1. Real Numbers

SEBA Class 10 Maths Chapter 1 Real Numbers

Chapter 1. Real Numbers

Class 10 Maths Chapter 1. Real Numbers Multiple Choice Questions and Solutions :

Question : Find the correct answer : is [SEBA 2015]

(a) an integer (b) a rational number (c) a prime number (d) an irrational number

Solution : (d) an irrational number . [ Since , is an irrational number. ]

Question : In the following real numbers, which one is non-terminating repeating decimal expansion ? [ SEBA 2015]

(a) (b) (c) (d)

Solution: (c) [ is non-terminating repeating decimal expansion . ]

Question : If LCM (91,26) = 182 , then HCF(91,26) = [SEBA 2016]

(a) 13 (b) 26 (c) 7 (d) 9

Solution : (a) 13

[ We have, ]

Question : According to Euclid’s Division lemma, given two positive integers a and b , there exist unique integers q and r such that - [ SEBA 2016]

(a)

(b)

(c)

(d)

Solution: (c)

Question : The smallest number by which should be multiplied so as to get rational number is : [ SEBA 2017]

(a) (b) (c) (d) 3

Solution: (c)

[ We have, ]

Question : The number of decimal places after which the decimal expansion of the rational number will terminate is : [SEBA 2017]

(a) 3 (b) 4 (c) 1 (d) 5

Solution: (b) 4

[ We have, ]

Question : Which one of the following is a rational number ? [SEBA 2018]

(a) (b) (c) (d)

Solution: (d)

[ We have, ]

Question : The decimal expansion of the rational number will terminate after

(A) one decimal place (B) two decimal places

Solution: (B) two decimal places .

[ We have, ]

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

(A) always irrational (B) always rational (C) rational or irrational (D) one

Solution: (A) always irrational .

[Example : is a irrational; number.]

Question : If the HCF of 65 and 117 is expressible in the form , then the value of is :

(A) 4 (B) 2 (C) 1 (D) 3

Solution: (B) 2

[ We have, 65 = 5 × 13 and 117 = 3 × 3 × 13

HCF (70 , 125) = 13

A/Q ,

]

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

(A) 13 (B) 65 (C) 875 (D) 1750

Solution: (A) 13 .

[Since, 70 – 5 = 65 = 5 × 13 and 125 – 8 = 117 = 3 × 3 × 13

HCF (70 , 125) = 13 ]

Question : If two positive integers and are written as and are prime numbers, then HCF is :

(A) (B) (C) (D)

Solution: (B) .

[Product of the smallest power of each common prime factor in the numbers .

Here, and

]

Question : If two positive integers and can be expressed as and are prime numbers, then HCF is :

(a) (b) (c) (d)

Solution : (a)

[ Product of the smallest power of each common prime factor in the numbers .

Here, and

HCF ]

Question : 120 can be expressed as a product of its prime factors as : [CBSE 2020 basic]

(a)

(b)

(c)

(d)

Solution : (d)

[ We have , ]

Question : The H.C.F of 8 , 9 , 25 is :

(a) 8 (b) 9 (c) 25 (d) 1

Solution: (d) 1

[ We have , ; ;

HCF (8 , 9 , 25) = 1 ]

Question : Which of the following is an irrational number ? [SEBA 2020]

(a) 0.142857142857142857……………

(b) (c) π (d)

Solution: (c) π [ π is an irrational number .]

Question : Which of the following is not irrational ?

(a)

(b)

(c)

(d)

Solution: (c)

[ We have, (c) is not irrational ]

Question : Which one of the following is a non-terminating repeating decimal ? [SEBA 2019]

(a) (b) (c) (d)

Solution: (c)

[ We have, is a non-terminating repeating decimal. ]

Question : Given three statement such as :

(i) The sum or difference of a rational and an irrational number is irrational .

(ii) The product and quotient of a non-zero rational and irrational number is irrational .

(iii) The product of the two numbers is not equal to the product of their HCF and LCM .

(iv) The LCM is equal to the product of the greatest power of each common prime factor in the numbers.

(a) (i) , (ii) and (iii) are correct .

(b) (i) , (ii) and (iv) are correct .

(d) (ii) , (iii) and (iv) are correct .

Solution: (d) (ii) , (iii) and (iv) are correct .

Question : The decimal expansion of the rational number will terminate after

(A) one decimal place (B) two decimal places

Solution: (D) four decimal places .

[ We have, ]

Question : Which number is not divisible by 11 ?

(a) 253 (b) 1771 (c) 286 (d) 91

Solution: (d) 196

[ We have, 253 = 11×23 ; 1771 = 7 × 11 × 23 , 286 = 2 × 11 × 13 ; 91 = 7 × 13 ]

Question : The largest number which divides 60 and 75 , leaving remainders 8 and 10 respectively,is

(a) 260 (b) 75 (c) 65 (d) 13

Solution: (a) 260

[ We have, 60 – 8= 52 = 2 × 2 × 13 ; 75 - 10 = 65 = 5 × 13

LCM (52 , 65) = 2 × 2 × 5 × 13 = 260 ]

Question : If LCM (91 , 26) = 182 , then HCF (91 , 26) is :[SEBA 2016]

(a) 13 (b) 26 (c) 7 (d) 9

Solution: (a) 13

[ We have, ]

Question : For some integers , every even integer is of the form

(A) (B) (C) (D)

Solution: (C) .

Question : The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is

(a) 840 (b) 2520 (c) 10 (d) 420

Solution: (b) 2520

[ We have , LCM (1, 2 , 3 , 4 , ………., 10)

]

Question : When a number is divided by 7 , its remainder is always :

(a) greater than 7

(b) at least 7

(d) at most 7

Solution: (a) greater than 7 .

Question : If HCF (16,y) = 8 and LCM(16, y) = 48 , then the value of y is :

(a) 24 (b) 16 (c) 8 (d) 48

Solution: (a) 24

[ A/Q , HCF (16 ,y) × LCM (16 , y) = 16 × y

⇒ 16 × y = 8 × 48

⇒ y = 8 × 3 = 24 ]

Question : Find the least number of 3 digits , that will gives us remainder of 9 when divided by 2 and 5 respectively .

(a) 121 (b) 141 (c) 110 (d) 109

Solution: (d) 109

[ We have , 109 = 10 × 10 + 9 ]

Question : The decimal expansion of the rational number will terminate after :

(a) One decimal place

(b) Two decimal places

(d) Four decimal places

Solution: (c) Three decimal places

[ We have, ]

Question : The ratio between the LCM and HCF of 5 , 15 , 20 is :

(a) 9 : 1 (b) 4 : 3 (c) 11 : 1 (d) 12 : 1

Solution: (d) 12 : 1

[ We have , 5 = 1 × 5 ; 15 = 3 × 5 ; 20 =

HCF (5 , 15 , 20) = 5

LCM (5 , 15 , 20) =

So, ]

Question : HCF of 52× 42 and 35 × 65 is :

(a) 52 × 35 (b) 5 × 33

Solution: (d) 7 × 13

[ We have, 52 × 42 = 2 × 2 × 13 × 2 × 3 × 7 =

and 35 × 65 = 5 × 7 × 5 × 13 = 52 × 7 × 13

HCF ( 52 × 42 , 35 × 65) = 7 × 13 ]

Question : What type of decimal form will have ?

(a) Terminating

(b) Non-terminating repeating

(d) None

Solution: (a) (ii) Non-terminating repeating .

[ The prime factorisation of the denominator is not of the form , where and are non-negative integers. ]

Question : The decimal expansion of the rational number will terminate after :

(a) one decimal place

(b) Two decimal places

(d) four decimal places .

Solution: (c) Three decimal places.

[ We have, ]

Question : If is written in the form , where are co-primes and is of the form , then values of and are :

(a) 0 , 3 (b) 3 , 0 (c) 2 , 3 (d) 2 , 2

Solution : (a) 0 , 3

[ We have, ]

Question : Which of the following a rational number lying between and ?

(a) 2.110111101111110………………

(b) 1.515785515…………….

(c)

(d) 1.14287514……………

Solution : (b) 1.515785515…………….

[We have, = 1.41421........ and = 1.73205..........

So, 1.515785515……………. is a rational number lying between and ]

Question : Which of the following rational numbers have terminating decimal ?

(a) (b) (c) (d)

Solution: (d)

[ Since, is a terminating decimal expansion . ]

Question : Given 36 , 72 and 120 are three number, then the HCF is ........... . .

(a) 10 (b) 15 (c) 18 (d) 12

Solution : (a) (i) 12 .

[ We have, ; and

]

Question : Given 29 and 53 are two number, then HCF (29 , 53) will be :

(a) 29 × 53

(b) 53

(d) 1

Solution :

Question : What is the LCM of 36 , 72 and 120 ?

(a) 620 (b) 720 (c) 360 (d) 260

Solution : (c) 360

[ We have, ; and

]

Question : The ratio of the two number is 3 : 7 . If the HCF is 13 , then the number are :

(a) 29 , 91

(b) 39 , 72

(d) 29 , 82

Solution: (c) 39 , 92

[ let the two number are and .

and

HCF

and ]

Question : If the LCM of and 18 is 36 and the HCF of and 18 is 2 , then

(a) 2 (b) 3 (c) 4 (d) 1

Solution : (c) 4

[Here, ,

We have,

Filled in the blanks :

Question : The HCF of two number is 27 and their LCM is 162 , if one of the number is 54 , then the other number is ......................................... .

Solution: 81

[ The other number ]

Question : If product of two numbers is 2366 and their LCM is 26 , then their HCF is ...............................

Solution: 91

[ We have, ]

Question : The HCF and LCM of two numbers are 33 and 264 respectively , When the first number is completely divided by 2 and the quotient is 33 , then other number is...................................... .

Solution: 132

[ We have ,

Other number ]

Question : If the prime factorisation of a natural number is , then number is.........................

Solution: 8232

[ We have , ]

Question : is ( irrational / a rational number ) .

Solution: a rational number .

[ We have , is a rational ]

Question : If is expressed in the form , then values of is .................................. .

Solution: 4

[ We have , ]

Answers following the question :

Q1. Find the LCM and HCF of 6 and 20 by prime factorization method .

Solution: We have, and

and .

Q2. Express the number 0.104 in the form of rational number .

Solution: We have ,

is the form of rational number .

Q3. Given that HCF(306 , 657) = 9 , find LCM (306 , 657) .

Solution: We have ,

Q4. The LCM of two number is 182 and their HCF is 13 . If one of the numbers is 26 , find the other . [CBSE 2020 standard]

Solution: We have , LCM × HCF one number × other number

The other number

Q5. Find HCF of 1001 and 385 .

Solution: We have,

and

Q6. Find the LCM of the two digit smallest prime and smallest odd composite natural number .

Solution: The two digit smallest prime number is 11 and the smallest odd composite number is 15 .

So,

Q7. Decompose 32760 into prime factors .

Solution: We have ,

Q8. What is the HCF of smallest prime number and the smallest composite number ? [CBSE 2018]

Solution: The smallest prime number is 2 and the smallest composite number is 4 .

So,

Q9. Find a rational number between and

Solution: We have , and

Thus , the rational number between and is .

Q10. Find one irrational number between and .

Solution: We have , and

Thus , the one irrational number between and is 1.21021002100021………….. .

Q11. The decimal expansion of the rational number will terminate after how many places of decimals ?

Solution: We have ,

The decimal expansion of the rational number will terminate after 4 places of decimals .

Q12. After how many decimal places will the rational number terminate ?

Solution: We have,

The decimal expansion of the rational number will terminate after 4 places of decimals .

Q13. If HCF(336 , 54) = 6 , find LCM (336 , 54) . [CBSE 2019]

Solution: We have ,

Class 10 Real Numbers SECTION = B

Q1. Find the LCM and HCF of 120 and 144 by using Fundamental theorem of Arithmetic . [CBSE2012]

Solution: We have ,

and

Q2. Check whether can end with the digit 0 for any natural number .

Solution: We have,

Prime factors of are only 2 and 3 .

So, the prime factors of does not contain , where and are positive integers .

Therefore, does not end with the digit 0 .

Q3. Prove that is irrational . [SEBA 2016]

Solution: let us assume , to the contrary that is rational .

We can find co-prime and ( ) such that

Since, 2 , and are integers ,is rational and so, is rational .

But this contradicts the fact that is irrational . So , is irrational .

Q4. Find the largest number that will divides 398 , 436 and 542 leaving remainders 7 , 11 and 15 respectively .

Solution: We have ,

398 - 7 = 391 = 17 × 23

436 - 11 = 425 = 5 × 5 × 17

542 - 15 = 527 = 17 × 31

HCF(391 , 425 , 527) = 17

Therefore, 17 is the largest number that will divide given numbers.

Q5. Express 5050 as product of its prime factors . Is it unique ?

Solution: We have,

It is not unique .

Q6. Check whether can end with the digit 0 for any natural number .

Solution: Since,

Prime factors of are only 2 and 3 .

So, the prime factors of does not contain , where and are positive integers .

Therefore, does not end with the digit 0 .

Q7. Find the HCF and LCM of 12 , 15 and 21 ,using the prime factorization method .

Solution: We have ,

and

Q8. Prove that is irrational .

Solution: let us assume , to the contrary that is rational .

We can find co-prime and ( ) such that,

Since and are integers , is rational and so, is rational .

But this contradicts the fact that is irrational . So, is irrational .

Q9. Use Euclid’s division algorithm to find the HCF of 867 and 255 .

Solution: We have , 867 > 255

We apply the division lemma ,

Q10. Use Euclid’s division algorithm to find the HCF of 4052 and 12576 .

Solution: We have , 12576 > 4052

We apply the division lemma ,

Thus , the HCF of 12576 and 4052 is 4 .

Q11. Using Euclid’s Algorithm , find the HCF of 2048 and 960 . [CBSE 2019]

Solution: We have , 2048 > 960

We apply the division lemma ,

Thus , the HCF of 2048 and 960 is 64 .

Q12. 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 ?

Solution: The maximum number of column HCF (616 , 32)

Using Euclid’s division algorithm , we have

The HCF (616 , 32) is 8 .

Thus , the maximum number of column is 8 .

Q13. Write the smallest number which is divisible by both 306 and 657 . [CBSE 2019]

Solution: We have , 306 = 2 × 3 × 3 × 17

and 657 = 3 × 3 × 73

HCF of 306 and 657 is 9 .

Class 10 Real Numbers SECTION = C

Q1. Find LCM and HCF of 6 , 72 and 120 by prime factorization method . Is HCF × LCM of three numbers equal to the product of the three numbers ?

Solution: We have,

LCM (6 , 72 , 120)

HCF (6 , 72 , 120)

LCM (6 , 72 , 120 ) × HCF (6 , 72 , 120 )

So, the product of three numbers is not equal to the product of their HCF and LCM .

Q2. Find the LCM and HCF of 144 , 112 and 418 by prime factorization .

Solution : We have,

LCM (144 , 112 , 418)

HCF (144 , 112 , 418)

Q3. Prove that is irrational .

Solution: let us assume , to the contrary that is rational .

So, we can find integers and () such that , [ and are co-prime]

Therefore , is divisible by 3 and so is also divisible by 3 .

let , , for some integer .

From and , we get

Therefore , is divisible by 3 and so is also divisible by 3 .

So, and have at least 3 as a common factor .

But this contradicts the fact that and are co-prime . So, is irrational .

Q4. Prove that is an irrational .

Solution: let us assume , to the contrary , that is rational .

We can find co-prime and ( ) such that

–

Since , 2 , and are integers , is rational and so, is rational . But this contradicts the fact that is irrational . So , is irrational .

Q5. Given that is an irrational , prove that is an irrational number . [CBSE 2018]

Solution: let us assume , to the contrary that is rational .

We can find co-prime and ( ) such that,

Since 3 , and are integers , is rational and so, is rational .

Given is irrational . So, is irrational .

Q6. Prove that is irrational and hence show that is also irrational .

Solution: let us assume , to the contrary that is rational .

So, we can find integers and () such that [ and are co-prime]

Therefore , is divisible by 5 and so is also divisible by 5 .

let , , for some integer .

From and , we get

Therefore , is divisible by 5 and so is also divisible by 5 .

So, and have at least 5 as a common factor .

But this contradicts the fact that and are co-prime . So, is irrational .

Since is an irrational number . Therefore , is also irrational .

Q7. Find HCF and LCM of 404 and 96 and verify that HCF × LCM Product of the two given numbers . [SEBA2019 , CBSE 2018]

Solution: We have, and

HCF (96 , 404)

LCM (96 , 404)

HCF (96 , 404) × LCM (96 , 404)

Verified.

Q8. Rajesh has two vessels containing 720 ml and 405 ml of milk respectively . Milk from these containers is poured into glasses of equal capacity to their brim . Find the minimum number of glasses that can be filled .

Solution: We have ,

Therefore, the number of glasses is 45 .

Q9. Example : Find the HCF of 455 and 42 , use Euclid’s division algorithm .

Solution : We have ,

Apply Euclid’s division algorithm ,

Therefore , the HCF of 455 and 42 is 7 .

Q10. Find the HCF of 455 and 42 , Use Euclid’s division algorithm .

Solution : We have , 455 > 42

Apply Euclid’s division algorithm ,

455 = 42 × 10 + 35

42 = 35 × 1 + 7

35 = 7 × 5 + 0

Therefore , the HCF of 455 and 42 is 7 .

Q11. Prove that is irrational .

Solution: let us assume , to the contrary that is rational .

We can find co-prime and ( ) such that ,

Since, 2 , and are integers , is rational and so, is rational .

But this contradicts the fact that is irrational . So, is irrational .

Q12. Use Euclid’s division lemma to show that the square of any positive integer is either of the form or for some integer .

Solution: let, be any positive integers and .

We apply the Euclid’s division algorithm ,

, then , or .

If then ,

where

If then ,

, where

If then ,

, where

Thus, the square of any positive integer is either of the form or for some integer .

Q13. Show that any positive odd integer is of the form or or,where is some integer.

Solution: let , be any positive odd integers and .

Using Euclid’s algorithm ,

, ,

So, 0 , 1 , 2 , 3 , 4 or 5 .

If then

is an even numbers .

If then is an odd numbers .

If then is an even numbers .

If then is an odd numbers .

If then is an even numbers .

If then is an odd numbers .

Therefore, any positive odd integer is of the form , or .