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Class 10th
CHAPTER 1
Real Numbers (SIQ 2020-21)
S3 = 3 ( S2 –S1).
1. Algorithm : An algorithm means a series of well defined step which gives a procedure for solving a
type of problem.
2. Lemma : A lemma is a proven statement used for proving another statement.
3. Fundamental Theorem of Arithmetic : Every composite number can be expressed (factorised) as
a product of primes and this factorisation is unique apart from the order in which the prime factors occur.
4. If p is prime number and p divides a2, then p divides a , where a is a positive integer.
5. If x be any rational number whose decimal expansion terminates, then we can express x in the form p/q , where p and q are co-prime and the prime factorisation of q is of the form 2ⁿ ×5m, where n and m are non-negative integers.
6. Let x =p/q be a rational number such that the prime factorisation of q is not of the form 2ⁿ×5m , where n
and m are non-negative integers, then x has a decimal expansion which terminates.
7. Let x=p/q, be a rational number such that the prime factorisation of q is not of the form 2ⁿ×5m, where
n and m are non-negative integers, then x has a decimal expansion which is non-terminating repeating
(recurring).
8. For any two positive integers p and q , HCF of p q, and LCM of p q, = p q .
9. For any three positive integers p q, and r ,
Test Practice Time
Q.1. The sum of exponents of prime factors in the prime-factorisation of 196 is
(a) 3 (b) 4 (c) 5 (d) 2
Sol.
Prime factors of 196,
196 = 4 × 49
= 2²× 7²
The sum of exponents of prime factor is 2+2 = 4.
Thus (b) is correct option.
Q.2 The total number of factors of prime number is
(a) 1 (b) 0 (c) 2 (d) 3
Sol.
There are only two factors (1 and number itself) of any prime number.
Thus (c) is correct option.
Q.3 The HCF and the LCM of 12, 21, 15 respectively are
(a) 3, 140 (b) 12, 420. (c) 3, 420
(d) 420, 3
Sol.
We have
12 = 2×2×3
21 = 3 × 7
15 = 3 × 5
HCF(12, 21, 15) = 3
LCM (12, 21, 15) = 2×2×3×5×7 = 420
Thus (c) is correct option.
Q.4 The least number that is divisible by all the numbers from 1 to 10 (both inclusive)
Sol.
Factor of 1 to 10 numbers
1 = 1
2 = 1 × 2
3 = 1 × 3
4 = 1 × 2 × 2
5 = 1 × 5
6 = 1 × 2 × 3
7 = 1 × 7
8 = 1 × 2 × 2 × 2
9 = 1 × 3 × 3
10 = 1 × 2 × 5
LCM(1 to 10) = LCM (1,2,3,4,5,6,7,8,9,10)
= 1 × 2 × 2 × 3 × 3 × 5 × 7
= 2520
Q.5 If p¹ and p² are two odd prime numbers such that p1> p2p, then p1²–p2² is
(a) an even number
(b) an odd number
(c) an odd prime number
(d) a prime number
Sol.
Let us take p1 = 5 and p2= 3
Then, p1²–p2² = 5² – 3²
= 25 –9
=16
16 is an even number.
Thus (a) is correct option.
Q.6. The rational form of .02 54 is in the form of p/q then p + q is
(a) 14 (b) 55 (c) 69 (d) 79
Sol.
1st Method
Let, x = .02 54,
then
x = .02545454 ..............(1)
Multiplying equation (1) by 100, we get
100x = 2.545454. ......... ...(2)
Multiplying equation (2) by 100, we get
10000x = 254.545454. ......... ...(3)
Subtracting equation (2) from equation. (3), we get
10000x = 254.545454. ......... ...(3)
100x = 2.545454. ......... ...(2)
9900x = 252
9900x = 252
x = 252/9900
x = 28/1100
x = 14/550
Comparing with p/q , we get
p = 14
and q = 550
Hence, p q + = 14 +550 + = 564
Q.7. Show that the number 3¹³– 3¹⁰ is divisible by 2, 3 and 13
Sol.
3¹³– 3¹⁰
= 3¹⁰(3³– 1)
= 3¹⁰(27– 1)
= 3¹⁰(26)
= 3¹⁰×2×13
Hence, 3¹³– 3¹⁰ is divisible by 2, 3 and 13.
Q.8. If a = 2³× 3 , b = 2×3×5, c = 3ⁿ×5 and LCM (a, b, c) 2³× 3²× 5, then n is
(a) 1 (b) 2 (c) 3 (d) 4
Sol.
Value of n must be 2.
Thus (b) is correct option.
Q.9 Explain why 13233343563715 is a composite number?
Sol. : [Board Term-1 2016]
The number 13233343563715 ends in 5.
Hence it is a multiple of 5. Therefore it is a composite number.
Q.10. Explain why ( 7× 13 ×11) + 11 and (7×6×5×4×3×2×1) + 3 are composite numbers.
Sol.
(7× 13× 11) + 11
= 11×(7×13 +1)
= 11 ×(91 +1)
= 11 × 92
and
(7×6×5×4×3×2×1) + 3
= 3×(7×6×5×4×2×1+ 1)
= 3 (1680+1)
= 3 (1681)
= 3× 41×41
Since given numbers have more than two prime factors, so that both numbers are composite.
Q.11. Check whether 4ⁿ can end with the digit 0 for any natural number n.
Sol.
If the number, 4ⁿ for any n, were to end with the digit zero, then it would be divisible by 5 and 2.
That is, the prime factorization of 4n would contain the prime 5 and 2. This is not possible because the
only prime in the factorization of 4ⁿ = 2²ⁿ is 2. So, the uniqueness of the Fundamental Theorem of Arithmetic
guarantees that there are no other primes in the factorization of 4ⁿ.
So, there is no natural number
n for which 4ⁿ ends with the digit zero. Hence 4ⁿcannot end with the digit zero.
Q.12 Show that 7ⁿ cannot end with the digit zero, for any natural number n.
Sol.
If the number, 7ⁿ for any n, were to end with the digit zero, then it would be divisible by 5 and 2.
That is, the prime factorization of 7ⁿ would contain the prime 5 and 2. This is not possible because the
only prime in the factorization of 7ⁿ= (1×7)ⁿ is 7.
So, the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorization of 7ⁿ . So, there is no natural number n for which 7ⁿ ends with the digit zero. Hence 7ⁿ cannot end with the digit zero.
Q.13. Check whether (15)ⁿ can end with digit 0 for any n € N ! .
Sol.
If the number (15)ⁿ for any n, were to end with the digit zero, then it would be divisible by 5 and 2.
That is, the prime factorization of (15)ⁿ would contain the prime 5 and 2.
This is not possible because the
only prime in the factorization of (15)ⁿ = (3×5)ⁿ are 3 and 5.
The uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorization of (15)ⁿ . Since there is no prime factor 2, (15)ⁿ cannot end with the digit zero.
Q.14. Write a rational number between 2 and 3 .
Sol.
Q.15. Show that 571 is a prime number.
Sol.
Let x = 571
√x = √571
Now 571 lies between the perfect squares of (23)² =529 and (24)= 576. Prime numbers less than 24 are 2, 3,
5, 7, 11, 13, 17, 19, 23.
Here 571 is not divisible by any of the above numbers, thus 571 is a prime number.
Q.16 An army contingent of 612 members is to march behind an army band of 48 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
Sol. [Board 2020 Delhi Basic]
Let the number of columns be x which is the largest number, which should divide both 612 and 48. It
means x should be HCF of 612 and 48.
We can write 612 and 48 as follows
612 = 2×2×3×3×5×17
= 2²×3²×5×17
48 = 2×2×2×2×3
= 2⁴×3
HCF(612, 28) = 2²×3
= 12
Q.17. Write the smallest number which is divisible by both 306 and 657.
Sol. [Board 2019 OD]
The smallest number that is divisible by two numbers is obtained by finding the LCM of these numbers
Here, the given numbers are 306 and 657.
306 = 2×3×3×17= 2×3²×17
657 = 3×3×73. = 3² ×73
LCM(306, 657) = 2×3²×17×73
= 22338
Hence, the smallest number which is divisible by 306 and 657 is 22338.
Q.18. 144 cartons of Coke cans and 90 cartons of Pepsi cans are to be stacked in a canteen. If each stack is of the same height and if it equal contain cartons of the same drink, what would be the greatest number of cartons each stack would have?
Sol. [Board Term-1 2011]
The required answer will be HCF of 144 and 90.
144 = 2×2×2×2×3×3 = 2⁴×3²
90 = 2×3×3×5 = 2×3²×5
HCF(144, 90) = 2×3²
= 2×9
= 18
Thus each stack would have 18 cartons.
Q.19. Three bells toll at intervals of 9, 12, 15 minutes respectively. If they start tolling together, after what time will they next toll together?
Sol. [Board Term-1 2011]
The required answer is the LCM of 9, 12, and 15 minutes.
Finding prime factor of given number we have,
9 = 3×3 = 3²
12= 2×2×3 =2²×3
15 = 3×5 = 3×5
LCM(9, 12, 15) = 2²×3²×5
= 4×9×5
= 180 minutes
The bells will toll next together after 180 minutes or 3 hr.
Q.20. Find HCF and LCM of 16 and 36 by prime factorization and check your answer.
Sol.
Finding prime factor of given number we have,
16= 2×2×2×2. = 2⁴
36 = 2×2×3×3 = 2²×3²
HCF(16, 36) = 2²
= 4
LCM (16, 36) =2⁴×3²
= 16×9
= 144
Check :
HCF(a, b)×LCM(a, b) = a×b
16 × 36 = 4 × 144
576 = 576
Thus LHS = RHS
Q.21. The HCF of 65 and 117 is expressible in the form 65 117– m . Find the value of m. Also find the LCM of 65 and 117 using prime factorization method.
Sol. [Board Term-1 2011]
Finding prime factor of given number we have,
117 = 2×3 ×13
65 = 5×13
HCF(117, 65) = 13
LCM(117, 65) = 2×3×5×13
= 585
Given that
HCF = 65m –117
13 = 65m –117
65m = 117 + 13
65m = 130
m = 130/65
m= 2
Q.22. Given that √5 is irrational, prove that (2 √5 – 3) is an irrational number.
Sol.
Assume that (2√5 – 3) is a rational number. Therefore, we can write it in the form of p/q where p and q are
co-prime integers and q ≠ 0.
Now (2√5 - 3)=p/q
where q ≠ 0 and p and q are co-prime integers.
Rewriting the above expression as,
2√5 = p/q +3
2√5 = (p+3q)/p
√5 = (p+3q)/2p
Here (p+3q)/2p is rational because p and q are co-prime integers, thus 5 should be a rational number. But
√5 is irrational.
This contradicts the given fact that √5 is irrational. Hence (2√5 - 3) is an irrational number.
Q.23. Prove that (2+√3)/5 is an irrational number, given that √3 is an irrational number.
Sol. [Board 2019 Delhi]
Assume that (2+√3)/5 is a rational number. Therefore, we can write it in the form of p/q where p and q are co-
prime integers and q ≠ 0.
Since, p and q are co-prime integers, then (2+√3)/5 is a rational number. But this contradicts the fact that
√3 is an irrational number. So, our assumption is wrong. Therefore (2+√3)/5 is an irrational number.
Q.24. Express (15/4+5/40) as a decimal fraction without actual
division.
Sol.
We have
Q.25. If p is prime number, then prove that √p is an irrational.
Sol. [Board Term-1 2013]
Let √p be a prime number and if possible, let p be rational
Thus √p = m / n ,
where m and n are co-primes and n ≠ 0.
Squaring on both sides, we get
p = m²/n²
or, pn²= m² ...(1)
Here p divides pn². Thus p divides m²
and in result p also divides m.
Let m = pq for some integer q and putting m = pq in eq. (1), we have
pn² = p² q²
or, n² = pq²
Here p divides pq².Thus p divides n²
and in result p also divides n.
[ Since p is prime and p divides n²=>p & divides n]
Thus p is a common factor of m and n but this contradicts the fact that m and n are primes. The contradiction arises by assuming that √p is rational.
Hence, √p is irrational.
Q.26. Prove that √2 is an irrational number.
Sol.
Since q² is divisible by 2, thus q is also divisible by 2. We have seen that p and q are divisible by 2, which contradicts the fact that p and q are co-primes. Hence, our assumption is false and √2 is irrational.
Q.27. Show that there is no positive integer n, for
is rational.
Sol. [Board Term-1 2012]
Let us assume that there is a positive integer n for which
is rational and equal to p/q , where p and q areare positive integers and q≠0 .
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