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CBSE Previous Year Questions: Real Numbers Notes | Study Mathematics (Maths) Class 10 - Class 10

Document Description: CBSE Previous Year Questions: Real Numbers for Class 10 2022 is part of Real Numbers for Mathematics (Maths) Class 10 preparation. The notes and questions for CBSE Previous Year Questions: Real Numbers have been prepared according to the Class 10 exam syllabus. Information about CBSE Previous Year Questions: Real Numbers covers topics like Short Answer Type Questions, Long Answer Type Questions and CBSE Previous Year Questions: Real Numbers Example, for Class 10 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for CBSE Previous Year Questions: Real Numbers.

Introduction of CBSE Previous Year Questions: Real Numbers in English is available as part of our Mathematics (Maths) Class 10 for Class 10 & CBSE Previous Year Questions: Real Numbers in Hindi for Mathematics (Maths) Class 10 course. Download more important topics related with Real Numbers, notes, lectures and mock test series for Class 10 Exam by signing up for free. Class 10: CBSE Previous Year Questions: Real Numbers Notes | Study Mathematics (Maths) Class 10 - Class 10
Table of contents
Short Answer Type Questions
Long Answer Type Questions
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Short Answer Type Questions

Q.1. Find a rational number between √2 and √3.    [Delhi 2019]
Ans: Let p be rational number between √2 and √3
√2 = 1.41
√3 = 1.73
∴ √2 < p < √3
On squaring throughout, we have
2 < p2 < 3
The perfect squares which lie between 2 and 3 are 2 < 2.25 < 2.56 < 2.89 < 3.
Taking square root throughout
The rational numbers between √2 and √3 are 1.5, 1.6, 1.7 and more.

Q.2. Prove that √2 is an irrational number.    [Delhi 2019]
Ans:  Let us assume that √2 is a rational number.
So it can be expressed in the form p/q where p, q are co-prime integers and q ≠ 0
√2 = p/q
Here p and q are coprime numbers and q ≠ 0
√2 = p/q
On squaring both the sides we get,
=> 2 = (p/q)2
=> 2q2 = p2……………………………..(1)
p2/2 = q2 
So 2 divides p and p is a multiple of 2.
⇒ p = 2m
⇒ p2 = 4m2 ………………………………..(2)
From equations (1) and (2), we get,
2q2 = 4m2
⇒ q2 = 2m2
⇒ q2 is a multiple of 2
⇒ q is a multiple of 2
Hence, p, q have a common factor 2. This contradicts our assumption that they are co-primes. Therefore, p/q is not a rational number
√2 is an irrational number.

Q.3. Prove that 2 + 5 √3 is an irrational number, given that √3 is an irrational number.    [CBSE, Allahabad 2019]
Ans: Let 2 + 5 √3 be a rational number. (p and q are co-prime positive integers, q ≠ 0)
CBSE Previous Year Questions: Real Numbers Notes | Study Mathematics (Maths) Class 10 - Class 10
CBSE Previous Year Questions: Real Numbers Notes | Study Mathematics (Maths) Class 10 - Class 10
which is a contradiction because √3 is an irrational number and CBSE Previous Year Questions: Real Numbers Notes | Study Mathematics (Maths) Class 10 - Class 10 is  a rational.

Q.4. The HCF of two numbers a and b is 5 and their LCM is 200. Find the product ab.    [CBSE 20 19 (30/5/2)]
Ans: The formula that shows the relation between LCM and HCF of a and b is: LCM (a, b) = (a × b) / HCF (a, b)
HCF (a,  b) = 5, and LCM (a, b) = 200
LCM (a, b) = (a × b) / HCF (a, b)
200 = (a × b) / 5
a × b = 200 × 5
a × b = 1000
Therefore, the product of 'a' and 'b' is 1000.

Q.5. A number N when divided by 14 gives the remainder 5. What is the remainder when the same number is divided by 7?    [CBSE 2019 (30/2/1)]
Ans: Let the number be X
Since dividing the number gives remainder as 5, X can be written as 14 * Y + 5
where Y will be the quotient on dividing X by 14.
X = 14 * Y + 5
X = 2 * 7 * Y + 5
Let's divide both sides by 7 (since we want to know the remainder when X is divided by 7)
X/7 = (2 * 7 * Y)/7 + 5/7
X/7 = 2 * Y + 5/7
Since Y is a whole number (quotient), thus 2 * Y will also be a whole number and what remains is 5/7.
Thus, the remainder is 5 when X is divided by 7.

Q.6. Find after how many places the decimal form o f the numberCBSE Previous Year Questions: Real Numbers Notes | Study Mathematics (Maths) Class 10 - Class 10will terminate.    [CBSE 2019 (30/3/3)
Ans: CBSE Previous Year Questions: Real Numbers Notes | Study Mathematics (Maths) Class 10 - Class 10
So, the decimal form will end after four decimal places.

Q.7. Express 429 as product of its prime factors.    [CBSE 2019 (30/3/3)]
Ans:
CBSE Previous Year Questions: Real Numbers Notes | Study Mathematics (Maths) Class 10 - Class 10
∴ 429 = 3 x 13 x 11

Q.8. Find the HCF of 612 and 1314 using prime factorisation.    [CBSE 2019 (30/5/3)]
Ans:
CBSE Previous Year Questions: Real Numbers Notes | Study Mathematics (Maths) Class 10 - Class 10
612 = 2 x 2 x 3 x 3 x 17
1314 = 2 x 3 x 3 x 73
HCF = 2 x 3 x 3 = 18

Q.9. Find the HCF o f 1260 and 7344 using Euclid’s algorithm.    

[NCERT, CBSE 2019 (30/1/1)]
OR
Use Euclid’s division algorithm to find the HCF of 255 and 867.    

[CBSE 2019 (30/4/1)]
Ans:
Since 7344 > 1260 and from Euclid’s algorithm
a = bq + r
CBSE Previous Year Questions: Real Numbers Notes | Study Mathematics (Maths) Class 10 - Class 10
OR
Solution is similar as above.
Only values are changed

Q.10. Write a rational number between √2 and √3.    [CBSE 2019 (30/1/2)]
Ans:
A rational number between √2 and √3 is √2.89 = 1.7 = 17/10.

Q.11. If HCF of 65 and 117 is expressible in the form 65n - 117, find the value of n.    [CBSE 2019 (30/3/3)]
Ans:
HCF (65, 117) is given by
65 = 5 x 13
117 = 13 x 3 x 3
⇒ HCF =13
According to question,
65n - 117 = 13
65n = 130
n = 2

Q.12. Write the smallest number which is divisible by both 306 and 657.    

[CBSE 2019 (30/2/1)]
Ans: 
Here, to find the required smallest number we will find LCM of 306 and 657.
CBSE Previous Year Questions: Real Numbers Notes | Study Mathematics (Maths) Class 10 - Class 10
306 = 2 x 32 x 17
657 = 32 x 73
LCM - 2 X 32 x 17 x 73 = 22338

Q.13. Show that any positive odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5 where q is some integer. 
[NCERT]
OR
Show that any positive odd integer is of the form 6 m + 1 or 6 m + 3 or 6 m + 5, where m is some integer.    [CBSE 2019(30/5/2)]
Ans: Let a be any positive odd integer and b = 6. Then, by Euclid's algorithm, a = 6q + r, for some integer q > 0 and 0 < r < 6.
i.e., the possible remainders are 0, 1, 2, 3, 4, 5.
Thus, a can be of the form 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5, where q is some quotient.
Since a is odd integer, so a cannot be of the form 6q or x 6q + 2 or 6q + 4 (since they are even).
Thus, a is of the form 6q + 1 or 6q + 3 or 6q + 5, where q is some integer.
Hence, any odd positive integer is of the form 6q + 1 or 6q + 3 or 6q + 5, where q is some integer.
OR
Let a be any positive odd integer and b = 6. Then, by Euclid’s algorithm, a = 6m + r, for some integer m > 0 and 0 < r < 6.
i.e., the possible remainders are 0, 1, 2, 3, 4, 5.
Thus, a can be of the form 6m, or 6m + 1 or 6m + 2 or 6m + 3 or 6m + 4 or 6m + 5, where m is some quotient.
Since a is odd integer, so a cannot be of the form 6m or 6m + 2 or 6m + 4 (since they are even). Thus, a is of the form 6m + 1, 6m + 3 or 6m + 5, where m is some integer.
Hence, any odd positive integer is of the form 6m + 1 or 6m + 3 or 6m + 5, where m is some integer.

Q.14. Find HCF and LCM of 404 and 96 and verify that HCF x LCM = Product of the two given numbers.    [CBSE 2018]
Ans:
CBSE Previous Year Questions: Real Numbers Notes | Study Mathematics (Maths) Class 10 - Class 10
HCF of 404 and 9
404 = 2 x 2 x 101
96 = 2 x 2 x 2 x 2 x 2 x 3
Common factor = 2 x 2 = 4
∴ HCF = 4
CBSE Previous Year Questions: Real Numbers Notes | Study Mathematics (Maths) Class 10 - Class 10
LCM of 404 and 96
= 2 x 2 x 101 x 2x 2x 2 x 3 = 9696
Verification:
HCF x LCM = 4 x 9696 = 38784
Product of two numbers = 404 x 96 = 38784
Clearly, HCF x LCM = Product of two numbers
Hence verified

Q.15. Write whether CBSE Previous Year Questions: Real Numbers Notes | Study Mathematics (Maths) Class 10 - Class 10 on simplification gives an irrational or a rational number.    [CBSE 2018 (C)]
Ans: CBSE Previous Year Questions: Real Numbers Notes | Study Mathematics (Maths) Class 10 - Class 10
CBSE Previous Year Questions: Real Numbers Notes | Study Mathematics (Maths) Class 10 - Class 10(rational number)

Q.16. What is the HCF of smallest prime number and the smallest composite number?    [CBSE 2018]
Ans: 
Prime number: A prime number is a number that has exactly two factors i.e. it can be divided by only the number ‘1’ and itself.
The smallest prime number is = 2
Composite number: A composite number has more than two factors, which means apart from getting divided by number 1 and itself, it can also be divided by at least one integer or number. We don’t consider the number ‘1’ as a composite number.
The smallest composite number is = 4
2 is the factor of 4
∴ H.C.F of the smallest prime number and the smallest composite number is 2.

Q.17. Given that √2 is irrational, prove that (5 + 3 √2 ) is an irrational number    [CBSE 2018]
Ans: 
Let 5 + 3 √2 = p/q  be a rational number, where p and q, have no common factor other than 1 and q ≠ 0.
CBSE Previous Year Questions: Real Numbers Notes | Study Mathematics (Maths) Class 10 - Class 10
For any values of p and q(q ≠ 0),  RHS = CBSE Previous Year Questions: Real Numbers Notes | Study Mathematics (Maths) Class 10 - Class 10 is rational, but LHS = √2 is s irrational.
This contradicts the fact. So, our assumption is wrong.
∴ 5 + 3 √2  is an irrational number.

Q.18. Given that √3 is an irrational number, prove that (2 + √3 ) is an irrational number.    [CBSE 2018 (C)]
Ans: Let (2 + √3 is not an irrational.
∴ it can be written in the form a/b, where b ≠ 0 and a and b are coprime.
So CBSE Previous Year Questions: Real Numbers Notes | Study Mathematics (Maths) Class 10 - Class 10
Subtracting 2 from both sides
CBSE Previous Year Questions: Real Numbers Notes | Study Mathematics (Maths) Class 10 - Class 10
CBSE Previous Year Questions: Real Numbers Notes | Study Mathematics (Maths) Class 10 - Class 10This contradicts.
LHS is an irrational number because 3 is an irrational number (given) and RHS is an rational number.
So, LHS is equal to RHS not possible.
Hence, our assumption is wrong.
∴ (2 + √3 is an irrational number.

Q.19. What is the HCF of the smallest composite number and the smallest prime number?    [CBSE 2018]
Ans: The smallest prime number is 2.
The smallest composite number is 4.
To find the HCF of 2 and 4 let us write their prime factorization.
2 = 2
4 = 2 × 2
HCF (2, 4) = 2
Thus, the HCF of the smallest composite number and the smallest prime number is 2.

Long Answer Type Questions

Q.1. Prove that √5 is an irrational number. [CBSE 2019 (30/5/1)]
Ans: Let us assume, to the contrary, that Vs is a rational number.
Then, there exist co-prime positive integers a and b such that
CBSE Previous Year Questions: Real Numbers Notes | Study Mathematics (Maths) Class 10 - Class 10
So, a = √5 b
Squaring both sides, we have
a2 = 5b2   ......(i)
⇒ 5 divides a⇒ 5 divides a
So, we can write
a = 5c (where c is any integer)
Putting the value of a = 5c in (i), we have
25c2 = 5b2 => 5c2 = b
It means 5 divides b2 and so 5 divides b.
So, 5 is a common factor of both a and b which is a contradiction.
So, our assumption that √5  is a rational number is wrong.
Hence, we conclude that √5  is an irrational number.

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