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## NCERT Exemplar Class 9 Maths Solutions Number Systems

**Exercise 1.1: Multiple Choice Questions (MCQs)**

**Question 1:**

Every rational number is

**(a)** a natural number ** (b)** an integer

**(c)** a real number ** (d)** a whole number

**Solution:**

**(c)** Since, real numbers are the combination of rational and irrational numbers.

Hence, every rational number is a real number.

**Question 2:**

Between two rational numbers

**(a)** there is no rational number

**(b)** there is exactly one rational number

**(c)** there are infinitely many rational numbers

**(d)** there are only rational numbers and no irrational numbers

**Solution:**

**(c)** Between two rational numbers, there are infinitely many rational numbers.

**Question 3:**

Decimal representation of a rational number cannot be

**(a)** terminating ** (b)** non-terminating

**(c)** non-terminating repeating ** (d)** non-terminating non-repeating

**Solution:**

**(d)** Decimal representation of a rational number cannot be non-terminating non-repeating because the decimal expansion of rational number is either terminating or non-terminating recurring (repeating).

**Question 4:**

The product of any two irrational numbers is

**(a)** always an irrational number **(b)** always a rational number

**(c)** always an integer ** (d)** sometimes rational, sometimes irrational

**Solution:**

**(d)** We know that, the product of any two irrational numbers is sometimes rational and sometimes irrational.

e.g., √2 x √2 = 2 (rational) and √2 x √3 = √6 (irrational)

**Question 5:**

The decimal expansion of the number √2 is

**(a)** a finite decimal ** (b)** 1.41421

**(c)** non-terminating recurring ** (d)** non-terminating non-recurring

**Solution:**

The decimal expansion of the number √2 is non-terminating non-recurring. Because √2 is an irrational number.

Also, we know that an irrational number is non-terminating non-recurring.

**Question 6:**

Which of the following is irrational?

**Solution:**

**(c)**

**Question 7:**

Which of the following is irrational?

Solution:

**(d)** An irrational number is non-terminating non-recurring which is 0.4014001400014….

**Question 8:**

**Solution:**

**(c)**

**Question 9:**

The value of 1.999… in the form of p/q, where p and q are integers and

**Solution:**

**(c)**

**Question 10:**

The value of 2√3 + √3 is

**(a)** 2√6 ** (b)** 6 ** (c)** 3√3 **(d)** 4√6

**Solution:**

2√3 + √3 = √3 (2 + 1) = 3√3

**Question 11:**

√10, √15 is equal to

**(a)** 6√5 **(b)** 5√6 ** (c)** 7√5 ** (d)** 10√5

Solution:

**(b)** √10,√15 = √2.5 √3.5 = √2 √5 √3 √5 = = 5 √6

**Question 12:**

**Solution:**

**Question 13:**

**Solution:**

**Question 14:**

**Solution:**

**Question 15:**

**Solution:**

**Question 16:**

**Solution:**

**Question 17:**

**Solution:**

**(c)**

**Question 18:**

Thinking Process

Take the LCM of indices of three irrational numbers. Then, convert all individually in the form whose index will be equal to LCM.

**Solution:**

**(b)**

**Question 19:**

**Solution:**

**Question 20:**

The value of (256)^{0.16} x (256)^{0.09} is

**(a)** 4 ** (b)** 16 ** (c)** 64 ** (d)** 256.25

**Solution:**

(a)

**Question 21:**

Which of the following is equal to X?

**Solution:**

**Exercise 1.2: Short Answer Type Questions**

**Question 1:**

Let x and y be rational and irrational numbers, respectively. Is x+y necessarily an irrational number? Give an example in support of your answer.

**Solution:**

Yes, (x + y) is necessarily an irrational number.

**Question 2:**

Let x be rational and y be irrational. Is xy necessarily irrational? Justify your answer by an example.

**Solution:**

No, (xy) is necessarily an irrational only when x ≠0.

Let x be a non-zero rational and y be an irrational. Then, we have to show that xy be an irrational. If possible, let xy be a rational number. Since, quotient of two non-zero rational number is a rational number.

So,(xy/x) is a rational number => y is a rational number.

But, this contradicts the fact that y is an irrational number. Thus, our supposition is wrong. Hence, xy is an irrational number. But, when x = 0, then xy = 0, a rational number.

**Question 3:**

State whether the following statements are true or false? Justify your answer.

**Solution:**

**(i)** False, here √2 is an irrational number and 3 is a rational number, we know that when we divide irrational number by non-zero rational number it will always give an irrational number.

**(ii)** False, because between two consecutive integers (likel and 2), there does not exist any other integer.

**(iii)** False, because between any two rational numbers there exist infinitely many rational numbers.

**(iv)** True, because there are infinitely many numbers which cannot be written in the form p/q, q ≠0. p,q both are integers and these numbers are called irrational numbers.

**Question 4:**

Classify the following numbers as rational or irrational with justification

Thinking Process

To classify, use the definition a rational number is in the form of p/q, where p and q are integers and q≠0 and otherwise it is an irrational number.

**Solution:**

**Exercise 1.3: Short Answer Type Questions**

**Question 1:**

Find which of the variables x, y, z and u represent rational numbers and which irrational numbers.

**(i)** x^{2} =5

**(ii)** y^{2} = 9

**(iii)** z^{2} = 0.04

**(iv)** u^{2} = 17/4

**Solution:**

**Question 2:**

Find three rational numbers between

**(i)** -1 and -2 ** (ii)** 0.1 and 0.11

**(iii)** 5/7 and 6/7 ** (iv)** 1/4 and 1/5

Thinking Process

Use the concept that three rational numbers between x and y are x+ d, x+2d and x+3d, where d =(y-x)/(n+1) , x<y and n=3.

**Solution:**

**Question 3:**

Insert a rational number and an irrational number between the following

**(i)** 2 and 3 ** (ii)** 0 and 0.1 ** (iii)** 1/3 and 1/2

**(iv)** -2/5 and -1/2 **(v)** 0.15 and 0.16 ** (vi)** √2 and √3

**(vii)** 2.357 and 3.121 ** (viii)** .0001 and .001

**(ix)** 3.623623 and 0.48484**8 (x)** 3.375289 and 6.375738

**Solution:**

We know that, there are infinitely many rational and irrational values between any two numbers.

**(i)** A rational number between 2 and 3 is 2.1.

To find an irrational number between 2 and 3. Find a number which is non-terminating non-recurring lying between them.

Such number will be 2.040040004…………..

**(ii)** A rational number between 0 and 0.1 is 0.03.

An irrational number between 0 and 0.1 is 0.007000700007……….

**(iii)** A rational number between 1/3 and 1/2 is 5/12. An irrational number between 1/3 and 1/2 i.e., between 0-3 and 0.5 is 0.4141141114………….

**(iv)** A rational number between -2/5 and 1/2 is 0. An irrational number between -2/5 and 1/2 i.e., between – 0.4 and 0.5 is 0.151151115………..

**(v)** A rational number between 0.15 and 0.16 is 0.151. An irrational number between 0.15 and 0.16 is 0.1515515551…….

**(vi)** A rational number between √2 and √3 i.e.,, between 1.4142…… and 1.7320…… is 1.5.

An irrational number between √2 and √3 is 1.585585558……….

**(vii)** A rational number between 2.357 and 3.121 is 3. An irrational number between 2.357 and 3.121 is 3.101101110……..

(viii) A rational number between 0.0001 and 0.001 is 0.00011. An irrational number between 0.0001 and 0.001 is 0.0001131331333………..

**(ix)** A rational number between 3.623623 and 0.484848 is 1. An irrational number between 3.623623 and 0.484848 is 1.909009000……….

**(x)** A rational number between 6.375289 and 6.375738 is 6.3753. An irrational number between 6.375289 and 6.375738 is 6.375414114111………

**Question 4:**

Represent the following numbers on the number line 7, 7.2, -3/2 and -12/5 .

**Solution:**

Firstly, we draw a number line whose mid-point is 0. Marks a positive numbers on right

hand side of 0 and negative numbers on left hand side of 0.

**(i)** Number 7 is a positive number. So we mark a number 7 on the right hand side of 0, which is a 7 units distance from zero.

**(ii)** Number 7.2 is a positive number. So, we mark a number 7.2 on the right hand side of 0, which is a 7.2 units distance from zero.

**(iii)** Number -3/2 or -1.5 is a negative number. So, we mark a number 1.5 on the left hand side of 0, which is a 1.5 units distance from zero.

**(iv)** Number – 12/5 or -2.4 is a negative number. So, we mark a number 2.4 on the left hand side of 0, which is a 2.4 units distance from zero.

**Question 5:**

Locate √5, √10 and √17 on the number line.

**Solution:**

**Question 6:**

Represent geometrically the following numbers on the number line

**(i)** √ 4.5 ** (ii)** √5.6

**(iii)** √ 8.1** (iv) **√2.3

Thinking Process

**(i)** Firstly, we draw a line segment AB of length equal to the number inside the root and extend it to C such that BC=1

**(ii)** Draw a semi-circle with centre O (O is the mid-point of AC) and radius OA

**(iii)** Now, draw a perpendicular line from B to cut the semi-circle atD.

(iv) Further, draw an arc with centre B and radius BD meeting AC produced at £

**Solution:**

**(i)** Firstly, we draw a line segment AB = 4.5 units and extend it to C such that BC = 1 unit. Let O be the mid-point of AC.

Now, draw a semi-circle with centre O and radius OA Let us draw BD perpendicular to AC passing through point B and intersecting the semi-circle at point D.

Hence, the distance BD is √4.5 units.

Draw an arc with centre B and radius BD, meeting AC produced at E, then BE = BD = √4.5 units.

**(ii)** Firstly, we draw a line segment AB = 5.6 units and extend it to C such that BC = 1 unit. Let O be the mid-point of AC.

Now, draw a semi-circte with centre O and radius OA. Let us draw BD perpendicular to AC passing through point B and intersecting the semi-circle at point D.

Hence, the distance BD is √5.6 units.

Draw an arc with centre B and radius BD, meeting AC produced at E, then BE = BD = √5.6 units.

**(iii)** Firstly, we draw a line segment AB = 8.1 units and extend it toC such that SC = 1 unit. Let O be the mid-point of AC.

Now, draw a semi-circle with centre 0 and radius OA

Let us draw BD perpendicular to AC passing through point 6 intersecting the semi-circle at point D:

Hence, the distance BD is √8.1 units .

Draw an arc with centre Sand radius BD, meeting AC produced at E, then BE = BD = √8.1 units.

**(iv)** Firstly, we draw a line segment AS = 2.3 units and extend it to C such that SC = 1 unit. Let O be the mid-point of AC.

Now, draw a semi-circle with centre 0 and radius OA.

Let us draw BD perpendicular to AC passing through point 6 and intersecting the semi-circle at point D.

Hence, the distance BD is √2.3 units.

Draw an arc with centre 6 and radius BD, meeting AC produced at E, then BE = BD = √2.3 units.

**Question 7:**

**Solution:**

**Question 8:**

Show that 0.142857142857… = 1/7.

**Solution:**

Let x = 0.142857142857 …………………..(i)

On multiplying both sides of Eq. (i) by 1000000, we get

1000000 x = 142857.142857…………………………(ii)

On subtracting Eq. (i) from Eq. (ii), we get

1000000 x – x = (142857.142857…) – (0.142857..) => 999999 x = 142857

∴ x = 142857/999999 = 1/7 Hence proved.

**Question 9:**

Simplify the following

**Solution:**

**Question 10:**

Rationalise the denominator of the following

**Solution:**

**Question 11:**

Find the values of a and b in each of the following

**Solution:**

**Question 12:**

**Solution:**

**Question 13:**

Rationalise the denominator in each of the following and hence evaluate by taking √2 = 1.414, √3 = 1.732 and √5 = 2.236 upto three places of decimal.

**Solution:**

**Question 14:**

Simplify

**Solution:**

**Exercise 1.4: Long Answer Type Questions**

**Question 1:**

**Solution:**

**Question 2:**

**Solution:**

**Question 3:**

**Solution:**

**Question 4:**

**Solution:**

**Question 5:**

**Solution:**

**Question 6:**

**Solution:**

**Question 7:**

**Solution:**

NCERT Exemplar ProblemsNCERT Exemplar MathsNCERT Exemplar Science

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give some extra ordinary question of exponent of real number

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