Well-designed AP Board Solutions Class 6 Maths Chapter 7 Fractions Exercise 7.6 offers step-by-step explanations to help students understand problem-solving strategies.
Fractions Class 6 Exercise 7.6 Solutions – 6th Class Maths 7.6 Exercise Solutions
Question 1.
Solve:
a) \(\frac { 2 }{ 3 }\) + \(\frac { 1 }{ 7 }\)
Solution:
\(\frac { 2 }{ 3 }\) + \(\frac { 1 }{ 7 }\) (LCM of 7 and 3 is 21)
\(\frac{2 \times 7}{3 \times 7}\) + \(\frac{1 \times 3}{7 \times 3}\) = \(\frac{14}{21}\) + \(\frac{3}{21}\) = \(\frac{14+3}{21}\) = \(\frac{17}{21}\)
b) \(\frac { 3 }{ 10 }\) + \(\frac { 7 }{ 15 }\)
Solution:
\(\frac { 3 }{ 10 }\) + \(\frac { 7 }{ 15 }\) (LCM of 10 and 15 is 30)
\(\frac{3 \times 3}{10 \times 3}\) + \(\frac{7 \times 2}{15 \times 2}\) = \(\frac{9}{30}\) + \(\frac{14}{30}\) = \(\frac{9+14}{30}\) = \(\frac{23}{30}\)
c) \(\frac { 4 }{ 9 }\) + \(\frac { 2 }{ 7 }\)
Solution:
\(\frac { 4 }{ 9 }\) + \(\frac { 2 }{ 7 }\) (LCM of 9 and 7 is 63)
\(\frac{4 \times 7}{9 \times 7}\) + \(\frac{2 \times 9}{7 \times 9}\) = \(\frac{28}{63}\) + \(\frac{18}{63}\) = \(\frac{28+18}{63}\) = \(\frac{46}{63}\)
d) \(\frac { 5 }{ 7 }\) + \(\frac { 1 }{ 3 }\)
Solution:
\(\frac { 5 }{ 7 }\) + \(\frac { 1 }{ 3 }\) (LCM of 7 and 3 is 21)
\(\frac{5 \times 3}{7 \times 3}\) + \(\frac{1 \times 7}{3 \times 7}\) = \(\frac{15}{21}\) + \(\frac{7}{21}\) = \(\frac{15+7}{21}\) = \(\frac{22}{21}\)
e) \(\frac { 2 }{5 }\) + \(\frac { 1 }{ 6 }\)
Solution:
\(\frac { 2 }{ 5 }\) + \(\frac { 1 }{ 6 }\) (LCM of 5 and 6 is 30)
\(\frac{2\times 6}{5 \times 6}\) + \(\frac{1 \times 5}{6 \times 5}\) = \(\frac{12}{30}\) + \(\frac{5}{30}\) = \(\frac{12+5}{30}\) = \(\frac{17}{30}\)
f) \(\frac { 4 }{ 5 }\) + \(\frac { 2 }{ 3 }\)
Solution:
\(\frac { 4 }{ 5 }\) + \(\frac { 2 }{ 3 }\) (LCM of 5 and 3 is 15)
\(\frac{4 \times 3}{5 \times 3}\) + \(\frac{2 \times 5}{3 \times 5}\) = \(\frac{12}{15}\) + \(\frac{10}{15}\) = \(\frac{12+10}{15}\) = \(\frac{22}{15}\)
g) \(\frac { 3 }{ 4 }\) – \(\frac { 1 }{ 3 }\)
Solution:
\(\frac { 3 }{ 4 }\) + \(\frac { 1 }{ 3 }\) (LCM of 4 and 3 is 12)
\(\frac{3 \times 3}{4 \times 3}\) + \(\frac{1 \times 4}{3 \times 4}\) = \(\frac{9}{12}\) – \(\frac{4}{12}\) = \(\frac{9-4}{12}\) = \(\frac{5}{12}\)
h) \(\frac { 5 }{ 6 }\) – \(\frac { 1 }{ 3 }\)
Solution:
\(\frac { 5 }{ 6 }\) – \(\frac { 1 }{ 3 }\) (LCM of 6 and 3 is 6)
\(\frac{5 \times 1}{6 \times 1}\) + \(\frac{1 \times 2}{3 \times 2}\) = \(\frac{5}{6}\) – \(\frac{2}{6}\) = \(\frac{5-2}{6}\) = \(\frac{3}{6}\) = \(\frac{3 \div 3}{6 \div 3}\) = \(\frac{1}{2}\)
i) \(\frac { 2 }{ 3 }\) + \(\frac { 3 }{ 4 }\) + \(\frac { 1 }{ 2 }\)
Solution:
\(\frac { 2 }{ 3 }\) + \(\frac { 3 }{ 4 }\) + \(\frac { 1 }{ 2 }\) (LCM of 3,4,2 is 12)
\(\frac{2 \times 4}{3 \times 4}\) + \(\frac{3 \times 3}{4 \times 3}\) + \(\frac{1 \times 6}{2 \times 6}\)
= \(\frac { 8 }{ 12 }\) + \(\frac { 9 }{ 12 }\) + \(\frac { 6 }{ 12 }\) = \(\frac { 8+9+6 }{ 12 }\) = \(\frac { 23 }{ 12 }\)
j) \(\frac { 1 }{ 2 }\) + \(\frac { 1 }{ 3 }\) + \(\frac { 1 }{ 6 }\)
Solution:
\(\frac { 1 }{ 2 }\) + \(\frac { 1 }{ 3 }\) + \(\frac { 1 }{ 6 }\) (LCM of 2,3 and 6 is 6)
\(\frac{1 \times 3}{2 \times 3}\) + \(\frac{1 \times 2}{3 \times 2}\) + \(\frac{1 \times 1}{6 \times 1}\)
= \(\frac { 3 }{ 6 }\) + \(\frac { 2 }{ 6 }\) + \(\frac { 1 }{ 6 }\) = \(\frac { 3+2+1}{ 6 }\) = \(\frac { 6 }{ 6 }\) = 1
k) 1\(\frac { 1 }{ 3 }\) + 3\(\frac { 2 }{ 3 }\)
Solution:
1\(\frac { 1 }{ 3 }\) + 3\(\frac { 2 }{ 3 }\)
1 + \(\frac { 1 }{ 3 }\) + 3 + \(\frac { 2 }{ 3 }\) = 4 + \(\frac { 1 }{ 3 }\) + \(\frac { 2 }{ 3 }\)
= 4 + \(\frac { 1 + 2}{ 3 }\) = 4 + \(\frac { 3 }{ 3 }\) = 4 + 1 = 5
l) 4\(\frac { 2 }{ 3 }\) + 3\(\frac { 1 }{ 4 }\)
Solution:
4\(\frac { 2 }{ 3 }\) + 3\(\frac { 1 }{ 4 }\) [LCM of 4 and 3 is 12]
4 + \(\frac { 2 }{ 3 }\) + 3 + \(\frac { 1 }{ 4 }\) = 4 + 3 + \(\frac { 2 }{ 3 }\) + \(\frac { 1 }{ 4 }\)
= 7 + \(\frac{2 \times 4}{3 \times 4}\) + \(\frac{1 \times 3}{4 \times 3}\) = 7 + \(\frac { 8 }{ 12 }\) + \(\frac { 3 }{ 12 }\)
= 7 + \(\frac { 8+3 }{ 12 }\) = 7 + \(\frac { 11 }{ 12 }\) = 7\(\frac { 11 }{ 12 }\)
m) \(\frac { 16 }{ 5 }\) – \(\frac { 7 }{ 5 }\)
Solution:
\(\frac { 16 }{ 5 }\) – \(\frac { 7 }{ 5 }\) = \(\frac { 16-7 }{ 5 }\) = \(\frac { 9 }{ 5 }\)
n) \(\frac { 4 }{ 3 }\) – \(\frac { 1 }{ 2 }\)
Solution:
\(\frac { 4 }{ 3 }\) – \(\frac { 1 }{ 2 }\) [LCM of 3 and 2 is 6]
\(\frac{4 \times 2}{3 \times 2}\) – \(\frac{1 \times 3}{2 \times 3}\) = \(\frac { 8 }{ 6 }\) – \(\frac { 3 }{ 6 }\) = \(\frac { 8 – 3 }{ 6 }\) = \(\frac { 5 }{ 6 }\)
Question 2.
Sarita 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?
Solution:
Length of ribbon bought by Sarita = \(\frac { 2 }{ 5 }\) metre
Length of ribbon bought by Lalita = \(\frac { 3 }{ 4 }\) metre
Length of ribbon bought by both Sarita and Lalita
= \(\frac { 2 }{ 5 }\) + \(\frac { 3 }{ 4 }\) = \(\frac{2 \times 4}{5 \times 4}\) + \(\frac{3 \times 5}{4 \times 5}\) = \(\frac { 8 }{ 20 }\) + \(\frac { 15 }{ 20 }\) = \(\frac { 23 }{ 20 }\)
Hence required length = \(\frac { 23 }{ 20 }\) metre.
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 was given to both of them.
Solution:
Piece of cake given to Naina = 1\(\frac { 1 }{ 2 }\)
Piece of cake given to Najma = 1\(\frac { 1 }{ 3 }\)
Total piece of cake given to Naina and Najma = 1\(\frac { 1 }{ 2 }\) + 1\(\frac { 1 }{ 3 }\)
= 1 + \(\frac { 1 }{ 2 }\) + 1 + \(\frac { 1 }{ 3 }\) = 2 + \(\frac { 1 }{ 2 }\) + \(\frac { 1 }{ 3 }\)
= 2 + \(\frac{1 \times 3}{2 \times 3}\) + \(\frac{1 \times 2}{3 \times 2}\) [LCM of 2 and 3 is 6 ]
= 2 + \(\frac { 3 }{ 6 }\) + \(\frac { 2 }{ 6 }\)
= 2 + \(\frac { 3+2 }{ 6 }\) = 2 + \(\frac { 5 }{ 6 }\) = 2\(\frac { 5 }{ 6 }\)
Hence the total amount of piece given to both = 2\(\frac { 5 }{ 6 }\)
Question 4.
Fill in the boxes :
a) ____ – \(\frac { 5 }{ 8 }\) = \(\frac { 1 }{ 4 }\)
Solution:
____ – \(\frac { 5 }{ 8 }\) = \(\frac { 1 }{ 4 }\)
[Missing number is \(\frac { 1 }{ 4 }\) more than \(\frac { 5 }{ 8 }\) ]
∴ ____ = \(\frac { 1 }{ 4 }\) + \(\frac { 5 }{ 8 }\) [LCM of 4 and 8 is 8]
____ = \(\frac{1 \times 2}{4 \times 2}\) + \(\frac{5 \times 1}{8 \times 1}\) = \(\frac { 2 }{ 8 }\) + \(\frac { 5 }{ 8 }\) = \(\frac { 7 }{ 8 }\)
(OR)
____ – \(\frac { 5 }{ 8 }\) = \(\frac { 1 }{ 4 }\)
____ – \(\frac { 5 }{ 8 }\) = \(\frac { 2 }{ 8 }\) [\(\frac { 1 }{ 4 }\) = \(\frac{1 \times 2}{4 \times 2}\) = \(\frac { 2 }{ 8 }\)]
\(\frac { 7 }{ 8 }\) – \(\frac { 5 }{ 8 }\) = \(\frac { 2 }{ 8 }\)
b) ____ – \(\frac { 1 }{ 5 }\) = \(\frac { 1 }{ 2 }\)
Solution:
____ – \(\frac { 1 }{ 5 }\) = \(\frac { 1 }{ 2 }\)
[Missing number is \(\frac { 1 }{ 2 }\) more than \(\frac { 1 }{ 5 }\)]
∴ ____ = \(\frac { 1 }{ 2 }\) + \(\frac { 1 }{ 5}\) [LCM of 2 and 5 is 10]
____ = \(\frac{1 \times 5}{2 \times 5}\) + \(\frac{1 \times 2}{5 \times 2}\) = \(\frac { 5 }{ 10 }\) – \(\frac { 2 }{ 10 }\) = \(\frac { 7 }{ 10 }\)
∴ ____ = \(\frac { 7 }{ 10 }\)
c) \(\frac { 1 }{ 2 }\) – ____ = \(\frac { 1 }{ 6 }\)
Solution:
\(\frac { 1 }{ 2 }\) – ____ = \(\frac { 1 }{ 6 }\)
[Missing number is \(\frac { 1 }{ 6 }\) more than \(\frac { 1 }{ 2 }\)]
∴ ____ = \(\frac { 1 }{ 2 }\) – \(\frac { 1 }{ 6 }\) [LCM of 2 and 6 is 6]
____ = \(\frac{1 \times 3}{2 \times 3}\) – \(\frac{1 \times 1}{6 \times 1}\) = \(\frac { 3 }{ 6 }\) – \(\frac { 1 }{ 6 }\) = \(\frac { 3-1 }{ 6 }\) = \(\frac { 2 }{ 6 }\) = \(\frac { 1 }{ 3 }\)
Question 5.
Complete the addition subtraction box.
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?
Solution:
Total length of the wire = \(\frac { 7 }{ 8 }\) metre
Length of on piece of wire = \(\frac { 1 }{ 4 }\) metre.
∴ Length of the piece = \(\frac { 7 }{ 8 }\) – \(\frac { 1 }{ 4 }\) [LCM of 8 and 4 is 8 ]
= \(\frac{7 \times 1}{8 \times 1}\) – \(\frac{1 \times 2}{4 \times 2}\) = \(\frac { 7 }{ 8 }\) – \(\frac { 2 }{ 8 }\) = \(\frac { 7-2 }{ 8 }\) = \(\frac { 5 }{ 8 }\)
Hence, the length of the other piece = \(\frac { 5 }{ 8 }\) metre.
Question 7.
Nandini’s 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?
Solution:
Total distance from Nandini’s house to school = \(\frac { 9 }{ 10 }\) km
Distance travelled by Nandini by bus = \(\frac { 1 }{ 2 }\) km
∴ Distance travelled by her on foot = \(\frac {9 }{ 10 }\) – \(\frac { 1 }{ 2 }\) [LCM of 10 and 2 is 10]
= \(\frac{9 \times 1}{10 \times 1}\) – \(\frac{1 \times 5}{2 \times 5}\)
= \(\frac { 9 }{ 10 }\) – \(\frac { 5 }{ 10 }\) = \(\frac { 9 – 5 }{ 10 }\) = \(\frac { 4 }{ 10 }\) = \(\frac { 2 }{ 5 }\)
Hence, the distance travelled by her on foot = \(\frac { 2 }{ 5 }\) 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 bookshelf is more full? By what fraction?
Solution:
Asha’s shelf is \(\frac { 5 }{ 6 }\)th full and Samule’s shelf is \(\frac { 2 }{ 5 }\)th full.
Comparing \(\frac { 5 }{ 6 }\) and \(\frac { 2 }{ 5 }\), LCM of 5 and 6 is 30 .
\(\frac{5 \times 5}{6 \times 5}\) and \(\frac{2 \times 6}{5 \times 6}\) = \(\frac { 25 }{ 30 }\) and \(\frac { 12 }{ 30 }\)
Hence 25 > 12
∴ \(\frac { 25 }{ 30 }\) > \(\frac { 12 }{ 30 }\)
So, \(\frac { 5 }{ 6 }\) is more than \(\frac { 2 }{ 5 }\)
Hence, Asha’s shelf is full more than Samuel’s shelf.
Now \(\frac { 5 }{ 6 }\) – \(\frac { 2 }{ 5 }\) = \(\frac { 25 }{ 30 }\) – \(\frac { 12 }{ 30 }\) = \(\frac { 25 – 12 }{ 30 }\) = \(\frac { 13 }{ 30 }\)
Hence, \(\frac { 13 }{ 30 }\) is more than full of Asha’s shelf.
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?
Solution:
Jaidev takes 2\(\frac { 1 }{ 5 }\) minutes
Rahul takes \(\frac { 7 }{ 4 }\) minutes
Comparing 2\(\frac { 1 }{ 5 }\) minutes and \(\frac { 7 }{ 4 }\) minutes.
2\(\frac { 1 }{ 5 }\) = 2 + \(\frac { 1 }{ 5 }\) = \(\frac{2 \times 5}{5}\) = \(\frac { 1 }{ 5 }\) = \(\frac { 10 }{ 5 }\) + \(\frac { 1 }{ 5 }\) = \(\frac { 11 }{ 5 }\)
(OR) 2\(\frac { 1 }{ 5 }\) and \(\frac { 7 }{ 4 }\)
Now given fractions
\(\frac { 11 }{ 5 }\) and \(\frac { 7 }{ 4 }\)
2\(\frac { 1 }{ 5 }\) and \(\frac { 7 }{ 4 }\)
[LCM of 5 and 4 is 20]
\(\frac{11 \times 4}{5 \times 4}\) and \(\frac{7 \times 5}{4 \times 5}\)
2\(\frac { 1 }{ 5 }\) and \(\frac { 7 }{ 4 }\)
\(\frac { 7 }{ 4 }\) = 1\(\frac { 3 }{ 4 }\)
= \(\frac { 44 }{ 20 }\) and \(\frac { 35 }{ 20 }\)
\(\frac {35 }{ 20 }\) < \(\frac { 44 }{ 20 }\) (∵ 35 < 44) 1 < 2
∴ \(\frac { 7 }{ 4 }\) < 2\(\frac { 1 }{ 5 }\)
∴ 1\(\frac { 3 }{ 4 }\) < 2\(\frac { 1 }{ 5 }\)
∴ \(\frac { 7 }{ 4 }\) < 2\(\frac { 1 }{ 5 }\)
So, the time take to cover the same distance by Rahul is less than that of Jaidev.
2\(\frac { 1 }{ 5 }\) – \(\frac { 7 }{ 4 }\) = \(\frac { 11 }{ 5 }\) – \(\frac { 7 }{ 4 }\)
\(\frac { 44 }{ 20 }\) – \(\frac { 35 }{ 20 }\) = \(\frac { 9 }{ 20 }\) minutes
Hence, Rahul takes \(\frac { 9 }{ 20 }\) minutes less to walk across the school ground.