PSEB 7th Class Maths Solutions Chapter 2 Fractions and Decimals Ex 2.1

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:
\(\frac{11}{16}-\frac{2}{5}+\frac{8}{10}\)
PSEB 7th Class Maths Solutions Chapter 2 Fractions and Decimals Ex 2.1 1a
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\(\frac {7}{8}\)

PSEB 7th Class Maths Solutions Chapter 2 Fractions and Decimals Ex 2.1

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:
\(\frac{1}{5}, \frac{3}{7}, \frac{7}{10}\)
PSEB 7th Class Maths Solutions Chapter 2 Fractions and Decimals Ex 2.1 2a
L.C.M of 5, 7 , 10 = 2 × 5 × 7
= 70
PSEB 7th Class Maths Solutions Chapter 2 Fractions and Decimals Ex 2.1 3a

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.
PSEB 7th Class Maths Solutions Chapter 2 Fractions and Decimals Ex 2.1 9
Answer:
Slides of ΔABC are
AB = \(\frac {5}{6}\) cm,
BC = \(\frac {2}{3}\)
CA = \(\frac {7}{10}\)
PSEB 7th Class Maths Solutions Chapter 2 Fractions and Decimals Ex 2.1 4a
L.C.M. (6, 3, 10) = 2 × 3 × 5 = 30
Perimeter of ΔABC = AB + BC + CA
PSEB 7th Class Maths Solutions Chapter 2 Fractions and Decimals Ex 2.1 5a

PSEB 7th Class Maths Solutions Chapter 2 Fractions and Decimals Ex 2.1

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
PSEB 7th Class Maths Solutions Chapter 2 Fractions and Decimals Ex 2.1 6a

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.
PSEB 7th Class Maths Solutions Chapter 2 Fractions and Decimals Ex 2.1 7a
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.

PSEB 7th Class Maths Solutions Chapter 2 Fractions and Decimals Ex 2.1

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) \(\frac {2}{3}\)

PSEB 7th Class Maths Solutions Chapter 2 Fractions and Decimals Ex 2.1 8a
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) \(\frac {29}{28}\)

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