teacher's Note:
1. Or: 12 ÷ 6 = 2, then 5 × 2 = 10.
2. Dividing first keeps the numbers small.

Question 8. What is the ancient Indian name for the process of reducing a fraction to its lowest terms?
Answer:
It is called apavartana. The Jaina scholar Umasvati (c. 150 CE) used it even in a philosophical work.

Teacher's Note:
1. Apavartana means cancelling common factors.
2. Remember the spelling.

Question 9. Find \( 1\div\frac{3}{4} \).
Answer:
\( 1\div\frac{3}{4}=1\times\frac{4}{3} \) = \( \frac{4}{3}=1\frac{1}{3} \), since \( \frac{3}{4}\times\frac{4}{3}=1 \).

Teacher's Note:
1. 1 divided by a number gives its reciprocal.
2. Check by multiplying back.

Question 10. In the division \( \frac{5}{8}\div\frac{2}{3} \), name the dividend and the divisor.
Answer:
Dividend (the number being divided) = \( \frac{5}{8} \); divisor (the number we divide by) = \( \frac{2}{3} \).

Teacher's Note:
1. The dividend comes before the ÷ sign.
2. The divisor comes after it.

Question 11. A tortoise walks \( \frac{1}{4} \) km in one hour. How far does it walk in 3 hours?
Answer:
Distance = \( 3\times\frac{1}{4}=\frac{3}{4} \) km. So it walks \( \frac{3}{4} \) km.

Teacher's Note:
1. Repeated addition: 1/4 + 1/4 + 1/4.
2. Multiply only the numerator.

Question 12. Find the area of a rectangle whose sides are \( \frac{2}{3} \) unit and \( \frac{3}{5} \) unit.
Answer:
Area = \( \frac{2}{3}\times\frac{3}{5}=\frac{6}{15}=\frac{2}{5} \), that is \( \frac{2}{5} \) square unit.

Teacher's Note:
1. Area = length × breadth, even for fractions.
2. Reduce to the lowest form.

Question 13. Multiply \( \frac{15}{8}\times\frac{4}{9} \) and write the answer in its lowest form.
Answer:
Cancel 4 with 8 (giving 1 and 2) and 15 with 9 (giving 5 and 3): \( \frac{5\times1}{2\times3} \) = \( \frac{5}{6} \).

Teacher's Note:
1. Cancel before multiplying to keep numbers small.
2. Check: 60/72 = 5/6.

Question 14. Find \( \frac{7}{10}\times\frac{2}{3} \) in its lowest form.
Answer:
\( \frac{7}{10}\times\frac{2}{3}=\frac{14}{30} \) = \( \frac{7}{15} \) (dividing the numerator and denominator by 2).

Teacher's Note:
1. 2 and 10 share the factor 2.
2. 7 and 15 have no common factor.

Question 15. Evaluate \( \frac{9}{4}\div\frac{3}{8} \).
Answer:
\( \frac{9}{4}\div\frac{3}{8}=\frac{9}{4}\times\frac{8}{3}=\frac{72}{12} \) = 6.

Teacher's Note:
1. Flip the divisor, then multiply.
2. Check: 6 × 3/8 = 18/8 = 9/4.

Question 16. How many pieces of ribbon, each \( \frac{1}{5} \) m long, can be cut from a ribbon of length 4 m?
Answer:
Number of pieces = \( 4\div\frac{1}{5}=4\times5 \) = 20 pieces.

Teacher's Note:
1. Each metre gives 5 pieces.
2. Dividing by 1/5 is multiplying by 5.

Question 17. \( \frac{1}{3} \) litre of juice is shared equally among 6 children. How much juice does each child get?
Answer:
Each child gets \( \frac{1}{3}\div6=\frac{1}{3}\times\frac{1}{6} \) = \( \frac{1}{18} \) litre.

Teacher's Note:
1. Sharing equally means dividing.
2. Dividing by 6 is multiplying by 1/6.

Question 18. Without multiplying, state whether \( \frac{7}{9}\times\frac{5}{8} \) is greater or less than \( \frac{7}{9} \), and give the reason.
Answer:
It is less than \( \frac{7}{9} \), because \( \frac{5}{8} \) lies between 0 and 1, and multiplying by such a number makes the other number smaller.

Teacher's Note:
1. Look at the second factor only.
2. 5/8 is less than 1.

Question 19. A car covers 24 km on one litre of petrol. How far will it go on \( 1\frac{1}{2} \) litres?
Answer:
\( 1\frac{1}{2}=\frac{3}{2} \), so distance = \( 24\times\frac{3}{2} \) = 36 km.

Teacher's Note:
1. Change the mixed number first.
2. Check: 24 + 12 = 36.

Question 20. Evaluate \( 2\frac{2}{3}\div1\frac{1}{3} \).
Answer:
\( \frac{8}{3}\div\frac{4}{3}=\frac{8}{3}\times\frac{3}{4}=\frac{24}{12} \) = 2.

Teacher's Note:
1. Convert both mixed numbers first.
2. Never divide whole parts and fraction parts separately.

Question 21. Without dividing, state whether \( \frac{3}{5}\div\frac{2}{7} \) is greater or less than \( \frac{3}{5} \), and give the reason.
Answer:
It is greater than \( \frac{3}{5} \), because the divisor \( \frac{2}{7} \) lies between 0 and 1, and dividing by such a number gives a quotient larger than the dividend.

Teacher's Note:
1. Small divisor, big quotient.
2. Look only at the divisor.

Question 22. Which is heavier: \( \frac{3}{4} \) of 800 g or \( \frac{1}{5} \) of 2 kg?
Answer:
\( \frac{3}{4}\times800 \) = 600 g, and 2 kg = 2000 g, so \( \frac{1}{5}\times2000 \) = 400 g.
So \( \frac{3}{4} \) of 800 g is heavier.

Teacher's Note:
1. Use the same unit before comparing.
2. 'Of' means multiply.

Question 23. Find the value of \( \left(1-\frac{1}{4}\right)\times\left(1-\frac{1}{3}\right) \).
Answer:
\( \frac{3}{4}\times\frac{2}{3}=\frac{6}{12} \) = \( \frac{1}{2} \).

Teacher's Note:
1. Work out each bracket first.
2. Then multiply and reduce.

Question 24. The product of two fractions is \( \frac{3}{10} \) and one of them is \( \frac{5}{6} \). Find the other fraction.
Answer:
Other fraction = \( \frac{3}{10}\div\frac{5}{6}=\frac{3}{10}\times\frac{6}{5}=\frac{18}{50} \) = \( \frac{9}{25} \).

Teacher's Note:
1. A missing factor is found by division.
2. Check: 5/6 × 9/25 = 45/150 = 3/10.

Question 25. A floor of area 6 square units is to be covered completely with square tiles of side \( \frac{1}{4} \) unit. How many tiles are needed?
Answer:
Area of one tile = \( \frac{1}{4}\times\frac{1}{4}=\frac{1}{16} \) square unit.
Number of tiles = \( 6\div\frac{1}{16}=6\times16 \) = 96 tiles.

Teacher's Note:
1. Find the tile's area, not just its side.
2. Then divide the floor area by it.

Question 26. Four taps can fill a tank alone in 1 day, \( \frac{1}{2} \) day, \( \frac{1}{3} \) day and \( \frac{1}{6} \) day respectively. If all four run together, in what part of a day will the tank be filled?
Answer:
In one day the taps fill the tank 1, 2, 3 and 6 times, a total of 12 times.
So together they fill it in \( \frac{1}{12} \) of a day.

Teacher's Note:
1. 1 ÷ (1/2) = 2 fillings a day, and so on.
2. This problem is from ancient Indian texts.

Question 27. The government takes \( \frac{1}{6} \) of Somu’s land for a road. She gives half of the remaining land to her daughter. What part of the original land does the daughter get?
Answer:
Land left = \( 1-\frac{1}{6}=\frac{5}{6} \). Daughter gets \( \frac{1}{2}\times\frac{5}{6} \) = \( \frac{5}{12} \) of the original land.

Teacher's Note:
1. Half of the remaining, not half of the whole.
2. Find what is left first.

Question 28. Evaluate \( \left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times\cdots\times\left(1-\frac{1}{10}\right) \).
Answer:
The product is \( \frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\cdots\times\frac{9}{10} \).
Each numerator cancels the previous denominator, leaving \( \frac{1}{10} \).

Teacher's Note:
1. Look for the cancelling pattern.
2. Only the first numerator and last denominator stay.

Question 29. Is the product \( \frac{565}{465}\times\frac{707}{676} \) greater or less than \( \frac{565}{465} \)? Give the reason without multiplying.
Answer:
It is greater than \( \frac{565}{465} \), because \( \frac{707}{676} \) is greater than 1 (707 > 676), and multiplying by a number greater than 1 makes a number larger.

Teacher's Note:
1. Compare the numerator and denominator of the other factor.
2. No long multiplication is needed.

Question 30. A trader gave away \( \frac{1}{4} \) of \( \frac{2}{3} \) of \( \frac{3}{5} \) of a dramma. If 1 dramma equals 1280 cowrie shells, how many cowrie shells did he give away?
Answer:
Amount = \( \frac{1}{4}\times\frac{2}{3}\times\frac{3}{5}=\frac{6}{60}=\frac{1}{10} \) of a dramma.
Cowrie shells = \( \frac{1}{10}\times1280 \) = 128 cowrie shells.

Teacher's Note:
1. Multiply all the 'of' fractions first.
2. Then multiply by 1280.