Q.1. The decimal representation of 6/1250 will terminate after how many places of decimal?
1. Simplify the fraction to its lowest form.
2. Now we can either convert the lowest fraction into decimals or a quicker way to do this is to convert the denominator to a multiple of 10 so that it is easier to convert it into decimals.
3. So we multiply with and divide by 2^{4 }to convert the denominator to 10^{4. }
24 2
Therefore, this representation will terminate after 4 decimal places.
Q.3. Consider the number 12^{n}, where n is a natural number. Check whether there is any value of n ∈ N for which 12^{n} ends with the digit zero or 5.
∵ Any number ending with the digit zero is always divisible by 5.
∴ If 12^{n} ends with the digit zero, then it must be divisible by 5.
⇒ Prime factorisation of 12^{n} must contain a prime factor 5.
Now,
12 = 2 × 2 × 3 = 2^{2} × 3
⇒ (12)^{n} = (2^{2} × 3)^{n} = 2^{2}^{n} × 3^{n}
∴ The prime factorisation of 12^{n} does not contain the prime factor 5.
⇒ There is no value of n ∈ N such that 12^{n} ends with the digit zero.
Q.4. If ‘p’ is prime, prove that √p is irrational.
Let √p be rational in the simplest form a/b, where p is prime.
∴ a and b are integers having no common factor other than 1 and b ≠ 0.
⇒ Now, √p = a/b
⇒ Squaring both sides, we have
⇒ pb^{2 }= a^{2} ...(1)
⇒ Since pb^{2} is divisible by p, a^{2} is also divisible by p.
⇒ a is also divisible by p ...(2)
Let a = pc for some integer c.
⇒ Substituting a = pc in (1), we have
pb^{2} = (pc)^{2}
⇒ pb^{2 }= p^{2}c^{2}
⇒ b^{2} = pc^{2}
pc^{2} is divisible by p,
∴ b^{2} is divisible by p
⇒ b is divisible by p ...(3)
From (2) and (3),
⇒ p is a common factor of ‘a’ and ‘b’. But this contradicts our assumption that a and b are coprime.
∴ Our assumption that √p is rational is wrong. Thus, √p is irrational if p is prime.
Q.5. Find the HCF of 18 and 24 using prime factorisation.
Using factor tree method, we have:
∴ 18 = 2 × 3 × 3 = 2 × 3^{2}
24 = 2 × 2 × 2 × 3 = 2^{3} × 3
HCF = Product of common prime factors with lowest powers.
⇒ HCF (18, 24) = 3 × 2 = 6
Q.6. Find the LCM of 10, 30 and 120.
∴ 10 = 2 × 5 = 2^{1} × 5^{1}
30 = 2 × 3 × 5 = 2^{1} × 3^{1} × 5^{1}
120 = 2 × 2 × 2 × 3 × 5 = 2^{3} × 3^{1} × 5^{1}
LCM = Product of each prime factor with highest powers
⇒ LCM of 10, 30 and 120 = 2^{3} × 3 × 5 = 120.
Q.7. Express as a rational number in the simplest form.
Let x = 0.666 ..... ...(1)
∴ 10x = 0.666 ..... × 10
= 6.666 .......(2)
Subtracting (1) from (2), we have:
⇒ 10x  x = 6.666 .....  0.666 .....
⇒ 9x = 6
⇒ x = 6/9 = 2/3
Thus,= 2/3
Q.8. Express as a rational number in the simplest form.
Let x = 1.161616 ........(1)
∴ 100x = (1.161616 .....) × 100
⇒ 100x = 116.161616 ........(2)
Subtracting (1) from (2), we have:
100x  x = [116.161616 .....] – [1.161616 .....]
⇒ 99x = 115
⇒ x = 115/99
Thus, =115/99
Q.9. Use Euclid's division algorithm to find HCF of 870 and 225.
We have 870 = 3 × 225 + 195
⇒ 225 = 1 × 195 + 30
⇒ 195 = 6 × 30 + 15
⇒ 30 = 2 × 15 + 0
∴ HCF (870, 225) is 15.Note:
LCM of two numbers = Product of the numbers
Q.10. Find the LCM and HCF of 1296 and 5040 by prime factorisation method.
and
∴ 5040 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 7
= 2^{4} × 3^{2} × 5 × 7 and
⇒ 1296 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3
= 2^{4} × 3^{4}
∴ LCM = Product of each prime factor with highest powers
= 2^{4} × 3^{4} × 5 × 7
= 16 × 81 × 5 × 7 = 45360
HCF = Product of common prime factors with lowest powers
= 2^{4} × 3^{2}
= 16 × 9 = 144
Q.11. Prove that √3 is irrational.
Let √3 be rational in the simplest form of P/q.
i.e., p and q are integers having no common factor other than 1 and q ≠ 0.
Now, √3 = p/q
Squaring both sides, we have
⇒ (√3)^{2} = (p/q)^{2}
⇒ 3 = p^{2}/q^{2}
⇒ 3q^{2} = p^{2 }............(1)
Since 3q^{2} is divisible by 3
∴ p^{2} is also divisible by 3
⇒ p is divisible by 3 ..........(2)
Let p = 3c for some integer ‘c’.
Substituting p = 3c in (1), we have:
⇒ 3q^{2 } = (3c)^{2}
⇒ 3q^{2} = 9c^{2}
⇒ q^{2} = 3c^{2}
3c^{2 }is divisible by 3
∴ q^{2} is divisible by 3
⇒ q is divisible by 3 ...(3)
From (2) and (3)
3 is a common factor of ‘p’ and ‘q’. But this contradicts our assumption that p and q are having no common factor other than 1.
∴ Our assumption that √3 is rational is wrong.
Thus, √3 is an irrational.
Q.12. Show that 3√2 is irrational.
Let 3√2 be a rational number
∴ p/q = 3√2 where p and q are prime to each other and q ≠ 0.
∴ p/3q = √2 ...(1)
Since, p is integer and 3q is also integer (3q≠ 0).
∴ p/3q is a rational number.
From (1), √2 is an integer
But this contradicts the fact that √2 is irrational.
⇒ 3√2 is irrational.
Q.13. Show that 2  √3 is an irrational number.
Let 2  √3 is rational.
∴ It can be expressed as p/q where p and q are integers (prime to each other) such that q ≠ 0.
∴ 2  √3 = p/q
⇒ ...(1)∵ p is an integer}
∴ q is an integer}
⇒ p/q is a rational number.
∴ is a rational number. ...(2)
From (1) and (2), √3 is a rational number. This contradicts the fact that √3 is an irrational number.
∴ Our assumption that (2 √3) is a rational number is not correct. Thus, (2  √3) is irrational.
Q.14. Using Euclid’s division algorithm, find the HCF of 56, 88 and 404.
Using Euclid’s division algorithm to 88 and 56, we have
⇒ 3q^{2} = (3c)^{2}
⇒ 3q^{2} = (3c)^{2}
⇒ 88 = 56 × 1 + 32
⇒ 56 = 32 × 1 + 24
⇒ 32 = 24 × 1 + 8
⇒ 24 = 8 × 3 + 0
∴ HCF of 88 and 56 is 8
Again, applying Euclid’s division algorithm to 8 and 404, we have:
⇒ 404 = 8 × 504 + 4
⇒ 8 = 4 × 2 + 0
∴ HCF of 404 and 8 is 4
Thus, HCF of 88, 56 and 404 is 4.
Q.15. Express in the decimal form.
We have
[multiplying and dividing by 5]
Q.16. Express in the p/q form.
Let x =
or x = 5.4178178178 .....
∴ 10x = 54.178178178 ..... ...(1)
Also 1000 (10x) = 54178.178178178 .....
⇒ 10000x = 54178.178178178 ..... ...(2)
Subtracting (1) from (2), we have:
⇒ 10000x  10x
= [54178.178178 .....]  [54.178178 .....]
⇒ 9990x = 54124⇒
Thus,
Q.17. State whether is a rational number or not.
= 1.23333 ..... is a nonterminating repeating decimal.
∴ is a rational number.
3/4 is in the form of p/q, where q ≠ 0 [Here 4 ≠ 0]
∴ 3/4 is a rational number.
Since the sum of two rational numbers is a rational number.
Therefore, is a rational number.
Q.18. The LCM of two numbers is 45 times their HCF. If one of the numbers is 225 and sum of their LCM and HCF is 1150, find the other number.
One of the numbers = 225
Let the other number = x
Also LCM = 45 (HCF) ...(1)
And LCM + HCF =1150
⇒ (45 HCF) + HCF = 1150
⇒ 46 HCF = 1150
⇒ HCF = 1150/46 = 25
From (1),
LCM = 45 × 25
∴ LCM × HCF = (45 × 25) × 25
Now, LCM × HCF = Product of the numbers
∴ x × 225 = (45 × 25) × 25
⇒= 125
Thus, the required number is 125.
Q.19. Three different containers contain 496 litres, 403 litres and 713 litres of a mixture. What is the capacity of the biggest container that can measure all the different quantities exactly?
For the capacity of the biggest container, we have to find the HCF.
HCF: By Long Division method
First find the HCF of two numbers, 496 and 403
The HCF of 496 and 403 = 31
Now find the HCF of 31 and 713
HCF of 713 and 31 is 31
So, the maximum capacity is 31 liters.
Q.20. Prove that 3+√2 is an irrational number.
Let 3 + √2 is a rational number.
∴ 3 + √2 = a/b such that ‘a’ and ‘b’ are coprime integers and b ≠ 0.
We have
Since a and b are integers,
∴ is rational.
⇒ √2 is a rational. This contradicts the fact that √2 is irrational.
∴ Our assumption that 3 + √2 is rational is not correct.
⇒ (3 + √2) is an irrational number.
126 videos477 docs105 tests

1. What are real numbers? 
2. How do you differentiate between rational and irrational numbers? 
3. Can real numbers be categorized into different subsets? 
4. What is the significance of real numbers in mathematics? 
5. How are real numbers used in reallife applications? 

Explore Courses for Class 10 exam
