Fractions Ex-7.6

NCERT solutions for Maths Fractions 

NCERT Solutions for Class 6 Maths Fractions

Class –VI Mathematics

(Ex. 7.6)

Question 1.Solve:

(a) $\frac{2}{3} + \frac{1}{7}$

(b) $\frac{3}{{10}} + \frac{7}{{15}}$

(c) $\frac{4}{9} + \frac{2}{7}$

(d) $\frac{5}{7} + \frac{1}{3}$

(e) $\frac{2}{5} + \frac{1}{6}$

(f) $\frac{4}{5} + \frac{2}{3}$

(g) $\frac{3}{4} – \frac{1}{3}$

(h) $\frac{5}{6} – \frac{1}{3}$

(i) $\frac{2}{3} + \frac{3}{4} + \frac{1}{2}$

(j) $\frac{1}{2} + \frac{1}{3} + \frac{1}{6}$

(k) $1\frac{1}{3} + 3\frac{2}{3}$

(l) $4\frac{2}{3} + 3\frac{1}{4}$

(m) $\frac{{16}}{5} – \frac{7}{5}$

(n) $\frac{4}{3} – \frac{1}{2}$

Answer:

(a) L.C.M. of 3 and 7 is 21

$\therefore$ $\frac{2}{3} + \frac{1}{7} = \frac{{2 \times 7 + 1 \times 3}}{{21}} = \frac{{14 + 3}}{{21}} = \frac{{17}}{{21}}$

(b) L.C.M. of 10 and 15 is 30

$\therefore$ $\frac{3}{{10}} + \frac{7}{{15}} = \frac{{3 \times 3 + 7 \times 2}}{{30}} = \frac{{9 + 14}}{{30}} = \frac{{23}}{{30}}$

(c) L.C.M. of 9 and 7 is 63

$\therefore$ $\frac{4}{9} + \frac{2}{7} = \frac{{4 \times 7 + 2 \times 9}}{{63}} = \frac{{28 + 18}}{{63}} = \frac{{46}}{{63}}$

(d)L.C.M. of 7 and 3 is 21

$\therefore$ $\frac{5}{7} + \frac{1}{3} = \frac{{5 \times 3 + 7 \times 1}}{{21}} = \frac{{15 + 7}}{{21}} = \frac{{22}}{{21}} = 1\frac{1}{{21}}$

(e)L.C.M. of 5 and 6 is 30

$\therefore$ $\frac{2}{5} + \frac{1}{6} = \frac{{2 \times 6 + 5 \times 1}}{{30}} = \frac{{12 + 5}}{{30}} = \frac{{17}}{{30}}$

(f)L.C.M. of 5 and 3 is 15

$\therefore$ $\frac{4}{5} + \frac{2}{3}$ = $\frac{{4 \times 3 + 2 \times 5}}{{15}} = \frac{{12 + 10}}{{15}} = \frac{{22}}{{15}} = 1\frac{7}{{15}}$

(g)L.C.M. of 4 and 3 is 12

$\therefore$ $\frac{3}{4} – \frac{1}{3} = \frac{{3 \times 3 – 4 \times 1}}{{12}} = \frac{{9 – 4}}{{12}} = \frac{5}{{12}}$

(h)L.C.M. of 6 and 3 is 6

$\therefore$

$\frac{5}{6}-\frac{1}{3}=\frac{5\times 1-2\times 1}{6}=\frac{5-2}{6}=\frac{3}{6}=\frac{1}{2}$

(i)L.C.M. of 3, 4 and 2 is 12

$\therefore$ $\frac{2}{3} + \frac{3}{4} + \frac{1}{2} = \frac{{2 \times 4 + 3 \times 3 + 1 \times 6}}{{12}} = \frac{{6 + 9 + 6}}{{12}} = \frac{{23}}{{12}} = 1\frac{{11}}{{12}}$

(j)L.C.M. of 2, 3, and 6 is 6

$\therefore$

$\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{1\times 3+1\times 2+1\times 1}{6}=\frac{3+2+1}{6}=\frac{6}{6}=1$

(k)L.C.M. of 3 and 3 is 3

$\therefore$

$\frac{4}{3}+\frac{11}{3}=\frac{4+11}{3}=\frac{15}{3}=5$

(l)L.C.M. of 3 and 4 is 12

$\therefore$ $\frac{{14}}{3} + \frac{{13}}{4} = \frac{{14 \times 4 + 13 \times 3}}{{12}} = \frac{{56 + 39}}{{12}} = \frac{{95}}{{12}} = 7\frac{{11}}{{12}}$

(m)L.C.M. of 5 and 5 is 5

$\therefore$ $\frac{{16}}{5} – \frac{7}{5} = \frac{{16 – 7}}{5} = \frac{9}{5} = 1\frac{4}{5}$

(n)L.C.M. of 3 and 2 is 6

$\therefore$ $\frac{4}{3} – \frac{1}{2} = \frac{{4 \times 2 – 1 \times 3}}{6} = \frac{{8 – 3}}{6} = \frac{5}{6}$

NCERT Solutions for Class 6 Maths Exercise 7.6

Question 2.Sarika bought $\frac{2}{5}$ meter of ribbon and Lalita $\frac{3}{4}$ meter of ribbon. What is the total length of the ribbon they bought?

Answer:

Ribbon bought by Sarita = $\frac{2}{5}$ m and Ribbon bought by Lalita = $\frac{3}{4}$ m

Total length of ribbon = $\frac{2}{5} + \frac{3}{4}$ = $\frac{{2 \times 4 + 5 \times 3}}{{20}}$ [$\because$ L.C.M. of 5 and 4 is 20]

= $\frac{{8 + 15}}{{20}} = \frac{{23}}{{20}} = 1\frac{3}{{20}}$ m

Therefore, they bought $1\frac{3}{{20}}$ m of ribbon.

NCERT Solutions for Class 6 Maths Exercise 7.6

Question 3.Naina was given $1\frac{1}{2}$ piece of cake and Najma was given $1\frac{1}{3}$ piece of cake. Find the total amount of cake given to both of them.

Answer:

Cake taken by Naina = $1\frac{1}{2}$ piece and Cake taken by Najma = $1\frac{1}{3}$ piece

Total cake taken = $1\frac{1}{2}$ + $1\frac{1}{3}$ = $\frac{3}{2} + \frac{4}{3}$ = $\frac{{3 \times 3 + 4 \times 2}}{6}$ [$\because$ L.C.M. of 2 and 3 is 6]

= $\frac{{9 + 8}}{6} = \frac{{17}}{6} = 2\frac{5}{6}$

Therefore total consumption of cake is $2\frac{5}{6}$.

Question 4.Fill in the boxes:

(a) $\boxed{{\text{ }}} – \frac{5}{8} = \frac{1}{4}$

(b) $\boxed{{\text{ }}} – \frac{1}{5} = \frac{1}{2}$

(c) $\frac{1}{2} – \boxed{{\text{ }}} = \frac{1}{6}$

Answer:

(a) $\frac{1}{4} + \frac{5}{8} = \frac{{2 + 5}}{8} = \frac{7}{8}$

(b) $\frac{1}{2} + \frac{1}{5} = \frac{{5 + 2}}{{10}} = \frac{7}{{10}}$

(c) $\frac{1}{2} – \frac{1}{6} = \frac{{3 – 1}}{6} = \frac{2}{6}$

NCERT Solutions for Class 6 Maths Exercise 7.6

Question 5.Complete the addition – subtraction box:

Answer:

Sol.

Question 6.A piece of wire $\frac{7}{8}$ meter long broke into two pieces. One piece was $\frac{1}{4}$ meter long. How long is the other piece?

Answer:

Total length of wire = $\frac{7}{8}$ meter

Length of first part = $\frac{1}{4}$ meter

Remaining part = $\frac{7}{8} – \frac{1}{4} = \frac{{7 \times 1 – 2 \times 1}}{8}$ [$\because$ L.C.M. of 8 and 4 is 8]

= $\frac{{7 – 2}}{8} = \frac{5}{8}$ meter

Therefore, the length of remaining part is $\frac{5}{8}$ meter.

NCERT Solutions for Class 6 Maths Exercise 7.6

Question 7.Nandini house is $\frac{9}{{10}}$ km from her school. She walked some distance and then took a bus for $\frac{1}{2}$ km to reach the school. How far did she walk?

Answer:

Total distance between school and house = $\frac{9}{{10}}$ km

Distance covered by bus = $\frac{1}{2}$ km

Remaining distance = $\frac{9}{{10}} – \frac{1}{2} = \frac{{9 \times 1 – 1 \times 5}}{{10}}$ [$\because$ L.C.M. of 10 and 2 is 10]

= $\frac{9-5}{10}=\frac{4}{10}=\frac{2}{5}$ km

Therefore, distance covered by walking us $\frac{2}{5}$ km.

Question 8.Ahsa and Samuel have bookshelves of the same size partly filled with books. Asha’s shelf is $\frac{5}{6}th$ full and Samuel’s shelf is $\frac{2}{5}th$ full. Whose bookshelf is more full? By what fraction?

Answer:

$\frac{5}{6}$ and $\frac{2}{5}$

$\Rightarrow$ $\frac{5}{6} \times \frac{5}{5} = \frac{{25}}{{30}}$ and $\frac{2}{5} \times \frac{6}{6} = \frac{{12}}{{30}}$ [$\because$ L.C.M. of 6 and 5 is 30]

$\because$ $\frac{{25}}{{30}} > \frac{{12}}{{30}}$ $\Rightarrow$ $\frac{5}{6} > \frac{2}{5}$

$\therefore$ Asha’s bookshelf is more covered than Samueal.

Difference = $\frac{{25}}{{30}} – \frac{{12}}{{30}} = \frac{{13}}{{30}}$

NCERT Solutions for Class 6 Maths Exercise 7.6

Question 9.Jaidev takes $2\frac{1}{5}$ minutes to walk across the school ground. Rahul takes $\frac{7}{4}$ minutes to do same. Who takes less time and by what fraction?

Answer:

Time taken by jaidev = $2\frac{1}{5}$ minutes = $\frac{{11}}{5}$ minutes

Time taken by Rahul = $\frac{7}{4}$ minutes

Difference = $\frac{{11}}{5} – \frac{7}{4}$ = $\frac{{11 \times 4 – 7 \times 5}}{{20}}$ [$\because$ L.C.M. of 5 and 4 is 20]

= $\frac{{44 – 35}}{{20}} = \frac{9}{{20}}$ minutes

Thus, Rahul takes less time, which is $\frac{9}{{20}}$ minutes.