NCERT Solutions for Class 9 Maths Chapter 3 — End-of-Chapter Exercises#

Bring down	÷ 50	Remainder
30	0	30
300	6	0

The remainder becomes 0, so the decimal stops:

$$\dfrac{3}{50} = 0.06$$

This is a terminating decimal.

Bring down	÷ 9	Remainder
20	2	2
20	2	2
20	2	2

The remainder 2 repeats at every step, so the digit 2 repeats forever:

$$\dfrac{2}{9} = 0.222\ldots = 0.\overline{2}$$

This is a non-terminating repeating decimal.

Step 1: Assumption. Assume $\sqrt{5}$ is rational. Then it can be written as $\dfrac{p}{q}$ in its simplest form, where $p$ and $q$ are integers, $q \neq 0$, and $p$ and $q$ have no common factor other than 1:

$$\sqrt{5} = \dfrac{p}{q}$$

Step 2: Squaring both sides:

$$5 = \dfrac{p^2}{q^2}$$

Step 3: Multiplying both sides by $q^2$:

$$5q^2 = p^2 \quad \cdots (1)$$

Step 4: Deduction for $p$. $p^2$ is 5 times an integer, so 5 divides $p^2$. Since 5 is a prime number, if 5 divides $p^2$ then 5 divides $p$. So $p = 5k$ for some integer $k$.

Step 5: Substituting $p = 5k$ in equation (1):

$$5q^2 = p^2$$

$$5q^2 = (5k)^2$$

$$5q^2 = 25k^2$$

Step 6: Dividing both sides by 5:

$$q^2 = 5k^2$$

Step 7: Deduction for $q$. $q^2$ is 5 times an integer, so 5 divides $q^2$, and therefore 5 divides $q$.

Step 8: Contradiction. Both $p$ and $q$ have 5 as a factor. But in Step 1 we said $\dfrac{p}{q}$ was in its simplest form, with no common factor other than 1. This is a contradiction.

So our assumption was wrong.

∴ $\sqrt{5}$ is irrational.

A terminating decimal with one decimal place is written over 10. We convert $12.6$:

$$= \dfrac{126}{10}$$

$$= \dfrac{63}{5}$$

The last 0 does not change the value, so $0.0120 = 0.012$, which has three decimal places. We convert $0.012$:

$$= \dfrac{12}{1000}$$

$$= \dfrac{3}{250}$$

Let $x = 3.0\overline{52} = 3.05252\ldots$.

There is 1 non-repeating digit after the decimal point, so we first multiply by $10^1 = 10$:

$$10x = 30.5252\ldots \quad \cdots (1)$$

Then we move one full repeating block, multiplying by $10^2 = 100$ again:

$$1000x = 3052.5252\ldots \quad \cdots (2)$$

Subtracting equation (1) from equation (2):

$$\begin{array}{rcl} 1000x &=& 3052.5252\ldots \\ 10x &=& 30.5252\ldots \\ \hline 990x &=& 3022 \end{array}$$

$$\Rightarrow x = \dfrac{3022}{990}$$

$$\Rightarrow x = \dfrac{1511}{495}$$

∴ $3.0\overline{52} = \dfrac{1511}{495}$.

Let $x = 1.2\overline{35} = 1.23535\ldots$.

There is 1 non-repeating digit after the decimal point, so we first multiply by $10^1 = 10$:

$$10x = 12.3535\ldots \quad \cdots (1)$$

Then we move one full repeating block, multiplying by $10^2 = 100$ again:

$$1000x = 1235.3535\ldots \quad \cdots (2)$$

Subtracting equation (1) from equation (2):

$$\begin{array}{rcl} 1000x &=& 1235.3535\ldots \\ 10x &=& 12.3535\ldots \\ \hline 990x &=& 1223 \end{array}$$

$$\Rightarrow x = \dfrac{1223}{990}$$

∴ $1.2\overline{35} = \dfrac{1223}{990}$.

Let $x = 0.\overline{23} = 0.2323\ldots$.

2 digits repeat, so we multiply by $10^2 = 100$:

$$x = 0.2323\ldots \quad \cdots (1)$$

$$100x = 23.2323\ldots \quad \cdots (2)$$

Subtracting equation (1) from equation (2):

$$\begin{array}{rcl} 100x &=& 23.2323\ldots \\ x &=& 0.2323\ldots \\ \hline 99x &=& 23 \end{array}$$

$$\Rightarrow x = \dfrac{23}{99}$$

∴ $0.\overline{23} = \dfrac{23}{99}$.

Let $x = 2.0\overline{5} = 2.0555\ldots$.

There is 1 non-repeating digit after the decimal point, so we first multiply by $10^1 = 10$:

$$10x = 20.555\ldots \quad \cdots (1)$$

Then we move one full repeating block, multiplying by $10^1 = 10$ again:

$$100x = 205.555\ldots \quad \cdots (2)$$

Subtracting equation (1) from equation (2):

$$\begin{array}{rcl} 100x &=& 205.555\ldots \\ 10x &=& 20.555\ldots \\ \hline 90x &=& 185 \end{array}$$

$$\Rightarrow x = \dfrac{185}{90}$$

$$\Rightarrow x = \dfrac{37}{18}$$

∴ $2.0\overline{5} = \dfrac{37}{18}$.

Let $x = 2.12\overline{5} = 2.12555\ldots$.

There are 2 non-repeating digits after the decimal point, so we first multiply by $10^2 = 100$:

$$100x = 212.555\ldots \quad \cdots (1)$$

Then we move one full repeating block, multiplying by $10^1 = 10$ again:

$$1000x = 2125.555\ldots \quad \cdots (2)$$

Subtracting equation (1) from equation (2):

$$\begin{array}{rcl} 1000x &=& 2125.555\ldots \\ 100x &=& 212.555\ldots \\ \hline 900x &=& 1913 \end{array}$$

$$\Rightarrow x = \dfrac{1913}{900}$$

∴ $2.12\overline{5} = \dfrac{1913}{900}$.

Let $x = 3.12\overline{5} = 3.12555\ldots$.

There are 2 non-repeating digits after the decimal point, so we first multiply by $10^2 = 100$:

$$100x = 312.555\ldots \quad \cdots (1)$$

Then we move one full repeating block, multiplying by $10^1 = 10$ again:

$$1000x = 3125.555\ldots \quad \cdots (2)$$

Subtracting equation (1) from equation (2):

$$\begin{array}{rcl} 1000x &=& 3125.555\ldots \\ 100x &=& 312.555\ldots \\ \hline 900x &=& 2813 \end{array}$$

$$\Rightarrow x = \dfrac{2813}{900}$$

∴ $3.12\overline{5} = \dfrac{2813}{900}$.

Let $x = 2.\overline{1625} = 2.16251625\ldots$.

4 digits repeat, so we multiply by $10^4 = 10000$:

$$x = 2.16251625\ldots \quad \cdots (1)$$

$$10000x = 21625.16251625\ldots \quad \cdots (2)$$

Subtracting equation (1) from equation (2):

$$\begin{array}{rcl} 10000x &=& 21625.16251625\ldots \\ x &=& 2.16251625\ldots \\ \hline 9999x &=& 21623 \end{array}$$

$$\Rightarrow x = \dfrac{21623}{9999}$$

∴ $2.\overline{1625} = \dfrac{21623}{9999}$.

We use successive magnification: we zoom in on one interval at a time.

1. $0.532$ lies between 0 and 1. Divide the unit into 10 parts; it lies between 0.5 and 0.6.
2. Divide the interval from 0.5 to 0.6 into 10 parts; it lies between 0.53 and 0.54.
3. Divide the interval from 0.53 to 0.54 into 10 parts; $0.532$ is the 2nd mark after 0.53.

$1.1\overline{5} = 1.1555\ldots$ We locate it to four decimal places by successive magnification.

1. It lies between 1.1 and 1.2.
2. Dividing 1.1 to 1.2 into 10 parts, it lies between 1.15 and 1.16.
3. Dividing 1.15 to 1.16 into 10 parts, it lies between 1.155 and 1.156.
4. Dividing 1.155 to 1.156 into 10 parts, it lies just past 1.1555 (between 1.1555 and 1.1556).

We can keep zooming in, and the point is always just past the 5th mark, because the digit 5 repeats forever.

We need at least 6 numerators between the two numbers, so we write both with denominator 7 (any denominator of 7 or more works):

$$3 = \dfrac{21}{7}$$

$$4 = \dfrac{28}{7}$$

The numerators between 21 and 28 are 22, 23, 24, 25, 26 and 27.

∴ Six rational numbers between 3 and 4 are $\dfrac{22}{7}, \dfrac{23}{7}, \dfrac{24}{7}, \dfrac{25}{7}, \dfrac{26}{7}, \dfrac{27}{7}$.

(Using decimals, 3.1, 3.2, 3.3, 3.4, 3.5 and 3.6 is another correct answer.)

The numerators 2 and 3 have no whole numbers between them, so we multiply the numerator and denominator by 6:

$$\dfrac{2}{5} = \dfrac{12}{30}$$

$$\dfrac{3}{5} = \dfrac{18}{30}$$

The numerators between 12 and 18 are 13, 14, 15, 16 and 17.

∴ Five rational numbers are $\dfrac{13}{30}, \dfrac{14}{30}, \dfrac{15}{30}, \dfrac{16}{30}, \dfrac{17}{30}$, that is, $\dfrac{13}{30}, \dfrac{7}{15}, \dfrac{1}{2}, \dfrac{8}{15}, \dfrac{17}{30}$.

The LCM of 6 and 5 is 30:

$$\dfrac{1}{6} = \dfrac{5}{30}$$

$$\dfrac{2}{5} = \dfrac{12}{30}$$

The numerators between 5 and 12 are 6, 7, 8, 9, 10 and 11. We choose any five of them.

∴ Five rational numbers are $\dfrac{6}{30}, \dfrac{7}{30}, \dfrac{8}{30}, \dfrac{9}{30}, \dfrac{10}{30}$, that is, $\dfrac{1}{5}, \dfrac{7}{30}, \dfrac{4}{15}, \dfrac{3}{10}, \dfrac{1}{3}$.

The LCM of 3 and 5 is 15. We solve $\dfrac{x}{3} + \dfrac{x}{5} = \dfrac{16}{15}$:

$$\Rightarrow \dfrac{5x}{15} + \dfrac{3x}{15} = \dfrac{16}{15}$$

$$\Rightarrow \dfrac{8x}{15} = \dfrac{16}{15}$$

$$\Rightarrow 8x = 16$$

$$\Rightarrow x = 2$$

∴ $x = 2$.

Check: $\dfrac{2}{3} + \dfrac{2}{5} = \dfrac{10}{15} + \dfrac{6}{15} = \dfrac{16}{15}$ ✓

We are given:

$$a + \dfrac{1}{b} = 0$$

Multiplying both sides by $b$ (allowed, since $b \neq 0$):

$$ab + 1 = 0$$

$$\Rightarrow ab = -1$$

∴ $ab = -1$, so $ab$ is negative.

This also makes sense with signs: $a = -\dfrac{1}{b}$, so $a$ and $b$ always have opposite signs, and the product of a positive and a negative number is negative.

Let the number be $x = n.d_1 d_2 d_3 d_4$, where $n$ is the whole-number part and $d_4 \neq 0$ (the last non-zero digit is in the 4th place).

Multiplying by $10^4$ moves the decimal point 4 places to the right, so $10^4 x$ is an integer $p$ (the digits of $n$ followed by $d_1 d_2 d_3 d_4$):

$$x = \dfrac{p}{10^4}$$

The last digit of $p$ is $d_4$, and $d_4 \neq 0$. A number divisible by 10 ends in 0, so $p$ is not divisible by 10.

For example, $0.0375 = \dfrac{375}{10^4}$, and 375 is not divisible by 10.

Yes — the lowest-form denominator must be divisible by at least one of $2^4$ and $5^4$, but not necessarily both.

Reason: $10^4 = 2^4 \times 5^4$, so

$$x = \dfrac{p}{2^4 \times 5^4}$$

To reach lowest form, we cancel the factors 2 and 5 that $p$ shares with the denominator. Since $p$ is not divisible by 10, $p$ cannot have both 2 and 5 as factors. So:

• if 2 does not divide $p$, nothing cancels from $2^4$, and the denominator keeps $2^4$;
• if 5 does not divide $p$, nothing cancels from $5^4$, and the denominator keeps $5^4$.

At least one of these is always true.

Examples:

$$0.0005 = \dfrac{5}{10^4} = \dfrac{1}{2000} = \dfrac{1}{2^4 \times 5^3}$$

Here the denominator is divisible by $2^4$ but not by $5^4$.

$$0.0002 = \dfrac{2}{10^4} = \dfrac{1}{5000} = \dfrac{1}{2^3 \times 5^4}$$

Here the denominator is divisible by $5^4$ but not by $2^4$.

$18 = 2 \times 3^2$ and $125 = 5^3$ have no common factor, so $\dfrac{18}{125}$ is in lowest terms.

The denominator $125 = 5^3$ has only the prime factor 5, so the decimal terminates.

To make the denominator a power of 10, we multiply the numerator and denominator by $2^3 = 8$:

$$\dfrac{18}{125} = \dfrac{18 \times 8}{125 \times 8}$$

$$\dfrac{18}{125} = \dfrac{144}{1000}$$

$$\dfrac{18}{125} = 0.144$$

∴ The decimal terminates after 3 decimal places.

Let the number be $\dfrac{p}{2^3 \times 5}$ in lowest form. Since it is in lowest form, $p$ has no factor 2 and no factor 5.

To make the denominator a power of 10, we need as many 5s as 2s, so we multiply the numerator and denominator by $5^2$:

$$\dfrac{p}{2^3 \times 5} = \dfrac{p \times 5^2}{2^3 \times 5^3}$$

$$\dfrac{p}{2^3 \times 5} = \dfrac{25p}{1000}$$

Dividing by 1000 gives 3 decimal places. The last digit is not 0, because $25p$ is not divisible by 10 ($p$ is odd, so $25p$ is odd).

∴ The decimal expansion has exactly 3 decimal places. In general, the number of places is the larger of the two powers of 2 and 5 in the denominator.

For example, $\dfrac{1}{40} = \dfrac{25}{1000} = 0.025$.

With the LCM 12 as the denominator:

$$a = \dfrac{7}{12}$$

$$b = \dfrac{10}{12}$$

Here $k_2 - k_1 = 10 - 7 = 3$, which is not more than 6. Multiplying the numerator and denominator by 3:

$$a = \dfrac{21}{36}$$

$$b = \dfrac{30}{36}$$

Now $k_1 = 21$, $k_2 = 30$ and $m = 36$:

$$\Rightarrow k_2 - k_1 = 9 > 6$$

The integers between 21 and 30 are 22, 23, …, 29. We take five of them.

∴ Five rational numbers between $a$ and $b$ are $\dfrac{22}{36}, \dfrac{23}{36}, \dfrac{24}{36}, \dfrac{25}{36}, \dfrac{26}{36}$.

Why the condition is needed: the numbers we can write are $\dfrac{k}{m}$ with $k$ an integer strictly between $k_1$ and $k_2$. These integers are $k_1 + 1, k_1 + 2, \ldots, k_2 - 1$, and there are $k_2 - k_1 - 1$ of them. To get $n$ numbers we need

$$k_2 - k_1 - 1 \geq n$$

$$\Rightarrow k_2 - k_1 \geq n + 1$$

If $k_2 - k_1$ is too small, there are not enough integer numerators, and we must first multiply the numerator and denominator by a larger number (as we did, going from 12 to 36). The condition $k_2 - k_1 > n + 1$ guarantees more than $n$ numerators in between, so $n$ distinct numbers can always be chosen. (Strictly, $k_2 - k_1 \geq n + 1$ is already enough.)

We are given:

$$x + y + z = 0 \quad \cdots (1)$$

$$xy + yz + zx = 0 \quad \cdots (2)$$

We use the identity

$$(x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx)$$

Substituting equation (1) and equation (2) in this identity:

$$0^2 = x^2 + y^2 + z^2 + 2 \times 0$$

$$\Rightarrow x^2 + y^2 + z^2 = 0$$

The square of any rational number is positive or zero. A sum of three numbers that are each $\geq 0$ can be 0 only if each of them is 0:

$$x^2 = 0, \quad y^2 = 0, \quad z^2 = 0$$

$$\Rightarrow x = 0, \quad y = 0, \quad z = 0$$

∴ $x$, $y$ and $z$ must all be zero.

Suppose $a < b$ (if $b < a$, swap the names). Then $b - a > 0$.

We compare $\dfrac{a + b}{2}$ with $a$ by subtracting:

$$\dfrac{a + b}{2} - a = \dfrac{a + b - 2a}{2}$$

$$\dfrac{a + b}{2} - a = \dfrac{b - a}{2}$$$$\Rightarrow \dfrac{b - a}{2} > 0$$

∴ $\dfrac{a + b}{2} > a$.

We compare $b$ with $\dfrac{a + b}{2}$ by subtracting:

$$b - \dfrac{a + b}{2} = \dfrac{2b - a - b}{2}$$

$$b - \dfrac{a + b}{2} = \dfrac{b - a}{2}$$$$\Rightarrow \dfrac{b - a}{2} > 0$$

∴ $b > \dfrac{a + b}{2}$.

Putting both together:

$$a < \dfrac{a + b}{2} < b$$

So the average of two rational numbers always lies between them (exactly halfway). It is also rational, because sums and quotients of rational numbers are rational.

The spiral has 10 right triangles. The first has both legs equal to 1. Each next triangle uses the hypotenuse of the previous one as a leg, with the other leg equal to 1.

By the Baudhāyana–Pythagoras theorem, for the first triangle:

$$h_1^2 = 1^2 + 1^2$$

$$h_1^2 = 2$$

$$\Rightarrow h_1 = \sqrt{2}$$

For the second triangle, the legs are $\sqrt{2}$ and 1:

$$h_2^2 = (\sqrt{2})^2 + 1^2$$

$$h_2^2 = 3$$

$$\Rightarrow h_2 = \sqrt{3}$$

In the same way, each hypotenuse squared is 1 more than the one before:

$$h_n^2 = h_{n-1}^2 + 1$$

Triangle	Hypotenuse
1	$\sqrt{2}$
2	$\sqrt{3}$
3	$\sqrt{4} = 2$
4	$\sqrt{5}$
5	$\sqrt{6}$
6	$\sqrt{7}$
7	$\sqrt{8} = 2\sqrt{2}$
8	$\sqrt{9} = 3$
9	$\sqrt{10}$
10	$\sqrt{11}$

∴ The hypotenuses are $\sqrt{2}, \sqrt{3}, 2, \sqrt{5}, \sqrt{6}, \sqrt{7}, 2\sqrt{2}, 3, \sqrt{10}$ and $\sqrt{11}$.