Real Numbers Class 10 Worksheet Maths Chapter 1

True and False

Q1: State whether the given statement is true or false :
(i) The product of two rationals is always rational
Ans: True

(ii) The product of two irrationals is an irrational
Ans: False 

(iii) The product of a rational and an irrational is irrational
Ans: True

(iv) The sum of two rationals is always rational
Ans: True

(v) The sum of two irrationals is an irrational.
Ans: False 

Very Short Questions

Q2: Classify the following numbers as rational or irrational:
(i) 3.1416
Ans: Rational

(ii) 3.142857
Ans: Rational

(iii) 2.040040004......
Ans: Irrational

(iv) 3.121221222...
Ans: Irrational

(v) 3√3
Ans: Irrational

Q.3. Find the prime factorization of 1152
Ans: 1152 = 2⁷ × 32

Short Questions

Q4: Show that the product of two numbers 60 and 84 is equal to the product of their HCF and LCM
Ans: Prime factorisation of 60 = 2 × 2 × 3 × 5
Prime factorisation of 84 = 2 × 2 × 3 × 7
Hence, LCM of 60 , 84 = 2 × 2 × 3 × 5 × 7 = 420
And HCF of 60 , 84 = 2 × 2 × 3 = 12
Now, LCM × HCF = 420 × 12 = 5040
Also,60 × 84 = 5040
i.e., HCF × LCM = Product of the two numbers

Q5: If p and q are two prime number, then what is their LCM?
Ans: It is given that p and q are two prime numbers; we have to find their LCM.
We know that the factors of any prime number are 1 and the prime number itself.
For example, let p = 2 and q = 3
Thus, the factors are as follows
p = 2 × 1
And
q = 3 × 1
Now, the LCM of 2 and 3 is 2 × 3 = 6.
Thus the HCF of p and q is p × q

Q6: Given that HCF (306, 657) = 9, find LCM (306, 657)..
Ans: HCF of two numbers 306 and 657 is 9
To find: LCM of number
We know that,
LCM × HCF = first number x second number
LCM × 9 = 306 × 657
LCM=306 × 657/9
= 22338

Q7: The H.C.F. and L.C.M. of two numbers are 12 and 240 respectively. If one of these numbers is 48; find the other numbers.
Ans: Since, the product of two numbers
= Their H.C.F. × Their L.C.M.
⇒ One no. × other no. = H.C.F. × L.C.M.
⇒ Other no. =12×240 / 48 = 60.

Long Questions

Q8: In a school, the duration of a period in junior section is 40 minutes and in senior section is 1 hour: If the first bell for each section ring at 9:00 a.m., when will the two bells ring together again? 
Ans:  1 hour = 60 minutes
40 = 2 × 2 × 2 × 5 = 2³ × 5
60 = 2 × 2 × 3 × 5 = 2² × 3 × 5
∴ LCM (40, 60) = 2³ × 3 × 5 = 120
120 minutes = 2 hours
Hence, the two bells will ring together again at 9:00 + 2:00 = 11:00 a.m.

Q9: The HCF of 408 and 1032 is expressible in the form 1032 m – 2040. Find the value of m. Also, find the LCM of 408 and 1032.
Ans: Let us find HCF of 408 and 1032.
Here, 1032 > 408
∴ 1032 = 2 × 408 + 216
408 = 1 × 216 + 192
216 = 1 × 192 + 24
192 = 8 × 24 + 0
Thus, HCF of 408 and 1032 is 24.
Now, HCF (408, 1032)
i.e., 24 = 1032 × m – 2040
⇒ 1032 × m = 24 + 2040
⇒ 1032 × m = 2064
⇒ m = 20641032 = 2
408 = 2³ × 3 × 17
1032 = 2³ × 3 × 43 ,
∴ LCM of 408 and 1032 = 2³ × 3 × 17 × 43 = 17544.

Q10: Prove that
(i) √2 is irrational number
(ii) √3 is irrational number
Similarly √5, √7, √11…... are irrational numbers.
Ans: (i) Let us assume, to the contrary, that 2 is rational.
So, we can find integers r and s (≠ 0) such that .√2= r/s
Suppose r and s not having a common factor other than 1. Then, we divide by the common factor to get ,√2=a/b
where a and b are coprime.
So, b √2= a.
Squaring on both sides and rearranging, we get 2b² = a². Therefore, 2 divides a². Now, by
Theorem it following that 2 divides a.
So, we can write a = 2c for some integer c.
Substituting for a, we get 2b² = 4c², that is,
b² = 2c².
This means that 2 divides b², and so 2 divides b (again using Theorem with p = 2).
Therefore, a and b have at least 2 as a common factor.
But this contradicts the fact that a and b have no common factors other than 1.
This contradiction has arisen because of our incorrect assumption that √2 is rational.
So, we conclude that √2 is irrational.

(ii) Let us assume, to contrary, that √3 is rational. That is, we can find integers a and b (≠ 0) such that √3=a/b
Suppose a and b not having a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.
So, b √3= a .
Squaring on both sides, and rearranging, we get 3b² = a².
Therefore, a² is divisible by 3, and by Theorem, it follows that a is also divisible by 3.
So, we can write a = 3c for some integer c.
Substituting for a, we get 3b² = 9c², that is,b² = 3c².
This means that b² is divisible by 3, and so b is
also divisible by 3 (using Theorem with p = 3).
Therefore, a and b have at least 3 as a common factor.
But this contradicts the fact that a and b are coprime.
This contradicts the fact that a and b are coprime.
This contradiction has arisen because of our incorrect assumption that √3 is rational.
So, we conclude that √3 is irrational.