PSEB 5th Class Maths Solutions Chapter 4 भिन्नात्मक संख्याएँ Ex 4.7

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 4 भिन्नात्मक संख्याएँ Ex 4.7 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 5 Maths Chapter 4 भिन्नात्मक संख्याएँ Exercise 4.7

प्रश्न 1.
निम्नलिखित भिन्नों का दशमलव रूप दर्शाओ:
(क) \(\frac{9}{10}\)
(ख) \(\frac{35}{100}\)
(ग) \(\frac{31}{1000}\)
(घ) \(\frac{117}{100}\)
(ङ) \(\frac{37}{10}\)
हल:
(क) \(\frac{9}{10}\) = 0.9
(ख) \(\frac{35}{100}\) = 0.35
(ग) \(\frac{31}{1000}\) = 0.031
(घ) \(\frac{117}{100}\) = 1.17
(ङ) \(\frac{37}{10}\) = 3.7

PSEB 5th Class Maths Solutions Chapter 4 भिन्नात्मक संख्याएँ Ex 4.7

प्रश्न 2.
निम्नलिखित भिन्नात्मक संख्याओं में प्रत्येक भिन्न को दशमलव रूप में बदलो :
(क) \(\frac{3}{5}\)
(ख) \(\frac{15}{20}\)
(ग) \(\frac{4}{25}\)
(घ) \(\frac{5}{4}\)
(ङ) \(\frac{7}{40}\)
हल:
(क) \(\frac{3}{5}\) = \(\frac{3×20}{5×20}\) = \(\frac{60}{100}\) = 0.60
(ख) \(\frac{15}{20}\) = \(\frac{15×5}{20×5}\) = \(\frac{75}{100}\) = 0.75
(ग) \(\frac{4}{25}\) = \(\frac{4×4}{25×4}\) = \(\frac{16}{100}\) = 0.16
(घ) \(\frac{5}{4}\) = \(\frac{5×25}{4×25}\) = \(\frac{125}{100}\) = 1.25
(ङ) \(\frac{7}{40}\) = \(\frac{7×25}{40×25}\) = \(\frac{175}{1000}\) = 0.175

PSEB 5th Class Maths Solutions Chapter 4 भिन्नात्मक संख्याएँ Ex 4.7

प्रश्न 3.
निम्नलिखित दशमलव संख्याओं को भिन्न रूप में लिखिए :
(क) 1.3
(ख) 1.75
(ग) 4.5
(घ). 0.35
(ङ) 0.8
(च) 3.84
(छ) 8.345
(ज) 0.024
(झ) 3.001
(ञ) 0.98
हल:
(क) 1.3 = \(\frac{13}{10}\)
(ख) 1.75 = \(\frac{175}{100}\)
(ग) 4.5 = \(\frac{45}{10}\)
(घ). 0.35 = \(\frac{35}{100}\)
(ङ) 0.8 = \(\frac{8}{10}\)
(च) 3.84 = \(\frac{384}{100}\)
(छ) 8.345 = \(\frac{8345}{1000}\)
(ज) 0.024 = \(\frac{24}{1000}\)
(झ) 3.001 = \(\frac{3001}{1000}\)
(ञ) 0.98 = \(\frac{98}{100}\)

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