Maharashtra Board Class 8 Maths Chapter 1 Rational and Irrational Numbers Set 1.2 Solutions

Official MSBSHSE Solutions for Class 8 Maths: Chapter 1 Rational and Irrational Numbers Set 1.2

Explore reliable textbook solutions for Chapter 1 Rational and Irrational Numbers Set 1.2 tailored for Class 8 learners. Utilizing these Maths answers ensures thorough preparation and strengthens foundational knowledge before final MSBSHSE evaluations.

Chapter-wise Solutions for Maths: Chapter 1 Rational and Irrational Numbers Set 1.2

View or download the dedicated Chapter 1 Rational and Irrational Numbers Set 1.2 solution resource below. Engaging with these textbook answers under focused study conditions ensures continuous academic progress and mastery of the 2026-27 curriculum for Maths.

Question 1. Compare the following numbers.
(i) 7, -2
(ii) 0, \( \frac{-9}{5} \)
(iii) \( \frac{7}{8}, 0 \)
(iv) \( \frac{-5}{4}, \frac{1}{4} \)
(v) \( \frac{40}{29}, \frac{141}{29} \)
(vi) \( \frac{-17}{20}, \frac{-13}{20} \)
(vii) \( \frac{15}{12}, \frac{7}{16} \)
(viii) \( \frac{-25}{8}, \frac{-9}{4} \)
(ix) \( \frac{12}{15}, \frac{3}{5} \)
(x) \( \frac{-7}{11}, \frac{-3}{4} \)
Answer:
Solution:
(i) 7, -2
If a and b are positive numbers such that a < b, then -a > -b.
Since, 2 < 7
\( \implies \) -2 > -7
(ii) 0, \( \frac{-9}{5} \)
On a number line, \( \frac{-9}{5} \) is to the left of zero.
\( \implies \) 0 > \( \frac{-9}{5} \)
(iii) \( \frac{7}{8}, 0 \)
On a number line, zero is to the left of \( \frac{7}{8} \).
\( \implies \frac{7}{8} \) > 0
(iv) \( \frac{-5}{4}, \frac{1}{4} \)
We know that, a negative number is always less than a positive number.
\( \implies \frac{-5}{4} < \frac{1}{4} \)
(v) \( \frac{40}{29}, \frac{141}{29} \)
Here, the denominators of the given numbers are the same.
Since, 40 < 141
\( \implies \frac{40}{29} < \frac{141}{29} \)
(vi) \( \frac{-17}{20}, \frac{-13}{20} \)
Here, the denominators of the given numbers are the same.
Since, -17 < -13
\( \implies \frac{-17}{20} < \frac{-13}{20} \)
(vii) \( \frac{15}{12}, \frac{7}{16} \)
Here, the denominators of the given numbers are not the same.
LCM of 12 and 16 = 48
\( \frac{15}{12} = \frac{15 \times 4}{12 \times 4} = \frac{60}{48} \),
\( \frac{7}{16} = \frac{7 \times 3}{16 \times 3} = \frac{21}{48} \)
Since, 60 > 21
\( \implies \frac{60}{48} > \frac{21}{48} \)
\( \implies \frac{15}{12} > \frac{7}{16} \)
Alternate method:
15 x 16 = 240
12 x 7 = 84
Since, 240 > 84
\( \implies \) 15 × 16 > 12 × 7
\( \implies \frac{15}{12} > \frac{7}{16} \)
\[ ... \text{If a \(\times\) d > b \(\times\) c, then } \frac{a}{b} > \frac{c}{d} \]
(viii) \( \frac{-25}{8}, \frac{-9}{4} \)
Here, the denominators of the given numbers are not the same.
LCM of 8 and 4 = 8
\( \frac{-9}{4} = \frac{-9 \times 2}{4 \times 2} = \frac{-18}{8} \)
Since, 25 > 18
\( \implies \frac{25}{8} > \frac{18}{8} \)
\( \implies \frac{-25}{8} < \frac{-18}{8} \)
\( \implies \frac{-25}{8} < \frac{-9}{4} \)
(ix) \( \frac{12}{15}, \frac{3}{5} \)
Here, the denominators of the given numbers are not the same.
LCM of 15 and 5 = 15
\( \frac{3}{5} = \frac{3 \times 3}{5 \times 3} = \frac{9}{15} \)
Since, 12 > 9
\( \implies \frac{12}{15} > \frac{9}{15} \)
\( \implies \frac{12}{15} > \frac{3}{5} \)
(x) \( \frac{-7}{11}, \frac{-3}{4} \)
Here, the denominators of the given numbers are not the same.
LCM of 11 and 4 = 44
\( \frac{-7}{11} = \frac{-7 \times 4}{11 \times 4} = \frac{-28}{44} \)
\( \frac{-3}{4} = \frac{-3 \times 11}{4 \times 11} = \frac{-33}{44} \)
Since, 28 < 33
\( \implies \frac{28}{44} < \frac{33}{44} \)
\( \implies \frac{-28}{44} > \frac{-33}{44} \)
\( \implies \frac{-7}{11} > \frac{-3}{4} \)
In simple words: To compare rational numbers, make their denominators the same by finding the LCM, then compare the numerators. For negative numbers, remember that a larger absolute value means a smaller number.

🎯 Exam Tip: Always pay close attention to the signs of the numbers. When comparing negative numbers, the number closer to zero is greater. Ensure correct LCM calculation for different denominators.

 

Maharashtra Board Class 8 Maths Solutions Chapter 1 Rational And Irrational Numbers Practice Set 1.2 Questions And Activities

 

Question 1. Verify the following comparisons using a number line.
(i) 2 < 3 but - 2 > - 3
(ii) \( \frac{5}{4} < \frac{7}{4} \) but \( \frac{-5}{4} < \frac{-7}{4} \)
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक संख्या रेखा है जिस पर पूर्णांक -3 से 3 तक अंकित हैं। इसमें भिन्नात्मक संख्याएँ जैसे -7/4, -5/4, 5/4 और 7/4 भी उनके सही स्थानों पर दर्शाई गई हैं, जिससे संख्याओं की तुलना करना आसान हो जाता है। बाईं ओर की संख्याएँ दाईं ओर की संख्याओं से छोटी होती हैं।
Solution:
We know that, on a number line the number to the left is smaller than the other.
\( \implies \) 2 < 3 and -3 < -2
i.e. 2 < 3 and -2 > -3
i.e. \( \frac{5}{4} < \frac{7}{4} \) and \( \frac{-7}{4} < \frac{-5}{4} \)
i.e. \( \frac{5}{4} < \frac{7}{4} \) and \( \frac{-5}{4} > \frac{-7}{4} \)
In simple words: The number line visually represents the order of numbers. Numbers to the left are smaller, and numbers to the right are larger. This helps in understanding comparisons, especially for negative numbers and fractions.

🎯 Exam Tip: When drawing a number line for verification, ensure equal spacing between integers and accurate placement of fractions. Clearly label all compared numbers on the line to avoid confusion.

MSBSHSE Solutions for Class 8 Maths Chapter 1 Rational and Irrational Numbers Set 1.2

Textbook Solutions for Class 8 Maths Chapter 1 Rational and Irrational Numbers Set 1.2

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