GANITA MANJARI | Chapter 3 - The World of Numbers
CHAPTER 3

The World of Numbers

Verbatim Questions & Solutions | GANITA MANJARI | Class 9 Maths | Part I

EXERCISE SET 3.1 (Page 43)

1.
A merchant in the port city of Lothal is exchanging bags of spices for copper ingots. He receives 15 ingots for every 2 bags of spices. If he brings 12 bags of spices to the market, how many copper ingots will he leave with?
Solution
For 2 bags → 15 ingots. For 1 bag → 15/2 ingots. For 12 bags → (15/2) × 12 = 15 × 6 = 90 ingots.
2.
Look at the sequence of numbers on one column of the Ishango bone: 11, 13, 17, 19. What do these numbers have in common? List the next three numbers that fit this pattern.
Solution
They are all prime numbers. The next three prime numbers are 23, 29, 31.
3.
We know that Natural Numbers are closed under addition (the sum of any two natural numbers is always a natural number). Are they closed under subtraction? Provide a couple of examples to justify your answer.
Solution
No, natural numbers are not closed under subtraction. Examples: 3 - 5 = -2 (not a natural number); 2 - 2 = 0 (not a natural number).
*4.
Ancient Indians used the joints of their fingers to count, a practice still seen today. Each finger has 3 joints, and the thumb is used to count them. How many can you count on one hand? How does this relate to the ancient base-12 counting systems?
Solution
On one hand (excluding thumb as counter), 4 fingers × 3 joints = 12 counts. This is the origin of base-12 counting systems.

EXERCISE SET 3.2 (Page 46)

1.
The temperature in the high-altitude desert of Ladakh is recorded as 4°C at noon. By midnight, it drops by 15°C. What is the midnight temperature?
Solution
4 - 15 = -11°C
2.
A spice trader takes a loan (debt) of 850. The next day, he makes a profit (fortune) of 1,200. The following week, he incurs a loss of 450. Write this sequence as an equation using integers and calculate his final financial standing.
Solution
-850 + 1200 - 450 = -100. Final standing = -100 (debt of 100).
3.
Calculate the following using Brahmagupta's laws:
(i) (-12) × 5
(ii) (-8) × (-7)
(iii) 0 - (-14)
(iv) (-20) ÷ 4
Solution
(i) -60    (ii) 56    (iii) 14    (iv) -5
4.
Explain, using a real-world example of debt, why subtracting a negative number is the same as adding a positive number (e.g., 10 - (-5) = 15).
Solution
If you have ₹10 and someone removes (cancels) a debt of ₹5 that you owed, you effectively gain ₹5, so 10 - (-5) = 15.

Think and Reflect

Page 46
Why does a negative times a negative equal a positive? Think of it in terms of action and debt. If a negative number represents a debt, then multiplying by a negative number represents the removal of that debt.
Answer
If someone removes (takes away) 4 debts of ₹3 each, you become ₹12 richer. Therefore, (-3) × (-4) = +12.

EXERCISE SET 3.3 (Pages 49-50)

1.
Prove that the following rational numbers are equal:
(i) \(\frac{2}{3}\) and \(\frac{4}{6}\)
Solution
(i) Cross multiply: 2×6 = 12, 3×4 = 12. Hence equal.
2.
Find the sum:
(i) \(\frac{2}{5} + \frac{3}{10}\)
(iii) \(-\frac{4}{7} + \frac{3}{14}\)
Solution
(i) \(\frac{4}{10} + \frac{3}{10} = \frac{7}{10}\)
(iii) \(-\frac{8}{14} + \frac{3}{14} = -\frac{5}{14}\)
3.
Find the difference:
(i) \(\frac{5}{6} - \frac{1}{4}\)
(iii) \(-\frac{7}{9} - \left(-\frac{2}{3}\right)\)
Solution
(i) \(\frac{10}{12} - \frac{3}{12} = \frac{7}{12}\)
(iii) \(-\frac{7}{9} + \frac{6}{9} = -\frac{1}{9}\)
4.
Find the product:
(i) \(\frac{2}{3} \times \frac{3}{10}\)
(iii) \(-\frac{4}{7} \times \frac{5}{14}\)
Solution
(i) \(\frac{6}{30} = \frac{1}{5}\)
(iii) \(-\frac{20}{98} = -\frac{10}{49}\)
5.
Find the quotient:
(i) \(\frac{2}{3} \div \frac{3}{10}\)
(ii) \(\frac{7}{11} \div \frac{5}{8}\)
(iii) \(-\frac{4}{7} \div \frac{5}{14}\)
Solution
(i) \(\frac{2}{3} \times \frac{10}{3} = \frac{20}{9}\)
(ii) \(\frac{7}{11} \times \frac{8}{5} = \frac{56}{55}\)
(iii) \(-\frac{4}{7} \times \frac{14}{5} = -\frac{56}{35} = -\frac{8}{5}\)
6.
Show that: \(\left(\frac{1}{2} + \frac{3}{4}\right) \times \frac{8}{3} = \frac{1}{2} \times \frac{8}{3} + \frac{3}{4} \times \frac{8}{3}\).
Solution
LHS = \(\left(\frac{2}{4} + \frac{3}{4}\right) \times \frac{8}{3} = \frac{5}{4} \times \frac{8}{3} = \frac{40}{12} = \frac{10}{3}\). RHS = \(\frac{8}{6} + \frac{24}{12} = \frac{4}{3} + 2 = \frac{4}{3} + \frac{6}{3} = \frac{10}{3}\). Hence proved.
7.
Simplify the following using the distributive property: \(\frac{7}{9}\left(\frac{6}{7} - \frac{3}{4}\right)\).
Solution
\(\frac{7}{9} \times \frac{6}{7} - \frac{7}{9} \times \frac{3}{4} = \frac{6}{9} - \frac{21}{36} = \frac{2}{3} - \frac{7}{12} = \frac{8}{12} - \frac{7}{12} = \frac{1}{12}\).
8.
Find the rational number \(x\) such that: \(\frac{5}{6}\left(x + \frac{3}{5}\right) = \frac{5}{6}x + \frac{1}{2}\).
Solution
LHS = \(\frac{5}{6}x + \frac{5}{6} \times \frac{3}{5} = \frac{5}{6}x + \frac{1}{2}\). RHS = \(\frac{5}{6}x + \frac{1}{2}\). The equation is true for all \(x\). So \(x\) can be any rational number.

Think and Reflect

Page 47
Can you explain why we need q ≠ 0 in the definition of a rational number?
Answer
Division by zero is undefined. There is no number that when multiplied by 0 gives a non-zero number.

Think and Reflect

Page 51
Try and represent \(\frac{8}{5}\) and \(-\frac{7}{4}\) on a number line.
Answer
\(\frac{8}{5} = 1.6\) lies between 1 and 2. \(-\frac{7}{4} = -1.75\) lies between -2 and -1.

EXERCISE SET 3.4 (Pages 52-53)

1.
Represent the rational numbers \(\frac{2}{3}, -\frac{5}{4}\) and \(1\frac{1}{2}\) on a single number line.
Solution
\(\frac{2}{3} \approx 0.67\) between 0 and 1. \(-\frac{5}{4} = -1.25\) between -2 and -1. \(1\frac{1}{2} = 1.5\) between 1 and 2.
2.
Find three distinct rational numbers that lie strictly between \(-\frac{1}{2}\) and \(\frac{1}{4}\).
Solution
Answers may vary. Example: \(-\frac{3}{8}, -\frac{1}{4}, -\frac{1}{8}\).
3.
Simplify the expression: \(\left(-\frac{1}{4}\right) + \left(\frac{5}{12}\right)\).
Solution
\(-\frac{3}{12} + \frac{5}{12} = \frac{2}{12} = \frac{1}{6}\).
4.
A tailor has \(15\frac{3}{4}\) metres of fine silk. If making one kurta requires \(2\frac{1}{4}\) metres of silk, exactly how many kurtas can he make?
Solution
\(15\frac{3}{4} = \frac{63}{4}\), \(2\frac{1}{4} = \frac{9}{4}\). Number of kurtas = \(\frac{63}{4} \div \frac{9}{4} = \frac{63}{4} \times \frac{4}{9} = 7\).
5.
Find three rational numbers between 3.1415 and 3.1416.
Solution
Answers may vary. Example: 3.14152, 3.14154, 3.14155.
*6.
Can you think of other way(s) to find a rational number between any two rational numbers?
Solution
Take the average: \(\frac{a+b}{2}\). Also, use the formula \(\frac{ma+nb}{m+n}\) for any positive integers m, n.

Think and Reflect

Page 53
Can \(\sqrt{2}\) be written as a rational number \(\frac{p}{q}\)?
Answer
No, \(\sqrt{2}\) is irrational. It cannot be written as a ratio of two integers.

Think and Reflect

Page 55
Try to prove the irrationality of \(\sqrt{3}\) using the approach of proof by contradiction. Will the same approach work for \(\sqrt{5}\), \(\sqrt{7}\), or \(\sqrt{10}\)?
Answer
Yes, the same method works for \(\sqrt{5}, \sqrt{7}\) but not for \(\sqrt{10}\) because 10 is not prime.

Think and Reflect

Page 56
We have seen how to obtain a line whose length is a rational number. How do we obtain lines whose lengths are irrational?
Answer
Use the Baudhayana-Pythagoras theorem: construct a right triangle with legs of rational lengths; the hypotenuse is often irrational.

Think and Reflect

Page 58
Try to find the decimal expansions of \(\frac{10}{3}\) and \(\frac{11}{12}\). What do you observe about the repetition of the digits after the decimal point?
Answer
\(\frac{10}{3} = 3.333...\) (repeating 3). \(\frac{11}{12} = 0.91666...\) (repeating 6).

EXERCISE SET 3.5 (Pages 61-62)

1.
Without performing long division, determine which of the following rational numbers will have terminating decimals and which will be repeating: \(\frac{7}{20}, \frac{4}{15}\) and \(\frac{13}{250}\). Then check your answers by explicitly performing the long divisions and expressing these rational numbers as decimals.
Solution
\(\frac{7}{20}\): denominator = \(2^2 \times 5\) → terminating = 0.35
\(\frac{4}{15}\): denominator = \(3 \times 5\) → repeating = 0.2666...
\(\frac{13}{250}\): denominator = \(2 \times 5^3\) → terminating = 0.052
2.
Perform the long division for \(\frac{1}{13}\). Identify the repeating block of digits. Does it show cyclic properties if you evaluate \(\frac{2}{13}\)? Now compute \(\frac{3}{13}, \frac{4}{13}\), etc. What do you notice?
Solution
\(\frac{1}{13} = 0.\overline{076923}\). Repeating block: 076923. \(\frac{2}{13} = 0.\overline{153846}\), \(\frac{3}{13} = 0.\overline{230769}\), etc. Each is a cyclic permutation of the digits of the same block.
3.
Classify the following numbers as rational or irrational:
(i) \(\sqrt{81}\)
(ii) \(\sqrt{12}\)
(iii) 0.33333...
(iv) 0.123451234512345...
(v) 1.01001000100001...
(vi) 23.560185612239874790120
Solution
(i) Rational (9)
(ii) Irrational
(iii) Rational (\(\frac{1}{3}\))
(iv) Rational (\(\frac{12345}{99999}\))
(v) Irrational (pattern not repeating)
(vi) Rational (terminating decimal)
4.
The number \(0.\overline{9}\) (which means 0.99999...) is a rational number. Using algebra (let \(x = 0.\overline{9}\), multiply by 10, and subtract), explain why \(0.\overline{9}\) is exactly equal to 1.
Solution
Let \(x = 0.\overline{9}\). Then \(10x = 9.\overline{9}\). Subtract: \(10x - x = 9.\overline{9} - 0.\overline{9} = 9\). So \(9x = 9\), hence \(x = 1\).
*5.
We have seen that the repeating block of \(\frac{1}{7}\) is a cyclic number. Try to find more numbers (n) whose reciprocals \(\frac{1}{n}\) produce decimals with repeating blocks that are cyclic.
Solution
Other examples: \(\frac{1}{17} = 0.\overline{0588235294117647}\) (16-digit cycle), \(\frac{1}{19}\) (18-digit cycle).

Think and Reflect

Page 64
Consider this puzzle: What is the square root of \(-1\)?
Answer
\(\sqrt{-1}\) is not a real number. It is denoted by \(i\) (imaginary unit).

END-OF-CHAPTER EXERCISES (Pages 64-66)

1.
Convert the following rational numbers in the form of a terminating decimal or non-terminating and repeating decimal, whichever the case may be, by the process of long division:
\(\frac{3}{50}\)    (ii) \(\frac{2}{9}\)
Solution
\(\frac{3}{50} = 0.06\) (terminating)
\(\frac{2}{9} = 0.\overline{2}\) (repeating)
2.
Prove that \(\sqrt{5}\) is an irrational number.
Solution
Assume \(\sqrt{5} = \frac{p}{q}\) in lowest terms. Then \(5q^2 = p^2\). So 5 divides \(p^2\), hence 5 divides p. Let p = 5k. Then \(5q^2 = 25k^2\) → \(q^2 = 5k^2\). So 5 divides q. Contradiction. Hence \(\sqrt{5}\) is irrational.
3.
Convert the following decimal numbers in the form of \(\frac{p}{q}\):
(i) 12.6   (ii) 0.0120   (iii) 3.052   (iv) 1.235   (v) 0.23   (vi) 2.05   (vii) 2.125   (viii) 3.125   (ix) 2.1625
Solution
(i) \(\frac{126}{10} = \frac{63}{5}\)
(ii) \(\frac{12}{1000} = \frac{3}{250}\)
(iii) \(\frac{3052}{1000} = \frac{763}{250}\)
(iv) \(\frac{1235}{1000} = \frac{247}{200}\)
(v) \(\frac{23}{100}\)
(vi) \(\frac{205}{100} = \frac{41}{20}\)
(vii) \(\frac{2125}{1000} = \frac{17}{8}\)
(viii) \(\frac{3125}{1000} = \frac{25}{8}\)
(ix) \(\frac{21625}{10000} = \frac{173}{80}\)
4.
Locate the following rational numbers on the number line:
(i) 0.532   (ii) 1.15
Solution
0.532 lies between 0.5 and 0.6. 1.15 lies between 1.1 and 1.2.
5.
Find 6 rational numbers between 3 and 4.
Solution
3.1, 3.2, 3.3, 3.4, 3.5, 3.6 (or any six rational numbers between 3 and 4)
6.
Find 5 rational numbers between \(\frac{2}{5}\) and \(\frac{3}{5}\).
Solution
\(\frac{21}{50}, \frac{22}{50}, \frac{23}{50}, \frac{24}{50}, \frac{25}{50}\) (or simplified forms)
7.
Find 5 rational numbers between \(\frac{1}{6}\) and \(\frac{2}{5}\).
Solution
\(\frac{1}{6} \approx 0.1667\), \(\frac{2}{5} = 0.4\). Examples: 0.2, 0.25, 0.3, 0.35, 0.38 = \(\frac{1}{5}, \frac{1}{4}, \frac{3}{10}, \frac{7}{20}, \frac{19}{50}\)
8.
If \(\frac{x}{3} + \frac{x}{5} = \frac{16}{15}\), find the rational number \(x\).
Solution
\(\frac{5x}{15} + \frac{3x}{15} = \frac{16}{15}\) → \(\frac{8x}{15} = \frac{16}{15}\) → \(8x = 16\) → \(x = 2\).
9.
Let \(a\) and \(b\) be two non-zero rational numbers such that \(a + \frac{1}{b} = 0\). Without assigning any numerical values, determine whether \(ab\) is positive or negative. Justify your answer.
Solution
\(a + \frac{1}{b} = 0\) ⇒ \(a = -\frac{1}{b}\) ⇒ \(ab = -1\). So \(ab\) is negative.
10.
A rational number has a terminating decimal expansion whose last non-zero digit occurs in the 4th decimal place. Show that such a number can be written in the form \(\frac{p}{10^4}\), where \(p\) is an integer not divisible by 10. Is it necessary that the denominator of this rational number, when written in the lowest form, is divisible by \(2^4\) or \(5^4\)? Give reasons.
Solution
The decimal has 4 decimal places, so it can be written as \(\frac{p}{10000}\) where p is an integer not ending with 0. When reduced to lowest terms, the denominator must be of the form \(2^m 5^n\) with m, n ≤ 4.
11.
Without performing division, determine whether the decimal expansion of \(\frac{18}{125}\) is terminating or non-terminating. If it terminates, state the number of decimal places.
Solution
\(125 = 5^3\). Denominator has only prime factor 5. So terminating. Number of decimal places = 3.
12.
A rational number in its lowest form has denominator \(2^3 \times 5\). How many decimal places will its decimal expansion have? Explain your answer.
Solution
It will have 3 decimal places (the maximum of the exponents of 2 and 5, which is 3).
*13.
Let \(a = \frac{7}{12}\) and \(b = \frac{5}{6}\). Express both \(a\) and \(b\) in the form \(\frac{k_1}{m}\) and \(\frac{k_2}{m}\) where \(k_1, k_2, m\) are integers and \(k_2 - k_1 > 6\). Using the same denominator \(m\), write exactly five distinct rational numbers lying between \(a\) and \(b\) keeping an integer numerator. Explain why the condition \(k_2 - k_1 > n + 1\) is necessary to find \(n\) such rational numbers between the two rational numbers \(a\) and \(b\) using this method.
Solution
\(\frac{7}{12} = \frac{14}{24}\), \(\frac{5}{6} = \frac{20}{24}\). Then integers between 14 and 20 are 15,16,17,18,19. So rational numbers: \(\frac{15}{24}, \frac{16}{24}, \frac{17}{24}, \frac{18}{24}, \frac{19}{24}\). Need \(k_2 - k_1 > n+1\) to have at least n integers between them.
*14.
Three rational numbers \(x, y, z\) satisfy \(x + y + z = 0\) and \(xy + yz + zx = 0\). Show that all the rational numbers \(x, y, z\) must be simultaneously zero.
Solution
Using identity: \(x^2 + y^2 + z^2 = (x+y+z)^2 - 2(xy+yz+zx) = 0\). So each is zero.
15.
Show that the rational number \(\frac{(a + b)}{2}\) lies between the rational numbers \(a\) and \(b\).
Solution
Without loss, assume \(a < b\). Then \(a + a < a + b < b + b\) ⇒ \(2a < a+b < 2b\) ⇒ \(a < \frac{a+b}{2} < b\).
16.
Find the lengths of the hypotenuse of all the right triangles in Fig. 3.14 which is referred to as the square root spiral.
Solution
The hypotenuses are \(\sqrt{2}, \sqrt{3}, \sqrt{4}, \sqrt{5}, \sqrt{6}, \ldots\)

Chapter Summary (Pages 66-67)

  • Natural Numbers (N) are the counting numbers {1, 2, 3, ...}
  • The Concept of Zero (Shunya) was formalised in India by Brahmagupta (628 CE)
  • Integers (Z) extend the number line to include zero and negative numbers
  • Rational Numbers (Q) are numbers that can be expressed as \(\frac{p}{q}\) where \(q \neq 0\)
  • Rational numbers are dense — a rational number always exists between any two other rational numbers
  • Irrational Numbers are values like \(\sqrt{2}\) and \(\pi\) that cannot be written as fractions
  • Real Numbers (R) represent the union of all rational and irrational numbers
  • Rational numbers have terminating or repeating decimal expansions
  • Irrational numbers have non-terminating, non-repeating decimal expansions
  • Cyclic Numbers, such as the repeating block of \(\frac{1}{7}\) (142857), reveal symmetrical internal patterns

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