1 : Real numbers
Real Numbers
1. REAL NUMBERS
SECTION = A
Q1. 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)
Q2. 120 can be expressed as a product of its prime factors as : [CBSE 2020 basic]
(a) 
(b) 
(c) 
(d) 
Solution : (d) 
[ We have , 
]
Q3. 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 ]
Q4. Which of the following is an irrational number ? [SEBA 2020]
(a) 0.142857142857142857……………
(b) 
(c) π
(d) 
Solution: (c) π
Q5. Which of the following is not irrational ?
(a) 
(b) 
(c) 
(d) 
Solution: (c)
[ We have, (c) 
is not irrational ]
Q6. Which one of the following is a non-terminating repeating decimal ? [SEBA 2019]
(a) 
(b) 
(c) 
(d) 
Solution: (c)
[ We have, 
]
Q7. 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 .
(c) (ii) , (ii) and (iii) are not correct .
(d) (ii) , (iii) and (iv) are correct .
Solution: (d) (ii) , (iii) and (iv) are correct .
Q8. In the following real numbers, which one is non-terminating repeating decimal expansion ? [ SEBA 2015]
(a) 
(b) 
(c) 
(d) 
Solution: (c)
Q9. 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 ]
Q10. 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 ]
Q11. 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, 
]
Q12. 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)
]
Q13. When a number is divided by 7 , its remainder is always :
(a) greater than 7
(b) at least 7
(c) less than 7
(d) at most 7
Solution: (a) greater than 7 .
Q14. 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 ]
Q15. 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 ]
Q16.The decimal expansion of the rational number 
will terminate after :
(a) One decimal place
(b) Two decimal places
(c) Three decimal places
(d) Four decimal places
Solution: (c) Three decimal places
[ We have, 
]
Q17. 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, 
]
Q18. HCF of 52× 42 and 35 × 65 is :
(a) 52 × 35
(b) 5 × 33
(c) 65 × 32
(d) 7 × 13
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 ]
Q19. Which one of the following is a rational number ? [SEBA 2018]
(a) 
(b) 
(c) 
(d) 
Solution: (d)
[ We have, 
]
Q20. 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, 
]
Q21. 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, 
]
Filled in the blanks :
Q1. 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 
]
Q2. If product of two numbers is 2366 and their LCM is 26 , then their HCF is ...............................
Solution: 91
[ We have, 
]
Q3. 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 
]
Q4. If the prime factorisation of a natural number is 
, then number is.........................
Solution: 8232
[ We have ,
]
Q5. 
is 
( irrational / a rational number ) .
Solution: a rational number .
[ We have , 
is a rational ]
Q6. 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 ,
Section = II
Case study based questions are compulsory . Attempt any four sub parts of each question .
Each subpart carries 1 marks.
Q1. let 
be a rational number such that the prime factorization of 
is not of the form 
, where 
are non-negative integers , then has a decimal expansion which is non-terminating repeating (recurring) .
(a) What type of decimal form 
will have ?
(i) Terminating
(ii) Non-terminating repeating
(iii) Non-terminating non-repeating
(iv) None
(b) The decimal expansion of the rational number will terminate after :
(i) one decimal place
(ii) Two decimal places
(iii) Three decimal places
(iv) four decimal places .
(c) If 
is written in the form 
, where 
are co-primes and 
is of the form 
, then values of 
and are :
(i) 0 , 3
(ii) 3 , 0
(iii) 2 , 3
(iv) 2 , 2
(d) Which of the following a rational number lying between 
and ?
(i) 2.110111101111110………………
(ii) 1.515785515…………….
(iii) 
(iv) 1.14287514……………
(e) Which of the following rational numbers have terminating decimal ?
(i) 
(ii) 
(iii) 
(iv) 
Solution: (a) (ii) Non-terminating repeating .
[ The prime factorisation of the denominator is not of the form 
, where 
and 
are non-negative integers. ]
(b) (iii) Three decimal places.
[ We have, 
]
(c) (i) 0 , 3
[ We have, 
]
(d) (ii) 1.515785515…………….
(e) (iv)
Q2. 36 ducks , 72 hens and 120 goats have to be taken across a river. There is only one boat which will have to make many trips in order to do so . The lazy boatman has his own conditions for transporting them . He insists that he will take the same number of animals in every trip and they have to be of the same kind. He will naturally like to take the largest possible number each time.
(a) Can you tell how many animals went in each trip ?
(i) 12
(ii) 13
(iii) 14
(iv) 15
(b) Given 36 , 72 and 120 are three number, then the HCF and LCM is ............................. .
(i) less than 63 × 72 × 12
(ii) greater than 63 × 72 × 12
(iii) equal to 63 × 72 × 12
(iv) not equal to 63 × 72 × 12
(c) Given 29 and 53 are two number, then HCF (29 , 53) will be :
(i) 29 × 53
(ii) 53
(iii) 29
(iv) 1
(d) What is the LCM of 36 , 72 and 120 ?
(i) 620
(ii) 720
(iii) 360
(iv) 260
(e) The ratio of the two number is 3 : 7 . If the HCF is 13 , then the number are :
(i) 29 , 91
(ii) 39 , 72
(iii) 39 , 92
(iv) 29 , 82
Solution: (a) (i) 12 .
(b) (iv) not equal to 63 × 72 × 12 .
(c) (iv) 1 .
(d) (iii) 360 .
(e) (iii) 39 , 92
[ let the two number are 
and 
.
and 
HCF 
and 
]
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 .
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 
.
Posted 5 years ago