Number System Class 10 Worksheets have been designed as per the latest pattern for CBSE, NCERT and KVS for Grade 10. Students are always suggested to solve printable worksheets for Mathematics Number System Grade 10 as they can be really helpful to clear their concepts and improve problem solving skills. We at worksheetsbag.com have provided here free PDF worksheets for students in standard 10 so that you can easily take print of these test sheets and use them daily for practice. All worksheets are easy to download and have been designed by teachers of Class 10 for benefit of students and is available for free download.

## Mathematics Number System Worksheets for Class 10

We have provided **chapter-wise worksheets for class 10 Mathematics Number System** which the students can download in Pdf format for free. This is the best collection of Mathematics Number System standard 10th worksheets with important questions and answers for each grade 10th Mathematics Number System chapter so that the students are able to properly practice and gain more marks in Class 10 Mathematics Number System class tests and exams.

### Chapter-wise Class 10 Mathematics Number System Worksheets Pdf Download

**1.Use Euclid’s division algorithm to find the HCF of:**

**(i) 135 and 225**

Ans. Start with the larger integer, that is, 225. Apply the division lemma to 225 and 135, we get

225 – 135 x 1 + 90

Since the remainder 90 ≠ 0!, we apply the division lemma to 135 and 90, we get

135 – 90 x 1 + 45

We consider the new divisor 90 and the new remainder 45, and apply the division lemma to get

90 – 45 x 2 + 0

The remainder has now become zero, so our procedure stops.

Since the divisor at this stage is 45, the HCF of 225 and 135 is 45.

**(ii) 196 and 38220**

Ans. Start with the larger integer, that is, 38220. Apply the division lemma to 38220 and 196, we get

38220 – 196 x 195 + 0

The remainder is zero, so our procedure stops.

Since the divisor is 196, the HCF of 38220 and 196 is 196.

**(iii) 867 and 255**

Ans. Start with the larger integer, that is, 867. Apply the division lemma, we get

867 – 255 x 3 + 102

Since the remainder 102 ≠ 0!, we apply the division lemma to 255 and 102, to get

255 – 102 x 2 + 51

We, consider the new divisor 102 and the new remainder 51, and apply the division lemma, we get

102 – 51 x 2 + 0

The remainder has now become zero, so our procedure stops.

Since the divisor at this stage is 51, the HCF of 867 and 255 is 51.

**2. Show that any positive odd integer is of the form q6 + 1, or q6 + 3, or q6 + 5, where q is some integer.**

Ans. Let us consider a positive odd integer as a. On dividing a by 6, let q be the quotient and r be the remainder.

Using Euclid’s lemma, we have

a – 6q + r where 0 ≤ r < 6 i.e., 0 1, ……6

a – 6q + 0 = 6

or a – 6q + 1

or a – 6q + 2

or a – 6q + 3

or a – 6q + 4

or a – 6q + 5

But, a 6q =, a – 6q + 2, a – 6q + 4 are even values of a.

[6q – 2 (3q) – 2m1, 6q + 2 – 2 ( 3q + 1) -2m_{2}, 6q + 4 – 2 (3q + 2) – 2m_{3}

But a being an odd integer, we have,

a – 6q + 1

or a – 6q + 3

or a – 6q + 5

**3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?**

Ans. By applying the Euclid’s division lemma, we can find the maximum number of columns in which an army contingent of 616 members can march behind an army band of 32 members in a parade. HCF of 616 and 32 is equal to maximum number of columns in which 616 and 32 members can march.

Since 616 > 32, we apply the division lemma to 616 and 32, we get

616 – 32 x 19 + 8

Since the remainder 80!, we apply the division lemma to 32 and 8, to get

32 – 8 x 4 + 0

The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 8, the HCF of 616 and 32 is 8.

Therefore, the maximum number of columns in which an army contingent of 616 members can march behind an army band of 32 members in a parade is 8.

a^{2} = 9q^{2} = 3 (3q^{2}) = 3m

where, m = 3q^{3} [From (i)]

Case-II r = 1, a^{2} – 9q^{2} + 6q + 1 -3 (3q^{2} + 2q) + 1

– 3m + 1

where, m – 3q^{2} + 2q [From (i)]

Case-III r = 1, a^{2} – 9q^{3} + 6q + 1 – 3 (3q^{2} + 2q) +1

– 3m + 1

where, m – (3q^{2} + 4q + 1) [From (i)]

Hence, square of any positive integer is either of the form m3 or 3m + 1 for some integer m.

**5.Use Euclid’s division lemma to show that the cube of any positive integer is of the form m9, 9m + 1 or 9m + 8.**

Ans. Let us consider an arbitrary positive integer a such that it is in the form of 3q, (3q + 1) or (3q + 2).

Case 1 : a = 3q,

For, a = 3q, a3 (3q)^{3} = q^{3 }= 9 (9q^{3})

= 9m …(1)

Here we have substituted 3q^{3} = m , where m is an integer.

Case 2 : a – 3q + 1

For, a – 3q + 1, a^{3} – (3q + 1)^{3}

– 27q^{3} + 27q^{3} + 9q + 1

– 9 (3q^{3 }+ 3q^{2} + q) + 1

9m + 1 …(2)

Here we have substituted (3q^{3 }+ 3q^{2} + q) – m, where m is an integer.

Case 3 : a – 3q + 2

For, a – 3q + 2, a^{3} – (3q + 2)^{3}

27q^{3} + 54q^{2 }+ 36q + 8

-9(3q^{3} +6q^{2} + 4q) + 8

9m + 8 …(3)

Here we have substituted (3q^{2} + 6q^{2} + 4q), where m is an integer.

From (1), (2), (3) we have :

a^{3} = 9m, (9m + 1) or (9m + 8)

Thus, cube of any positive integer can be in the form

9m, (9m + 1) or (9m + 8) for some integer m.

**1. Express each number as a product of its prime factors:**

**(i) 140****Using factor there method we have,**

Ans. 140 = 2 x 2 x 5 x 7 = 2^{2} x 5 x 7

**(ii) 156**

Ans. 156 = 2 x 2 x 3 x 13

= 2^{2} x 3 x 13

**(iii) 3825**

Ans. 3825 = 3 x 3 x 5 x 17

= 2^{2} x 5^{2} x 17

**(iv) 5005**

Ans. 5006 = 5 x 7 x 11 x 13

**(v) 7429**

Ans. 7429 = 17 x 19 x 23

**2. Find the LCM and HCF of the following pairs of integers and verify that LCM X HCF = product of the two numbers.**

**(i) 26 and 91**

Ans. 26 = 2 x 13

Ans. 91 = 7 x 13

Therefore,

LCM (26,91) = 2 x 7 x 13

HCF (26, 91) = 13

Verification:

LCM X HCF = 182 X 13 = 2366

and 26 X 91 = 2366

i.e. LCM X HCF = Product of two numbers.

**(ii) 510 and 92**

Ans. 510 = 2 X 5 X 3 X 17

Ans. 92 = 2 X 2 X 23 X 17

Therefore,

LCM (510,92) = 2 X 2 X 5 X 3 X 17 X 23

= 23460

HCF (510, 92) = 2

Verification:

LCM X HCF = 23460 X 2 = 46920

and 510 X 92 = 46920

i.e., LCM X HCF = Product of two numbers

**(iii) 336 and 54**

Ans. 336 = 2 X 2 X 2 X2 3 X 7

= 24 X 3 X 7

Ans. 54= 2 X 3 X 3 X3 = 2 X 33

Therefore,

LCM (336, 52) = 24 X 33 X 7 = 3024

HCF (336, 52) = 2 X 3 = 6

Verification:

LCM X HCF 3024 X 6 = 18144

and 336 X 54 = 18144

i.e., LCM X HCF = Product of two numbers.

**3. Find the LCM and HCF of the following integers by applying the prime factorisation method.**

**(i) 12, 15 and 21**

Ans. 12 = 2 X 2 X 3 = 2^{2} X 3

So, 21 = 3 X 7

Therefore,

HCF (12, 15, 21) = 3

LCM (12, 15, 21) = 22 X 3 X 5 X 7 X = 420

**(ii) 17, 23 and 29**

Ans. 17 = 17

23 = 23

29 = 29

Therefore,

HCF (17, 23, 29) = 1

LCM (17, 23, 29) = 17 X 23 X 19 = 11339

**(iii) 8, 9 and 25**

Ans. 8 = 2 X 2 X 2 = 2^{3}

Ans. 9 = 3 X 3 = 3^{2}

Ans. 25 = 5 X 5 = 5^{2}

Therefore,

HCF (8, 9, 25) = 1

LCM (8, 9, 25) 2^{3} X 3^{2} X 5^{2} = 1800

**4. Given that HCF (306, 657) = 9 , find LCM (306, 657)**

Ans. As we know that,

LCM X HCF = Product of numbers

**5. Check whether 6 ^{n} can end with the digit 0 for any natural number n.**

Ans. If the number 6n for any natural number n, end with digit 0, then it would be divisible by 5. That is, the prime factorisation of 6n would contain the prime 5. This is not possible because 6n = (2 X 3)

^{n}= 2

^{n}X 3

^{n}so the only primes in the factorisation of 6n are 2 and 3 and the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorisation of 6

^{n}. So, there is no natural number n for which 6n ends

**6. Explain why 7 X 11 X 13 + 13 and 7 X 6 X 5 X 4 X 3 X 2 X 1 + 5 are composite numbers.**

**(i) 7 X 11 X 13 + 13**

Ans. 7 X 11 X 13 + 13 – (7 X 11 X 1) X 13

– (77 + 1) X 13 = 78 X 13

= (2 X 3 X 13) X 13

78 = 2 X 3 X 13

2 X 3 X 13^{2}

Since, 7 X 11 X 13 + 13 can be expressed as a product of primes, therefore, it is a composite number.

**(ii) 7 X 6 X 5 X 4 X 3 X 2 X 1 + 5**

Ans. 7 X 6 X 5 X 4 X 3 X 2 X 1 + 5

(7 X 6 X 4 X 3 X 2 X 1 + 1) X 5

(1008 + 1) X 5 = 1009 X 5

= 5 X 1009

Since, 7 X 6 X 5 X 4 X 3 X 2 X 1 + 5 can be expressed as a product of primes, therefore, it is a composite number.

**7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?**

Ans. By taking LCM of time taken (in minutes) by Sonia and Ravi, we can get the actual number of minutes after which they meet again at the starting point after both start as same point and at the same time, and go in the same direction

18 = 2 X 3 X 3 = 2 X 3^{2}

12 = 2 X 2 X 3 = 22 X 3

= 22 X 3

LCM(18, 12) = 22 X 32 = 36

Therefore, both Sonia and Ravi will meet again at the starting point after 36 minutes.

**1. Prove that 5 is irrational.**

Ans. Let √5 be a rational number.

So, we can find co-prime integers a and b (≠ 0) such that

√5 = a/b

√5b = a

Squaring on both sides, we get

5B^{2} = a^{2}

Therefore, 5 divides a^{2}

Therefore, 5 divides a

So, we can write

a = 5c for some integer c.

Substituting for a, we get

5b^{2} = 25c^{2}

b^{2} = 5c^{2}

This means that 5 divides b^{2}, and so 5 divides b. Therefore, a and b have at least 5 as a common factor. But this contradicts the fact that a and b have no common factor other than 1. This contradiction arose because of our incorrect assumption that √5 is rational. So, we conclude that √5 is irrational

**2. Prove that 3 + 2√5 is irrational.**

Ans. Let, 3 + 2√5 is rational number.

That is, we can find co-prime integers a and b (b ≠ 0)

Such that, 3 + 2√5 = a/b where b ≠ 0

Therefore, a/b – 3 = 2 √5

a – 3b/b = 2√5

a – 3b/2b = √5

a/2b – 3/2 = √5

Since a and b are integers, we get a/2b – 3/2 is rational, and so √5 is rational.

But this contradicts the fact that √5 is irrational.

This contradiction has arisen because of our incorrect assumption that 3 + 2√5+ is rational.

So, we conclude that 3 + 2√5+ is irrational

CBSE Class 10 Mathematics Number Systems Worksheet Set A |

CBSE Class 10 Mathematics Number Systems Worksheet Set B |

CBSE Class 10 Mathematics Number Systems Worksheet |

## Mathematics Number System Worksheets for Class 10 as per CBSE NCERT pattern

Parents and students are welcome to download as many worksheets as they want as we have provided all free. As you can see we have covered all topics which are there in your Class 10 Mathematics Number System book designed as per CBSE, NCERT and KVS syllabus and examination pattern. These test papers have been used in various schools and have helped students to practice and improve their grades in school and have also helped them to appear in other school level exams. You can take printout of these chapter wise test sheets having questions relating to each topic and practice them daily so that you can thoroughly understand each concept and get better marks. As Mathematics Number System for Class 10 is a very scoring subject, if you download and do these questions and answers on daily basis, this will help you to become master in this subject.

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- CBSE Class 10 Mathematics Number System Workbook will surely help to improve knowledge of this subject

These Printable practice worksheets are available for free download for Class 10 Mathematics Number System. As the teachers have done extensive research for all topics and have then made these worksheets for you so that you can use them for your benefit and have also provided to you for each chapter in your ebook. The Chapter wise question bank and revision worksheets can be accessed free and anywhere. Go ahead and click on the links above to download free CBSE Class 10 Mathematics Number System Worksheets PDF.

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