Chapter Test ML Aggarwal Class 9 Rational and Irrational Numbers Full Solutions » True Answer

Exercise 1.1 Exercise 1.2 Exercise 1.3 Exercise 1.4 Exercise 1.5 Chapter Test-1

Chapter Test ML Aggarwal Class 9 of chapter 1 rational and irrational numbers from Mathematics ICSE Textbook contains a total of 19 questions. The questions are based on all the concepts given in the chapter.

The main topics of each exercise are rational numbers and their representation on the number-line, irrational numbers and their representation on the number line, surds, rationalisation of surds.

Chapter Test ML Aggarwal Class 9 Mathematics Solutions

Q.1 Q.2 Q.3 Q.4 Q.5 Q.6 Q.7 Q.8 Q.9 Q.10 Q.11 Q.12 Q.13 Q.14 Q.15 Q.16 Q.17 Q.18 Q.19

The first question of the Chapter Test ML Aggarwal Class 9 asks us to find the given numbers for terminating decimals or recurring decimals.

1. Without actual division, find whether the following rational numbers are terminating decimals or recurring decimals:
In case of terminating decimals, write their decimal expansions.

(i) $\frac{13}{45}$

Answer

A rational number is terminating if its denominator only has factors of 2 or 5, or both. It’s non-terminating if its denominator has at least one factor other than 2 and 5.

The prime factors of the denominator of $\frac{13}{45}$ i.e. $45$ = $3 \times 3 \times 5$ .
Since $45$ contains $3$ as its one of prime factors other than 5, so $\frac{13}{45}$ is a recurring decimal.

(ii) $- \frac{5}{56}$

Answer

A rational number is terminating if its denominator only has factors of 2 or 5, or both. It’s non-terminating if its denominator has at least one factor other than 2 and 5.

The prime factors of the denominator of $- \frac{5}{56}$ i.e. $56$ = $2 \times 2 \times 2 \times 7$ .
Since $56$ contains $7$ as its one of prime factors other than 2, so $- \frac{5}{56}$ is a recurring decimal.

(iii) $\frac{7}{125}$

Answer

A rational number is terminating if its denominator only has factors of 2 or 5, or both. It’s non-terminating if its denominator has at least one factor other than 2 and 5.

The prime factors of the denominator of $\frac{7}{125}$ i.e. $125$ = $5 \times 5 \times 5$ .
Since $125$ only contains $5$ as its prime factor, $\frac{7}{125}$ is a terminating decimal.

Now,

$\frac{7}{125} = \frac{7}{5^{3}} \times \frac{2^{3}}{2^{3}} = \frac{7 \times 8}{{(5 \times 2)}^{3}}$ $= \frac{56}{10^{3}} = \frac{56}{1000} =$ $0.0 56$

(iv) $- \frac{23}{80}$

Answer

A rational number is terminating if its denominator only has factors of 2 or 5, or both. It’s non-terminating if its denominator has at least one factor other than 2 and 5.

The prime factors of the denominator of $- \frac{23}{80}$ i.e. $80$ = $2 \times 2 \times 2 \times 2 \times 5$ .
Since $125$ only contains $5$ as its prime factor, $\frac{7}{125}$ is a terminating decimal.

Now,

$- \frac{23}{80} = - \frac{23}{2^{4} \times 5} = - \frac{23}{2^{4} \times 5} \times \frac{5^{3}}{5^{3}}$ $= - \frac{2875}{{(2 \times 5)}^{4}} = - \frac{2875}{10^{4}} = - \frac{2875}{10000} =$ $- 0. 2875$ .

(v) $- \frac{15}{66}$

Answer

A rational number is terminating if its denominator only has factors of 2 or 5, or both. It’s non-terminating if its denominator has at least one factor other than 2 and 5.

The prime factors of the denominator of $- \frac{15}{66}$ i.e. $66$ = $3 \times 2 \times 11$ .
Since $66$ contains $3$ and $11$ as its two of the prime factors other than 2, so $- \frac{15}{66}$ is a recurring decimal.

The second question of the Chapter Test ML Aggarwal Class 9 asks us to express the recurring decimals as vulgar fractions.

2. Express the following recurring decimals as vulgar fractions:

(i) 1.345

Answer

Let $x =$ 1.345.
$\Rightarrow x = 1.3454545 . . .$ –––––––––––––– (i)
Since there are two digits containing a bar, we multiply both sides by 100. We get

$100 x = 134.54545 . . .$ –––––––––––––– (ii)

Subtracting equation (i) from (ii), we get.

$100 x - x = (134.5454545 . . .) - (1.3454545 . . .)$

$\Rightarrow 99 x = 133.2$

$\Rightarrow x = \frac{133.2}{99} \times \frac{10}{10} = \frac{1332}{990} = \frac{74}{55}$ .

Hence, 1.345 = $\frac{74}{55}$ .

(ii) 2.357

Answer

Let $x =$ 2.357.
$\Rightarrow x = 2.357357 . . .$ –––––––––––––– (i)
Since there are three digits containing a bar, we multiply both sides by 1000. We get

$1000 x = 2357.357 . . .$ –––––––––––––– (ii)

Subtracting equation (i) from (ii), we get.

$1000 x - x = (2357.357357 . . .) - (2.357357 . . .)$

$\Rightarrow 999 x = 2355$

$\Rightarrow x = \frac{2355}{999} = \frac{785}{333}$ .

Hence, 2.357 = $\frac{785}{333}$ .

The third question of the Chapter Test ML Aggarwal Class 9 asks us to insert a rational number between two rational numbers and arrange them in ascending order.

3. Insert a rational number between $\frac{5}{9}$ and $\frac{7}{13}$ , and arrange in ascending order.

Answer

A rational number between $\frac{5}{9}$ and $\frac{7}{13}$ is

$\frac{\frac{5}{9} + \frac{7}{13}}{2}$ = $\frac{\frac{65 + 63}{117}}{2}$ = $\frac{\frac{128}{117}}{2}$ = $\frac{128}{2 \times 117}$ = $\frac{64}{117}$ .

Since $\frac{64}{117}$ lies between $\frac{5}{9}$ and $\frac{7}{13}$ . So, $\frac{5}{9} < \frac{64}{117} < \frac{7}{13}$ .

Hence, the ascending order of the rational numbers is

$\frac{5}{9}, \frac{64}{117}, \frac{7}{13}$ .

The fourth question of exercise 1.5 ML Aggarwal Class 9 asks us to insert four rational numbers between two given rational numbers.

4. Insert four rational numbers between $\frac{4}{5}$ and $\frac{5}{6}$ .

Answer

To find four rational numbers between $\frac{4}{5}$ and $\frac{5}{6}$ , we follow the following steps:

Firstly, we make the denominators of both the rational numbers equal by taking their L.C.M.
The L.C.M. of 5 and 6 is 30.
∴ $\frac{4}{5} = \frac{4 \times 6}{5 \times 6} = \frac{24}{30}$ , and $\frac{5}{6} = \frac{5 \times 5}{6 \times 5} = \frac{25}{30}$ .

Secondly, we multiply by 4 + 1 = 5 in both the numerators and denominators of both the rational numbers, as we need to find only four rational numbers in between them.
$\frac{24}{30} \times \frac{5}{5} = \frac{120}{150}$ , and $\frac{25}{30} \times \frac{5}{5} = \frac{125}{150}$ .

Thirdly, we write the rational numbers lying between $\frac{120}{150}$ and $\frac{125}{150}$ , which are the required four rational numbers.

$\frac{4}{5} = \frac{120}{150} < \frac{121}{150} < \frac{122}{150} < \frac{123}{150} < \frac{124}{150} < \frac{125}{150} = \frac{5}{6}$

Hence, the four rational numbers between $\frac{4}{5}$ and $\frac{5}{6}$ are

$\frac{121}{150}, \frac{122}{150}, \frac{123}{150}, \frac{124}{150}$ = $\frac{121}{150}, \frac{61}{75}, \frac{41}{50}, \frac{62}{75}$ .

The fifth question of the Chapter Test ML Aggarwal Class 9 asks us to prove a proposal.

5. Prove that the reciprocal of an irrational number is irrational.

Answer

To prove: The reciprocal of an irrational number is irrational.

Proof: Let 'b' be an irrational number. Now, we have to prove that $\frac{1}{b}$ is also irrational.

Let us assume that $\frac{1}{b}$ is a rational number.
Then, we can write $\frac{1}{b} = \frac{p}{q}$ , where p, q are integers and q ≠ 0.

$\frac{1}{b} = \frac{p}{q}$ $\Rightarrow b \times p = q \times 1$ $\Rightarrow b = \frac{q}{p}$

$\Rightarrow b$ is a rational number.
This contradicts the fact that b is an irrational number. Thus, our assumption that $\frac{1}{b}$ is a rational number is wrong. So, $\frac{1}{b}$ is an irrational number.

Hence, the reciprocal of an irrational number is also irrational. We proved it.

The sixth question of the Chapter Test ML Aggarwal Class 9 asks us to prove the given numbers to be irrational.

6. Prove that the following numbers are irrational:

(i) $\sqrt{8}$

Answer

We have
$\sqrt{8} = 2 \sqrt{2}$
We know that the product of a rational number and an irrational number is an irrational number.
Here, 2 is a rational number and $\sqrt{2}$ is an irrational number.
Their product must be an irrational number.
Hence, $\sqrt{8}$ is an irrational number.

(i) $\sqrt{14}$

Answer

Suppose that $\sqrt{14}$ is a rational number, then

$\sqrt{14} = \frac{p}{q},$ where $p, q$ are integers, $q \neq 0$ and $p, q$ have no common factors other than 1.

$\Rightarrow 14 = \frac{p^{2}}{q^{2}}$ ... [∵ Squaring on both sides]

$\Rightarrow p^{2} = 14 q^{2}$ ... $(i)$

As $14 q^{2}$ is a multiple of 2 and 7 both,

$\Rightarrow$ 2 divides $14 q^{2}$ .

$\Rightarrow$ 2 divides $p^{2}$ ... $[∵ p^{2} = 14 q^{2}]$

According to the theorem,

If $a$ is any natural number and $p$ is a prime number such that $p$ divides $a^{2}$ , then $p$ divides $a$ .

$\Rightarrow$ 2 divides $p$ ... [∵ 2 is prime.]

Let $p = 2 m$ , where m is an integer.

Substituting this value of $p$ in $(i)$ , we get

${(2 m)}^{2} = 14 q^{2}$
$\Rightarrow 4 m^{2} = 14 q^{2}$
$\Rightarrow 2 m^{2} = 7 q^{2}$

As $2$ divides $2 m^{2}$ ,

$\Rightarrow 2$ divides $7 q^{2}$ ... $[∵ 2 m^{2} = 7 q^{2}]$

$\Rightarrow 2$ divides $7$ or $2$ divides $q^{2}$

But, we know that 2 doesn't divide 7, so $2$ divides $q^{2}$ .

$\Rightarrow 2$ divides $q$ ... [From the above theorem]

∵ 2 divides both $p$ and $q$ , 2 is a common factor of $p$ and $q$ .

This contradicts the initial supposition that $p$ and $q$ have no common factors other than 1.

Hence, our supposition is wrong. Therefore, $\sqrt{14}$ is an irrational number.

(i) $\sqrt[3]{2}$

Answer

To prove: $\sqrt[3]{2}$ is an irrational number.

Proof: Let us assume that $\sqrt[3]{2}$ is a rational number.
$\Rightarrow \sqrt[3]{2} = \frac{p}{q}$ , where $p$ , $q$ are integers, $q \neq 0$ , $p$ and $q$ have no common factor other than 1.
$\Rightarrow {(\sqrt[3]{2})}^{3} = {(\frac{p}{q})}^{3}$ ... [Taking cube on both sides]

$\Rightarrow 2 = \frac{p^{3}}{q^{3}}$

$\Rightarrow p^{3} = 2 q^{3}$ .... (i)

We have a theorem which states that

If $a$ is any natural number and $p$ is a prime number such that $p$ divides $a^{n}$ , then $p$ divides $a$ . In this case, we choose $n = 3$ .

Using the above theorem and equation (i), we get
As $2$ divides $2 q^{3}$
$\Rightarrow$ $2$ divides $p^{3}$
$\Rightarrow 2$ divides $p$ ....... [Using above theorem]

Now, let $p = 2 k$ , where k is an integer.
Substituting this value of $p$ in (i), we get
${(2 k)}^{3} = 2 q^{3}$
$\Rightarrow 8 k^{3} = 2 q^{3}$
$\Rightarrow 4 k^{3} = q^{3}$ ......... [Dividing by 2 on both sides]

As $2$ divides $4 k^{3}$ ,
$\Rightarrow 2$ divides $q^{3}$
$\Rightarrow 2$ divides $q$ .......... [Using above theorem]

Thus, $2$ divides $p$ and $q$ both, i.e. $p$ and $q$ have a common factor other than 1. This contradicts the fact that $p$ and $q$ have no common factors other than 1.

Hence, our assumption is wrong. It follows that $\sqrt[3]{2}$ cannot be expressed as $\frac{p}{q}$ , where $p$ , $q$ are integers, $q$ ≠ $0$ , $p$ and $q$ have no common factors other than 1. Therefore, $\sqrt[3]{2}$ is an irrational number.
Hence, proved.

The seventh question of the Chapter Test ML Aggarwal Class 9 asks us a given surd irrational when another irrational is given.

7. Prove that $\sqrt{3}$ is an irrational number. Hence, show that $5 - \sqrt{3}$ is an irrational number.

Answer

The eighth question of the Chapter Test ML Aggarwal Class 9 asks us to prove the given surds irrational.

8. Prove that the following numbers are irrational:

(i) $3 + \sqrt{5}$

Answer

(ii) $15 - 2 \sqrt{7}$

Answer

(iii) $\frac{1}{3 - \sqrt{5}}$

Answer

The ninth question of the Chapter Test ML Aggarwal Class 9 asks us to rationalise the denominator of the following.

9. Rationalise the denominator of the following:

(i) $\frac{10}{2 \sqrt{2} + \sqrt{3}}$

Answer

(ii) $\frac{7 \sqrt{3} - 5 \sqrt{2}}{\sqrt{48} + \sqrt{18}}$

Answer

(ii) $\frac{1}{\sqrt{3} - \sqrt{2} + 1}$

Answer

The tenth question of the Chapter Test ML Aggarwal Class 9 asks us to find the value of $p$ and $q$ from the given equation.

10. If $p, q$ are rational numbers and $p - \sqrt{15} q = \frac{2 \sqrt{3} - \sqrt{5}}{4 \sqrt{3} - 3 \sqrt{5}}$ , find the values of $p$ and $q$ .

Answer

The eleventh question of the Chapter Test ML Aggarwal Class 9 asks us find the value of an expression.

11. If $x = \frac{1}{3 + 2 \sqrt{2}}$ , then find the value of $x - \frac{1}{x}$ .

Answer

The twelfth question of the Chapter Test ML Aggarwal Class 9 asks us to find the value of expressions

12.

(i) If $x = \frac{7 + 3 \sqrt{5}}{7 - 3 \sqrt{5}}$ , find the value of $x - \frac{1}{x}$ .

Answer

(ii) If $x = \frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} + \sqrt{2}}$ and $y = \frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}}$ , then find the value of $x^{2} + x y + y^{2}$ .

Answer

(iii) If $x = \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}$ and $y = \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}$ , find the value of $x^{3} + y^{3}$ .

Answer

The 13th question of the Chapter Test ML Aggarwal Class 9 asks us to arrange the given real numbers in descending order.

13. Write the following real numbers in descending order:

$\sqrt{2}, 3.5, \sqrt{10}, - \frac{5}{\sqrt{2}}, \frac{5}{2} \sqrt{3}$

Answer

The 14th question of the Chapter Test ML Aggarwal Class 9 asks us to find a rational number and an irrational number between two irrational numbers.

14. Find a rational number and an irrational number between $\sqrt{3}$ and $\sqrt{5}$ .

Answer

The 15th question of the Chapter Test ML Aggarwal Class 9 asks us to insert three rational numbers between two irrational numbers.

15. Insert three irrational numbers between $2 \sqrt{3}$ and $2 \sqrt{5}$ , arrange in descending order.

Answer

The 16th question of the Chapter Test ML Aggarwal Class 9 asks us to find examples of irrational numbers.

16. Give an example each of the two different irrational numbers, whose

(i) sum is an irrational number.
(ii) product is an irrational number.

Answer

The 17th question of the Chapter Test ML Aggarwal Class 9 asks us to find examples of irrational numbers.

17. Give an example each of the two different irrational numbers, $a$ and $b$ , where $\frac{a}{b}$ is a rational number.

Answer

The 18th question of the Chapter Test ML Aggarwal Class 9 asks us to find examples of irrational numbers.

18. If 34.0356 is expressed in the form $\frac{p}{q}$ , where $p$ and $q$ are co-prime integers, then what can you say about the factorisation of $q$ ?

Answer

The 19th question of the Chapter Test ML Aggarwal Class 9 asks us to find the criteria for q.

19. In each case, state whether the following numbers are rational or irrational. If they are rational and expressed in the form $\frac{p}{q}$ , where $p$ and $q$ are co-prime integers, then what can you say about the prime factors of $q$ ?

(i) 279.034
(ii) 76.17893
(iii) 3.010010001...
(iv) 39.546782
(v) 2.3476817681...
(vi) 59.120120012000...

Answer

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