RBSE Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.6

Rajasthan Board RBSE Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.6 Textbook Exercise Questions and Answers.

RBSE Class 6 Maths Solutions Chapter 7 Fractions Ex 7.6

Question 1.
Solve

Answer:

Question 2.
Santa bought \(\frac{2}{5}\) metre of ribbon and Lalita \(\frac{3}{4}\) metre of ribbon. What is the total length of the ribbon they. bought?
Answer:
Length of ribbon bought by Sarita = \(\frac{2}{5}\) metre
Length of ribbon bought by Lauta = \(\frac{3}{4}\) metre
∴ Total length of ribbon bought
= \(\frac{2}{5}\) metre + \(\frac{3}{4}\) metre
∵ LCM of 5 and 4 = 20, then

Question 3.
Nalna 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 w given to both of them.
Answer:
Amount of cake given to Naina
= 1 \(\frac{1}{2}\) piece = \(\frac{3}{2}\) piece
Amount of cake given to Najma
= 1\(\frac{1}{3}\) piece = \(\frac{4}{3}\) piece

Question 4.
(a)

Answer:

(b)

Answer:

(c)

Answer:

Question 5.
Complete the addition-subtraction box:
(a)

Answer:

(b)

Answer:

Question 6.
A piece of wire \(\frac{7}{8}\) metre long broke into two pieces. One piece was \(\frac{1}{4}\) metre long. How long is the other piece?
Answer:
Total length of wire = \(\frac{7}{8}\)m
Length of one piece = \(\frac{1}{4}\)m
∴ Length of other piece = \(\frac{7}{8}\)m - \(\frac{1}{4}\)m
= \(\frac{7}{8}-\frac{1 \times 2}{4 \times 2}=\frac{7}{8}-\frac{2}{8}=\frac{7-2}{8}=\frac{5}{8} \mathrm{~m}\)
Hence, length of other piece = \(\frac{5}{8}\)m

Question 7.
Nandini’s house is \(\frac{9}{10}\)km far 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:
Distance between Nandini’s house and school = \(\frac{9}{10}\)km
Distance covered by bus = \(\frac{1}{2}\)km
∴ Distance covered by walking = \(\left(\frac{9}{10}-\frac{1}{2}\right)\) km
= \(\frac{9}{10}-\frac{1 \times 5}{2 \times 5}=\frac{9}{10}-\frac{5}{10}\)
= \(\frac{9-5}{10}=\frac{4}{10}=\frac{2}{5}\) km

Question 8.
Asha 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 bookself is more full? By what fraction?
Answer:
Portion of Asha’s shelf full of books = \(\frac{5}{6}\)
Portion of Samuel’s shelf full of books = \(\frac{2}{5}\)
By converting them into like fraction, we have
\(\frac{5}{6}=\frac{5 \times 5}{6 \times 5}=\frac{25}{30}\) and \(\frac{2}{5}=\frac{2 \times 6}{5 \times 6}=\frac{12}{30}\)
∵ 25 > 12
Hence, portion of Asha’s shelf is more full.
Now \(\frac{5}{6}-\frac{2}{5}=\frac{25}{30}-\frac{12}{30}=\frac{25-12}{30}=\frac{12}{30}\)
∴ Fraction by which Asha's shelf is more full = \(\frac{13}{30}\)

Question 9.
Jaidev takes 2\(\frac{1}{5}\) minutes to walk across the school ground. Rahul takes \(\frac{7}{4}\) minutes to do the same. Who takes less time and by what fraction?
Answer:
Time taken by Jaidev = 2\(\frac{1}{5}\) min = \(\frac{11}{5}\)min
Time taken by Rahul = \(\frac{7}{4}\)min
By converting into like fractions,
\(\frac{11}{5}=\frac{11 \times 4}{5 \times 4}=\frac{44}{20}\) and \(\frac{7}{4}=\frac{7 \times 4}{4 \times 5}=\frac{35}{20}\)
Clearly, Rahul takes less time to walk across the ground.
Now \(\frac{11}{5}-\frac{7}{4}=\frac{44}{20}-\frac{35}{20}=\frac{44-35}{20}=\frac{9}{20}\)
Thus, Rahul takes less time by \(\frac{9}{20}\) minutes less.