Important Questions Class 10 Maths Chapter 1

Important Questions Class 10 Mathematics Chapter 1 – Real numbers

Mathematics plays an important role in building the students’ careers as it is included as a core subject in almost every field. In order to be good in Mathematics, students need to practice consistently. Hence, Extramarks has included all the vital questions to boost students’ performance in their examinations in  Class 10 Mathematics Chapter 1 available on the Extramarks official website.

You may recall all the concepts covered in the chapter on real numbers in Classes 8, 9 and 10. You would also know more about Euclid’s Geometry which you already read in the lower classes. The theorems covered in this chapter will help you better understand the concepts of rational and irrational numbers. The important topics covered in the chapter include Euclid’s division lemma, a theorem on Euclid’s division lemma, fundamental theorems of arithmetics, more information about irrational numbers, and advanced level concepts of rational numbers and their decimal expansion and all the important theorems associated with it.

For students to score well, it is very necessary that none of the topics covered in the chapter remains untouched. As a result, in the Important Questions Class 10 Mathematics Chapter 1, we have included different sets of questions from various references and sources, which will aid students in understanding every concept included in the chapter in the form of questions and getting a hold on the subject.

One can find questions from the NCERT textbook, NCERT Exemplar and various other reference sources included in it. The questions are selected by experienced subject matter experts. After a detailed understanding of the questions, the solutions are given by the Mathematics experts, ensuring students that they are studying from the authentic and reliable study material. No wonder students and teachers swear by Extramarks because of the absolute trust and confidence it has built over the years. 

Extramarks website provides NCERT solutions along with the important questions Mathematics Class 10 Chapter 1. You can also get the NCERT revision notes and mock test by registering on the official website.

Important Questions Class 10 Mathematics Chapter 1 – With Solutions

We have provided the students with all the important questions along with their detailed solutions in  Class 10 Mathematics Chapter 1. The questions cover important topics like Euclid’s division lemma, a theorem on Euclid’s division lemma, advanced concepts of rational numbers and their decimal expansion, and more about the irrational numbers and all the vital theories associated with the chapter. 

After referring to this, students will be able to recall all the interrelated concepts of real numbers in an efficient manner.

The following important questions and their solutions: 

Question 1. For some integer x, every even integer is of the form.

(a) x

(b) x + 1

(c) 2x

(d) 2x + 1

Answer 1:  (C) 2x

Explanation: An integer is called even if it is divided by 2. Let m be an integer. 

We get, x = …,-6, −5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5, 6 … 

so x can be even or odd.

Therefore, ‘x’ cannot always be even.

Now, x + 1 = …, -5,−4, −3, −2, −1, 0, 1, 2, 3, 4, 7 …

We observe that ‘x+1’ cannot always be even.

Now consider 2x =…, -12,−10, −8, −6, −4, −2, 0, 2, 4, 6, 8, 10, …

We observe that ‘2x’ is always even.

Consider 2x + 1 = …, -11, −9, −7, −5, −3, −1, 1, 3, 5, 7, 9, 11, ….

We observe that ‘2x+1’ is always an odd integer. 

So, ‘2x’ is the answer.  

 Therefore, the correct answer is option C.

Question 2. The decimal form of 129/225775 is

(a) terminating

(b) non-terminating

(c) non-terminating non-repeating

(d) none of the above

Answer 2 : (c) non-terminating non-repeating

Question 3. The decimal expansion of 22/7  is

(a) Terminating

(b) Non-terminating and repeating

(c) Non-terminating and Non-repeating

(d) None of the above

Answer 3 : (b) Non-terminating and repeating

Explanation: 22/7 = 3.14285714286.

Question 4. For some integer m, every odd integer is of the form

(a) m

(b) m + 1

(c) 2m

(d) 2m + 1

Answer 4 : (D) 2m + 1

Explanation: An integer is called odd if it is not divided by 2. Let m be an integer i.e m = …, −2, −1, 0, 1, 2, …

Multiplication of both sides by 2 gives, 2m = …,-6, −4, −2, 0, 2, 4, 6 …

Adding 1 on both sides, 2m + 1 = …,-5, −3, −1, 1, 3, 5, 7 …

Hence, for any integer m, (2m +1) is always odd.

Therefore, the correct answer is option  D.

 

Question 5. For some integer a, the odd integer is represented in the form of:

(a) a

(b) a+ 1

(c) 2a + 1

(d) 2a

Answer 5 : (c) 2a + 1

Explanation: Since 2a represents even numbers, hence 2a + 1 will always represent odd numbers. Suppose if n = 2, then 2a = 4 and 2a + 1 = 5.

Question 6. n 2 – 1 is divisible by 8, if n is

(a) an integer

(b) a natural number

(c) an odd integer

(d) an even integer

Answer 6 : (C) an odd integer

Explanation: Let a = n 2 − 1. Here, n can be odd or even.

Case 1: When n = even. Let n = 4k, where k is an integer.

⇒a=(4k) 2-1

⇒a=16k 2-1

which is not divisible by 8.

Case 2: When n = odd. Let n = 4k + 1, where k is an integer.

⇒a=(4k + 1) 2-1

⇒a=16k 2+1+8k-1⇒a=16k 2+8k

⇒a=8(2k 2+k)

which is always divisible by 8. Thus, if n is odd, then n 2-1 is divisible by 8.

Hence, the correct answer is option C.

 

Question 7. HCF of 26 and 91 are:

(a) 15

(b) 13

(c) 19

(d) 11

Answer 7 : (b) 13

Explanation: 

Let us do the prime factorisation of 26 and 91;

26 = 2 x 13

91 = 7 x 13

Hence, HCF (26, 91) = 13

Question 8. If HCF of 65 and 117 is expressible in form 65m– 117, then the value of m is

(a) 4

(b) 2

(c) 1

(d) 3

Answer 8 :  (b) 2

Explanation: By Euclid’s division lemma, b = aq + r, here 0 ≤r<a

117=65×1+52

65=52×1+13

52=13×4+0

∴ HCF(65,117) = 13         …..(1)

We observe that HCF (65,117) = 65m − 117        …..(2)

From (1) and (2), we get:

65m−117=13

65m=130

m=2

Therefore, the right answer is option (b).

 

Question 9. Which of the following is not irrational?

(a) (3 + √7)

(b) (3 – √7)

(c) (3 + √7) (3 – √7)

(d) 3√7

Answer 9 : (c)  (3 + √7) (3 – √7)

Explanation: When we solve (3 + √7) (3 – √7), we would get;

(3 + √7) (3 – √7) = 3*3 – (√7)*(√7) = 9 – 7 = 2 [using a2 – b2 = (a – b) (a + b)]

 

Question 10: The largest number, which divides 70 and 125, leaving remainders 5 and 8 respectively are

(a) 13

(b) 65

(c) 875

(d) 1750

Answer 10 :  (a) 13

Explanation: Firstly, we subtract the remainder 5 and 8, respectively, from corresponding numbers and then get the HCF of the resulting numbers using Euclid’s division lemma to get the associated number. After removing these remainders from the numbers, we have:

(70 − 5) = 65

(125 − 8) = 117

Now, the required number is HCF(65,117).

Using Euclid’s division algorithm,

117=65×1+52

65=52×1+13

52=13×4+0

∴  HCF = 13. Thus, 13 is the largest number which is divisible by 70 and 125 leaving remainder 5 and 8. 

Hence, the correct answer is option (a) .

Question 11. The addition of a rational number and an irrational number is equal to:

(a) a rational number

(b) Irrational number

(c) Both

(d) None of the above

Answer 11 : (b) Irrational number

Explanation: The addition of a rational number and an irrational number equals an irrational number.

Question 12. If two positive integers ‘a’ as well as ‘b’ are written as a = x3y2 and b = xy3; x, y are prime numbers, then HCF (a, b) is

(a) xy

(b) xy2

(c) x3y3

(d) x2y2

Answer 12 :  (b) xy2

Explanation: Given: a= x3y2 = x×x×x×y×y b= xy3= x×y×y×y

HCF has full-form as the highest common factor of two or more related numbers.

Here, HCF(a, b) = HCF(x3y3, xy3) = x×y×y = xy2

Hence, the correct answer is option B.

Question 13. The multiplication of two irrational numbers is:

(a) irrational number

(b) rational number

(c) Maybe rational or irrational

(d) None

Answer 13 : (c) Maybe rational or irrational

Explanation: The multiplication of two irrational numbers is maybe rational or irrational.

Question 14. If two positive integers ‘p’ as well as ‘q’ can be expressed as p = ab2 and q = a3b; a, b being prime numbers, then LCM (p, q) is

(a) ab

(b) a2b3

(c) a3b2

(d) a3b3

Answer 14 : (c) a3b2

Explanation: Given that,

p=ab2=a×b×b

q=a3b=a×a×a×b

The LCM has the full-form as the smallest common multiple that can be divided by both the numbers.

Thus,

LCM(p, q) 

= LCM (ab2, a3b) 

Hence,

=a×b×b×a×a =a3b2

Therefore, the right answer is option (c)

 

Question 15. If the set A = {1, 2, 3, 4, 5, 6, 7…} is given, then it represents:

(a) Whole numbers

(b) Rational Numbers

(c) Natural numbers

(d) Complex numbers

Answer 15 : (c) Natural numbers

Explanation: If set A = {1, 2, 3, 4, 5, 6, 7…} is given, then it represents natural numbers.

Question 16. The product of an irrational number and a non-zero rational is

(a) it’s always irrational

(b) it’s always rational

(c) rational or irrational

(d) one

Answer 16 : (a) always irrational

Explanation: Consider the rational number as 7/2 and the irrational number as √5/2

Their product is given as:

7/2×√5/2=7√5/4, which is irrational.

Hence, the correct answer is option A.

Question 17. If a and b are integers and is represented in the form of a/b, then it is a:

(a) Whole number

(b) Rational number

(c) Natural number

(d) Even number

Answer 17 : (b) Rational number

Explanation: If a and b are integers and are represented in the form of a/b, then it is a rational number.

 

Question 18. The least number which is divisible by all the numbers from 1 to 10 (both 1 and 10 inclusive) is

(a) 10

(b) 100

(c) 504

(d) 2520

Answer 18 : (d) 2520

Explanation: The LCM is the smallest positive integer that is divisible by the given numbers.

Here, we need to find the LCM that is the lowest common multiple of the numbers starting from 1 to 10 (both inclusive). Factors of 1 to 10 numbers:

1=1

2=1×2

3=1×3

4=1×2×2

5=1×5

6=1×2×3

7=1×7

8=1×2×2×2

9=1×3×3

10=1×2×5

Therefore, LCM of numbers from 1 to 10 = LCM (1, 2, 3, 4, 5, 6, 7, 8, 9, 10)

Hence, LCM = 1×2×2×2×3×3×5×7=2520

Therefore, the right answer is option (d)

Question 19. The largest number that divides 70 and 125, and which leaves the remainders 5 and 8, is:

(a) 65

(b) 15

(c) 13

(d) 25

Answer 19 : (c) 13

Explanation: 

Step 1: 70 – 5 = 65 and 125 – 8 = 117

Step 2: HCF (65, 117) is the largest number that divides 70 and 125 and leaves the remainder 5 and 8.

HCF (65, 117) = 13

Question 20. The decimal expansion of rational number 14587/1250 will terminate after:

(a) one decimal place

(b) two decimal places

(c) three decimal places

(d) four decimal places

Answer 20 : (d) four decimal places

Explanation: For terminating decimal expansion of a rational number, the denominator is of the form 2m×5n.

Here, 14587/1250=14587/21×54

⇒[14587/10×53]×[23/23\ = [14587×8]/[10×1000] = 116696/10000=11.6696

Hence, the decimal expansion of the given rational number terminates after four decimal places.

Thus, the right answer is option (d).

Question 21. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = the product of the two numbers.

(i) 26 and 91

(ii) 510 and 92

(iii) 336 and 54

Answer 21:

(i) 26 and 91

Explanation: 

Expressing 26 and 91 as products of the prime factors, we observe,

26 = 2 × 13 × 1

91 = 7 × 13 × 1

Hence, LCM (26, 91) = 2 × 7 × 13 × 1 = 182

And HCF (26, 91) = 13

Verification

Now, multiplication of 26 and 91 = 26 × 91 = 2366

And multiplication of LCM and HCF = 182 × 13 = 2366

Therefore, LCM × HCF = product of the 26 and 91.

(ii) 510 and 92

Expressing 510 and 92 as products of the prime factors, we observe,

510 = 2 × 3 × 17 × 5 × 1

92 = 2 × 2 × 23 × 1

Hence, LCM(510, 92) = 2 × 2 × 3 × 5 × 17 × 23 = 23460

And HCF (510, 92) = 2

Verification

Now, multiplication of 510 and 92 = 510 × 92 = 46920

And multiplication of LCM and HCF = 23460 × 2 = 46920

Therefore, LCM × HCF = product of the 510 and 92.

(iii) 336 and 54

Expressing 336 and 54 as products of the prime factors,  

we observe, 336 = 2 × 2 × 2 × 2 × 7 × 3 × 1 and, 54 = 2 × 3 × 3 × 3 × 1

Hence, LCM(336, 54) =  3024

And HCF (336, 54) = 2×3 = 6

Verification

Now, multiplication of 336 and 54 = 336 × 54 = 18,144

And multiplication of LCM and HCF = 3024 × 6 = 18,144

Therefore, LCM × HCF = product of  336 and 54.

Question 22: Express each number as a product of its prime factors:

(i) 140

(ii) 156

(iii) 3825

(iv) 5005

(v) 7429

Answer 22:

(i) 140

Explanation:

With the help of division of the number by the prime numbers method, we observe the multiple  of prime factors of 140

hence,

 140 = 2 × 2 × 5 × 7 × 1 = 22 × 5 × 7

(ii) 156

With the help of division of the number by the prime numbers method, we observe the multiple of prime factors of 156

here, 156 = 2 × 2 × 13 × 3 = 22 × 13 × 3

(iii) 3825

With the help of division of the number by the prime numbers method, we observe the multiple  of prime factors of 3825

here, 3825 = 3 × 3 × 5 × 5 × 17 = 32 × 52 × 17

(iv) 5005

With the help of division of the number by the prime numbers method, we observe the multiple of prime factors of 5005

here, 5005 = 5 × 7 × 11 × 13 = 5 × 7 × 11 × 13

(v) 7429

With the help of division of the number by the prime numbers method, we observe the multiple of prime factors of 7429

here, 7429 = 17 × 19 × 23 = 17 × 19 × 23

Question 23. Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:

(i) 13/3125 

(ii) 17/8 

(iii) 64/455 

(iv) 15/1600 

(v) 29/343 

(vi) 23/(2352) 

(vii) 129/(225775) 

(viii) 6/15 

(ix) 35/50 

(x) 77/210

Answer 23:

Note: If the denominator of a number has only factors of 2 and 5 in the form of 2m ×5n, then it shows terminating decimal expansion.

If the denominator of a number has factors not 2 and 5 then it shows a non-terminating decimal expansion.

(i) 13/3125

When we factorise the denominator, we observe,

3125 = 5 × 5 × 5 × 5 × 5 = 55

Then, the denominator has only 5 as its factor, 13/3125 has a terminating decimal expansion.

(ii) 17/8

 When we factorise the denominator, we observe,

8 = 2×2×2 = 23

then, the denominator has only 2 as its factor, 17/8 has a terminating decimal expansion.

(iii) 64/455

When we factorise the denominator, we observe,

455 = 5×7×13

then, the denominator is not in the form of 2m × 5n, thus 64/455 has a non-terminating decimal expansion.

(iv) 15/ 1600

When we factorise the denominator, we observe,

1600 = 26×52

then, the denominator is in the form of 2m × 5n, hence 15/1600 shows a terminating decimal expansion.

(v) 29/343

When we factorise the denominator, we observe,

343 = 7×7×7 = 73 

then, the denominator is not in the form of 2m × 5n thus 29/343 has a non-terminating decimal expansion.

(vi)23/(2352)

It is observed that the denominator is in the form of 2m × 5n.

So, 23/ (2352) shows a terminating decimal expansion.

(vii) 129/(225775)

As you can observe, the denominator is not in the form of 2m × 5n.

Therefore, 129/ (225775) shows a non-terminating decimal expansion.

(viii) 6/15

6/15 = ⅖

Then, the denominator has only 5 as its factor, thus, 6/15 shows a terminating decimal expansion.

(ix) 35/50

35/50 = 7/10

When we factorise the denominator, we observe

10 = 2 × 5

Then, the denominator is in the form of 2m × 5n thus, 35/50 shows a terminating decimal expansion.

(x) 77/210

77/210 

Then,

= (7× 11)/ (30 × 7) 

= 11/30

When we factorise the denominator, we observe,

30 = 2 × 3 × 5

As you can observe, the denominator is not in the form of 2m × 5n. Therefore, 77/210 has a non-terminating decimal expansion.

Question 24. Given that HCF (306, 657) = 9, find LCM (306, 657).

Answer 24:

You are aware about,

HCF × LCM = Multiplication of the two given numbers

So,

9 × LCM = 306 × 657

LCM = (306 × 657)/9 = 22338

Hence, LCM (306,657) = 22338

Question 25. Prove that the following are irrationals:

(i) 1/√2

(ii) 7√5

(iii) 6 + √2

Answer 25:

(i) 1/√2

Consider, 1/√2 is rational.

Then we can get co-prime x and y (y ≠ 0) such that 1/√2 = x/y

Rearranging, we observe,

√2 = y/x

Then, x and y are integers, so √2 is a rational number, which tells the fact that √2 is irrational.

Therefore, we can conclude that 1/√2 is irrational.

(ii) 7√5

Consider 7√5 is a rational number.

Then we can get co-prime a and b (b ≠ 0) such that 7√5 = x/y

Rearranging, we observe,

√5 = x/7y

Then, x and y are integers, thus, √5 is a rational number, which tells the fact that √5 is irrational.

Therefore, we can conclude that 7√5 is irrational.

(iii) 6 +√2

Consider 6 +√2 is a rational number.

Then we can get co-primes x and y (y ≠ 0) such that 6 +√2 = x/y⋅

Rearranging, we observe,

√2 = (x/y) – 6

Then, x and y are integers, thus (x/y) – 6 is a rational number and therefore, √2 is rational. This tells the fact that √2 is an irrational number.

Therefore, we can conclude that 6 +√2 is irrational

Question 26. Show that any positive odd integer is in the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.

Answer 26:

Consider any positive integer and b = 6. Hence, by Euclid’s algorithm, a = 6q + r, if some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5, due to 0≤r<6.

Now substituting the value of r, we observe,

If r = 0, then a = 6q

if, for r= 1, 2, 3, 4 and 5, we consider the value of a is 6q+1, 6q+2, 6q+3, 6q+4 and 6q+5, respectively.

for a = 6q, 6q+2, 6q+4, then a is even number and divided by 2. A positive integer can be even or not odd, hence, any positive odd integer is of the form of 6q+1, 6q+3 and 6q+5, where q is some integer.

Question 27. “The product of three consecutive positive integers is divisible by 6”. Is this statement true or false? Justify your answer.

Answer 27:

Yes, the statement is true “the product of three consecutive positive integers is divisible by 6”.

Justification:

we can consider the 3 consecutive numbers 2, 3, 4

(2 × 3 × 4)/6 = 24/6 = 4

Now, we can consider another 3 consecutive numbers 4, 5, 6

(4 × 5 × 6)/6 = 120/6 = 20

Now, we can consider another 3 consecutive numbers 7, 8, 9

(7 × 8 × 9)/6 = 504/6 = 84

Therefore, the statement “product of three consecutive positive integers is divisible by 6” is true.

Question 28. Check whether 6n can end with the digit 0 for any natural number n.

Answer 28:

If any number 6n ends with the digit zero (0), it will be divisible by 5, as we know any number with a unit place as 0 or 5 can be divided by 5.

Prime factorization of 6n = (2 × 3)n

Hence, the prime factorization of 6n doesn’t include the prime number 5.

Therefore, it is clear that for any natural number n, 6n is not divisible by 5 and thus it proves that 6n cannot end with the digit 0 for any natural number n.

 

Question 29. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

Answer 29: 

As per the definition of a composite number, we know that if the number is composite, it means it has factors one and itself. hence, for the expression;

7 × 11 × 13 + 13

Take 13 we observe as a common factor, we get,

=13(7×11×1+1) = 13(77+1) = 13×78 = 13×3×2×13

Hence, 7 × 11 × 13 + 13 is a composite number.

Now let’s take another number,

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5

Take 5 we observe as a common factor, we get,

=5(7×6×4×3×2×1+1) = 5(1008+1) = 5×1009

so, 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 is a composite number.

 

Question 30. Write whether every positive integer can be of the form 4q + 2, where q is an integer. Justify your answer.

Answer 30:

No, each positive integer cannot take the form 4q + 2, where q is the integer.

clarification:

All the related numbers of the form 4q + 2, where ‘q’ is the integer, are even numbers which we can not divide by ‘4’.

As given example,

if q=1,

4q+2 = 4(1) + 2= 6.

if q=2,

4q+2 = 4(2) + 2= 10

if q=0,

4q+2 = 4(0) + 2= 2 etc.

Therefore, any number which has the form 4q+2 will give only even numbers which can not be multiplied by 4.

Therefore, every positive integer cannot be written in the form 4q+2

 

Question 31. Find HCF and LCM of 96 and 404 and then verify that HCF × LCM = Product of the two given numbers.

Answer 31:

Prime factors of 96 = 2 × 2 × 2 × 2 × 2 × 3 = 25 × 3

Prime factors of 404 = 2 × 2 × 101

HCF = 2 × 2 = 4

LCM = 25 × 3 × 101 = 9696

HCF × LCM = 4 × 9696 = 38784

Multiplication of the given two numbers = 404 × 96 = 38784

Therefore, verified that LCM × HCF = Product of the given two numbers.

 

Question 32. Prove that √5 is irrational.

Answer 32: 

Consider that √5 is a rational number.

i.e. √5 = x/y (as well as, x and y are co-primes)

y√5= x

Squaring the both the sides, we observe,

(y√5)2 = x2

⇒5y2 = x2……………………………….. (1)

Therefore, x2 is divided by 5, so x is also divided by 5.

Consider, x = 5k, for some value of k and putting the value of x in equation (1), we observe,

5y2 = (5k)2

⇒y2 = 5k2

is divisible by 5 it means y is divisible by 5.

Clearly, x and y are not co-primes. Thus, our assumption about √5 is rational is incorrect.

Hence, √5 is an irrational number.

Question 33. A positive integer is of the form 3q + 1, q being a natural number. Can you write its square in any form other than 3m + 1, i.e., 3m or 3m + 2 for some integer m? Justify your answer.

Answer 33:

Answer is No.

clarification:

Let the positive integer 3q + 1, where q is the natural number.

(3q + 1)2 = 9q2 + 6q + 1

= 3(3q2 + 2q) + 1

= 3m + 1, (here, m is an integer which is equal to 3q2 + 2q.

hence (3q + 1)2 cannot be expressed in any other form apart from 3m + 1.

Question 34. Prove that 3 + 2√5 is irrational.

Answer 34:

Let us assume  3 + 2√5 to be a rational number.

if the co-primes x and y of the required rational number where (y ≠ 0) is such that:

3 + 2√5 = x/y

By rearranging, we get,

2√5 = (x/y) – 3

√5 = 1/2[(x/y) – 3]

Now x and y are integers, thus, 1/2[(x/y) – 3] is a rational number.

Therefore, √5 is also a rational number. But this confronts the fact that √5 is irrational.

Thus, our assumption that 3 + 2√5 is a rational number is wrong.

Hence, 3 + 2√5 is irrational.

Question 35. There is a circular path around a sports field. Shilpa takes 18 minutes to drive one round of the field, while Rahul takes 12 minutes for the same path. Suppose both of them start at the same point and at the same time and go in the same direction. After how many total minutes will both meet again at the starting point?

Answer 35: 

Since, if Shilpa and Rahul move in the same direction as well as at the same time, the method to get the time when they will  meet again at the initial point is LCM of 18 and 12.

hence, LCM(18,12) = 2×3×3×2×1=36

We conclude, Shilpa and Rahul will meet again at the starting point after 36 minutes.

 

Question 36. If n is an odd integer, then show that n2 – 1 is divisible by 8.

Answer 36:

We are aware that any odd positive integer n can be written in the form 4q + 1 or 4q + 3.

thus, according to the question,

if n = 4q + 1,

then we get n2 – 1 = (4q + 1)2 – 1 = 16q2 + 8q + 1 – 1 = 8q(2q + 1), divided by 8.

if n = 4q + 3,

then we get n2 – 1 = (4q + 3)2 – 1 = 16q2 + 24q + 9 – 1 = 8(2q2 + 3q + 1), divided by 8.

thus, from the above equations, it is clear that, if n is an odd positive integer

n2 – 1 is divisible by 8.

Hence Proved.

Question 37. An army contingent consisting of 616 members is to march behind an army band of 32 members in a parade. Both the groups are required to march in the same number of columns. What is the maximum number of columns for which they could march?

Answer 37:

We have  given,

Number of army contingent members=616

Number of army band members = 32

Suppose the two groups meet to march in the same column, we meet to get the highest common factor between the two groups. HCF(616, 32), which gives the maximum number of columns for which they can march.

By Using Euclid’s lemma to find their HCF, we get,

then, 616>32, thus,

616 = 32 × 19 + 8

then, 8 ≠ 0, thus, take 32 as new divisor, we have,

32 = 8 × 4 + 0

Now we have got the remainder as 0, thus, HCF (616, 32) = 8.

Thus, the maximum number of columns in which they can march is 8.

Question 38. Prove that if x and y are both odd positive integers, then x2 + y2 is even but not divisible by 4.

Answer 38:

Assume the two odd positive numbers x and y be 2r + 1 and 2s + 1, that is

 = x2 + y2 = (2r + 1)2 +(2s + 1)2

= 4r2 + 4r + 1 + 4s2 + 4s + 1

= 4r2 + 4s2 + 4r + 4s + 2

= 4 (r2 + s2 + r + s) + 2

we conclude , the sum of square is even the number is not divided by 4

Thus, if x and y are odd positive integers, then x2 + y2 is even but not divisible by four.

Hence proved

Question 39. The below real numbers have decimal expansions as given in the below options. In each case, decide whether they are rational or not. If they are rational numbers, and of the form, p/q what can you say about the prime factors of q?

(i) 43.123456789

(ii) 0.120120012000120000. . .

Answer 39:

(i) 43.123456789

then shows a terminating decimal expansion, it is the rational number of the form of p/q and q have factors of 2 and 5 only.

(ii) 0.120120012000120000. . .

then show a non-terminating and non-repeating decimal expansion, it is an irrational number.

Question 40. Show that the square of any odd integer is of the form 4q + 1, for some integer q.

Answer 40:

Assume a be any odd integer and b = 4.

As per Euclid’s algorithm,

a = 4m + r , heresome integer m ≥ 0

also r = 0,1,2,3 if 0 ≤ r < 4.

hence, we find here,

a = 4m or 4m + 1 or 4m + 2 or 4m + 3 So, a = 4m + 1 or 4m + 3

We already know that, a cannot be 4m or 4m + 2, as they are divided by 2.

(4m + 1)2 = 16m2 + 8m + 1

= 4(4m2 + 2m) + 1

= 4q + 1, here q is some integer and q = 4m2 + 2m.

(4m + 3)2 = 16m2 + 24m + 9

= 4(4m2 + 6m + 2) + 1

= 4q + 1, here q is some integer and q = 4m2 + 6m + 2

Hence, Square of any odd integer is of the form 4q + 1, for some integer q.

Hence proved.

 

Question 41. Use Euclid’s division method to show that the square of any positive integer is either in the form 3m or 3m + 1 for some integer m.

Answer 41:

Assume x be any positive integer and y = 3.

By Euclid’s division algorithm;

x =3q + r (for some integer q ≥ 0 and r = 0, 1, 2 as r ≥ 0 and r < 3)

therefore,

x = 3q, 3q + 1 and 3q + 2

As per in given question, if we take the square on both the sides, we get;

x2 = (3q)2 = 9q2 = 3.3q2

Let 3q2 = m

hence,

x2 = 3m ………………….(1)

x2 = (3q + 1)2

= (3q)2 + 12 + 2 × 3q × 1

= 9q2 + 1 + 6q

= 3(3q2 + 2q) + 1

Substituting 3q2+2q = m we observe,

x2 = 3m + 1 ……………………………. (2)

x2 = (3q + 2)2

= (3q)2 + 22 + 2 × 3q × 2

= 9q2 + 4 + 12q

= 3(3q2 + 4q + 1) + 1

Again, substituting 3q2 + 4q + 1 = m, we observe,

x2 = 3m + 1…………………………… (3)

Therefore, from eq. 1, 2 and 3, we conclude that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

 

Question 42. Show that the cube of any positive integer is of the form 4m, 4m + 1 or 4m + 3, for some integer m.

Answer 42:

Assume x be any positive integer and y = 4.

According to Euclid Division Lemma,

x = yq + r [0 ≤ r < y]

x = 3q + r [0 ≤ r < 4]

In the question, the required values for r,

Here,

r = 0, r = 1, r = 2 and, r = 3

if r = 0,

x = 4q + 0

x = 4q

Take cubes on LHS and RHS,

We observe,

x³ = (4q)³

Then

x³ = 4 (16q³)

we get,

x³ = 4m         [here m is an integer = 16q³]

if r = 1,

x = 4q + 1

Take cubes on LHS and RHS,

We observe,

x³ = (4q + 1)³

x³ = 64q³ + 1³ + 3 × 4q × 1 (4q + 1)

x³ = 64q³ + 1 + 48q² + 12q

x³ = 4 (16q³ + 12q² + 3q) + 1

x³ = 4m + 1        [here m is an integer = 16q³ + 12q² + 3q]

if r = 2,

x = 4q + 2

Take cubes on LHS and RHS,

We observe,

x³ = (4q + 2)³

x³ = 64q³ + 2³ + 3 × 4q × 2 (4q + 2)

x³ = 64q³ + 8 + 96q² + 48q

x³ = 4 (16q³ + 2 + 24q² + 12q)

x³ = 4m   [here m is an integer =16q³ + 2 + 24q² + 12q]

if r = 3,

x = 4q + 3

Take cubes on LHS and RHS,

We observe,

x³ = (4q + 3)³

x³ = 64q³ + 27 + 3 × 4q × 3 (4q + 3)

x³ = 64q³ + 24 + 3 + 144q² + 108q

x³ = 4 (16q³ + 36q² + 27q + 6) + 3

x³ = 4m + 3 [here m is an integer =16q³ + 36q² + 27q + 6]

Thus, the cube of any positive integer is in the form of 4m, 4m+1 or 4m+3.

 

Question 43. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

Answer 43:

Assume x be any positive integer and y = 3.

As per Euclid’s division algorithm, for,

x = 3q+r, here q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.

Therefore, putting the value of r, we observe,

x = 3q

or

x = 3q + 1

or

x = 3q + 2

Now, take the cube of all the three above expressions, we observe,

Case (i): if r = 0, then,

x2= (3q)3 = 27q3= 9(3q3)= 9m; where m = 3q3

Case (ii): if r = 1, then,

x3 = (3q+1)3 = (3q)3 +13+3×3q×1(3q+1) = 27q3+1+27q2+9q

Take 9 as common factor, we observe,

x3 = 9(3q3+3q2+q)+1

substituting = m, we observe,

substituting (3q3+3q2+q) = m, we observe ,

x3 = 9m+1

Case (iii): if r = 2, then,

x3 = (3q+2)3= (3q)3+23+3×3q×2(3q+2) = 27q3+54q2+36q+8

Take 9 as common factor, we observe,

x3=9(3q3+6q2+4q)+8

substituting (3q3+6q2+4q) = m, we observe ,

x3= 9m+8

Thus, from all the three cases explained above, it is proved that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

 

Question 44. Show that the square of any positive integer cannot be of the form 6m + 2 or 6m + 5 for any integer m.

Answer 44:

Assume the positive integer = a

As per Euclid’s division algorithm,

a = 6q + r, here 0 ≤ r < 6

a2 = (6q + r)2 = 36q2 + r2 + 12qr [∵(a+b)2 = a2 + 2ab + b2]

a2 = 6(6q2 + 2qr) + r2   …(i), here,0 ≤ r < 6

If r = 0, putting r = 0 in Eq.(i), we observe

a2 = 6 (6q2) = 6m, here, m = 6q2 is an integer.

If r = 1, pitting r = 1 in Eq.(i), we observe

a2 = 6 (6q2 + 2q) + 1 = 6m + 1, here, m = (6q2 + 2q) is an integer.

If r = 2, putting r = 2 in Eq(i), we observe

a2 = 6(6q2 + 4q) + 4 = 6m + 4, where, m = (6q2 + 4q) is an integer.

If r = 3, putting r = 3 in Eq.(i), we observe

a2 = 6(6q2 + 6q) + 9 = 6(6q2 + 6a) + 6 + 3

a2 = 6(6q2 + 6q + 1) + 3 = 6m + 3, here, m = (6q + 6q + 1) is integer.

If r = 4, putting r = 4 in Eq.(i) we observe

a2 = 6(6q2 + 8q) + 16

= 6(6q2 + 8q) + 12 + 4

⇒ a2 = 6(6q2 + 8q + 2) + 4 = 6m + 4, here, m = (6q2 + 8q + 2) is integer.

If r = 5, substituting r = 5 in Eq.(i), we observe

a2 = 6 (6q2 + 10q) + 25 = 6(6q2 + 10q) + 24 + 1

a2= 6(6q2 + 10q + 4) + 1 = 6m + 1, here, m = (6q2 + 10q + 4) is integer.

Therefore, the square of any positive integer cannot be of the form 6m + 2 or 6m + 5 for any integer m.

Hence proved

 

Question 45. Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

Answer 45:

Assume x be any positive integer and y = 3.

As per Euclid’s division algorithm, then,

x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.

Hence , x = 3q, 3q+1 and 3q+2

Now as per the question given, by squaring both the sides, we observe,

x2 = (3q)2 = 9q2 = 3 × 3q2

Let 3q2 = m

Hence , x2= 3m ……………………..(1)

x2 = (3q + 1)2 = (3q)2+12+2×3q×1 = 9q2 + 1 +6q = 3(3q2+2q) +1

putting, 3q2+2q = m, to get,

x2= 3m + 1 ……………………………. (2)

x2= (3q + 2)2 = (3q)2+22+2×3q×2 = 9q2 + 4 + 12q = 3 (3q2 + 4q + 1)+1

Again, putting, 3q2+4q+1 = m, to get,

x2= 3m + 1…………………………… (3)

Therefore, from equations 1, 2 and 3, we can say that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

 

Question 46. Show that the cube of a positive integer of the form 6q + r, q is an integer and r = 0, 1, 2, 3, 4, 5 is also of the form 6m + r.

Answer 46:

6q + r is the positive integer, where q is an integer and r = 0, 1, 2, 3, 4, 5

Now, the positive integers are of the form 6q, 6q+1, 6q+2, 6q+3, 6q+4 and 6q+5.

Take cube on L.H.S and R.H.S,

If 6q,

(6q)³ = 216 q³ = 6(36q)³ + 0

= 6m + 0, (here m is an integer = (36q)³)

If 6q+1,

(6q+1)³ = 216q³ + 108q2 + 18q + 1

= 6(36q³ + 18q2 + 3q) + 1

= 6m + 1, (here m is an integer = 36q³ + 18q2 + 3q)

If 6q+2,

(6q+2)³ = 216q³ + 216q2 + 72q + 8

= 6(36q³ + 36q2 + 12q + 1) +2

= 6m + 2, (here m is an integer = 36q³ + 36q2 + 12q + 1)

For 6q+3,

(6q+3)³ = 216q³ + 324q2 + 162q + 27

= 6(36q³ + 54q2 + 27q + 4) + 3

= 6m + 3, (here m is an integer = 36q³ + 54q2 + 27q + 4)

If 6q+4,

(6q+4)³ = 216q³ + 432q2 + 288q + 64

= 6(36q³ + 72q2 + 48q + 10) + 4

= 6m + 4, (here m is an integer = 36q³ + 72q2 + 48q + 10)

If 6q+5,

(6q+5)³ = 216q³ + 540q2 + 450q + 125

= 6(36q³ + 90q2 + 75q + 20) + 5

= 6m + 5, (here m is an integer = 36q³ + 90q2 + 75q + 20)

Thus, the cube of a positive integer of the form 6q + r, q is an integer and r = 0, 1, 2, 3, 4, 5 is also of the form 6m + r.

Question 47. There are 104 students in  Class X  and 96 students in  Class IX  in a school. In a house examination, students are to be evenly seated in parallel rows such that no two adjacent rows are of the same class.  

 (I) Find the maximum number of parallel rows of each class for the seating arrangement.

 (II) Also, find the number of students of Class IX and also of Class X in a row.

 (III) What is the objective of the school administration behind such an arrangement?

 Answer 47:

 104 = 23 × 13

 96 = 25 × 3

 HCF = 23 = 8

(I) Number of rows of students of Class X = 1048 = 13

 Number maximum of rows Class IX = 968 = 12

 Total number of rows = 13 + 12 = 25

 (II) No. of students of Class IX  in a row = 8

 No. of students of Class X  in a row = 8

 (III) the objective of our school administration behind such an arrangement is a fair and clean examination hence that no student can take help from another student.

 

Question 48. Dudhnath has two vessels, which contain 720 ml and 405 ml of the milk, respectively. Milk from these containers is poured into the glasses of equal capacity to their brim. Thus, Find the minimum number of glasses that can be filled.  

Answer 48: we have, 1st vessel = 720 ml and 2nd vessel = 405 ml

Thus,  we have to find the HCF of 720 and 405 to find the highest quantity of milk to be filled in a single glass.

we observe, , 

 405 = 9 x 9 × 5

 720 = 5 x 9 x 16

 HCF = 9 × 5 = 45 ml = Capacity of glass

 No. of glasses filled from 1st vessel = 720/45 = 16

 No. of glasses filled from 2nd vessel = 405/45 = 9

 so, Total number of the glasses is = 25

 

Question 49. Ravi, Seema, and Raju start preparing cards for all persons in an old age home. They complete one card in 20, 16 and 10 minutes. If all of them started together at the same time, after what time will they start preparing a new card together?  

Answer 49:

 To find the least time, they will start preparing a new card together, we have to find the LCM of 20, 16 and 10.

 20 = 2 x 2 × 5

 16 = 2 x 2 x 4

 10 = 2 × 5

 LCM = 16 × 5 = 16 × 5 = 80 minutes

Thus, They will start preparing a new card together after 80 minutes.

Question 50. Find out HCF of the numbers 134791, 6341 and 6339 by Euclid’s division algorithm.  

Answer 50: First, we have to find HCF of 6339 and 6341 by the method of Euclid’s division.

 6341 > 6339

 6341 = 6339 × 1 + 2

 6339 = 2 × 3169 + 1

 2 = 1 × 2 + 0

Thus, HCF of 6341 and 6339 is 1.

 here, we find the HCF of 134791 and 1

 134791 = 1 × 134791 + 0

thus, HCF of 134791 and 1 is 1.

 Therefore, the HCF of the given three numbers is 1.

 

Question 51. If x and y are two positive integers expressible in the terms of primes as x = p2q3 and y = p3q, what can you say about their LCM and HCF. Is LCM a multiple of HCF? Explain.  

 Answer 51: x = p2q3 and y = p3q

 LCM = p3q3

 HCF = p2q …..(i)

 Now, LCM = p3q3

 ⇒ LCM = pq2 (p2q)

 ⇒ LCM = pq2 (HCF)

 Yes, LCM is a multiple of HCF.

 Explanation:

 Let a = 12 = 22 × 3

 b = 18 = 2 × 32

 HCF = 2 × 3 = 6 …(ii)

 LCM = 22 × 32 = 36

 LCM = 6 × 6

 LCM = 6 (HCF) …[From (ii)]

 Here LCM is 6 times HCF.

Question 52. Show that one as well as only one out of n, (n + 1) and (n + 2) is divisible by 3, where n is any positive integer.  

Answer 52: here, n, n + 1, n + 2 be three consecutive positive integers.

 We observe that n is in the form 3q, 3q + 1, and 3q + 2.

 Case I. if n = 3q,

 then, n is divisible by 3,

 but n + 1 and n + 2 are  indivisible by 3.

 Case II. if n = 3q + 1,

 we get that n + 2 = (3q + 1) + 2

 = 3q + 3

 = 3(q + 1 ), (n + 2) is divisible by 3,

 but n and n + 1 are  indivisible by 3.

 Case III.

 if n = 3q + 2, in this case,

 n + 1 = (3q + 2) + 1

 = 3q + 3 = 3 (q + 1 ), (n + 1) is divisible by 3,

 but n and n + 2 are not divisible by 3.

 Thus, one and only one out of n, n + 1 and n + 2 is divisible by 3.

Question 53. Find out the HCF and LCM of 306 and 657 and verify that LCM × HCF = Product of the two numbers. 

Answer 53: 306 = 2 × 32 × 17

 657 = 32 × 73

 HCF = 32 = 9

 LCM = 2 × 32 × 17 × 73 = 22338

 L.H.S. = LCM × HCF = 22338 × 9 = 201042

 R.H.S. = Product of two numbers = 306 × 657 = 201042

 L.H.S. = R.H.S.

Question 54. Show that any positive odd integer is in form 41 + 1 or 4q + 3, where q is a positive integer. 

Answer 54: Let a is a positive odd integer

 By the Euclid’s Division algorithm:

 a = 4q + r …[here q, r are positive integers thus, 0 ≤ r < 4]

 a = 4q

 or 4q + 1

 or 4q + 2

 or 4q + 3

 But 4q and 4q + 2 are both even

 a is in the form 4q + 1 or 4q + 3.

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