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Important Questions for CBSE Class 10 Maths Chapter 1 - Real Numbers 2024-25

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Boost Your Performance in CBSE Class 10 Maths Exam Chapter 1 with These Important Questions

Important questions for Class 10 Maths Chapter 1, Real Numbers is prepared by experts of Vedantu with a purpose of creating important questions for the chapter is to enable the students to pivotal concepts of the chapter that has been introduced in Maths NCERT Solutions Class 10. The important questions pdf version is prepared to give a better conceptual understanding to the students and help them to perceive what questions they can expect from this chapter in exams. There are pdf versions of important questions  for other subjects also. You can download them anytime on any device and practice them at your convenient time. The important questions will surely give an insight about what questions can be expected in exams. 

Chapter 1, Real numbers for class 10, continues with the real number operations that you studied in earlier grades and also introduces two important properties of positive integers namely Euclid's Division Lemma and algorithm and the fundamental theorem of arithmetic. The important questions are based on the topics that are discussed in this chapter. Let us have a quick glance at the summary of the chapter so that you can solve the important questions for Maths Chapter 1, Real Numbers for Class 10. 

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Important Questions for CBSE Class 10 Maths Chapter 1 - Real Numbers 2024-25
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Study Important Questions for Class 10 Mathematics Chapter 1 - Real Numbers

Very Short Answer Questions                                                                        (1 Mark)

1. Use Euclid’s division lemma to show that the square of any positive integer is either of form 3m or 3m+1 for some integer m.

(Hint: Let x be any positive integer then it is of the form 3q,3q+1 or 3q+2. Now square each of these and show that they can be rewritten in the form 3m or 3m+1.)

Ans: Let a be any positive integer, then we can write a=3q+r for some integer qβ‰₯0  …..(1)

Clearly, in the expression (1) we are dividing a by 3 with quotient q and remainder r, r=0,1,2 because 0≀r<3.    ..…(2)

Therefore from (1) and (2) we can get, a=3q or 3q+1 or 3q+2.    ……(3)

Squaring both sides of equation (3) we get,

a2=(3q)2 or (3q+1)2 or (3q+2)2 

β‡’a2=9q2 or 9q2+6q+1 or 9q2+12q+4  …..(4)

Taking 3 common from LHS of equation (4) we get,

a2=3Γ—(3q2) or 3(3q2+2q)+1 or 3(3q2+4q+1)+1.   …..(5)

Equation (5) could be written as 

a2=3k1 or 3k2+1 or (3k3+1) for some positive integers k1,k2 and k3.

Hence, it can be said that the square of any positive integer is either of the form 3m or 3m+1.


2. Express each number as product of its prime factors.

(i). 140 

Ans: The product of prime factors of number 140 is 140=2β‹…2β‹…5β‹…7=22β‹…5β‹…7.

(ii). 156 

Ans: The product of prime factors of number 156 is 156=2β‹…2β‹…3β‹…13=22β‹…3β‹…13.

(iii). 3825 

Ans: The product of prime factors of number 3825 is 3825=3β‹…3β‹…5β‹…5β‹…17=32β‹…52β‹…17.

(iv). 5005 

Ans: The product of prime factors of number 5005 is 5005=5β‹…7β‹…11β‹…13.

(v). 7429 

Ans: The product of prime factors of number 7429 is 7429=17β‹…19β‹…23.


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

Ans: It is given that the HCF of the two numbers 306,657 is 9. We have to find its LCM.

We know that, LCMΓ—HCF = Product of two numbers. Therefore,

LCMΓ—HCF =306Γ—657 

β‡’LCM=306Γ—657HCF

β‡’LCM=306Γ—6579

∴LCM=22338


4. Check whether 6ncan end with the digit 0 for any natural number n.

Ans: If any number ends with the digit 0, it should be divisible by 10.

Let us check this for natural number n=2

Clearly, 62=36, which is not divisible by 10. Therefore, 6n cannot end with the digit 0 for any natural number n.


5. Prove that (3+25) is irrational.    

Ans: We will prove this by contradiction. To solve this problem, suppose that (3+25) is rational i.e., βˆƒ two co-prime integers a and b (bβ‰ 0) such that ab=3+25     …..(1)

Subtracting 3 from both sides of the equation (1) and simplifying it further we get,

abβˆ’3=25

β‡’aβˆ’3bb=25 

β‡’aβˆ’3b2b=5        …..(2)

Since we know that 5 is irrational. Therefore, from (2) aβˆ’3b2b should be irrational but it could be written in pq form which means it is rational. It is not possible and hence our assumption is wrong and (3+25) cannot be rational.

Hence, (3+25) is irrational.


6. The number 7Γ—11Γ—13Γ—15+15 is a

  1. Composite Number

  2. Whole Number

  3. Prime Number

  4. None of these

Ans: (a) and (b) both


7. For what least value of β€²nβ€² a natural number, (24)n is divisible by 8?

  1.  0 

  2. βˆ’1 

  3. 1 

  4. No value of β€²nβ€² is possible

Ans: (c) 1


8. The sum of a rational and an irrational number is

  1. Rational

  2. Irrational

  3. Both (a) & (b)

  4. Either (a) & (b)

Ans: (b) Irrational


9. HCF of two numbers is 113, their LCM is 56952. If one number is 904, then other number is:

  1. 7719

  2. 7119

  3. 7791

  4. 7911

Ans: (b) 7119


10. A lemma is an axiom used for providing

  1. other statement

  2. no statement

  3. contradictory statement

  4. none of these

Ans: (a) other statement


11. If HCF of two numbers is 1, the two numbers are called relatively ______ or _______.

  1. prime, co-prime

  2. composite, prime

  3. Both (a) and (b)

  4. None of these

Ans: (a) prime, co-prime


12. The number 2.35― is

  1. a terminating decimal number

  2. a rational number 

  3. an irrational number

  4. Both (a) and (b)

Ans: (b) a rational number


13. The number 2.13113111311113...... is        

  1. a rational number

  2. a non-terminating decimal number

  3. an irrational number

  4. Both (a) & (c)

Ans: (b) a non-terminating decimal number


14. The smallest composite number is      

  1.  1

  2. 2

  3. 3

  4. 4

Ans: (d) 4 


15. The number 1.2348― is

  1. an integer

  2. an irrational number

  3. a rational number

  4. None if these

Ans: (c) a rational number


16. The number Ο€ is

  1. a rational number

  2. an irrational number

  3. Both (a) and (b)

  4. neither rational nor irrational

Ans: (b) an irrational number


17. The number (2+5) is

  1. a rational number

  2. an irrational number

  3. an integer

  4. not real number

Ans: (b) an irrational number


Short Answer Questions                                                                               (2 Marks)

1. Show that any positive odd integer is of the form 6q+1, or 6q+3 or 6q+5, where q is some integer.

Ans: Let a be any odd positive integer, then by Euclid’s algorithm we can write a=6q+r for some integer qβ‰₯0.        …..(1)

Clearly, in the expression (1) we are dividing a by 6 with quotient q and remainder r

r=1,3,5 because 0≀r<6 and a is odd.    …..(2)

Therefore from (1) and (2) we can get, 

a=6q+1 or 6q+3 or 6q+5 

And therefore, any odd integer can be expressed in the form 6q+1,6q+3, or 6q+5.


2. An army contingent of 616 members are to march behind a 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: To solve this problem, we have to find the HCF of two numbers (616,32).

To find the HCF, we can use Euclid’s algorithm which states that if there are any two integers a and b, then there exists q and r such that it satisfies the condition a=bq+r where 0≀r<b.

For a=616, 616=32Γ—19+8       …..(1)

For a=32, 32=8Γ—4+0       …..(2) 

Therefore, from (1) and (2), the HCF (616,32) is 8 and hence they can march in 8 columns each.


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

Ans: Let a be any positive integer, then by Euclid’s algorithm we can write a=3q+r, for some integer qβ‰₯0.    …..(1)

Clearly, in the expression (1) we are dividing a by 3 with quotient q and remainder r, r=0,1,2 because 0≀r<3.     …..(2)

Therefore from (1) and (2) we can get, 

a=3q or 3q+1 or 3q+2    …..(3)

Cubing both sides of equation (3) we get,

Case 1: When a=3q,

a3=27q3

β‡’a3=9(3q3) 

β‡’a3=9m 

Where m is an integer such that m=3q3

Case 2: When a=3q+1

a3=(3q+1)3

β‡’a3=9q3+9q2+9q+1 

β‡’a3=9(q3+q2+q)+1β‡’a3=9m+1

Where m is an integer such that m=(3q3+3q2+q)

Case 3: When a=3q+2

a3=(3q+2)3

β‡’a3=27q3+54q2+36q+8 

β‡’a3=9(3q3+6q2+4q)+8 

β‡’a3=9m+8 

Where m is an integer such that m=(3q3+6q2+4q)

Therefore, from case 1,2 and 3 we conclude that the cube of any positive integer is of the form 9m,9m+1 or 9m+8.


4. Find the LCM and HCF of the following pairs of integers and verify that LCMΓ—HCF=product of two numbers.                       

(i). 26 and 91

Ans: Let us first find the HCF of the two given numbers.

Writing the prime factorization of both of the numbers we get, 

26=2Γ—13 and 91=7Γ—13.       …..(1) 

Since 13 is common in the prime factorization of both of the numbers we get HCF=13.      …..(2)

From (1), LCM = 2Γ—7Γ—13=182     …..(3) 

Product of two numbers, 26Γ—91=2366   …..(4)

From (2) and (3), HCFΓ—LCM =13Γ—182=2366.    …..(5) 

Hence from (4) and (5), product of two numbers =HCFΓ—LCM

(ii). 510 and 92

Ans: Let us first find the HCF of the two given numbers.

Writing the prime factorization of both of the numbers we get, 

510=2Γ—3Γ—5Γ—17 and 92=2Γ—2Γ—23.   …..(1) 

Since 2 is common in the prime factorization of both of the numbers we get HCF=2.         …..(2)

From (1),LCM=2Γ—2Γ—3Γ—5Γ—17Γ—23=23460    …..(3) 

Product of two numbers, 510Γ—92=46920    …..(4)

From (2) and (3),HCFΓ—LCM=2Γ—23460=46920.  …..(5) 

Hence from (4) and (5), product of two numbers =HCFΓ—LCM.

(iii). 336 and 54

Ans: Let us first find the HCF of the two given numbers.

Writing the prime factorization of both of the numbers we get, 

336=2Γ—2Γ—2Γ—2Γ—3Γ—7=24Γ—3Γ—7and54=2Γ—3Γ—3Γ—3=2Γ—33.     …..(1) 

Since 2,3 are common in the prime factorization of the numbers, HCF=2Γ—3=6.      .....(2)

From (1),LCM=24Γ—33Γ—7=3024      …..(3) 

Product of two numbers =336Γ—54=18144      …..(4)

From (2) and (3), HCFΓ—LCM=6Γ—3024=18144.      …..(5) 

Hence from (4) and (5), product of two numbers =HCFΓ—LCM.


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

(i). 12,15 and 221

Ans: Let us first find the HCF of the two given numbers.

Writing the prime factorization of all the numbers we get, 

12=22Γ—3, 15=3Γ—5 and21=3Γ—7.    …..(1) 

Since 3 is common in the prime factorization of all the numbers we get HCF=3.                                   

From (1), LCM=22Γ—3Γ—5Γ—7=420

(ii). 17,23 and 29

Ans: Let us first find the HCF of the two given numbers.

Writing the prime factorization of all the numbers we get, 

17=1Γ—17, 23=1Γ—23 and 29=1Γ—29.   …..(1) 

Since 1 is common in the prime factorization of all the numbers we get HCF=1.                                   

From (1), LCM=17Γ—23Γ—29=11339

(iii). 8,9 and 25

Ans: Let us first find the HCF of the two given numbers.

Writing the prime factorization of all the numbers we get, 

8=2Γ—2Γ—2=23, 9=3Γ—3=32 and 25=5Γ—5=52.    …..(1) 

Since 1 is common in the prime factorization of all the numbers we get HCF=1.                                   

From (1), LCM=23Γ—32Γ—52=1800


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

Ans: Numbers are of two types – prime and composite.                  

Prime numbers can be divided by 1 and only itself, whereas composite numbers have factors other than 1 and itself.

It can be observed that 7Γ—11Γ—13+13=13Γ—(7Γ—11+1)

The given expression has 13 as one of the factors. Therefore, it is a composite number.

Also, 7Γ—6Γ—5Γ—4Γ—3Γ—2Γ—1+5=5Γ—(7Γ—6Γ—4Γ—3Γ—2Γ—1+1).

The given expression has 5 as one of the factors. 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: It can be observed that Ravi takes lesser time than Sonia for completing one round of the circular path. As they are going in the same direction, they will meet again at the time that will be the LCM of time taken by Sonia and Ravi for completing 1 round of circular path respectively i.e., LCM of 18 minutes and 12 minutes.

Let us now calculate LCM of 18 minutes and 12 minutes.

18=2Γ—3Γ—3 and 12=2Γ—2Γ—3    …..(1)

LCM(12,18)=2Γ—2Γ—3Γ—3=36

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


8. Prove that 5 is irrational.

Ans: We will prove this by contradiction.

Let us suppose that 5 is rational. It means that we have co-prime integers a and b (b≠0) such that 5=ab

β‡’b5=a   …..(1)

Squaring both sides of (1), we get 5b=a2    …..(2)

From (2) we can conclude that 5 is factor of a2, this means that 5 is also factor of a.     …..(3)

From (3) we can write a=5c for some integer c.   …..(4)

Substituting value of (4) in (2) we get,

5b2=25c2β‡’b2=5c2

It means that 5 is factor of b2, this means that 5 is also factor of b.   …..(5)

From (3),(5), we can say that 5 is factor of both a,b, which is a contradiction to our assumption. Therefore, our assumption was wrong and 5 is irrational.


9. Write down the decimal expansion of those rational numbers in Question 1 which have terminated decimal expansion.        

Ans: 

(i).The decimal expansion of 133125 is 1355=13Γ—2555Γ—25=13Γ—25105=416105=0.00416 

(ii) The decimal expansion of 178 is 1723=17Γ—5323Γ—53=17Γ—53103=2125103=2.215 

(iii) The decimal expansion of 151600 is 1526Γ—52=15Γ—542Γ—52Γ—54=15Γ—54106=9375106=0.009375 

(iv) The decimal expansion of 2323Γ—52 is 23Γ—5123Γ—52Γ—51=23Γ—51103=115103=0.115 

(v) The decimal expansion of 615 is 25=2Γ—25Γ—2=410=0.4 

(vi) The decimal expansion of 3550 is 710=0.7


10. The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If, they are rational, and of form pq, what can you say about the prime factors of q?                                                                               

(i). Number - 43.123456789

Ans: Since the decimal expansion is terminating so it is a rational number and therefore it can be expressed in pq form.

 43.123456789=43123456789109

Hence, the factors of q are of the form 2nΓ—5m where n and m are non-negative integers.

(ii). Number - 0.1201120012000120000...

Ans: Since the decimal expansion is neither terminating nor non-terminating repeating so it is an irrational number.

(iii). Number - 43.123456789―

Ans: Since the decimal expansion is non-terminating repeating so it is a rational number and therefore it can be expressed in pq form.

43.123456789―=43β‹…(123456789109)β‹…(123456789109)β‹…(123456789109)β‹…...

Hence, the factors of q are of the form 2nΓ—5m where n and m are non-negative integers.


11. Show that every positive integer is of the form 2q and that every positive odd integer is of the form 2q+1 for some integer q.                               

Ans: Let a be any positive integer, then by Euclid’s algorithm we can write a=2q+r, for some integer qβ‰₯0.   …..(1)

Clearly, in the expression (1) we are dividing a by 2 with quotient q and remainder r, r=0,1 because 0≀r<2.  …..(2)

Therefore from (1) and (2) we can get, 

a=2q or 2q+1   …..(3)

If a=2q (which is even)

If a=2q+1 (which is odd)

So, every positive even integer is of the form 2q and odd integer is of the form 2q+1.


12. Show that any number of the form 4n, n∈N can never end with the digit 0.  

Ans: The prime factorization of 10 is 2Γ—5. Therefore, for a number to end with 0 it should have both 2 and 5 as a factor.

For the given number of the form 4n=[22]n=22n, we can observe that it has only 2 as a factor and not 5. So 4n,n∈N can never end with the digit 0.


13. Use Euclid’s Division Algorithm to find the HCF of 4052 and 12576.

Ans: Step 1: Since12576>4052 , we will apply Euclid’s division lemma such that 12576=4052Γ—3+420   …..(1)

Step 2: Since the remainder in (1) is not zero so we have to apply the Euclid’s division lemma to 4052 and 420, such that 4052=420Γ—9+272   …..(2)

Step 3: Again, the remainder in (2) is not zero so we have to apply the Euclid’s division lemma to 420 and 272, such that 420=272Γ—1+148.   …..(3) 

Step 4: Again, the remainder in (3) is not zero so we have to apply the Euclid’s division lemma to 148 and 272, such that 272=148Γ—1+124.   …..(4) 

Step 5: Again, the remainder in (4) is not zero so we have to apply the Euclid’s division lemma to 148 and 124, such that 148=124Γ—1+24.  …..(5) 

Step 6: Again, the remainder in (5) is not zero so we have to apply the Euclid’s division lemma to 24 and 124, such that 124=24Γ—5+4.  …..(6) 

Step 7: Again, the remainder in (6) is not zero so we have to apply the Euclid’s division lemma to 24 and 4, such that 24=4Γ—6+0.  …..(7) 

Now the remainder in equation (7) is zero. Therefore, HCF of 12576 and 4052 is β€²4β€².


14. Given that HCF of two numbers is 23 and there LCM is 1449. If one of the numbers is 161, find the other.                             

Ans: It is given that the HCF of the two numbers is 23 and their LCM is 1449. We have to find the other number if one of the numbers is 161. Let the other number be a

We know that, LCMΓ—HCF = Product of two numbers. Therefore,

1449Γ—23=aΓ—161 

β‡’a=1449Γ—23161

∴a=207


15. Show that every positive odd integer is of the form (4q+1) or (4q+3) for some integer q.

Ans: Let a be any positive integer, then by Euclid’s algorithm we can write a=4q+r, for some integer qβ‰₯0. …..(1)

Clearly, in the expression (1) we are dividing a by 4 with quotient q and remainder r, r=0,1,2,3 because 0≀r<4.     …..(2)

Therefore from (1) and (2) we can get, 

a=4q or 4q+1 or 4q+2 or 4q+3     …..(3)

If a=4q=2(2q) (which is even)

If a=4q+1=2(2q)+1 (which is odd)

If a=4q+2=2(2q+1) (which is even)

If a=4q+3=2(2q+1)+1 (which is odd) 

So, every positive odd integer is of the form (4q+1) or (4q+3).


16. Show that any number of the form 6x,x∈N can never end with the digit 0.       

Ans: The prime factorization of 10 is 2Γ—5. Therefore, for a number to end with 0 it should have both 2 and 5 as a factor.

For the given number of the form 6n=(2Γ—3)n=2nΓ—3n., we can observe that it has only 2 as a factor and not 5. So 6x,x∈N can never end with the digit 0.


17. Find HCF and LCM of 18 and 24 by the prime factorization method.   

Ans: The prime factorization of 18 is 18=2Γ—3Γ—3=2Γ—32  …..(1)

The prime factorization of 24 is 24=2Γ—2Γ—2Γ—3=23Γ—3   …..(2)

From (1) and (2) we can see that 18,24 has prime factors 2,3 in common. 

∴HCF=2Γ—3=6

From (1) and (2) we can see that 3 appears twice in prime factorization of 18 and 2 appears thrice in prime factorization of 24.

 βˆ΄LCM=(32)Γ—(23)=72


18. The HCF of two numbers is 23 and their LCM is 1449. If one of the numbers is 161, find the other.

Ans: It is given that the HCF of the two numbers is 23 and their LCM is 1449. We have to find the other number if one of the numbers is 161. Let the other number be a

We know that, LCMΓ—HCF = Product of two numbers. Therefore,

1449Γ—23=aΓ—161 

β‡’a=1449Γ—23161

∴a=207


19. Prove that the square of any positive integer of the form 5g+1 is of the same form.

Ans: Let a be any positive integer, of the form a=5g+1      …..(1)

Squaring both sides of equation (1) we get,

a2=(5g+1)2

β‡’a2=25g2+10g+1 

β‡’a2=5(5g2+2g)+1

β‡’a2=5m+1 

Hence, the square of any positive integer of the form 5g+1 is of the same form.


20. Use Euclid’s Division Algorithm to find the HCF of 4052 and 12576.      

Ans: Step 1: Since12576>4052 , we will apply Euclid’s division lemma such that 12576=4052Γ—3+420    …..(1)

Step 2: Since the remainder in (1) is not zero so we have to apply the Euclid’s division lemma to 4052 and 420, such that 4052=420Γ—9+272   …..(2)

Step 3: Again, the remainder in (2) is not zero so we have to apply the Euclid’s division lemma to 420 and 272, such that 420=272Γ—1+148.    …..(3) 

Step 4: Again, the remainder in (3) is not zero so we have to apply the Euclid’s division lemma to 148 and 272, such that 272=148Γ—1+124.   …..(4) 

Step 5: Again, the remainder in (4) is not zero so we have to apply the Euclid’s division lemma to 148 and 124, such that 148=124Γ—1+24.     …..(5) 

Step 6: Again, the remainder in (5) is not zero so we have to apply the Euclid’s division lemma to 24 and 124, such that 124=24Γ—5+4.    …..(6) 

Step 7: Again, the remainder in (6) is not zero so we have to apply the Euclid’s division lemma to 24 and 4, such that 24=4Γ—6+0.   …..(7) 

Now the remainder equation (7) is zero. Therefore, HCF of 12576 and 4052 is β€²4β€².


21. Find the largest number which divides 245 and 1029 leaving remainder 5 in each case.

Ans. Since we want the remainder to be 5, therefore the required number is the HCF of (245βˆ’5) and (1029βˆ’5) i.e., 240 and 1024.

Step 1: Since1024>240 , we will apply Euclid’s division lemma such that 1024=240Γ—4+64                     …..(1)

Step 2: Since the remainder in (1) is not zero so we have to apply the Euclid’s division lemma to 240 and 64, such that 240=64Γ—3+48  …..(2)

Step 3: Again, the remainder in (2) is not zero so we have to apply the Euclid’s division lemma to 64 and 48, such that64=48Γ—1+16.   …..(3) 

Step 4: Again, the remainder in (3) is not zero so we have to apply the Euclid’s division lemma to 48 and 16, such that48=16Γ—3+0.    …..(4) 

Now the remainder in equation (4) is zero. Therefore, HCF of 1024 and 240 is β€²16β€².

The largest number which divides 245 and 1029 leaving remainder 5 in each case is 16


22. A shopkeeper has 120 litres of petrol, 180 litres of diesel and 240 litres of kerosene. He wants to sell oil by filling the three kinds of oils in tins of equal capacity. What should be the greatest capacity of such a tin?                                                                         

Ans: The required greatest capacity is the HCF of 120,180 and 240. First let us find the HCF of 240,180 by Euclid’s division algorithm.

Step 1: Since240>180 , we will apply Euclid’s division lemma such that 240=180Γ—1+60                     …..(1)

Step 2: Since the remainder in (1) is not zero so we have to apply the Euclid’s division lemma to 180 and 60, such that 180=60Γ—3+0     …..(2)

Now the remainder in equation (2) is zero. Therefore, HCF of 180 and 240 is 60.

Let us now find the HCF of 60,120. Clearly, 120=60Γ—2+0.

∴HCF of 120,180 and 240 is 60 and hence the required capacity is 60 litres.


Short Answer Questions                                                                               (3 Marks)

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

(i). 135 and 225

Ans: Step 1: Since 225>135, we will apply Euclid’s division lemma such that 225=135Γ—1+90                     …..(1)

Step 2: Since the remainder in (1) is not zero so we have to apply the Euclid’s division lemma to 135 and 90, such that 135=90Γ—1+45 …..(2)

Step 3: Again, the remainder in (2) is not zero so we have to apply the Euclid’s division lemma to 90 and 45, such that 90=2Γ—45+0.   …..(3) 

Now the remainder in equation (3) is zero. Therefore, the HCF of 135 and 225 is 45.

(ii). 196 and 38220

Ans: Since 38220>196, we will apply Euclid’s division lemma such that 38220=196Γ—195+0                     …..(1)

Since, the remainder in (1) is zero. Therefore, the HCF of 38220 and 196 is 196.

(iii). 867 and 255

Ans: Step 1: Since 867>255,, we will apply Euclid’s division lemma such that 867=255Γ—3+102                     …..(1)

Step 2: Since the remainder in (1) is not zero so we have to apply the Euclid’s division lemma to 255 and 102, such that 255=102Γ—2+51   …..(2)

Step 3: Again, the remainder in (2) is not zero so we have to apply the Euclid’s division lemma to 102 and 51, such that 102=51Γ—2+0.   …..(3) 

Now the remainder is zero. Therefore, the HCF of 867 and 255 is 51.


2. Find the greatest number of 6 digits exactly divisible by 24,15 and 36.

Ans: To find the greatest number of 6 digits exactly divisible by 24,15 and 36, first we have to find the LCM of 24,15 and 36.

The prime factorization of 24,15 and 36 are:

24=2Γ—2Γ—2Γ—3    …..(1)

15=5Γ—3    …..(2)

36=2Γ—2Γ—3Γ—3   …..(3)

Hence, from (1), (2) and (3), LCM(15,24,36)=2Γ—2Γ—2Γ—5Γ—3Γ—3=360.  .….(4)

Now, we know that the greatest number of 6 digits is 999999.  …..(5)

From (4) and (5), using Euclid’s division lemma we get, 999999=360Γ—2777+279.

β‡’999999βˆ’279=360Γ—2777 

β‡’999720=360Γ—2777

Hence, the greatest number of 6 digits exactly divisible by 24,15 and 36 is 999720.  


3. Prove that the square of any positive integer is of the form 4m or 4m+1 for some integer m.

Ans: Let a be any positive integer, then by Euclid’s algorithm we can write a=4q+r, for some integer qβ‰₯0.      …..(1)

Clearly, in the expression (1) we are dividing a by 4 with quotient q and remainder r, r=0,1,2,3 because 0≀r<4.   …..(2)

Therefore from (1) and (2) we can get, 

a=4q or 4q+1 or 4q+2 or 4q+3   …..(3)

Case 1: If a=4q then, squaring both sides we get,

a2=16q2

β‡’a2=4(4q2) …..(4)

Equation (4) could be written as a2=4m for some integer m ….. (5)

Case 2: If a=4q+1 then, squaring both sides we get,

a2=16q2+8q+1

β‡’a2=4(4q2+2q)+1 …..(6)

Equation (6) could be written as a2=4m+1 for some integer m ….. (7)

Case 3: If a=4q+2 then, squaring both sides we get,

a2=16q2+16q+4

β‡’a2=4(4q2+4q+1) …..(8)

Equation (8) could be written as a2=4m for some integer m ….. (9)

Case 4: If a=4q+3 then, squaring both sides we get,

a2=16q2+24q+9

a2=4(4q2+6q+2)+1 …..(10)

Equation (10) could be written as a2=4m+1 for some integer m ….. (11)

Therefore, from equations (5), (7), (9) and (11) it can be said that the square of any positive integer is either of the form 4m or 4m+1 for some integer m.


4. There are 144 cartons of coke can and 90 cartons of Pepsi can to be stacked in a canteen. If each stack is of the same height and is to contain cartoons of the same drink. What would be the greater number of cartons each stack would have?     

Ans: Given 144 cartons of coke can and 90 cartons of Pepsi can are to be stacked in a canteen. For stack to be of the same height and is to contain cartoons of the same drink, we have to find the HCF of 144 and 90.

Using Euclid’s division algorithm,

Step 1: Since 144>90,, we will apply Euclid’s division lemma such that 144=90Γ—1+54                     …..(1)

Step 2: Since the remainder in (1) is not zero so we have to apply the Euclid’s division lemma to 90 and 54, such that 90=54Γ—1+36  …..(2)

Step 3: Again, the remainder in (2) is not zero so we have to apply the Euclid’s division lemma to 54 and 36, such that 54=36Γ—1+18   …..(3) 

Step 4: Again, the remainder in (3) is not zero so we have to apply the Euclid’s division lemma to 18 and 36, such that36=18Γ—2+0.      …..(4)

Now the remainder in (4) is zero. Therefore, the HCF of 144 and 90 is 18.

So, greatest number of cartoons is 18.


5. Prove that product of three consecutive positive integers is divisible by 6.   

Ans: Let us consider three consecutive positive integers x,(x+1) and (x+2).

Using Euclid’s lemma, let us write x=6q+r   …..(1)

Clearly, in the expression (1) we are dividing a by 6 with quotient q and remainder r, r=0,1,2,3,4,5 because 0≀r<6.     …..(2)

Therefore from (1) and (2) we can get,                                                                                       

x=6q,6q+1,6q+2,6q+3,6q+4,6q+5

Case 1: If x=6q then product 

x(x+1)(x+2)=6q(6q+1)(6q+2), which is divisible by 6.   …..(3)

Case 2: If x=6q+1 then product 

x(x+1)(x+2)=(6q+1)(6q+2)(6q+3)

β‡’x(x+1)(x+2)=2(3q+1).3(2q+1)(6q+1) 

β‡’x(x+1)(x+2)=6(3q+1).(2q+1)(6q+1) 

 which is divisible by 6.    …..(4)

Case 3: If x=6q+2 then product 

x(x+1)(x+2)=(6q+2)(6q+3)(6q+4)

β‡’x(x+1)(x+2)=3(2q+1).2(3q+1)(6q+4) 

β‡’x(x+1)(x+2)=6(2q+1)(3q+1)(6q+4) 

 which is divisible by 6.  …..(5)

Case 4: If x=6q+3 then product 

x(x+1)(x+2)=(6q+3)(6q+4)(6q+5)

β‡’x(x+1)(x+2)=6(2q+1)(3q+2)(6q+5) 

 which is divisible by 6.   …..(6)

Case 5: If x=6q+4 then product 

x(x+1)(x+2)=(6q+4)(6q+5)(6q+6)

β‡’x(x+1)(x+2)=6(6q+4)(6q+5)(q+1) 

 which is divisible by 6.     …..(7)

Case 6: If x=6q+5 then product 

x(x+1)(x+2)=(6q+5)(6q+6)(6q+7)

β‡’x(x+1)(x+2)=6(6q+5)(q+1)(6q+7) 

 which is divisible by 6.   …..(8)

From equations (3) to (8) we can conclude that the product of three consecutive positive integers is divisible by 6.


6. Prove that (3βˆ’5) is an irrational number.

Ans: We will prove this by contradiction. To solve this problem, suppose that (3βˆ’5) is rational i.e., βˆƒ two co-prime integers a and b (bβ‰ 0) such that ab=3βˆ’5      …..(1)

Subtracting 3 from both sides of the equation (1) and simplifying it further we get,

abβˆ’3=βˆ’5

β‡’3bβˆ’ab=5      …..(2)

Since we know that 5 is irrational. Therefore, from (2) 3bβˆ’abshould be irrational but it could be written in pq form which means it is rational. It is not possible and hence our assumption is wrong and (3βˆ’5) cannot be rational.


7. Prove that if x and y are odd positive integers, then x2+y2 is even but not divisible by 4.

Ans: Let x=2p+1 and y=2q+1 be two odd positive integers βˆ€p,q∈N.       …..(1)

Squaring x,y from (1) and adding them we get, 

x2+y2=(2p+1)2+(2q+1)2

β‡’x2+y2=(4p2+4p+1)+(4q2+4q+1)

β‡’x2+y2=2(2p2+2q2+2p+2q+1)

β‡’x2+y2=2m    …..(2)

Where m=(2p2+2q2+2p+2q+1) 

From (2), it is clear that x2+y2 is even but not divisible by 4.


8. Show that one and only one out of n,(n+2) or (n+4) is divisible by 3, where n∈N.

Ans: Let n be a number such that n=3q+r.   …..(1)

Clearly, in the expression (1) we are dividing n by 3 with quotient q and remainder r, r=0,1,2 because 0≀r<3.    …..(2)

Therefore from (1) and (2) we can get, n=3q,3q+1,3q+2

Case 1: If n=3q then, numbers are 3q,(3q+2)(3q+4) out of these 3q is divisible by 3.     …..(1)

Case 2: If n=3q+1 then, numbers are (3q+1),(3q+3),(3q+5) out of these (3q+3) is divisible by 3.     …..(2)

Case 3: If n=3q+2 then, numbers are (3q+2),(3q+4),(3q+6) out of these (3q+6) is divisible by 3.     …..(3)

∴ from (1), (2) and (3) we can conclude that out of n,(n+2) and (n+4) only one is divisible by 3.


9. Use Euclid’s Division Lemma to show that the square of any positive integer of the form 3m or (3m+1) for some integer q.

Ans: Let a be any positive integer, then we can write a=3q+r for some integer qβ‰₯0    …..(1)

Clearly, in the expression (1) we are dividing a by 3 with quotient q and remainder r, r=0,1,2 because 0≀r<3.     ..…(2)

Therefore from (1) and (2) we can get, a=3q or 3q+1 or 3q+2.     ……(3)

Squaring both sides of equation (3) we get,

a2=(3q)2 or (3q+1)2 or (3q+2)2 

β‡’a2=9q2 or 9q2+6q+1 or 9q2+12q+4    …..(4)

Taking 3 common from LHS of equation (4) we get,

a2=3Γ—(3q2) or 3(3q2+2q)+1 or 3(3q2+4q+1)+1.    …..(5)

Equation (5) could be written as 

a2=3k1 or 3k2+1 or (3k3+1) for some positive integers k1,k2 and k3.

Hence, it can be said that the square of any positive integer is either of the form 3m or 3m+1.


10. Prove that if n is not a rational number, if n is not a perfect square. 

Ans: We will prove this by contradiction.

Let us suppose that n is rational. It means that we have co-prime integers a and b (b≠0) such that n=ab

β‡’bn=a   …..(1)

Squaring both sides of (1), we get nb2=a2   …..(2)

From (2) we can conclude that n is factor of a2, this means that n is also factor of a.    …..(3)

From (3) we can write a=nc for some integer c.   …..(4)

Substituting value of (4) in (2) we get,

nb2=n2c2β‡’b2=nc2

It means that n is factor of b2, this means that n is also factor of b.    …..(5)

From (3),(5), we can say that n is factor of both a,b, which is a contradiction to our assumption. Therefore, our assumption was wrong and n is irrational.


11. Prove that the difference and quotient of (3+23) and (3βˆ’23) are irrational. 

Ans: Difference of (3+23) and (3βˆ’23) is (3+23)βˆ’(3βˆ’23)=43.

We will prove this by contradiction.

Let us suppose that 43 is rational. It means that we have co-prime integers a and b (b≠0) such that 43=ab

β‡’4b3=a   …..(1)

Squaring both sides of (1), we get 48b2=a2  …..(2)

From (2) we can conclude that 48 is factor of a2, this means that 48 is also factor of a.   …..(3)

From (3) we can write a=48c for some integer c.   …..(4)

Substituting value of (4) in (2) we get,

43b2=48c2β‡’b2=43c2

It means that 48 is factor of b2, this means that 48 is also factor of b.   …..(5)

From (3),(5), we can say that 48 is factor of both a,b, which is a contradiction to our assumption. Therefore, our assumption was wrong and 43 is irrational.

Now, the quotient of (3+23),(3βˆ’23) is

3+233βˆ’23=3+233βˆ’23Γ—3+233+23

β‡’3+233βˆ’23=βˆ’(5+43) 

Similarly, using the approach used above, we can prove that βˆ’(5+43) is also irrational. 


12. Show that (n2βˆ’1) is divisible by 8, if n is an odd positive integer.

Ans: Let n=4q+1 be an odd positive integer.   …..(1)

Then from (1),

n2βˆ’1=(4q+1)2βˆ’1

β‡’n2βˆ’1=16q2+8q    …..(2)

Taking 8 common from RHS of equation (2) we get,

n2βˆ’1=8(2q2+q) 

β‡’n2βˆ’1=8m where m=2q2+q.   …..(3)

Hence, from (3) we can conclude that (n2βˆ’1) is divisible by 8, if n is an odd positive integer.


13. Use Euclid Division Lemma to show that cube of any positive integer is either of the form 9m,(9m+1) or (9m+8).

Ans: Let a be any positive integer, then we can write a=3q+r for some integer qβ‰₯0    …..(1)

Clearly, in the expression (1) we are dividing a by 3 with quotient q and remainder r, r=0,1,2 because 0≀r<3.    ..…(2)

Therefore from (1) and (2) we can get, a=3q or 3q+1 or 3q+2.    ……(3)

Cubing both sides of equation (3) we get,

a3=(3q)3 or (3q+1)3 or (3q+2)3 

β‡’a3=27q3 or 27q3+27q2+9q+1 or 27q3+54q2+36q+8    …..(4)

Taking 9 common from RHS of equation (4) we get,

a3=9(3q3) or 9(3q3+3q2+q)+1 or 9(3q3+6q2+4q)+8    …..(5)

Equation (5) could be written as 

a3=9k1 or 9k2+1 or (9k3+1) for some positive integers k1,k2 and k3.

Hence, it can be said that the square of any positive integer is either of the form 9m,9m+1 or 3m+8.


Long Answer Questions                                                                                (4 Marks)

1. Prove that the following are irrationals.       

(i). 12

Ans: (i) We will prove this by contradiction. Let us suppose that 12 is rational. It means that we have co-prime integers a and b (b≠0) such that 12=ab

β‡’ba=2      …..(1)

Since we know that 2 is irrational. Therefore, from (1) ba should be irrational but it could be written in pq form which means it is rational. It is not possible and hence our assumption is wrong and 12 cannot be rational.

(ii). 75

Ans: We will prove this by contradiction. Let us suppose that 75 is rational. It means that we have co-prime integers a and b (bβ‰ 0) such that 75=ab  

β‡’5=a7b      …..(1)

Since we know that 5 is irrational. Therefore, from (1) a7b should be irrational but it could be written in pq form which means it is rational. It is not possible and hence our assumption is wrong and 75 cannot be rational.

(iii). 6+2

Ans: We will prove this by contradiction. To solve this problem, suppose that 6+2 is rational i.e., βˆƒ two co-prime integers a and b (bβ‰ 0) such that ab=6+2        …..(1)

Subtracting 6 from both sides of the equation (1) and simplifying it further we get,

abβˆ’6=2

β‡’aβˆ’6bb=2       …..(2)

Since we know that 2 is irrational. Therefore, from (2) aβˆ’6bbshould be irrational but it could be written in pq form which means it is rational. It is not possible and hence our assumption is wrong and 6+2 cannot be rational.


2. Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating decimal expansion.

(i). 133125

Ans: Theorem: Let x be a rational number expressed in the form , pq where p and qare coprime, and the prime factorisation of q is of the form 2nΓ—5m, where m and n are non-negative integers. Then x has a decimal expansion which terminates.            …..(1)

For rational number pq=133125, the prime factorization of q is q=3125=5Γ—5Γ—5Γ—5Γ—5=55.                       …..(2)

From (2) we can see that the denominator q=3125 is of the form 2nΓ—5m, where m=5 and n=0.

Therefore, it follows from theorem (1), the rational number 133125 has a terminating decimal expansion.

(ii). 178

Ans: Theorem: Let x be a rational number expressed in the form , pq where p and qare coprime, and the prime factorisation of q is of the form 2nΓ—5m, where m and n are non-negative integers. Then x has a decimal expansion which terminates.            …..(1)

For rational number pq=178, the prime factorization of q is 

q=8=2Γ—2Γ—2=23.    …..(2)

From (2) we can see that the denominator q=8 is of the form 2nΓ—5m, where m=0 and n=3.

Therefore, it follows from theorem (1), the rational number 178 has a terminating decimal expansion.

(iii). 64455

Ans: Theorem: Let x be a rational number expressed in the form , pq where p and qare coprime, and the prime factorisation of q is of the form 2nΓ—5m, where m and n are non-negative integers. Then x has a decimal expansion which terminates.            …..(1)

For rational number pq=64455, the prime factorization of q is 

q=455=5Γ—91.   …..(2)

From (2) we can see that the denominator q=455 is not of the form 2nΓ—5m.

Therefore, it follows from theorem (1), the rational number 64455 has a non-terminating repeating decimal expansion.

(iv). 151600

Ans: Theorem: Let x be a rational number expressed in the form , pq where p and qare coprime, and the prime factorisation of q is of the form 2nΓ—5m, where m and n are non-negative integers. Then x has a decimal expansion which terminates.            …..(1)

For rational number pq=151600, the prime factorization of q is 

q=320=2Γ—2Γ—2Γ—2Γ—2Γ—2Γ—5=26Γ—5.    …..(2)

From (2) we can see that the denominator q=8 is of the form 2nΓ—5m, where m=1 and n=6.

Therefore, it follows from theorem (1), the rational number 151600 has a terminating decimal expansion.

(v). 29343

Ans: Theorem: Let x be a rational number expressed in the form , pq where p and qare coprime, and the prime factorisation of q is of the form 2nΓ—5m, where m and n are non-negative integers. Then x has a decimal expansion which terminates.            …..(1)

For rational number pq=29343, the prime factorization of q is 

q=343=7Γ—7Γ—7.   …..(2)

From (2) we can see that the denominator q=343 is not of the form 2nΓ—5m.

Therefore, it follows from theorem (1), the rational number 29343 has a non-terminating repeating decimal expansion.

(vi). 2323Γ—52

Ans: Theorem: Let x be a rational number expressed in the form , pq where p and qare coprime, and the prime factorisation of q is of the form 2nΓ—5m, where m and n are non-negative integers. Then x has a decimal expansion which terminates.            …..(1)

For rational number 2323Γ—52, the prime factorization of q is 

q=23Γ—52.   …..(2)

From (2) we can see that the denominator q=23Γ—52 is of the form 2nΓ—5m, where m=2 and n=3.

Therefore, it follows from theorem (1), the rational number 2323Γ—52 has a terminating decimal expansion.

(vii). 12922Γ—57Γ—75

Ans: Theorem: Let x be a rational number expressed in the form , pq where p and qare coprime, and the prime factorisation of q is of the form 2nΓ—5m, where m and n are non-negative integers. Then x has a decimal expansion which terminates.            …..(1)

For rational number 12922Γ—57Γ—75, the prime factorization of q is 

q=22Γ—57Γ—75.    …..(2)

From (2) we can see that the denominator q=23Γ—52 is not of the form 2nΓ—5m.

Therefore, it follows from theorem (1), the rational number 12922Γ—57Γ—75 has a non-terminating repeating decimal expansion.

(viii). 615=25

Ans: Theorem: Let x be a rational number expressed in the form , pq where p and qare coprime, and the prime factorisation of q is of the form 2nΓ—5m, where m and n are non-negative  integers. Then x has a decimal expansion which terminates.            …..(1)

For rational number pq=615=25, the prime factorization of q is q=5=51.    …..(2)

From (2) we can see that the denominator q=5 is of the form 2nΓ—5m, where m=1 and n=0. Therefore, it follows from theorem (1), the rational number 615=25 has a terminating decimal expansion.

(ix). 3550=710

Ans: Theorem: Let x be a rational number expressed in the form , pq where p and qare coprime, and the prime factorisation of q is of the form 2nΓ—5m, where m and n are non-negative integers. Then x has a decimal expansion which terminates.   …..(1)

For rational number 3550=710, the prime factorization of q is q=10=2Γ—5=21Γ—51.  …..(2)

From (2) we can see that the denominator q=5 is of the form 2nΓ—5m, where m=1 and n=1. Therefore, it follows from theorem (1), the rational number 3550 has a terminating decimal expansion.

(x). 77210=1130

Ans: Theorem: Let x be a rational number expressed in the form , pq where p and qare coprime, and the prime factorisation of q is of the form 2nΓ—5m, where m and n are non-negative integers. Then x has a decimal expansion which terminates.            …..(1)

For rational number 77210=1130, the prime factorization of q is 

q=30=2Γ—3Γ—5.  …..(2)

From (2) we can see that the denominator q=23Γ—52 is not of the form 2nΓ—5m.

Therefore, it follows from theorem (1), the rational number 77210=1130 has a non-terminating repeating decimal expansion.


Maths Chapter 1- Real Numbers

Introduction

All numbers can be classified into two categories -- Real and Imaginary. A real number can be actually perceived and can be represented in a number line. It includes all integers -- positive and negative, all natural numbers (1, 2, 3, ….., ) all whole numbers (0, 1, 2, 3, ….., ), fractions, rational and irrational numbers.

  • An imaginary number on the other hand cannot be perceived. Square root of all negative numbers is imaginary.

  • A real number can be represented as an infinite decimal expansion. Ex: 5 = 5.0000….., β…“ = 0.3333……

  • A real number can be rational or irrational.

  • All rational numbers and all irrational numbers together make a group of real numbers.

(i) A group of rational numbers is denoted by Q.

(ii) A group of irrational numbers is denoted by Q’.

(iii) A group of real numbers is denoted by R.

  • A rational number an be written in the form pq, both p and q being integers (q 0) while an irrational number can not be impressed in such a way.

  • The decimal expansion of a rational number say x can terminate after a finite number of digits.

Ex: ⅝ = 0.625 or eventually begins to repeat the same finite sequence of digits continuously. 

Ex: 56/99 = 0.565656……….

  • On the other hand in case of an irrational number, the decimal expansion continues without repeating. Ex: 1.7320508….,  πœ‹= 3.141592….)


Euclid’s Division Lemma and Algorithm

First, let us understand the meaning of Lemma and Algorithms.

A Lemma is a minor proven statement which is used to prove other statements. The Euclid’s division Lemma is actually another statement of the common long division process.

For example, on dividing 36 by 5, we get quotient = 7 and remainder = 1 which is less than 5.

If there are positive integers a and b, then there are unique integers existing q and r that satisfy a = bq + r, 0 rb.

Example: If a =47 and b = 9, then we can write 

  47 = 9 x 5 + 2, where q = 4 and r = 15 b

Note: 

(i) The unique characteristics of qand r are nothing but the quotient and remainder respectively.

(ii) Although Euclid’s division algorithm is expressed for only positive integers but it can be extended for all integers except 0.

(iii) When aand bare positive integers, then qand rcan take values only from whole numbers.

An algorithm means a series of well defined steps, which provides  a procedure of calculations repeatedly successfully on the results of earlier steps. With the help of defined procedure the desired result can be obtained. 


Euclid’s Division Algorithm

The Euclid’s Division Algorithm is also an algorithm to determine the H.C.F (Highest Common Factor) of the given positive integers. 

Ordinarily, if we want to find the HCF of two positive integers, we first find the factors of the two integers. 

For ex: For two integers 24 and 42, the common factors are 1, 2, 3 and 6. As 6 is the highest of other numbers, 6 is the HCF of 24 and 42. 

In Euclid’s Division Algorithm, the largest integer is divided by the smaller one and a remainder is obtained. 

Next the smaller integer is divided by the above remainder and the process is repeated till zero is obtained. The remainder in the last but one division is the required HCF.


The Fundamental Theorem of Arithmetic

According to this theorem every positive integer is either a prime number or it can be expressed as the product of primes. It is also known as Unique Factorization Theorem. Thus, any integer that is greater than 1 can be either a prime number or can be expressed as a product of prime factors.

Example: 8 = 2 x 2 x 2, where 2 is a prime factor. 

Note: 

(i) When a number has no factors other than 1 and the number itself, the number is called a prime number.

(ii) 1 is neither prime nor composite.

(iii) A number is a composite number if it has at least one factor other than 1 and the number itself.

(iv) 2 is the smallest prime number and an even number. It is the only number that is both prime and even.

(v) Two numbers are co-prime if they have no common factors other than 1, i.e, their HCF = 1.

This theorem is termed as β€˜Fundamental’ because of its importance in the development of number theory. 


Factor Tree

A chain of factors that is demonstrated in the form of a tree, is called a factor tree. 

HCF and LCM using Prime Factorization

HCF is the product of the smallest power of each common prime factor present in the numbers.

The product of the largest power of each prime factor present in the numbers is the LCM. 

Important Note: 

(i) LCM of two or more numbers is the smallest number which is the smallest number and divisible by all the given numbers.

(ii) If there is no common prime factor, then the HCF of the given number is 1.


Relationship Between Numbers and Their HCF and LCM

  • For given positive integers aand b, the relation between these numbers and their HCF and LCM is 

HCF (a, b) = aΓ—bLCM(a,b) or LCM (a, b) = aΓ—bHCF(a,b)

  • For given three positive integers a, b and c, the relation between these numbers and their HCF and LCM is  HCF (a, b, c) = aΓ—bΓ—cΓ—LCM(a,b,c)LCM(a,b)Γ—LCM(b,c)Γ—LCM(c,a) or LCM (a, b,c) = aΓ—bΓ—cΓ—HCF(a,b,c)HCF(a,b)Γ—HCF(b,c)Γ—HCF(c,a)


Method of Proving Irrationality of Numbers

We use the method of contradiction in order to prove irrationality of the numbers. i.e., first we assume that the given number is rational and after reaching a contradiction we prove that the given number is irrational. 


Theorem

If a prime number  p divides a2, then p divides a, where ais a positive integer.

Let us understand how to write the decimal expansion of those rational numbers with terminating decimal expansion

Let a rational number in the lowest form be  p/q such that the prime factorization of q is of the form 2mx 5n, where m, n are non-negative integers. To write decimal expansion of p/q, convert p/q to an equivalent rational number of the form c/d, where dis a power of 10. 

The above notes will be a huge benefit for solving the Important Questions Of Chapter 1 of Maths for Class 10, Real Numbers. You can now master the Euclid’s Division Lemma and Algorithm. The important questions for this chapter is basically a compilation of higher and more advanced techniques a student can expect in Class 10 examinations. The questions of the chapter and the notes related to the topic provided by Vedantu will not only help you to understand the concept better but also solve the questions successfully. If you harbour any doubts then you can get answers to all your queries with our experienced teachers by registering with Vedantu and gain expertise in the subject. 


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FAQs on Important Questions for CBSE Class 10 Maths Chapter 1 - Real Numbers 2024-25

1. What do you mean by HCF and LCM?

Ans: HCF stands for "Highest Common Factor," also known as the Greatest Common Divisor (GCD). It refers to the largest number that divides two or more integers without leaving a remainder. In other words, the HCF is the largest common factor shared by a set of numbers.


LCM stands for "Least Common Multiple." It refers to the smallest positive integer that is divisible by two or more integers without leaving a remainder. In other words, the LCM is the smallest common multiple shared by a set of numbers.

2. How will Important Questions for Chapter 1 for Class 10 help in exams?

Ans: Important Questions for Chapter 1 in Class 10 will help in exams by focusing on the key concepts, topics, and problem-solving techniques. These questions are carefully selected to cover the important aspects of the chapter, allowing students to practice and gain confidence in their understanding. They serve as a valuable tool for exam preparation and improving overall performance.

3. What does Chapter 1, real number for Class 10 discuss?

Ans: Chapter 1, "Real Numbers," for Class 10 discusses the concept of real numbers and their properties. It covers topics such as rational numbers, irrational numbers, decimal representation of rational numbers, fundamental theorem of arithmetic, Euclid's division algorithm, and the concept of prime and composite numbers. The chapter also introduces the concept of HCF (Highest Common Factor) and LCM (Least Common Multiple) of numbers. It provides a foundation for understanding number systems and lays the groundwork for further mathematical concepts in higher grades.

4. Which is the most important chapter in maths class 10?

Ans: The most important chapter in maths are:

Chapter 3 - Linear Equations

Chapter 5 - Arithmetic Progression

Chapter 6 - Triangles

Chapter 7 - Coordinate Geometry

Chapter 8 - Introduction to Trigonometry

Chapter 9 - Applications of Trigonometry

Chapter 13 - Surface Areas & Volumes

Chapter 14 - Statistics

5. What is the weightage of marks in class 10 maths 2024-25?

Ans: CBSE Class 10 Mathematics in the academic year 2024-25 includes a theory paper worth 80 marks and internal assessment carrying 20 marks. The theory paper consists of 30 questions divided into four sections, with different marks assigned to each question. The total duration of the examination is three hours.