Class 10 Exam  >  Class 10 Notes  >  Mathematics (Maths) Class 10  >  CBSE Previous Year Questions: Real Numbers

Class 10 Maths Chapter 1 Previous Year Questions - Real Numbers

Previous Year Questions 2024

Q1: The smallest irrational number by which √20 should be multipled so as to get a rational number, is:    (2024)
(a) √20
(b) 2
(c) 5
(d) √5

Class 10 Maths Chapter 1 Previous Year Questions - Real Numbers  View Answer

Ans: (d)
(a) Class 10 Maths Chapter 1 Previous Year Questions - Real Numbers
But Class 10 Maths Chapter 1 Previous Year Questions - Real Numbers is not the smallest among all options. 
(b) Class 10 Maths Chapter 1 Previous Year Questions - Real Numbers is irrational 
(c) Class 10 Maths Chapter 1 Previous Year Questions - Real Numbers is irrational 
(d) Class 10 Maths Chapter 1 Previous Year Questions - Real Numbers
Class 10 Maths Chapter 1 Previous Year Questions - Real Numbers
Hence, option (d) is correct.

Q2: The LCM of two prime numbers p and q (p > q) is 221. Then the value of 3p – q is:     (2024)
(a) 4
(b) 28
(c) 38
(d) 48

Class 10 Maths Chapter 1 Previous Year Questions - Real Numbers  View Answer

Ans: (c)
The numbers p and q are prime numbers, 
∴ HCF (p, q) = 1 
Here, LCM(p, q) = 221 
∴ As, p > q
p = 17, q = 13 
(As p × q = 221) 
Now, 3p – q = 3 × 17 – 13 
= 51 – 13
= 38

Q3: A pair of irrational numbers whose product is a rational number is     (2024)
(a) (√16, √4)
(b) (√5, √2)
(c) (√3, √27)
(d) (√36, √2)

Class 10 Maths Chapter 1 Previous Year Questions - Real Numbers  View Answer

Ans:(c)
Here √3 and √27 both are irrational numbers.
The product of Class 10 Maths Chapter 1 Previous Year Questions - Real Numbers
Class 10 Maths Chapter 1 Previous Year Questions - Real Numbers
∴ 9 is a rational number.

Q4: Given HCF (2520, 6600) = 40, LCM (2520, 6600) = 252 × k, then the value of k is:     (2024)
(a) 1650
(b) 1600
(c) 165
(d) 1625

Class 10 Maths Chapter 1 Previous Year Questions - Real Numbers  View Answer

Ans:(a)
HCF(2520, 6600) = 40
LCM(2520, 6600) = 252 × k 
∴ HCF × LCM = Ist No. × IInd No. 
∴ 40 × 252 × k = 2520 × 6600
Class 10 Maths Chapter 1 Previous Year Questions - Real Numbers
⇒ k = 1650

Q5: Teaching Mathematics through activities is a powerful approach that enhances students' understanding and engagement. Keeping this in mind, Ms. Mukta planned a prime number game for class 5 students. She announced the number 2 in her class and asked the first student to multiply it by a prime number and then pass it to the second student. The second student also multiplied it by a prime number and passed it to the third student. In this way by multiplying by a prime number, the last student got 173250.
Now, Mukta asked some questions as given below to the students:        (2024)
(A) What is the least prime number used by students?
(B) How many students are in the class?
OR
What is the highest prime number used by students?
(C) Which prime number has been used maximum times?

Class 10 Maths Chapter 1 Previous Year Questions - Real Numbers  View Answer

Ans:
(A)
Class 10 Maths Chapter 1 Previous Year Questions - Real Numbers
So least prime no. used by students = 3(because 2 is announced by the teacher, so the least number used by the students is 3)
(B)As the last student got  173250 = 2 × 3 × 3 × 5 × 5 × 5 × 7 × 11
there are 7 factors other than 2, which is announced by teacher. So, Number of student = 7
OR
Highest prime number used by student = 11
(C)Prime number 5 is used maximum times i.e., 3 times.

Previous Year Questions 2023

Q6: 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  

Class 10 Maths Chapter 1 Previous Year Questions - Real Numbers  View Answer
Ans:(a)

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

Q7: Find the least number which when divided by 12, 16, and 24 leaves the remainder 7 in each case.  (2023)

Class 10 Maths Chapter 1 Previous Year Questions - Real Numbers  View Answer

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

Q8: Two numbers are in the ratio 2 : 3 and their LCM is 180. What is the HCF of these numbers? (2023)

Class 10 Maths Chapter 1 Previous Year Questions - Real Numbers  View Answer

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

Q9: Prove that √3 is an irrational number. (2023)

Class 10 Maths Chapter 1 Previous Year Questions - Real Numbers  View Answer

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

Q10: 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

Class 10 Maths Chapter 1 Previous Year Questions - Real Numbers  View Answer

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 22 = 4x  11
Here, we will take a = 1 and b = 44 ; a = 4 and b = 11.
We do not take ab = 2  x 22 because 2 and 22 are not co-prime 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).

Q11: If 'n' is any natural number, then (12)n cannot end with the digit    (2022)
(a) 2
(b) 4
(c) 8
(d) 0

Class 10 Maths Chapter 1 Previous Year Questions - Real Numbers  View Answer

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

Q12: The number 385 can be expressed as the product of prime factors as   (2022)
(a) 5 x 11 x 13
(b) 5 x  7 x 11
(c) 5 x 7 x 13
(d) 5  x  11 x 17

Class 10 Maths Chapter 1 Previous Year Questions - Real Numbers  View Answer

Ans: (b)
Sol:We have,

Class 10 Maths Chapter 1 Previous Year Questions - Real Numbers
∴ Prime factorisation of 385 = 5 x 7 x 11

Previous Year Questions 2021

Q13: Explain why 2 x 3 x 5 + 5 and 5 x 7 x 11 + 7 x 5 are composite numbers.   (2021)

Class 10 Maths Chapter 1 Previous Year Questions - Real Numbers  View Answer

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

Q14: 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

Class 10 Maths Chapter 1 Previous Year Questions - Real Numbers  View Answer

Ans: (c)
Sol:We have, 
12 = 2 x 2 x 3 = 2x 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

Q15: The LCM of two numbers is 182 and their HCF is 13. If one of the numbers is 26. find the other.   (2020)

Class 10 Maths Chapter 1 Previous Year Questions - Real Numbers  View Answer

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.

Q16: Given that HCF (135, 225) = 45, find the LCM (135, 225). (CBSE 2020)

Class 10 Maths Chapter 1 Previous Year Questions - Real Numbers  View Answer

Ans: We know that
LCM × HCF = Product of two numbers
∴ LCM (135, 225) = Product of 135 and 225 / HCF(135, 225)
= 135 x 225 / 45
= 675
So, LCM (135, 225) = 675

Previous Year Questions 2019

Q17: If HCF (336, 54) = 6. find LCM (336, 54).   (2019)

Class 10 Maths Chapter 1 Previous Year Questions - Real Numbers  View Answer

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

Q18: The HCF of two numbers a and b is 5 and their LCM is 200. Find the product of ab.  (2019)

Class 10 Maths Chapter 1 Previous Year Questions - Real Numbers  View Answer

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

Q19: 1f HCF of 65 and 117 is expressible in the form 65n - 117, then find the value of n.   (2019)

Class 10 Maths Chapter 1 Previous Year Questions - Real Numbers  View Answer

Ans: 2

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

Q20: Find the HCF of 612 and 1314 using prime factorization.   (2019)

Class 10 Maths Chapter 1 Previous Year Questions - Real Numbers  View Answer

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

Q21: Prove that √5 is an irrational number.   (2019)

Class 10 Maths Chapter 1 Previous Year Questions - Real Numbers  View Answer

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
Class 10 Maths Chapter 1 Previous Year Questions - Real Numbers
⇒ 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.

Q22: Prove that √2 is an irrational number.   (2019)

Class 10 Maths Chapter 1 Previous Year Questions - Real Numbers  View Answer

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
Class 10 Maths Chapter 1 Previous Year Questions - Real Numbers ------------(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.

Q23: Prove that 2 + 5√3 is an irrational number given that √3 is an irrational number.   (2019)

Class 10 Maths Chapter 1 Previous Year Questions - Real Numbers  View Answer

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.
Class 10 Maths Chapter 1 Previous Year Questions - Real Numbers
3 is a rational number.

[ ∵ a, b are integers, so Class 10 Maths Chapter 1 Previous Year Questions - Real Numbers 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.

Q24: Write the smallest number which is divisible by both 306 and 657. (CBSE 2019)

Class 10 Maths Chapter 1 Previous Year Questions - Real Numbers  View Answer

Ans:Given numbers are 306 and 657. 
The smallest number divisible by 306 and 657 = LCM(306, 657) 
Prime factors of 306 = 2 × 3 × 3 × 17 
Prime factors of 657 = 3 × 3 × 73 
LCM of (306, 657) = 2 × 3 × 3 × 17 × 73 
= 22338 
Hence, the smallest number divisible by 306 and 657 is 22,338.

The document Class 10 Maths Chapter 1 Previous Year Questions - Real Numbers is a part of the Class 10 Course 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 mathematics?
Ans.Real numbers include all the numbers on the number line. This includes rational numbers (like integers and fractions) and irrational numbers (like the square root of 2 or pi). They can be positive, negative, or zero.
2. How can I represent real numbers on a number line?
Ans.To represent real numbers on a number line, you draw a horizontal line and mark points at equal intervals. Each point corresponds to a real number, with positive numbers to the right of zero and negative numbers to the left.
3. What is the difference between rational and irrational numbers?
Ans.Rational numbers can be expressed as the quotient of two integers (like 1/2 or 3), while irrational numbers cannot be expressed in this form (like √2 or π). Rational numbers have terminating or repeating decimal expansions, while irrational numbers have non-repeating, non-terminating decimals.
4. How do I simplify a fraction involving real numbers?
Ans.To simplify a fraction involving real numbers, you divide both the numerator and the denominator by their greatest common divisor (GCD). For example, to simplify 4/8, divide both by 4 to get 1/2.
5. What are some common properties of real numbers?
Ans.Common properties of real numbers include the commutative property (a + b = b + a), associative property ((a + b) + c = a + (b + c)), and distributive property (a(b + c) = ab + ac). These properties help in performing operations with real numbers.
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