CBSE Previous Year Questions: Real Numbers

# Class 10 Maths Chapter 1 Previous Year Questions - Real Numbers

 Table of contents Previous Year Questions 2023 Previous Year Questions 2022 Previous Year Questions 2021 Previous Year Questions 2020 Previous Year Questions 2019

## Previous Year Questions 2023

Q1: The ratio of HCF to LCM of the least composite number and the least prime number is  (2023)
(a) 1 : 2
(b) 2 : 1
(c) 1 : 1
(d) 1 : 3   [2023, 1 Mark]
Ans: (a)

Sol: Least composite number = 4
Least prime number = 2
∴ HCF = 2, LCM = 4
∴ Required ratio = 2/4
i.e. 1 : 2

Q2: Find the least number which when divided by 12, 16 and 24 leaves remainder 7 in each case.  (2023)

Ans: 55

Given, least number which when divided by 12, 16 and 24 leaves remainder 7 in each case
∴  Least number = LCM( 12, 16, 24) + 7
= 48 + 7
= 55

Q3: Two numbers are in the ratio 2 : 3 and their LCM is 180. What is the HCF of these numbers?  (2023)

Ans: 30

Let the two numbers be 2x and 3x
LCM of 2x and 3x = 6x, HCF(2x, 3x) = x
Now, 6x = 180
⇒ x = 180/6
x = 30

Q4: Prove that √3 is an irrational number.   (2023)

Ans:  Let us assume that √3 is a rational number.

Then √3 = a/b; where a and b ( ≠ 0) are co-prime positive integers.
Squaring on both sides, we get
3 = a2/b2 ⇒ a2 = 3b
⇒ 3 divides a2
⇒ 3 divides a    _________(i)
= a = 3c, where c is an integer
Again, squaring on both sides, we get
a2 = 9c2
3b2 = 9c2 b2 = 3c2 ⇒ 3 divides b2
⇒ 3 divides b     _________(ii)
From (i) and (ii), we get 3 divides both a and b.
⇒ a and b are not co- prime integers.
This contradicts the fact that a and b are co-primes.
Hence,  √3 is an irrational number.

## Previous Year Questions 2022

Q5: Two positive numbers have their HCF as 12 and their product as 6336. The number of pairs possible for the numbers, is   (2022)
(a) 2
(b) 3
(c) 4
(d) 1

Ans: (a)
Sol: Given, HCF = 12
Let two numbers be 12a and 12b
So. 12a x 12b = 6336 ⇒ ab = 44
We can write 44 as product of two numbers in these ways:
ab = 1 x 44 = 2 x 2 2 = 4x  11
Here, we will take a = 1 and b = 44 ; a = 4 and b = 11.
We do not take ab = 2 x22 because 2 and 22 are not co-priine to each other.

For a = 1 and b = 44, 1st no. = 12a = 12, 2nd no. = 12b = 528
For a = 4 and b = 11, 1st no. = 12a = 48, 2nd no. = 12b = 132
Hence, we get two pairs of numbers, (12, 528) and (48, 132).

Q6: If 'n' is any natural number, then (12)n cannot end with the digit    (2022)

(a) 2
(b) 4
(c) 8
(d) 0

Ans: (d)
Sol: for n = 1, 2,  3, 4...
(12)n cannot end with 0.

Q7: The number 385 can be expressed as the product of prime factors as   (2022)
(a) 5 x 11 x 1 3
(b) 5 x  7 x 11
(c) 5 x 7 x 1 3
(d) 5  x  11 x 17

Ans: (b)
Sol: We have,

∴ Prime factorisation of 385 = 5 x 7 x 11

## Previous Year Questions 2021

Q8: Explain why 2 x 3 x 5 + 5 and 5 x 7 x 11 + 7 x 5 are composite numbers.   (2021)

Ans: We have, 2 x 3 x 5 + 5 and 5 x 7 x 11 + 7 x 5.
We can write these numbers as:
2 x 3 x 5+5 = 5(2 x 3 + 1)
=1 x 5 x 7
and 5 x 7 x 11 + 7 x 5 = 5 x 7(11 + 1)
= 5 x 7 x 12 = 1 x 5 x 7 x 12
Since, on simplifying. we find that both the numbers have more than two factors. So. these are composite numbers.

## Previous Year Questions 2020

Q9: The HCF and the LCM of 12, 21 and 15 respectively, are   (2020)
(a) 3, 140
(b) 12, 420
(c) 3, 420
(d) 420, 3

Ans: (c)
Sol: We have, 12 =  2 x 2 x 3 = 22x 3
21= 3 x 7
15 = 3 x 5
∴ HCF (12, 21, 15) = 3
and LCM (12, 21 ,15 ) = 22 x  3 x  5  x  7
= 420

Q10: The LCM of two numbers is 182 and their HCF is 13. If one of the numbers is 26. find the other.   (2020)

Ans: 91
Let the other number be x
As, HCF (a, b) x  LCM [a, b) = a  x  b
⇒ 13 x  182= 26x
⇒ x = 13 x 182 / 26
= 91
Hence, other number is 91.

## Previous Year Questions 2019

Q11: If HCF (336, 54) = 6. find LCM (336, 54).   (2019)

Ans: 3024
Since. HCF (a, b) x LCM (a, b) = a x b
∴ HCF (336, 54) x  LCM {336, 54) = 336 x 54
⇒ 6 x LCM(336, 54) = 18144
⇒ LCM (336, 54) = 18144 / 6
= 3024

Q12: The HCF of two numbers a and b is 5 and their LCM is 200. Find the product of ab.   (2019)

Ans: 1000
We know that HCF (a, b) x  LCM (a, b)=a x b
⇒ 5 x 200 = ab
⇒  ab = 1000

Q13: 1f HCF of 65 and 117 is expressible in the form 65n-117, then find the value of n.    (2019)

Ans:  2

Since,  HCF (65 ,117) = 13
Given HCF ( 65, 117 ) = 65n - 117
13 = 65n - 117
⇒  65n = 13 +117
⇒  n = 2

Q14: Find the HCF of 612 and 1314 using prime factorisation.    (2019)

Ans: 18

Prime factorisation of 612 and 1314 are
612 = 2 x 2 x 3 x 3 x 17
1314 = 2 x 3 x 3 x 73
∴ HCF (612, 1314) = 2 x 3 x 3
= 18

Q15: Prove that √5 is an irrational number.    (2019)

Ans: Let us assume that 5 is a rational number.
Then 5 = a/b where a and b (≠ 0} are co-prime integers,
if Squaring on both sides, we get

⇒ 5 divides a2
⇒ 5 divides a  ----------(i)
⇒ a = 5c, where c is an integer
Again, squaring on both sides, we get
a2 = 25c2
⇒ 5b2 = 25c2  ⇒  b2 = 5c2
⇒  5 divides b2   ----------(ii)
⇒ 5 divides b
From (i) and {ii), we get 5 divides both a and b.
⇒  a and b are not co-prime integers.
Hence, our supposition is wrong.
Thus, 5 is an irrational number.

Q16: Prove that √2 is an irrational number.    (2019)

Ans: Let us assume 2 be a rational number.
Then, 2 = p/q where p, q (q ≠ 0) are integers and co-prime. ;
On squaring both sides. we get
------------(i)
⇒ 2 divides p2
⇒ 2 divides p -----------(ii)
So, p = 2a, where a is some integer.
Again squaring on both sides, we get
p2 = 4a2
⇒ 2q2 = 4a2               (using (i))
⇒ q2 = 2a2
⇒  2 divides q2
⇒ 2 divides q       -----------(iii)
From (ii) and (iii), we get
2 divides both p and q.
∴ p and q are not co-prime integers.
Hence, our assumption is wrong.
Thus 2 is an irrational number.

Q17: Prove that 2 + 5√3 is an irrational number given that √3 is an irrational number.    (2019)

Ans: Suppose 2 + 53 is a rational number.
We can find two integers a, b (b ≠ 0) such that
2 + 53 = a/b, where a and b are co -prime integers.

3 is a rational number.

[ ∵ a, b are integers, so  is a rational number]
But this contradicts the fact that 3 is an irrational number.
Hence, our assumption is wrong.
Thus, 2 + 53 is an irrational number.

The document Class 10 Maths Chapter 1 Previous Year Questions - Real Numbers is a part of the Class 10 Course Mathematics (Maths) Class 10.
All you need of Class 10 at this link: Class 10

## Mathematics (Maths) Class 10

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## FAQs on Class 10 Maths Chapter 1 Previous Year Questions - Real Numbers

 1. What are real numbers in Class 9 Mathematics?
Ans. Real numbers in Class 9 Mathematics include all rational and irrational numbers, which can be represented on a number line. They can be positive, negative, or zero, and can also be expressed in decimal form.
 2. How are real numbers classified in Class 9 Mathematics?
Ans. Real numbers in Class 9 Mathematics are classified as rational numbers and irrational numbers. Rational numbers can be expressed as a ratio of two integers, while irrational numbers cannot be expressed as a simple fraction.
 3. How do you represent real numbers on a number line in Class 9 Mathematics?
Ans. To represent real numbers on a number line in Class 9 Mathematics, plot each number at its corresponding position on the line. Rational numbers are represented as points on the line, while irrational numbers are represented as points between rational numbers.
 4. What is the importance of real numbers in Class 9 Mathematics?
Ans. Real numbers are essential in Class 9 Mathematics as they form the foundation for various mathematical concepts and calculations. They are used in algebra, geometry, and other branches of mathematics to solve problems and make accurate calculations.
 5. How do you perform operations on real numbers in Class 9 Mathematics?
Ans. In Class 9 Mathematics, operations such as addition, subtraction, multiplication, and division can be performed on real numbers following the rules of arithmetic. These operations help in solving equations, simplifying expressions, and solving mathematical problems.

## Mathematics (Maths) Class 10

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