Chapter 4 – Operations on Fractions
Practice set 9
1. Convert into improper fractions.
(i) 7 \(\large \frac {2}{5}\)
Solution:
7 \(\large \frac {2}{5}\) = \(\large \frac {7 \,×\, 5 \,+\, 2}{5}\)
∴ 7 \(\large \frac {2}{5}\) = \(\large \frac {37}{5}\)
(ii) 5 \(\large \frac {1}{6}\)
Solution:
5 \(\large \frac {1}{6}\) = \(\large \frac {5 \,×\, 6 \,+\, 1}{6}\)
∴ 5 \(\large \frac {1}{6}\) = \(\large \frac {31}{6}\)
(iii) 4 \(\large \frac {3}{4}\)
Solution:
4 \(\large \frac {3}{4}\) = \(\large \frac {4 \,×\, 4 \,+\, 3}{4}\)
∴ 4 \(\large \frac {3}{4}\) = \(\large \frac {19}{4}\)
(iv) 2 \(\large \frac {5}{9}\)
Solution:
2 \(\large \frac {5}{9}\) = \(\large \frac {2 \,×\, 9 \,+\, 5}{9}\)
∴ 2 \(\large \frac {5}{9}\) = \(\large \frac {23}{9}\)
(v) 1 \(\large \frac {5}{7}\)
Solution:
1 \(\large \frac {5}{7}\) = \(\large \frac {1 \,×\, 7 \,+\, 5}{7}\)
∴ 1 \(\large \frac {5}{7}\) = \(\large \frac {12}{7}\)
2. Convert into mixed numbers.
(i) \(\large \frac {30}{7}\)
Solution:
∴ \(\large \frac {30}{7}\) = 4 \(\large \frac {2}{7}\)
(ii) \(\large \frac {7}{4}\)
Solution:
∴ \(\large \frac {7}{4}\) = 1 \(\large \frac {3}{4}\)
(iii) \(\large \frac {15}{12}\)
Solution:
\(\large \frac {15}{12}\) = \(\large \frac {3 \,×\, 5}{3 \,×\, 4}\) = \(\large \frac {5}{4}\)
∴ \(\large \frac {15}{12}\) = 1 \(\large \frac {1}{4}\)
(iv) \(\large \frac {11}{8}\)
Solution:
∴ \(\large \frac {11}{8}\) = 1 \(\large \frac {3}{8}\)
(v) \(\large \frac {21}{4}\)
Solution:
∴ \(\large \frac {21}{4}\) = 5 \(\large \frac {1}{4}\)
(vi) \(\large \frac {20}{7}\)
Solution:
∴ \(\large \frac {20}{7}\) = 2 \(\large \frac {6}{7}\)
3. Write the following examples using fractions.
(i) If 9 kg rice is shared amongst 5 people, how many kilograms of rice does each person get?
Solution:
Total quantity of rice = 9 kg
Number of people = 5
∴ Each person will get \(\large \frac {9}{5}\) kg rice.
(ii) To make 5 shirts of the same size, 11 metres of cloth is needed. How much cloth is needed for one shirt?
Solution:
Total meters of cloth = 11 meters
Number of shirts to be made = 5
∴ Cloth needed to make 1 shirt is \(\large \frac {11}{5}\) meters.
Practice set 10
1. Add.
(i) 6 \(\large \frac {1}{3}\) + 2 \(\large \frac {1}{3}\)
Solution:
6 \(\large \frac {1}{3}\) + 2 \(\large \frac {1}{3}\)
= \(\large \frac {6 \,×\, 3 \,+\, 1}{3}\) + \(\large \frac {2 \,×\, 3 \,+\, 1}{3}\)
= \(\large \frac {19}{3}\) + \(\large \frac {7}{3}\)
= \(\large \frac {19 \,+\,7}{3}\)
= \(\large \frac {26}{3}\)
= 8 \(\large \frac {2}{3}\)
∴ 6 \(\large \frac {1}{3}\) + 2 \(\large \frac {1}{3}\) = 8 \(\large \frac {2}{3}\)
(ii) 1 \(\large \frac {1}{4}\) + 3 \(\large \frac {1}{2}\)
Solution:
1 \(\large \frac {1}{4}\) + 3 \(\large \frac {1}{2}\)
= \(\large \frac {1 \,×\, 4 \,+\, 1}{4}\) + \(\large \frac {3 \,×\, 2 \,+\, 1}{2}\)
= \(\large \frac {5}{4}\) + \(\large \frac {7}{2}\)
= \(\large \frac {5}{4}\) + \(\large \frac {7\,×\,2}{2\,×\,2}\)
= \(\large \frac {5}{4}\) + \(\large \frac {14}{4}\)
= \(\large \frac {5\,+\,14}{4}\)
= \(\large \frac {19}{4}\)
= 4 \(\large \frac {3}{4}\)
∴ 1 \(\large \frac {1}{4}\) + 3 \(\large \frac {1}{2}\) = 4 \(\large \frac {3}{4}\)
(iii) 5 \(\large \frac {1}{5}\) + 2 \(\large \frac {1}{7}\)
Solution:
5 \(\large \frac {1}{5}\) + 2 \(\large \frac {1}{7}\)
= \(\large \frac {5\,×\,5\,+\,1}{5}\) + \(\large \frac {2\,×\,7\,+\,1}{7}\)
= \(\large \frac {26}{5}\) + \(\large \frac {15}{7}\)
= \(\large \frac {26\,×\,7}{5\,×\,7}\) + \(\large \frac {15\,×\,5}{7\,×\,5}\)
= \(\large \frac {182}{35}\) + \(\large \frac {75}{35}\)
= \(\large \frac {182\,+\,75}{35}\)
= \(\large \frac {257}{35}\)
= 7 \(\large \frac {12}{35}\)
∴ 5 \(\large \frac {1}{5}\) + 2 \(\large \frac {1}{7}\) = 7 \(\large \frac {12}{35}\)
(iv) 3 \(\large \frac {1}{5}\) + 2 \(\large \frac {1}{3}\)
Solution:
3 \(\large \frac {1}{5}\) + 2 \(\large \frac {1}{3}\)
= \(\large \frac {3 \,×\, 5 \,+\, 1}{5}\) + \(\large \frac {2 \,×\, 3 \,+\, 1}{3}\)
= \(\large \frac {16}{5}\) + \(\large \frac {7}{3}\)
= \(\large \frac {16\,×\,3}{5\,×\,3}\) + \(\large \frac {7\,×\,5}{3\,×\,5}\)
= \(\large \frac {48}{15}\) + \(\large \frac {35}{15}\)
= \(\large \frac {48\,+\,35}{15}\)
= \(\large \frac {83}{15}\)
= 5 \(\large \frac {8}{15}\)
∴ 3 \(\large \frac {1}{5}\) + 2 \(\large \frac {1}{3}\) = 5 \(\large \frac {8}{15}\)
2. Subtract.
(i) 3 \(\large \frac {1}{3}\) – 1 \(\large \frac {1}{4}\)
Solution:
3 \(\large \frac {1}{3}\) – 1 \(\large \frac {1}{4}\)
= \(\large \frac {3 \,×\, 3 \,+\, 1}{3}\) – \(\large \frac {1 \,×\, 4 \,+\, 1}{4}\)
= \(\large \frac {10}{3}\) – \(\large \frac {5}{4}\)
= \(\large \frac {10\,×\,4}{3\,×\,4}\) – \(\large \frac {5\,×\,3}{4\,×\,3}\)
= \(\large \frac {40}{12}\) – \(\large \frac {15}{12}\)
= \(\large \frac {40\,–\,15}{12}\)
= \(\large \frac {25}{12}\)
= 2 \(\large \frac {1}{12}\)
∴ 3 \(\large \frac {1}{3}\) – 1 \(\large \frac {1}{4}\) = 2 \(\large \frac {1}{12}\)
(ii) 5 \(\large \frac {1}{2}\) – 3 \(\large \frac {1}{3}\)
Solution:
5 \(\large \frac {1}{2}\) – 3 \(\large \frac {1}{3}\)
= \(\large \frac {5 \,×\, 2 \,+\, 1}{2}\) – \(\large \frac {3 \,×\, 3 \,+\, 1}{3}\)
= \(\large \frac {11}{2}\) – \(\large \frac {10}{3}\)
= \(\large \frac {11\,×\,3}{2\,×\,3}\) – \(\large \frac {10\,×\,2}{3\,×\,2}\)
= \(\large \frac {33}{6}\) – \(\large \frac {20}{6}\)
= \(\large \frac {33\,–\,20}{6}\)
= \(\large \frac {13}{6}\)
= 2 \(\large \frac {1}{6}\)
∴ 5 \(\large \frac {1}{2}\) – 3 \(\large \frac {1}{3}\) = 2 \(\large \frac {1}{6}\)
(iii) 7 \(\large \frac {1}{8}\) – 6 \(\large \frac {1}{10}\)
Solution:
7 \(\large \frac {1}{8}\) – 6 \(\large \frac {1}{10}\)
= \(\large \frac {7 \,×\, 8 \,+\, 1}{8}\) – \(\large \frac {6 \,×\, 10 \,+\, 1}{10}\)
= \(\large \frac {57}{8}\) – \(\large \frac {61}{10}\)
= \(\large \frac {57\,×\,5}{8\,×\,5}\) – \(\large \frac {61\,×\,4}{10\,×\,4}\)
= \(\large \frac {285}{40}\) – \(\large \frac {244}{40}\)
= \(\large \frac {285\,–\,244}{40}\)
= \(\large \frac {41}{40}\)
= 1 \(\large \frac {1}{40}\)
∴ 7 \(\large \frac {1}{8}\) – 6 \(\large \frac {1}{10}\) = 1 \(\large \frac {1}{40}\)
(iv) 7 \(\large \frac {1}{2}\) – 3 \(\large \frac {1}{5}\)
Solution:
7 \(\large \frac {1}{2}\) – 3 \(\large \frac {1}{5}\)
= \(\large \frac {7 \,×\, 2 \,+\, 1}{2}\) – \(\large \frac {3 \,×\, 5 \,+\, 1}{5}\)
= \(\large \frac {15}{2}\) – \(\large \frac {16}{5}\)
= \(\large \frac {15\,×\,5}{2\,×\,5}\) – \(\large \frac {16\,×\,2}{5\,×\,2}\)
= \(\large \frac {75}{10}\) – \(\large \frac {32}{10}\)
= \(\large \frac {75\,–\,32}{10}\)
= \(\large \frac {43}{10}\)
= 4 \(\large \frac {3}{10}\)
∴ 7 \(\large \frac {1}{2}\) – 3 \(\large \frac {1}{5}\) = 4 \(\large \frac {3}{10}\)
3. Solve.
(1) Suyash bought 2 \(\large \frac {1}{2}\) kg of sugar and Ashish bought 3 \(\large \frac {1}{2}\) kg. How much sugar did they buy altogether? If sugar costs 32 rupees per kg, how much did they spend on the sugar they bought?
Solution:
Sugar bought by Suyash = 2 \(\large \frac {1}{2}\) kg
Sugar bought by Ashish = 3 \(\large \frac {1}{2}\) kg
∴ Total sugar bought by them
= 2 \(\large \frac {1}{2}\) + 3 \(\large \frac {1}{2}\) kg
= \(\large \frac {2 \,×\, 2 \,+\, 1}{2}\) + \(\large \frac {3\,×\, 2 \,+\, 1}{2}\)
= \(\large \frac {5}{2}\) + \(\large \frac {7}{2}\)
= \(\large \frac {5\,+\,7}{2}\)
= \(\large \frac {12}{2}\)
= 6 kg
Cost of 1 kg of sugar = ₹ 32
∴ Cost of 6 kg of sugar
= 32 × 6
= ₹ 192
Ans: They bought 6 kg sugar altogether and the total money spent on sugar is ₹ 192.
(2) Aradhana grows potatoes in \(\large \frac {2}{5}\) part of her garden, greens in \(\large \frac {1}{3}\) part and brinjals in the remaining part. On how much of her plot did she plant brinjals?
Solution:
Part of garden occupied by potatoes = \(\large \frac {2}{5}\) part
Part of garden occupied by greens = \(\large \frac {1}{3}\) part
Remaining part of garden is occupied by brinjals
∴ (Part occupied by potatoes) + (Part occupied by greens) + (Part occupied by brinjals) = 1 (i.e. The entire garden)
∴ Part of garden occupied by brinjals
= 1 – (Part of garden occupied by potatoes) – (Part of garden occupied by greens)
= 1 – \(\large \frac {2}{5}\) – \(\large \frac {1}{3}\)
= 1 – \(\large (\frac {2}{5}\) + \(\large \frac {1}{3})\)
= 1 – \(\large (\frac {2\,×\,3}{5\,×\,3}\) + \(\large \frac {1\,×\,5}{3\,×\,5})\)
= 1 – \(\large (\frac {6}{15}\) + \(\large \frac {5}{15})\)
= 1 – \(\large (\frac {6\,+\,5}{15})\)
= 1 – \(\large \frac {11}{15}\)
= \(\large \frac {1\,×\,15}{1\,×\,15}\) – \(\large \frac {11}{15}\)
= \(\large \frac {15}{15}\) – \(\large \frac {11}{15}\)
= \(\large (\frac {15\,–\,11}{15})\)
= \(\large \frac {4}{15}\) part
Ans: Aradhana planted brinjals on \(\large \frac {4}{15}\) part of her plot.
(3) Sandeep filled water in \(\large \frac {4}{7}\) of an empty tank. After that, Ramakant filled \(\large \frac {1}{4}\) part more of the same tank. Then Umesh used \(\large \frac {3}{14}\) part of the tank to water the garden. If the tank has a maximum capacity of 560 litres, how many litres of water will be left in the tank?
Solution:
Part of tank filled by Sandeep = \(\large \frac {4}{7}\)
Part of tank filled by Ramakant = \(\large \frac {1}{4}\)
Part of tank filled by both of them
= \(\large \frac {4}{7}\) + \(\large \frac {1}{4}\)
= \(\large \frac {4\,×\,4}{7\,×\,4}\) + \(\large \frac {1\,×\,7}{4\,×\,7}\)
= \(\large \frac {16}{28}\) + \(\large \frac {7}{28}\)
= \(\large (\frac {16\,+\,7}{28})\)
= \(\large \frac {23}{28}\) part
Part of tank used by Umesh = \(\large \frac {3}{14}\) part
∴ Part of tank still filled with water
= \(\large \frac {23}{28}\) – \(\large \frac {3}{14}\)
= \(\large \frac {23}{28}\) – \(\large \frac {3\,×\,2}{14\,×\,2}\)
= \(\large \frac {23}{28}\) – \(\large \frac {6}{28}\)
= \(\large \frac {23\,–\,6}{28}\)
= \(\large \frac {17}{28}\)
Capacity of the tank = 560 litres
∴ Quantity of water left in tank
= \(\large \frac {17}{28}\) × 560
= 340 litres
Ans: The quantity of water left in the tank is 340 litres.
Practice set 11
1. What fractions do the points A and B show on the number lines below?
Solution:
Every unit is divided in 6 parts
A is 5th division from 0
∴ A = \(\large \frac {5}{6}\)
B is 10th division from 0
∴ B = \(\large \frac {10}{6}\)
Solution:
Every unit is divided in 5 parts
A is 3rd division from 0
∴ A = \(\large \frac {3}{5}\)
B is 7th division from 0
∴ B = \(\large \frac {7}{5}\)
Solution:
Every unit is divided in 7 parts
A is 10th division from 0
∴ A = \(\large \frac {7}{10}\)
B is 3rd division from 0
∴ B = \(\large \frac {3}{10}\)
2. Show the following fractions on the number line.
(1) \(\large \frac {3}{5}\), \(\large \frac {6}{5}\), 2 \(\large \frac {3}{5}\)
Solution:
(2) \(\large \frac {3}{4}\), \(\large \frac {5}{4}\), 2 \(\large \frac {1}{4}\)
Solution:
Practice set 12
1. Multiply.
(i) \(\large \frac {7}{5}\) × \(\large \frac {1}{4}\)
Solution:
\(\large \frac {7}{5}\) × \(\large \frac {1}{4}\)
= \(\large \frac {7 \,×\, 1}{5 \,×\,4 }\)
= \(\large \frac {7}{5}\)
∴ \(\large \frac {7}{5}\) × \(\large \frac {1}{4}\) = \(\large \frac {7}{5}\)
(ii) \(\large \frac {6}{7}\) × \(\large \frac {2}{5}\)
Solution:
\(\large \frac {6}{7}\) × \(\large \frac {2}{5}\)
= \(\large \frac {6 \,×\,2 }{7\,×\, 5}\)
= \(\large \frac {12}{35}\)
∴ \(\large \frac {6}{7}\) × \(\large \frac {2}{5}\) = \(\large \frac {12}{35}\)
(iii) \(\large \frac {5}{9}\) × \(\large \frac {4}{9}\)
Solution:
\(\large \frac {5}{9}\) × \(\large \frac {4}{9}\)
= \(\large \frac {5 \,×\, 4}{9 \,×\,9}\)
= \(\large \frac {20}{81}\)
∴ \(\large \frac {5}{9}\) × \(\large \frac {4}{9}\) = \(\large \frac {20}{81}\)
(iv) \(\large \frac {4}{11}\) × \(\large \frac {2}{7}\)
Solution:
\(\large \frac {4}{11}\) × \(\large \frac {2}{7}\)
= \(\large \frac {4 \,×\, 2}{11 \,×\,7}\)
= \(\large \frac {8}{77}\)
∴ \(\large \frac {4}{11}\) × \(\large \frac {2}{7}\) = \(\large \frac {8}{77}\)
(v) \(\large \frac {1}{5}\) × \(\large \frac {7}{2}\)
Solution:
\(\large \frac {1}{5}\) × \(\large \frac {7}{2}\)
= \(\large \frac {1 \,×\, 7}{5 \,×\, 2}\)
= \(\large \frac {7}{10}\)
∴ \(\large \frac {1}{5}\) × \(\large \frac {7}{2}\) = \(\large \frac {7}{10}\)
(vi) \(\large \frac {9}{7}\) × \(\large \frac {7}{8}\)
Solution:
\(\large \frac {9}{7}\) × \(\large \frac {7}{8}\)
= \(\large \frac {9 \,×\, 7}{7 \,×\, 8}\)
= \(\large \frac {63}{56}\)
∴ \(\large \frac {9}{7}\) × \(\large \frac {7}{8}\) = \(\large \frac {63}{56}\)
(vii) \(\large \frac {5}{6}\) × \(\large \frac {26}{5}\)
Solution:
\(\large \frac {5}{6}\) × \(\large \frac {26}{5}\)
= \(\large \frac {5\,×\, 2}{6\,×\, 5}\)
= \(\large \frac {10}{30}\)
∴ \(\large \frac {5}{6}\) × \(\large \frac {26}{5}\) = \(\large \frac {10}{30}\)
(viii) \(\large \frac {6}{17}\) × \(\large \frac {3}{2}\)
Solution:
\(\large \frac {6}{17}\) × \(\large \frac {3}{2}\)
= \(\large \frac {6 \,×\, 3}{17 \,×\, 2}\)
= \(\large \frac {18}{34}\)
= \(\large \frac {9}{17}\)
∴ \(\large \frac {6}{17}\) × \(\large \frac {3}{2}\) = \(\large \frac {9}{17}\)
2. Ashokrao planted bananas on \(\large \frac {2}{7}\) of his field of 21 acres. What is the area of the banana plantation?
Solution:
Area of banana plantation is \(\large \frac {2}{7}\) of 21 acres field
∴ Area of banana plantation
= \(\large \frac {2}{7}\) × 21
= \(\large \frac {2\,×\,21}{7}\)
= \(\large \frac {42}{7}\)
= 6 acres
Ans: Area of banana plantation is 6 acres.
3. Of the total number of soldiers in our army, \(\large \frac {4}{9}\) are posted on the northern border and one‑third of them on the north‑eastern border. If the number of soldiers in the north is 540000, how many are posted in the north‑east?
Solution:
Number of soldiers posted on northern border = 5,40,000
Number of soldiers in north-east is \(\large \frac {1}{3}\) of the northern border
∴ Number of soldiers in the north-east
= \(\large \frac {1}{3}\) × 540000
= \(\large \frac {1\,×\,540000}{3}\)
= \(\large \frac {540000}{3}\)
= 1,80,000 soldiers
Ans: The number of soldiers in the north-east is 1,80,000.
Practice set 13
1. Write the reciprocals of the following numbers.
(i) 7
Solution:
Reciprocal of 7 = \(\large \frac {1}{7}\)
(ii) \(\large \frac {11}{3}\)
Solution:
Reciprocal of \(\large \frac {11}{3}\) = \(\large \frac {3}{11}\)
(iii) \(\large \frac {5}{13}\)
Solution:
Reciprocal of \(\large \frac {5}{13}\) = \(\large \frac {13}{5}\)
(iv) 2
Solution:
Reciprocal of 2 = \(\large \frac {1}{2}\)
(v) \(\large \frac {6}{7}\)
Solution:
Reciprocal of \(\large \frac {6}{7}\) = \(\large \frac {7}{6}\)
2. Carry out the following divisions.
(i) \(\large \frac {11}{3}\) ÷ \(\large \frac {11}{3}\)
Solution:
\(\large \frac {11}{3}\) ÷ \(\large \frac {11}{3}\)
= \(\large \frac {11}{3}\) × \(\large \frac {3}{11}\)
= \(\large \frac {11 \,×\, 3}{3 \,×\, 11}\)
= \(\large \frac {33}{33}\)
= 1
∴ \(\large \frac {11}{3}\) ÷ \(\large \frac {11}{3}\) = 1
(ii) \(\large \frac {5}{9}\) ÷ \(\large \frac {3}{2}\)
Solution:
\(\large \frac {5}{9}\) ÷ \(\large \frac {3}{2}\)
= \(\large \frac {5}{8}\) × \(\large \frac {2}{3}\)
= \(\large \frac {5 \,×\, 2}{8 \,×\, 3}\)
= \(\large \frac {10}{24}\)
= \(\large \frac {5}{12}\)
∴ \(\large \frac {5}{9}\) ÷ \(\large \frac {3}{2}\) = \(\large \frac {5}{12}\)
(iii) \(\large \frac {3}{7}\) ÷ \(\large \frac {5}{11}\)
Solution:
\(\large \frac {3}{7}\) ÷ \(\large \frac {5}{11}\)
= \(\large \frac {3}{7}\) × \(\large \frac {11}{5}\)
= \(\large \frac {3 \,×\, 11}{7 \,×\, 5}\)
= \(\large \frac {33}{35}\)
∴ \(\large \frac {3}{7}\) ÷ \(\large \frac {5}{11}\) = \(\large \frac {33}{35}\)
(iv) \(\large \frac {11}{12}\) ÷ \(\large \frac {4}{7}\)
Solution:
\(\large \frac {11}{12}\) ÷ \(\large \frac {4}{7}\)
= \(\large \frac {11}{12}\) × \(\large \frac {7}{4}\)
= \(\large \frac {11 \,×\, 7}{12 \,×\, 4}\)
= \(\large \frac {77}{47}\)
∴ \(\large \frac {11}{12}\) ÷ \(\large \frac {4}{7}\) = \(\large \frac {77}{47}\)
3. There were 420 students participating in the Swachh Bharat campaign. They cleaned \(\large \frac {42}{75}\) part of the town, Sevagram. What part of Sevagram did each student clean if the work was equally shared by all?
Solution:
Total number of students = 420
Part of town cleaned by all the students = \(\large \frac {42}{75}\) part
∴ Part of town cleaned by one student
= \(\large \frac {42}{75}\) ÷ 420
= \(\large \frac {42}{75}\) × \(\large \frac {1}{420}\)
= \(\large \frac {42 \,×\, 1}{75 \,×\, 420}\)
= \(\large \frac {1 \,×\, 1}{75 \,×\, 10}\)
= \(\large \frac {1}{750}\)
Ans: Part of town cleaned by one student is \(\large \frac {1}{750}\).