Class 9 Maths Chapter 3 End of Chapter Exercise Solutions Ganita Manjari NCERT

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Our expert-written solutions are structured precisely the way high-achieving students write their answers in school notebooks—utilizing clean mathematical steps, logical proofs, and concise explanations without unnecessary fluff.

Whether you are converting complex repeating decimals into fractional forms, locating precise decimal values on a number line, or proving algebraic identities involving rational numbers, this comprehensive walkthrough will guide you smoothly through every single question with absolute clarity.

🔑 Key Concepts to Remember:

Decimal Conversions: Terminating decimals have denominators with prime factors of only 2 and 5, while repeating decimals involve other prime factors.
Algebraic Fractions: Non-terminating repeating decimals can be converted into $\frac{p}{q}$ form using algebraic multiplication equations.
Pythagorean Spiral: Successive right triangles build side lengths using the Pythagorean theorem, generating square roots like $\sqrt{2}, \sqrt{3}, \sqrt{4}, \dots$.

NCERT Class 9 Maths Ganita Manjari Chapter 3 End of Chapter Exercise Solutions

Question 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: (i) $\frac{3}{50}$ (ii) $\frac{2}{9}$.

Solution:

(i) $\frac{3}{50}$: By long division ($3 \div 50$), we get $\mathbf{0.06}$ (Terminating decimal).
(ii) $\frac{2}{9}$: By long division ($2 \div 9$), we get $0.222... = \mathbf{0.\bar{2}}$ (Non-terminating and repeating decimal).

Question 2

Prove that $\sqrt{5}$ is an irrational number.

Solution:

Let us assume to the contrary that $\sqrt{5}$ is a rational number. Thus, it can be expressed in the form $\frac{a}{b}$, where $a$ and $b$ are coprime integers and $b \neq 0$.

$\sqrt{5} = \frac{a}{b} \implies a = b\sqrt{5}$

Squaring both sides:
$a^2 = 5b^2$

This means $5$ divides $a^2$, so $5$ also divides $a$. Let $a = 5c$ for some integer $c$.

Substituting $a$:
$(5c)^2 = 5b^2 \implies 25c^2 = 5b^2 \implies b^2 = 5c^2$

This means $5$ divides $b^2$, so $5$ also divides $b$. Therefore, both $a$ and $b$ have a common factor of $5$, which contradicts our assumption that $a$ and $b$ are coprime. Hence, $\sqrt{5}$ is an irrational number.

Question 3

Convert the following decimal numbers in the form of $\frac{p}{q}$:
(i) $12.6$ (ii) $0.0120$ (iii) $3.05\bar{2}$
(iv) $1.2\bar{35}$ (v) $0.\bar{23}$ (vi) $2.0\bar{5}$
(vii) $2.12\bar{5}$ (viii) $3.12\bar{5}$ (ix) $2.1\bar{1625}$

Solution:

(i) $12.6$

$\displaystyle 12.6 = \frac{126}{10}$
Dividing numerator and denominator by 2:
$\displaystyle = \mathbf{\frac{63}{5}}$

(ii) $0.0120$

$\displaystyle 0.0120 = \frac{120}{10000}$
Cancelling out zeros and dividing by 40:
$\displaystyle = \mathbf{\frac{3}{250}}$

(iii) $3.05\bar{2}$

Let $x = 3.05222... \quad \rightarrow (1)$
Multiply both sides by 10:
$10x = 30.5222... \quad \rightarrow (2)$
Multiply equation (1) by 100:
$100x = 305.222... \quad \rightarrow (3)$
Subtracting equation (2) from equation (3):
$100x - 10x = 305.222... - 30.5222...$
$90x = 274.7$
$x = \frac{274.7}{90} = \mathbf{\frac{2747}{900}}$

(iv) $1.2\bar{35}$

Let $x = 1.2353535... \quad \rightarrow (1)$
Multiply by 10:
$10x = 12.353535... \quad \rightarrow (2)$
Multiply equation (1) by 1000:
$1000x = 1235.353535... \quad \rightarrow (3)$
Subtracting equation (2) from equation (3):
$1000x - 10x = 1235.3535... - 12.3535...$
$990x = 1223$
$x = \mathbf{\frac{1223}{990}}$

(v) $0.\bar{23}$

Let $x = 0.232323... \quad \rightarrow (1)$
Multiply both sides by 100:
$100x = 23.232323... \quad \rightarrow (2)$
Subtracting equation (1) from equation (2):
$100x - x = 23.2323... - 0.2323...$
$99x = 23$
$x = \mathbf{\frac{23}{99}}$

(vi) $2.0\bar{5}$

Let $x = 2.0555... \quad \rightarrow (1)$
Multiply by 10:
$10x = 20.555... \quad \rightarrow (2)$
Multiply equation (1) by 100:
$100x = 205.555... \quad \rightarrow (3)$
Subtracting equation (2) from equation (3):
$100x - 10x = 205.555... - 20.555...$
$90x = 185$
$x = \frac{185}{90} = \mathbf{\frac{37}{18}}$

(vii) $2.12\bar{5}$

Let $x = 2.12555... \quad \rightarrow (1)$
Multiply by 100:
$100x = 212.555... \quad \rightarrow (2)$
Multiply equation (1) by 1000:
$1000x = 2125.555... \quad \rightarrow (3)$
Subtracting equation (2) from equation (3):
$1000x - 100x = 2125.555... - 212.555...$
$900x = 1913$
$x = \mathbf{\frac{1913}{900}}$

(viii) $3.12\bar{5}$

Let $x = 3.12555... \quad \rightarrow (1)$
Multiply by 100:
$100x = 312.555... \quad \rightarrow (2)$
Multiply equation (1) by 1000:
$1000x = 3125.555... \quad \rightarrow (3)$
Subtracting equation (2) from equation (3):
$1000x - 100x = 3125.555... - 312.555...$
$900x = 2813$
$x = \mathbf{\frac{2813}{900}}$

(ix) $2.1\bar{1625}$

Let $x = 2.116251625... \quad \rightarrow (1)$
Multiply by 10:
$10x = 21.16251625... \quad \rightarrow (2)$
Multiply equation (1) by 100000:
$100000x = 211625.16251625... \quad \rightarrow (3)$
Subtracting equation (2) from equation (3):
$100000x - 10x = 211625.1625... - 21.1625...$
$99990x = 211604$
$x = \frac{211604}{99990} = \mathbf{\frac{21604}{9999}}$

Question 4

Locate the following rational numbers on the number line: (i) $0.532$ (ii) $1.1\bar{5}$.

Solution:

(i) $0.532$: Located successively using magnification between $0.5$ and $0.6$, then between $0.53$ and $0.54$, narrowing down to $0.532$.
(ii) $1.1\bar{5}$: Since $1.1\bar{5} = 1.1555...$, it is located via successive magnification between $1.1$ and $1.2$, further magnified between $1.15$ and $1.16$.

Question 5

Find 6 rational numbers between 3 and 4.

Solution:

Multiply and divide both numbers by $6 + 1 = 7$:
$3 = \frac{21}{7}$ and $4 = \frac{28}{7}$

Six rational numbers between them are:
$$\mathbf{\frac{22}{7}, \frac{23}{7}, \frac{24}{7}, \frac{25}{7}, \frac{26}{7}, \frac{27}{7}}$$

Question 6

Find 5 rational numbers between $\frac{2}{5}$ and $\frac{3}{5}$.

Solution:

Multiply numerator and denominator by $5 + 1 = 6$:
$\frac{2}{5} = \frac{12}{30}$ and $\frac{3}{5} = \frac{18}{30}$

Five rational numbers between them are:
$$\mathbf{\frac{13}{30}, \frac{14}{30}, \frac{15}{30}, \frac{16}{30}, \frac{17}{30}}$$

Question 7

Find 5 rational numbers between $\frac{1}{6}$ and $\frac{2}{5}$.

Solution:

Make denominators equal using LCM of 6 and 5, which is 30:
$\frac{1}{6} = \frac{5}{30}$ and $\frac{2}{5} = \frac{12}{30}$

Five rational numbers between $\frac{5}{30}$ and $\frac{12}{30}$ are:
$$\mathbf{\frac{6}{30}, \frac{7}{30}, \frac{8}{30}, \frac{9}{30}, \frac{10}{30}}$$

Question 8

If $\frac{x}{3} + \frac{x}{5} = \frac{16}{15}$, find the rational number $x$.

Solution:

Taking LCM on the left side (LCM of 3 and 5 is 15):
$$\frac{5x + 3x}{15} = \frac{16}{15}$$ $$\frac{8x}{15} = \frac{16}{15}$$ $8x = 16 \implies \mathbf{x = 2}$

Question 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:

Given: $a + \frac{1}{b} = 0 \implies a = -\frac{1}{b}$

Multiply both sides by $b$:
$ab = -1$

Answer: $ab$ is negative because its value is exactly $-1$.

Question 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:

Since the decimal terminates at the 4th decimal place, it can be expressed with a denominator of $10^4$, written as $\frac{p}{10^4}$ where $p$ is an integer not ending in zero (meaning it is not divisible by 10). When reduced to its lowest form, factors of 10 ($2 \times 5$) might cancel out with factors in $p$, so the denominator in lowest form is not necessarily divisible by both $2^4$ or $5^4$.

Question 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:

Denominator $125 = 5^3 = 2^0 \times 5^3$. Since the prime factors contain only powers of 5, the decimal expansion is terminating. It terminates after 3 decimal places ($0.144$).

Question 12

A rational number in its lowest form has a denominator $2^3 \times 5$. How many decimal places will its decimal expansion have? Explain your answer.

Solution:

The denominator has factors $2^3 \times 5^1$. To make powers of 2 and 5 equal (forming powers of 10), we multiply numerator and denominator by $5^2$. The highest exponent between 2 and 3 is 3. Therefore, its decimal expansion will have 3 decimal places.

Question 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 $k_1$, $k_2$ and $m$ 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:

Convert $a$ and $b$ to a common denominator $m = 12$:
$a = \frac{7}{12} \implies k_1 = 7$
$b = \frac{5}{6} = \frac{10}{12} \implies k_2 = 10$

Since $k_2 - k_1 = 10 - 7 = 3$ (which does not allow 5 numbers), we scale up by multiplying numerator and denominator by 2 (so $m = 24$):
$a = \frac{14}{24} \implies k_1 = 14$
$b = \frac{20}{24} \implies k_2 = 20$

Now, $k_2 - k_1 = 20 - 14 = 6 > 5$. Five rational numbers between them are:
$$\mathbf{\frac{15}{24}, \frac{16}{24}, \frac{17}{24}, \frac{18}{24}, \frac{19}{24}}$$

Question 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:

We know the algebraic identity:
$x^2 + y^2 + z^2 = (x + y + z)^2 - 2(xy + yz + zx)$

Substituting the given values ($x + y + z = 0$ and $xy + yz + zx = 0$):
$x^2 + y^2 + z^2 = (0)^2 - 2(0) = 0$

Since the sum of squares of real numbers equals zero, each term must individually be zero: $x = 0, y = 0, z = 0$.

Question 15

Show that the rational number $\frac{a+b}{2}$ lies between the rational numbers $a$ and $b$.

Solution:

Assume $a < b$. Adding $a$ to both sides: $a + a < a + b \implies 2a < a + b \implies a < \frac{a+b}{2}$.

Similarly, adding $b$ to both sides: $a + b < b + b \implies a + b < 2b \implies \frac{a+b}{2} < b$.

Combining both inequalities: $a < \frac{a+b}{2} < b$, proving it lies strictly between $a$ and $b$.

Question 16

Find the lengths of the hypotenuses of all the right triangles in Fig. 3.14 which is referred to as the square root spiral.

Square root spiral geometric construction diagram for NCERT Class 9 Maths Ganita Manjari Chapter 3

Solution:

Using the Pythagorean theorem successively where each base is 1 and perpendicular is the previous hypotenuse:

First triangle hypotenuse = $\sqrt{1^2 + 1^2} = \mathbf{\sqrt{2}}$
Second triangle hypotenuse = $\sqrt{(\sqrt{2})^2 + 1^2} = \mathbf{\sqrt{3}}$
Third triangle hypotenuse = $\sqrt{(\sqrt{3})^2 + 1^2} = \mathbf{\sqrt{4} = 2}$
Continuing this pattern, subsequent hypotenuses measure $\mathbf{\sqrt{5}, \sqrt{6}, \sqrt{7}, \sqrt{8}, \sqrt{9} (=3), \dots}$

Frequently Asked Questions (FAQs)

Q1: How do you prove that a square root like $\sqrt{5}$ is irrational?
A: By contradiction, assuming it is rational in the form $\frac{a}{b}$, squaring both sides, and showing that $a$ and $b$ share a common factor which violates the coprime property.

Q2: How do you convert repeating decimals into fraction form?
A: By setting the repeating decimal equal to $x$, multiplying by powers of 10 corresponding to the length of the repeating block, and subtracting equations to eliminate the infinite decimal part.

Q3: What textbook do these solutions follow?
A: These answers strictly follow the official Ganita Manjari Part I textbook for NCERT Class 9 Mathematics.

Q4: How do you find the number of decimal places for a terminating rational number?
A: Look at the prime factorization of its denominator in lowest form ($2^n 5^m$); the maximum exponent between $n$ and $m$ determines the exact count of decimal places.

Q5: What is the square root spiral?
A: A geometric construction of connected right-angled triangles that visually represents the square roots of natural numbers ($\sqrt{2}, \sqrt{3}, \sqrt{4}, \dots$).

Q6: Can rational numbers have non-terminating decimals?
A: Yes, rational numbers can have non-terminating decimals as long as the digits repeat in a fixed recurring pattern.

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Last updated according to the latest NCERT Ganita Manjari syllabus.