Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 2 Fractions and Decimals Ex 2.1 Textbook Exercise Questions and Answers.
PSEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1
1. Solve the following fractions :
Question (i).
4 + \frac {7}{8}
Answer:
4 + \frac {7}{8}
= \frac{4 \times 8+7}{8}
= \frac{32+7}{8}
= \frac {39}{8}
= 4\frac {7}{8}
Question (ii).
\frac{9}{11}-\frac{4}{15}
Answer:
\frac{9}{11}-\frac{4}{15}
= \frac{9 \times 15-4 \times 11}{11 \times 15}
= \frac{135-44}{165}
= \frac {91}{165}
Question (iii).
\frac{11}{16}-\frac{2}{5}+\frac{8}{10}
Answer:
\frac{11}{16}-\frac{2}{5}+\frac{8}{10}
LCM of 16, 5 and 10
= 2 × 2 × 2 × 2 × 5
= 80
= \frac{11 \times 5-2 \times 16+8 \times 8}{80}
= \frac{55-32+64}{80}
= \frac {87}{80}
= 1\frac {7}{80}
Question (iv).
2 \frac{1}{5}+6 \frac{1}{2}
Answer:
2 \frac{1}{5}+6 \frac{1}{2}
= \frac{11}{5}+\frac{13}{2}
= \frac{11 \times 2+13 \times 5}{5 \times 2}
= \frac{22+65}{10}
= \frac {87}{10}
= 8\frac {7}{10}
Question (v).
8 \frac{1}{2}-3 \frac{5}{8}
Answer:
8 \frac{1}{2}-3 \frac{5}{8}
= \frac{17}{2}-\frac{29}{8}
= \frac{17 \times 4-29}{8}
= \frac{68-29}{8}
= \frac {39}{8}
= 4\frac {7}{8}
Question (vi).
\frac{9}{10}-\frac{9}{100}+\frac{9}{1000}
Answer:
\frac{9}{10}-\frac{9}{100}+\frac{9}{1000}
= \frac{9 \times 100-9 \times 10+9}{1000}
= \frac{900-90+9}{1000}
= \frac {810}{1000}
2. Arrange the following in ascending order :
Question (i).
\frac{2}{17}, \frac{10}{17}, \frac{3}{17}, \frac{16}{17}, \frac{5}{17}, \frac{8}{17}
Answer:
Ascending order of \frac{2}{17}, \frac{10}{17}, \frac{3}{17}, \frac{16}{17}, \frac{5}{17}, \frac{8}{17} is:
\frac{2}{17}, \frac{3}{17}, \frac{5}{17}, \frac{8}{17}, \frac{10}{17}, \frac{16}{17}
Question (ii).
\frac{1}{5}, \frac{3}{7}, \frac{7}{10}
Answer:
\frac{1}{5}, \frac{3}{7}, \frac{7}{10}
L.C.M of 5, 7 , 10 = 2 × 5 × 7
= 70
3. The three sides AB, BC and CA of a triangle ΔABC are \frac {5}{6}cm, \frac {2}{3}cm and \frac {7}{10} cm respectively. Find the perimeter of the triangle.
Answer:
Slides of ΔABC are
AB = \frac {5}{6} cm,
BC = \frac {2}{3}
CA = \frac {7}{10}
L.C.M. (6, 3, 10) = 2 × 3 × 5 = 30
Perimeter of ΔABC = AB + BC + CA
4. Ramesh studies for 5\frac {2}{3} hours daily. He devotes 2\frac {4}{5} hours of his time for science devotes for other subjects ?
Answer:
Total daily time for all subjects
= 5\frac {2}{3} hours = \frac {17}{3} hours
Time for science and mathematics
= 2\frac {4}{5} hours = \frac {14}{5} hours
Time for other subjects
5. Sonia jogs once around the rectangular park of sides 10\frac {2}{3}m and 12\frac {1}{2}m. Find the total distance covered by the Sonia.
Answer:
Length of rectangular park
= 12\frac {1}{2}m = \frac {25}{2}m
Breadth of rectangular park
= 10\frac {2}{3}m = \frac {32}{3}m
Total distance covered by Sonia = 2 [Length + Breadth]
= 2\left(\frac{32}{3}+\frac{25}{3}\right) \mathrm{m}
= 2\left(\frac{32 \times 2+25 \times 3}{3 \times 2}\right) \mathrm{m}
= 2\left(\frac{65+75}{6}\right) \mathrm{m}
= \frac {278}{6} m
= \frac {139}{3} m
= 46\frac {1}{3} m
6. Ritu coloured a picture in \frac {7}{12} hours. Vaibhav coloured the same picture in \frac {3}{4} hours. Who worked for a longer time and by what fraction ?
Answer:
Time taken by Ritu to colour
= \frac {7}{12} hours
Time taken by Vaibhav = \frac {3}{4} hours
= \frac {3}{4} × \frac {3}{3}
= \frac {9}{12} hours
Since 9 > 7
∴ \frac {9}{12} > \frac {7}{12}
∴ Vaibhav worked for more time.
Difference between time taken by
Vaibhav and Ritu = \frac{3}{4}-\frac{7}{12}
= \frac{3 \times 3-7}{12}
= \frac{9-7}{12}=\frac{2}{12}
= \frac {1}{6} of an hour.
7. Multiple Choice Questions :
Question (i).
Fraction \frac {2}{5}, \frac {7}{5} are :
(a) Like fractions
(b) Unlike fractions
(c) Equivalent fractions
(d) None of these
Answer:
(a) Like fractions
Question (ii).
What fraction do 8 hours of a day represents ?
(a) \frac {1}{2}
(b) \frac {1}{3}
(c) \frac {8}{60}
(d) \frac {2}{3}
Answer:
(b) \frac {1}{3}
Question (iii).
Equivalent fraction of \frac {3}{5} is :
(a) \frac {13}{15}
(b) \frac {5}{3}
(c) \frac {9}{15}
(d) \frac {5}{13}
Answer:
(c) \frac {9}{15}
Question (iv).
Shaded area of given triangle represents the fractions:
(a) \frac {1}{3}
(b) \frac {3}{4}
(c) \frac {1}{4}
(d) \frac {2}{3}
Answer:
(b) \frac {3}{4}
Question (v).
Sum of fractions \frac {2}{7}, \frac {3}{4} is equal to :
(a) \frac {5}{28}
(b) \frac {1}{3}
(c) \frac {5}{11}
(d) \frac {29}{28}
Answer:
(d) \frac {29}{28}