NCERT Solutions Class 6 Maths Chapter 7 fractions Exercise 7.8

Exercise 7.8 of NCERT Class 6 Maths Chapter 7 – Fractions is the final practice exercise that allows students to consolidate all aspects of the chapter. This exercise consists of various application-based questions that consolidate students' understanding of how to compare fractions, add fractions, subtract fractions, and simplify both like and unlike fractions in various contexts. This Exercise really is a fantastic wrap-up exercise that enables students to test their learning outcomes from the entire chapter on fractions.

ALLEN's NCERT Solutions Class 6 Maths for Exercise 7.8 provides comprehensive solutions, simplifying explanations, and step-by-step instructions for solving each question. This way students can improve their academic performance as well as acquire the real-world understanding of fractions that they must master, if they want a chance to progress in higher-level Maths.

1.0Download NCERT Solutions Class 6 Maths Chapter 7 fractions Exercise 7.8: Free PDF

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2.0NCERT Solutions Class 6 Chapter 7 Fractions: All Exercises

3.0NCERT Class 6 Maths Chapter 7 fractions Exercise 7.8: Detailed Solutions

• Add the following fractions using Brahmagupta's method: (a) $\frac{2}{7}+\frac{5}{7}+\frac{6}{7}$ (b) $\frac{3}{4}+\frac{1}{3}$ (c) $\frac{2}{3}+\frac{5}{6}$ (d) $\frac{2}{3}+\frac{2}{7}$ (e) $\frac{2}{3}+\frac{4}{5}$ (g) $\frac{4}{5}+\frac{2}{3}$ (h) $\frac{3}{5}+\frac{5}{8}$ (i) $\frac{9}{2}+\frac{5}{4}$ (j) $\frac{8}{3}+\frac{2}{7}$ (k) $\frac{3}{4}+\frac{1}{3}+\frac{1}{5}$ (i) $\frac{2}{5}+\frac{4}{5}+\frac{3}{7}$ (m) $\frac{9}{2}+\frac{5}{4}+\frac{7}{6}$ Sol. (a) $\frac{2}{7}+\frac{5}{7}+\frac{6}{7}=\frac{2+5+6}{7}=\frac{13}{7}$ (b) $\frac{3}{4}\times \frac{3}{3}=\frac{9}{12};\frac{1}{3}\times \frac{4}{4}=\frac{4}{12}$ $\frac{3}{4}+\frac{1}{3}=\frac{9}{12}+\frac{4}{12}=\frac{9+4}{12}=\frac{13}{12}$ (c) $\frac{2}{3}\times \frac{2}{2}=\frac{4}{6};\frac{5\times 1}{6\times 1}=\frac{5}{6}$ $\frac{2}{3}+\frac{5}{6}=\frac{4}{6}+\frac{5}{6}=\frac{4+5}{6}=\frac{10}{6}=\frac{10÷2}{6÷2}=\frac{5}{3}$ (d) $\frac{2\times 7}{3\times 7}=\frac{14}{21};\frac{2\times 3}{7\times 3}=\frac{6}{21}$ $\frac{2}{3}+\frac{2}{7}=\frac{14}{21}+\frac{6}{21}=\frac{14+6}{21}=\frac{20}{21}$ (e) $\frac{3}{4}+\frac{1}{3}+\frac{1}{5}$ The smallest common multiple of $3,4,5$ is 60 . $\frac{3\times 15}{4\times 15}=\frac{45}{60};\frac{1\times 20}{3\times 20}=\frac{20}{60};\frac{1\times 12}{5\times 12}=\frac{12}{60}$ $\frac{3}{4}+\frac{1}{3}+\frac{1}{5}=\frac{45}{60}+\frac{20}{60}+\frac{12}{60}=\frac{45+20+12}{60}=\frac{77}{60}$ (f) $\frac{2\times 5}{3\times 5}=\frac{10}{15};\frac{4\times 3}{5\times 3}=\frac{12}{15}$ $\frac{2}{3}+\frac{4}{5}=\frac{10}{15}+\frac{12}{15}=\frac{10+12}{15}=\frac{22}{15}$ (g) $\frac{4\times 3}{5\times 3}=\frac{12}{15};\frac{2\times 5}{3\times 5}=\frac{10}{15}$ $\frac{4}{5}+\frac{2}{3}=\frac{12}{15}+\frac{10}{15}=\frac{12+10}{15}=\frac{22}{15}$ (h) The smallest common multiple of 5 and 8 is 40 $\frac{3\times 8}{5\times 8}=\frac{24}{40};\frac{5\times 5}{8\times 5}=\frac{25}{40}$ $\frac{24}{40}+\frac{25}{40}=\frac{24+25}{40}=\frac{49}{40}$ (i) $\frac{9\times 2}{2\times 2}=\frac{18}{4};\frac{5\times 1}{4\times 1}=\frac{5}{4}$ $\frac{18}{4}+\frac{5}{4}=\frac{18+5}{4}=\frac{23}{4}$ (j) $\frac{8\times 7}{3\times 7}=\frac{56}{21};\frac{2\times 3}{7\times 3}=\frac{6}{21}$ $\frac{56}{21}+\frac{6}{21}=\frac{56+6}{21}=\frac{62}{21}$ (k) The smallest common multiple of $3,4,5$ is 60 . $\frac{3\times 15}{4\times 15}=\frac{45}{60};\frac{1\times 20}{3\times 20}=\frac{20}{60};\frac{1\times 12}{8\times 12}=\frac{12}{60}$ $\frac{45}{60}+\frac{20}{60}+\frac{12}{60}=\frac{45+20+12}{60}=\frac{77}{60}$ (1) The smallest common multiple of $3,5,7$ is 105 . $\frac{2\times 35}{3\times 35}=\frac{70}{105};\frac{4\times 21}{5\times 21}=\frac{84}{105};\frac{3\times 15}{7\times 15}=\frac{45}{105}$ $\frac{70}{105}+\frac{84}{105}+\frac{45}{105}=\frac{70+84+45}{105}=\frac{199}{105}$ (m) The smallest common multiple of $2,4,6$, is 12 . $\begin{array}{rl}& \frac{9\times 6}{2\times 6}=\frac{54}{12};\frac{5\times 3}{4\times 3}=\frac{15}{12};\frac{7\times 2}{6\times 2}=\frac{14}{12}\\ & \frac{54}{12}+\frac{15}{12}+\frac{14}{12}=\frac{54+15+14}{12}=\frac{83}{12}\end{array}$
• Rahim mixes $\frac{2}{3}$ litres of yellow paint with $\frac{3}{4}$ litres of blue paint to make green paint. What is the volume of green paint he has made? Sol. Volume of green paint $=\frac{2}{3}+\frac{3}{4}$ $\frac{2\times 4}{3\times 4}=\frac{8}{12};\frac{3\times 3}{4\times 3}=\frac{9}{12}$ $\frac{2}{3}+\frac{3}{4}=\frac{8}{12}+\frac{9}{12}=\frac{8+9}{12}=\frac{17}{12}$.
• Geeta bought $\frac{2}{5}$ meter of lace and Shamim bought $\frac{3}{4}$ meter of the same lace to put a complete border on a table cloth whose perimeter is 1 meter long. Find the total length of the lace they both have bought. Will the lace be sufficient to cover the whole border? Sol. Total length of the lace Geeta and Shamim have bought $=\frac{2}{5}+\frac{3}{4}$ The smallest common multiple of 5 and 4 is 20 . $\frac{2\times 4}{5\times 4}=\frac{8}{20};\frac{3\times 5}{4\times 5}=\frac{15}{20}$ $\frac{8}{20}+\frac{15}{20}=\frac{23}{20}\mathrm{m}=1\frac{3}{20}\mathrm{m}$ $=1\mathrm{m}+\frac{3}{20}\times 100\mathrm{cm}$ $=1\mathrm{m}15\mathrm{cm}$. Since perimeter of border on table cloth as 1 m . Here, length of lack is greater than perimeter, So, the required length of lace will be sufficient to cover the whole border.
• $\frac{5}{8}-\frac{3}{8}$ Sol. $\frac{5}{8}-\frac{3}{8}=\frac{5-3}{8}=\frac{2}{8}=\frac{2÷2}{8÷2}=\frac{1}{4}$
• $\frac{7}{9}-\frac{5}{9}$ Sol. $\frac{7}{9}-\frac{5}{9}=\frac{7-5}{9}=\frac{2}{9}$
• $\frac{10}{27}-\frac{1}{27}$ Sol. $\frac{10}{27}-\frac{1}{27}=\frac{10-1}{27}=\frac{9}{27}=\frac{9÷9}{27÷9}=\frac{1}{3}$
• Carry out the following subtractions using Brahmagupta's method: (a) $\frac{8}{15}-\frac{3}{15}$ (b) $\frac{2}{5}-\frac{4}{15}$ (c) $\frac{5}{6}-\frac{4}{9}$ (d) $\frac{2}{3}-\frac{1}{2}$ Sol. (a) $\frac{8}{15}-\frac{3}{15}$ $\Rightarrow \frac{8-3}{15}\{\because$ same denominators $\}$ $\Rightarrow \frac{5}{15}$ (b) $\frac{2}{5}-\frac{4}{15}$ Converting given fraction having same denominator into equivalent fraction, $\frac{2\times 3}{5\times 3}=\frac{6}{15};\frac{4\times 1}{15\times 1}=\frac{4}{15}$ $\Rightarrow \frac{6}{15}-\frac{4}{15}=\frac{6-4}{15}=\frac{2}{15}$. (c) $\frac{5}{6}-\frac{4}{9}$ Converting given fraction into equivalent fraction having same denominator, $\frac{5\times 3}{6\times 3}=\frac{15}{18};\frac{4\times 2}{9\times 2}=\frac{8}{18}$ $\Rightarrow \frac{15}{18}-\frac{8}{18}=\frac{15-8}{18}=\frac{7}{18}$. (d) $\frac{2}{3}-\frac{1}{2}$ Converting given fraction into equivalent fraction having same denominator, $\frac{2\times 2}{3\times 2}=\frac{4}{6};\frac{1}{2}\times \frac{3}{3}=\frac{3}{6}$ $\Rightarrow \frac{4}{6}-\frac{3}{6}=\frac{4-3}{6}=\frac{1}{6}$
• Subtract as indicated: (a) $\frac{13}{4}$ from $\frac{10}{3}$ (b) $\frac{18}{5}$ from $\frac{23}{3}$ (c) $\frac{29}{7}$ from $\frac{45}{7}$ Sol. (a) $\frac{13}{4}$ from $\frac{10}{3}\Rightarrow \frac{10}{3}-\frac{13}{4}$ $\begin{array}{rl}& \frac{10\times 4}{3\times 4}=\frac{40}{12};\frac{13\times 3}{4\times 3}=\frac{39}{12}\\ & \Rightarrow \frac{40}{12}-\frac{39}{2}=\frac{1}{12}\end{array}$ (b) $\frac{18}{5}$ from $\frac{23}{3}\Rightarrow \frac{23}{3}-\frac{18}{5}$ $\begin{array}{rl}& \frac{23\times 5}{3\times 5}=\frac{115}{15};\frac{18\times 3}{5\times 3}=\frac{54}{15}\\ & \Rightarrow \frac{115}{15}-\frac{54}{15}=\frac{115-54}{15}=\frac{61}{15}\end{array}$ (c) $\frac{29}{7}$ from $\frac{45}{7}\Rightarrow \frac{45}{7}-\frac{29}{7}$ $\frac{45-29}{7}=\frac{16}{7}$
• Solve the following problems: (a) Jaya's school is $\frac{7}{10}\mathrm{km}$ from her home. She takes an auto for $\frac{1}{2}\mathrm{km}$ from her home daily, and then walks the remaining distance to reach her school. How much does she walk daily to reach the school? (b) Jeevika takes $\frac{10}{3}$ minutes to take a complete round of the park and her friend Namit takes $\frac{13}{4}$ minutes to do the same. Who takes less time and by how much? Sol. (a)
  
  Distance between home and school $=\frac{7}{10}\mathrm{km}$. Distance travelled by auto $=\frac{1}{2}\mathrm{km}$. So, remaining distance travelled by walk. $\Rightarrow \frac{7}{10}-\frac{1}{2}$ $\frac{7\times 1}{10\times 1}=\frac{7}{10};\frac{1}{2}\times \frac{5}{5}=\frac{5}{10}$ $\Rightarrow \frac{7}{10}-\frac{1}{2}=\frac{7}{10}-\frac{5}{10}=\frac{7-5}{10}=\frac{2}{10}$ $=\frac{2÷2}{10÷2}=\frac{1}{5}\mathrm{km}$. (b) Time taken by Jeevika $=\frac{10}{3}$ minutes. Time taken by Namit $=\frac{13}{4}$ minutes. Now, $\frac{10\times 4}{3\times 4}=\frac{40}{12};\frac{13}{4}\times \frac{3}{3}=\frac{39}{12}$ Comparing $\frac{40}{12}$ and $\frac{39}{12}$, $\frac{40}{12}>\frac{39}{12}$ {For same denominator, fraction having greater numerator is greater} So, $\frac{10}{3}>\frac{13}{4}$ Or we can say Namit takes less time. Also, $\frac{10}{3}-\frac{13}{4}=\frac{40}{12}-\frac{39}{12}$ $=\frac{40-39}{12}=\frac{1}{12}$ minutes Namit takes less time by $\frac{1}{12}$ minutes.

4.0Key Features and benefits for Class 6 Maths Chapter 7 Exercise 7.8

Step-by-Step Explanations: Each solution is meticulously broken down into easy-to-understand steps, making challenging problems, especially those with unlike denominators, simple to follow.

Accuracy and Reliability: Developed by experienced subject matter experts, these solutions are highly accurate and strictly adhere to the latest NCERT guidelines for the 2025-2026 academic year.

Conceptual Clarity: Beyond just providing answers, our solutions aim to build a strong conceptual foundation in performing arithmetic operations with fractions.

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