real numbers - snd public school

Quick Review

z A number is called a rational number, if it can be written in the form pq

, where p and q are integers and q ≠ 0.

For example, 2 23

911

, , ,− etc.

z A number which cannot be expressed in the form pq

, where p and q are integers and q ≠ 0, is called an irrational

number. For example, 2 3 23

, , , ,π − etc.

z All rational numbers and all irrational numbers together make the collection of real numbers. (i) The collection of rational numbers is denoted by Q. (ii) The collection of irrational numbers is denoted by Q′.

(iii) The collection of real numbers is denoted by R.

z Decimal expansion of a rational number is either terminating or non-terminating recurring, while the decimal expansion of an irrational number is non-terminating and non-recurring.

For example, 0.375 ; 0.2222 ... = 0.2 are rational numbers while 1.05005000500005 ... is an irrational number.

z If r is a rational number and s is an irrational number, then r + s and r – s are irrational numbers. Further, if r is

a non-zero rational number, then rs and rs

are irrational numbers.

For example, 2 3 3 1 3 15

+ − −, , , , etc.

Real Numbers

1CHAPTER

TOPICS Euclid's Division Lemma and

Algorithm

The Fundamental Theorem of Arithmetic

Revisiting Irrational Numbers

Revisiting Rational Numbers and their Decimal Expansions

In previous classes, we have learnt about real numbers, operations on real numbers (addition, subtraction,

multiplication and division), their decimal expansions and laws of exponents for real numbers. In this chapter, we will continue our study on real numbers and begin with two important properties of positive integers namely Euclid's Division Lemma and the Fundamental Theorem of Arithmetic.

π = 3.1415926535...Irrational Number

π = 22__7 Rational

Number

100PERCENTMathematics Class-10

TOPIC 1 : EUCLID'S DIVISION LEMMA AND ALGORITHMLet us first understand the meaning of lemma and algorithm.Lemma : A lemma is a proven statement used for proving another statement.Algorithm : An algorithm is a series of well defined steps, which gives a procedure for solving a problem.

z Euclid’s Division LemmaIt says that any positive integer a can be divided by another positive integer b in such a way that it leaves a remainder r which is smaller than b.Note that this is nothing but a restatement of the long division process.For example, On dividing 37 by 5, we get quotient = 7 and remainder = 2,which is less than 5.

z Formal Statement of Above Lemma“Given positive integers a and b, there exist unique integers q and r satisfying a = bq + r, 0 ≤ r < b.”For example,(i) Let a = 47 and b = 9, then we can write 47 = 9 × 5 + 2, where q = 5 and r = 2 < b(ii) Let a = 83 and b = 17, then we can write 83 = 17 × 4 + 15, where q = 4 and r = 15 < b

Note : (i) The unique integers q and r are nothing but the quotient and remainder respectively.(ii) Although Euclid’s division algorithm is stated for only positive integers but it can be extended for all

integers except zero.(iii) When a and b are positive integers, then q and r can take values only from whole numbers.

z An Important Fact about the Quotient and the RemainderIf a < b, then in this case quotient, q = 0 and remainder, r = aFor example, if a = 5 and b = 8, then we can write 5 = 8 × 0 + 5, where q = 0 and r = 5.Let us now discuss some applications of Euclid’s division lemma with the help of some illustrations.

ILLUSTRATIONS Show that every positive odd integer is of

the form (4q + 1) or (4q + 3), where q is some integer. (NCERT, Delhi 2019)Sol. Let a be any positive odd integer and let b = 4.By Euclid’s division lemma, there exists integers q and r, such thata = 4q + r, where 0 ≤ r < 4 i.e., r = 0, 1, 2 or 3.\ a = 4q or 4q + 1 or 4q + 2 or + 3But a is an odd integer\ a ≠ 4q and a ≠ 4q + 2⇒ a = 4q + 1 or a = 4q + 3 Hence, any odd integer is of the form 4q + 1 or 4q + 3.

Show that the square of any odd integer is of the form 4q + 1, for some integer q. (NCERT Exemplar)Sol. By Euclid’s division lemma, we havea = bm + r, where 0 ≤ r < b ... (i)On putting b = 4 in eq. (i), we geta = 4m + r, where 0 ≤ r < 4, i.e., r = 0, 1, 2, 3If r = 0, then a = 4m which is divisible by 2⇒ 4m is even.

If r = 1, then a = 4m + 1 is not divisible by 2.If r = 2, then a = 4m + 2 = 2(2m + 1) which is divisible by 2⇒ 2(2m + 1) is even.If r = 3, then a = 4m + 3 which is not divisible by 2.So, for any positive integer m, (4m + 1) and (4m + 3) are odd integers.Now, a2 = (4m + 1)2 = 16m2 + 1 + 8m = 4(4m2 + 2m) + 1 = 4q + 1It is a square which is of the form 4q + 1, where q = (4m2 + 2m) is an integer.and a2 = (4m + 3)2 = 16m2 + 9 + 24m = 4(4m2 + 6m + 2) + 1 = 4q + 1It is a square which is of the form 4q + 1, where q = (4m2 + 6m + 2) is an integer.Hence, for some integer q, the square of any odd integer is of the form 4q + 1.

Show that the cube of any positive integer is of the form 4m, 4m + 1, 4m + 3, for some integer m. (NCERT Exemplar)Sol. Let a be an arbitrary positive integer. Then, by Euclid's division lemma, corresponding to the positive integer a and 4, there exist non-negative integers q and r such thata = 4q + r, where 0 ≤ r < 4

5 37–35

2

7Divisor

Dividend

Quotient

Remainder

2

Now, a3 = (4q + r)3 = 64q3 + r3 + 48q2r + 12qr2

⇒ a3 = 64q3 + 48q2r + 12qr2 + r3 ...(i)where 0 ≤ r < 4 If r = 0, then from (i), we get a3 = 64q3 = 4(16q3)⇒ a3 = 4m, where m = 16q3 is an integer.If r = 1, then from (i), we geta3 = 64 q3 + 48q2 + 12q + 1 = 4(16q3 + 12q2 + 3q) + 1= 4m + 1, where m = (16q3 + 12q2 + 3q) is an integer.If r = 2, then from (i), we get

a3 = 64q3 + 96q2 + 48q + 8 = 4(16q3 + 24q2 + 12q + 2)= 4m, where m = (16q3 + 24q2 + 12q + 2) is an integer.If r = 3, then from (i), we geta3 = 64q3 + 144q2 + 108q + 27= 64q3 + 144q2 + 108q + 24 + 3= 4(16q3 + 36q2 + 27q + 6) + 3= 4m + 3 where, m = (16q3 + 36q2 + 27q + 6) is an integer.Hence, the cube of any positive integer is of the form 4m, 4m + 1 or 4m + 3 for some integer m.

YOURSELF1. Prove that the square of any positive integer of the form 5q + 1, is of the same form.2. Show that one and only one out of n, (n + 1) and (n + 2) is divisible by 3, where n is any

positive integer.

For YOURSELF Solutions visit http://bit.ly/mtg-100-mt-tryyourself

Euclid’s Division AlgorithmEuclid’s division algorithm is a technique to compute the Highest Common Factor (HCF) of two given positive integers. According to this, the HCF of any two positive integers a and b, with a > b, is obtained as follows :Step 1 : Apply Euclid’s division Lemma to a and b, i.e., find whole numbers, q and r such that a = bq + r,where 0 ≤ r < b.Step 2 : If r = 0 then b is the HCF of a and b. If r ≠ 0, apply the division lemma to b and r.Step 3 : Continue the process till the remainder is zero. The divisor at this stage will be the required HCF.For example, we can calculate the HCF of 595 and 721 as follows:721 = 595 × 1 + 126 (Applying division lemma to 721 and 595)595 = 126 × 4 + 91 (Applying division lemma to 595 and 126)126 = 91 × 1 + 35 (Applying division lemma to 126 and 91)91 = 35 × 2 + 21 (Applying division lemma to 91 and 35)35 = 21 × 1 + 14 (Applying division lemma to 35 and 21)21 = 14 × 1 + 7 (Applying division lemma to 21 and 14)14 = 7 × 2 + 0 (Applying division lemma to 14 and 7)Since, the remainder becomes zero when divisor is 7, therefore HCF (595, 721) = 7.Note : (i) HCF of two positive integers a and b is the largest positive integer d that divides both a and b.(ii) To find HCF of three numbers say a > b > c by using Euclid’s Division Lemma, first find HCF (a, b) = d(say), then

find HCF (d, c), which will be required HCF (a, b, c).(iii) Euclid’s division lemma and algorithm are so closely interlinked that people often call former as the division

algorithm also.

ILLUSTRATIONS Find the HCF of 37728 and 12156 by

Euclid’s division algorithm.Sol. Here, 37728 > 12156\ Applying Euclid's division lemma, we get37728 = 12156 × 3 + 1260Since, remainder, 1260 ≠ 0, therefore applying Euclid's division lemma for 12156 and 1260, we get12156 = 1260 × 9 + 816

Similarly, we have 1260 = 816 × 1 + 444,816 = 444 × 1 + 372, 444 = 372 × 1 + 72,372 = 72 × 5 + 12 and 72 = 12 × 6 + 0Since, the remainder becomes 0, when divisor is 12.\ By Euclid’s division algorithm,HCF (37728, 12156) = 12

ByusingEuclid’sdivisionalgorithm,findthelargest number which divides 650 and 1170.Sol. The required largest number which divides 650 and 1170 is HCF (1170, 650)

3Real Numbers


1. Use Euclid’s division algorithm to find the HCF of(i) 135 and 225 (ii) 196 and 38220(iii) 867 and 2552. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.3. An army contingent of 616 members is to march behind an army band of 32 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?

4. Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.[Hint : Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m +1.]5. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

For SolutionsVisit http://bit.ly/mtg-100-mt-ncertfocus

Now, by Euclid’s division lemma, we have1170 = 650 × 1 + 520650 = 520 × 1 + 130520 = 130 × 4 + 0Since, the remainder becomes 0, when divisor is 130.\ By Euclid’s division algorithm,HCF (1170, 650) = 130Hence, the required largest number is 130.

Find the HCF of 180, 252 and 324 by using Euclid's division algorithm.Sol. Here, 324 > 252 > 180\ By applying Euclid's division lemma to 324 and 252, we get324 = (252 × 1) + 72252 = (72 × 3) + 3672 = (36 × 2) + 0Now, the remainder becomes 0, when divisor is 36.So, by Euclid's division algorithm, HCF (324, 252) is 36.Now, applying Euclid's division lemma to 180

and 36, we get180 = (36 × 5) + 0Here, remainder becomes 0, when divisor is 36.So, HCF (180, 36) is 36.Hence, HCF of 180, 252 and 324 is 36.

Two tankers contain 850L and 680L of petrol respectively. Find the maximum capacity of a container which can measure the petrol of either tanker, in exact number of times.Sol. Clearly, the maximum capacity of the container is the HCF of 850 and 680.Here, 850 > 680\ Applying Euclid’s division lemma, we get 850 = 680 × 1 + 170 680 = 170 × 4 + 0Here, remainder becomes zero, when divisor is 170.So, by Euclid’s division algorithm, HCF (850, 680) is 170.Hence, the maximum capacity of the required container is 170 L.

YOURSELF3. Find the HCF of 250 and 30 using Euclid's division algorithm. (Ans. 10)

4. A trader has 612 Dettol soaps and 342 Pears soaps. He packs them in boxes and each box containsexactlyonetypeofsoap.Ifeveryboxcontainsthesamenumberofsoaps,thenfindthe number of soaps in each box such that the number of boxes is the least. (Ans. 18)

5. Find the HCF of the numbers 1305, 1365 and 1530 by Euclid’s division algorithm. (Ans. 15)


EXERCISE - 1.1

4

Objective Type QuestionsMCQs (1 Mark)1. Euclid’s division lemma states that for any positive integers a and b, there exist unique integers q and r such that a = bq + r, where r must satisfy(a) 1 < r < b (b) 0 < r ≤ b(c) 0 ≤ r < b (d) 0 < r < b2. A number when divided by 143 leaves 31 as remainder. What will be the remainder, when the same number is divided by 13?(a) 0 (b) 1 (c) 3 (d) 53. The remainder when the square of any prime number greater than 3 is divided by 6, is(a) 1 (b) 3 (c) 2 (d) 4

Fill in the Blanks (1 Mark)4. _______ is divided by 53 gives 34 as quotient and 21 as remainder.5. 38220 = 176 × _______ + 28.

VSA Type Questions (1 Mark)6. For some integer q, what will be the form of every odd integer? (NCERT Exemplar)7. If the HCF of 65 and 117 is expressible in the form 65 m – 117, then find the value of m.

(NCERT Exemplar)

Short Answer Type QuestionsSA Type I Questions (2 Marks)8. Prove that if x and y are odd positive integers,

then x2 + y2 is even but not divisible by 4.(NCERT Exemplar)

9. “The product of three consecutive positive integers is divisible by 6”. Is this statement true or false? Justify your answer.

(NCERT Exemplar)

SA Type II Questions (3 Marks)10. Find the greatest number that will divide 445, 572 and 699 leaving remainders 4, 5 and 6 respectively.11. For what value of n, n2 – 1 is divisible by 8?

(NCERT Exemplar)12. Show that every positive even integer is of the form 2q and every positive odd integer is of the form 2q + 1 for some integer q.13. Prove that (n2 – n) is divisible by 2 for every positive integer n.

Long Answer Type QuestionsLA Type Questions (4 Marks)14. Show that the square of any positive integer cannot be of the form 5q + 2 or 5q + 3 for any integer q. (NCERT Exemplar)15. There are 156, 208 and 260 students in group A, B and C, respectively. Buses are to be hired to take them for a field trip. Find the minimum number of buses to be hired, if the same number of students should be accommodated in each bus.

1. (c) : On dividing a by b, let q be the quotient and r be the remainder. Then, a = bq + r, where 0 ≤ r < b.2. (d) : Let the given number when divided by 143 gives q as quotient and 31 as remainder.Then, number = 143q + 31 = 13 × 11q + 13 × 2 + 5 = 13 (11q + 2) + 5.So, the same number when divided by 13 gives 5 as remainder.

3. (a) : Any prime number greater than 3 is of the form 6k ± 1, where k is a natural number.\ (6k ± 1)2 = 36k2 ± 12k + 1 = 6(6k2 ± 2k) + 1\ Required remainder is 1.4. If a is divided by b gives q as quotient and r as remainder. Then, by Euclid’s division lemma, we have a = bq + r, 0 ≤ r < bHere, b = 53, q = 34, r = 21 \ a = 53 × 34 + 21 = 1823

5Real Numbers


5. Let the required number be x.\ 38220 = 176 × x + 28Now, 38220 = 176 × 217 + 28 \ x = 2176. We know that, odd integers are 1, 3, 5, ...So, it can be written in the form of 2q + 1, where q is an integer i.e., q = ...., –2, –1, 0, 1, 2, 3, ....\ 2q + 1 = ...., –3, –1, 1, 3, 5, ....7. By Euclid's division lemma, we havea = bq + r, 0 ≤ r < bWe have, 117 = 65 × 1 + 5265 = 52 × 1 + 13 ; 52 = 13 × 4 + 0\ HCF (65, 117) = 13 ...(i)Also, given that, HCF (65, 117) = 65 m – 117 ...(ii)From (i) and (ii), we get 65 m – 117 = 13 ⇒ 65 m = 130 ⇒ m = 28. We know that any odd positive integer is of the form 2q + 1 for some integer q. So, let x = 2m + 1 and y = 2n + 1 for some integers m and n.\ x2 + y2 = (2m + 1)2 + (2n + 1)2

= 4(m2 + n2) + 4(m + n) + 2 = 4{(m2 + n2) + (m + n)} + 2= 4q + 2, where q = (m2 + n2) + (m + n)⇒ x2 + y2 is even and leaves remainder 2, when divided by 4.⇒ x2 + y2 is even but not divisible by 4.9. The given statement is true.Three consecutive integers can be n, (n + 1) and (n + 2).So, one integer out of three must be divisible by 2 and another one must be divisible by 3. Hence, product of three consecutive integers is divisible by 6.10. Since the remainders are 4, 5 and 6 respectively.\ Wehave to find theHCF of 445 – 4, 572 – 5 and 699 – 6 i.e., 441, 567 and 693. Here, 693 > 567 > 441By applying Euclid's division lemma to 693 and 567, we get, 693 = 567 × 1 + 126, 567 = 126 × 4 + 63 ; 126 = 63 × 2 + 0Since, remainder is 0, when divisor is 63.So, HCF (693, 567) is 63.Again, applying Euclid's division lemma to 441 and 63 we get, 441 = 63 × 7 + 0.Here, remainder is 0, when divisor is 63.So, HCF (441, 63) is 63. Thus, HCF of 441, 567 and 693 = 63\ The required number is 63.11. Let a = n2 – 1. Here n can be even or odd.Case I : If n is even, i.e., n = 2k, where k is an integer.⇒ a = (2k)2 – 1 ⇒ a = 4k2 – 1At k = –1, a = 4(–1)2 – 1 = 4 – 1 = 3, which is not divisible by 8.At k = 0, a = 4(0)2 – 1 = 0 – 1 = –1, which is not divisible by 8.At k = 1, a = 4(1)2 – 1 = 3, which is not divisible by 8.Case II : If n is odd, i.e., n = 2k + 1, where k is an integer.⇒ a = (2k + 1)2 – 1 = 4k2 + 4k + 1 – 1 = 4k2 + 4k = 4k(k + 1)At k = –1, a = 4(–1)(–1 + 1) = 0, which is divisible by 8.At k = 0, a = 4(0)(0 + 1) = 0 , which is divisible by 8

At k = 1, a = 4(1)(1 + 1) = 8, which is divisible by 8.Hence, we can conclude from above two cases that if n is odd, then n2 – 1 is divisible by 8.12. Let a be any positive integer. On dividing a by 2, let q be the quotient and r be the remainder. Then, by Euclid’s Lemma, we have a = 2q + r, where 0 ≤ r < 2 ⇒ a = 2q + r, where r = 0 or r = 1 ⇒ a = 2q or a = 2q + 1When a = 2q for some integer q, then clearly a is even.Hence, an odd integer is of the form 2q + 1 for some integer q. Thus, every positive even integer is of the form 2q and every positive odd integer is of the form 2q + 1.13. We know that any positive integer is of the form 2q or 2q + 1, for some integer q. So, two cases arises :Case I : When n = 2q, thenn2 – n = (2q)2 – 2q = 4q2 – 2q = 2q(2q – 1)i.e., n2 – n = 2r, where r = q(2q – 1) ⇒ n2 – n is divisible by 2Case II : When n = 2q + 1, thenn2 – n =(2q + 1)2 – (2q + 1) = 4q2 + 1 + 4q – 2q – 1= 4q2 + 2q = 2q(2q + 1) = 2r, where r = q(2q + 1)⇒ n2 – n is divisible by 2.Hence, n2 – n is divisible by 2 for every positive integer n.14. Let x be any positive integer such that it is in the form of 5m, 5m + 1, 5m + 2, 5m + 3 or 5m + 4Case I : When x = 5mx2 = 25m2 = 5(5m2) = 5q, where q = 5m2

Case II : When x = 5m + 1x2 = (5m + 1)2 = 25m2 + 10m + 1 = 5(5m2 + 2m) + 1 = 5q + 1, where q = 5m2 + 2mCase III : When x = 5m + 2x2 = (5m + 2)2 = 25m2 + 20m + 4 = 5(5m2 + 4m) + 4 = 5q + 4, where q = 5m2 + 4mCase IV : When x = 5m + 3x2 = (5m + 3)2 = 25m2 + 30m + 9 = 5(5m2 + 6m + 1) + 4 = 5q + 4, where q = 5m2 + 6m + 1Case V : When x = 5m + 4x2 = (5m + 4)2 = 25m2 + 40m + 16 = 5(5m2 + 8m + 3) + 1 = 5q + 1, where q = 5m2 + 8m + 3Hence, x2 is of the form 5q or 5q + 1 or 5q + 4 but it cannot be of the form 5q + 2 or 5q + 3.15. Given numbers are 156, 208 and 260.Here, 260 > 208 > 156By applying Euclid's division lemma to 260 and 208, we get ,260 = (208 × 1) + 52 ; 208 = (52 × 4) + 0Here, the remainder is zero when divisor is 52.So, HCF (260, 208) is 52.Similarly, by applying Euclid's division lemma to 156 and 52, we get 156 = (52 × 3) + 0Here, the remainder is zero when divisor is 52.So, HCF (156, 52) is 52. Thus, HCF of 156, 208 and 260 is 52.Hence, the minimum number of buses required

= + +15652

20852

26052 = 3 + 4 + 5 = 12

6

TOPIC 2 : THE FUNDAMENTAL THEOREM OF ARITHMETICThe fundamental theorem of Arithmetic states that - ‘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’. It is also known as ‘Unique Factorisation Theorem’.Thus, any integer greater than 1, can either be a prime number or can be written as a product of prime factors.For example,(i) 8 = 2 × 2 × 2, where 2 is a prime number.(ii) 130 = 2 × 5 × 13, where 2, 5 and 13 are prime numbers.Note : (i) A number is called a prime number, if it has no factors other than 1 and the number itself.(ii) A number is called a composite number, if it has atleast one factor other than 1 and the number itself.(iii) 1 is neither prime nor composite.(iv) 2 is the smallest prime number. It is the only even prime number.(v) Two numbers are said to be co-prime if they have no common factors other than 1 i.e., their HCF = 1.

Factor TreeA chain of factors which is represented in the form of a tree, is called factor tree.

ILLUSTRATIONS Express each of the following positive

integers as the product of its prime factors: (i) 260 (ii) 196

Sol. (i) Using factor tree method, we have260

2 130

2 65

5 13

\ 260 = 2 × 2 × 5 × 13 = 22 × 5 × 13(ii) Using factor tree method, we have

196

2 98

2 49

7 7

\ 196 = 2 × 2 × 7 × 7 = 22 × 72

Explain why 3 × 5 × 7 + 7 is a composite number. (NCERT Exemplar)Sol. We have, 3 × 5 × 7 + 7 = 7 [3 × 5 + 1], which is not a prime number because it has 7 as a factor, other than 1 and the number itself i.e., (3 × 5 + 1).Hence, it is a composite number.

Consider the numbers 4n, where n is a natural number. Check whether there is any value of n for which 4n ends with the digit zero. (NCERT)Sol. If the number 4n, for any n, has to end with the digit zero, then it would be divisible by 5 i.e., the prime factorisation of 4n must contain the prime number 5. This is not possible because 4n = (2)2n and so the only prime number in the factorisation of 4n is 2. Also, 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.

YOURSELF6. Express each of the following positive integers as the product of its prime factors : (i) 4095 (ii) 1001 (Ans. (i) 32 × 5 × 7 × 13 (ii) 7 × 11 × 13)

7. Check whether 9 × 13 × 17 + 17 and 5 × 6 × 7 × 8 × 9 + 7 are composite numbers, or not. If yes, then justify your answer.

(Ans. Yes, both are composite numbers.)

8. Check whether 12n can end with the digit 0 or 5, or not for any natural number n.(Ans. No)


7Real Numbers


HCF and LCM using Prime FactorisationHCF = Product of the smallest power of each common prime factor involved in the numbers.LCM = Product of the greatest power of each prime factor involved in the numbers.For example, if a = 231 = 3 × 7 × 11 and b = 396 = 2 × 2 × 3 × 3 × 11 = 22 × 32 × 11,then (i) HCF (a, b) = 3 × 11 = 33 [ Only 3 and 11 are common factors](ii) LCM (a, b) = 22 × 32 × 7 × 11 = 2772 [ Prime numbers involved in a and b are 2, 3, 7, 11]Note : (i) LCM (Least Common Multiple) of two or more numbers is the smallest number which is divisible by all the given numbers.(ii) If there is no common prime factor, then HCF of given numbers is 1.

ILLUSTRATIONS Find the HCF and LCM of 404 and 96 by

the prime factorisation method. (NCERT)Sol. The prime factorisation of 404 and 96 is,404 = 2 × 2 × 101 and 96 = 2 × 2 × 2 × 2 × 2 × 3\ HCF (404, 96) = 2 × 2 = 4 andLCM (404, 96) = 2 × 2 × 2 × 2 × 2 × 3 × 101 = 9696

Find the LCM and HCF of 9, 117 and 729 by using prime factorisation method.Sol. The prime factorisation of 9, 117 and 729 is,9 = 3 × 3, 117 = 3 × 3 × 13,729 = 3 × 3 × 3 × 3 × 3 × 3⇒ 9 = 32, 117 = 32 × 13, 729 = 36

\ HCF (9, 117, 729) = 32 = 9

and LCM (9, 117, 729) = 36 × 13 = 729 × 13 = 9477.

Amita, Suneha and Rajiv start preparing cards. In order to complete one card, they take 10, 16 and 20 minutes respectively. If all of them started together, after what time will they start preparing a new card together?Sol. Given, Amita, Suneha and Rajiv takes 10, 16 and 20 minutes respectively to complete one card.Prime factorisation of 10, 16 and 20 is10 = 2 × 5, 16 = 2 × 2 × 2 × 2 = 24 and20 = 2 × 2 × 5 = 22 × 5\ Time after which they start preparing a new card together = LCM (10, 16 and 20)= 2 × 2 × 2 × 2 × 5 = 24 × 5 = 80 minutes.

YOURSELF9. Find the HCF and LCM of 144, 180 and 192 by prime factorisation method. (Ans. HCF = 12, LCM = 2880)

10. In a school there are two sections – section A and section B of class X. There are 32 students in section A and 36 students in section B. Determine the minimum number of books required for their class library so that they can be distributed equally among students of section A or section B. (Ans. 288)


Relationship between Numbers and their HCF and LCMz For any two positive integers a and b, the relation between these numbers and their HCF and LCM is

HCF (a, b) × LCM (a, b) = a × b

⇒ HCF (a, b) = a ba b

×LCM( , )

or LCM (a, b) = a ba b

×HCF ( , )

z For any three positive integers a, b and c, the relation between these numbers and their HCF and LCM is

HCF (a, b, c) = a b c a b ca b b c c a

× × ×× ×

LCMLCM LCM LCM

( , , )( , ) ( , ) ( , )

or LCM (a, b, c) = a b c a b ca b b c c a

× × ×× ×

HCFHCF HCF HCF

( , , )( , ) ( , ) ( , )

8

ILLUSTRATIONS The HCF of two numbers is 27 and their

LCM is 162. If one of the numbers is 54, thenfindtheothernumber.Sol. Let the other number be x.As, HCF (a, b) × LCM (a, b) = a × b

⇒ 27 × 162 = 54x ⇒ x = 27 16254

81× =Hence, other number is 81.

If the HCF of 35 and 45 is 5 and LCM of 35 and 45 is 63 × a,thenfindthevalueofa.Sol. We know that,HCF × LCM = Product of two numbers

⇒ 5 × 63 × a = 35 × 45 ⇒ a = 35 455 63

5××

=\ a = 5

IfHCF(28,35,343)=7,thenfindtheLCM(28, 35, 343).Sol. Given numbers are 28, 35 and 343. Also, it is given that HCF (28, 35, 343) = 7Clearly, HCF (28, 35) = 7,HCF (35, 343) = 7 and HCF (343, 28) = 7Now, LCM (a, b, c)

=× × ×

× ×a b c a b ca b b c c a

HCFHCF HCF HCF

( , , )( , ) ( , ) ( , )

\ LCM (28, 35, 343)

= × × ×× ×

28 35 343 28 35 34328 35 35 343 343 28

HCF ,HCF HCF HCF

( , )( , ) ( , ) ( , ))

= × × ×× ×

28 35 343 77 7 7

= 4 × 5 × 343 = 6860

YOURSELF11. Find the LCM of a and b, if product of a and b is 7623 and HCF of a and b is 11. (Ans. 693)12. Given that HCF (2520, 6600) = 120, LCM (2520, 6600) = 252 × k, thenfind the value

of k. (Ans. 550)13. If the product of two numbers a and b is 1152 and HCF of a and bis12,thenfindLCMof

a and b. (Ans. 96)14. If two positive integers x and y are expressible in terms of primes as x = p2q3 and y = p3q,

thenfindtheirLCMandHCF.IsLCMamultipleofHCF?Explain. (Ans. LCM = p3q3, HCF = p2q, Yes)


1. Express each number as a product of its prime factors:(i) 140 (ii) 156 (iii) 3825 (iv) 5005(v) 74292. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.(i) 26 and 91 (ii) 510 and 92(iii) 336 and 543. Find the LCM and HCF of the following integers by applying the prime factorisation method. (i) 12, 15 and 21 (ii) 17, 23 and 29(iii) 8, 9 and 25

4. Given that HCF(306, 657) = 9, find LCM (306, 657).5. Check whether 6n can end with the digit 0 for any natural number n.6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?

EXERCISE - 1.2

For Solutions visit http://bit.ly/mtg-100-mt-ncertfocus

9Real Numbers


Objective Type QuestionsMCQs (1 Mark)1. LCM of (23 × 3 × 5) and (24 × 5 × 7) is(a) 1680 (b) 560(c) 1240 (d) 11202. What is the LCM of smallest prime number and the smallest composite number?(a) 2 (b) 4(c) 6 (d) 83. Four persons P, Q, R and S start running around a circular track simultaneously. If they complete one round in 10, 8, 12 and 18 minutes respectively, after how much time will they next meet at the starting point?(a) 180 minutes (b) 270 minutes(c) 360 minutes (d) 450 minutes

Fill in the Blanks (1 Mark)4. If a = HCF (24, 72), then a = _______.5. The prime factorisation of 945 is _______.

VSA Type Questions (1 Mark)6. Find the LCM and HCF of x3y2 and xy.7. Why 5 × 7 × 11 + 7 is a composite number?8. If the product of two numbers is 980 and their HCF is 2, then find their LCM.

Short Answer Type QuestionsSA Type I Questions (2 Marks)9. On a morning walk, three persons steps off together and their steps measure 40 cm, 42 cm and 45 cm respectively. What is the minimum distance each should walk so that each can cover same distance in complete steps?

(NCERT Exemplar)10. If sum of two numbers is 1215 and their HCF is 81, then find the possible number of pairs.

SA Type II Questions (3 Marks)

11. Find the HCF of 90 and 126 by Euclid's division algorithm. Also, find their LCM and verify that LCM × HCF = Product of two numbers.

12. Three sets of English, Hindi and Mathematics books have to be stacked in such a way that all the books are stored topicwise and the height of each stack is the same. The number of English books is 96, the number of Hindi books is 240 and the number of Mathematics books is 336. Each book has the same size. Determine the number of stacks of English, Hindi and Mathematics books.

13. The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds respectively. If they change simultaneously at 8 a.m., at what time will they change together again?

Long Answer Type QuestionsLA Type Questions (4 Marks)

14. A circular track around a sports ground has circumference of 1080 m. Two cyclists Rohan and Sumeet start together and cycled at constant speed of 6 m/s and 9 m/s respectively around the circular track. After how many minutes, will they meet again at the starting point?

15. A, B and C start cycling around a circular path in the same direction and at the same time. Circumference of the path is 1980 m. If the speed of A is 330 m/min, speed of B is 198 m/min and C is 220 m/min and they start from the same point, then after how much time will they meet again?

10

1. (a) : LCM = 24 × 3 × 5 × 7 = 16 × 3 × 5 × 7 = 16802. (b) : Smallest prime number = 2Smallest composite number = 4\ LCM (2, 4) = 43. (c) : The time interval for them to meet at the starting point for the next time will be LCM (10, 8, 12 and 18).Thus, the prime factorisation of 10, 8, 12 and 18 is10 = 2 × 5, 8 = 2 × 2 × 2, 12 = 2 × 2 × 3 and 18 = 2 × 3 × 3\ LCM (10, 8, 12, 18) = 23 × 32 × 5 = 360 minutes4. The prime factorisation of 24 and 72 is, 24 = 2 × 2 × 2 × 3 72 = 2 × 2 × 2 × 3 × 3\ HCF (24, 72) = 2 × 2 × 2 × 3 = 24⇒ a = 245. The prime factorisation of 945 is, 945 = 3 × 3 × 3 × 5 × 7 = 33 × 5 × 76. We have, x3y2 = x × x × x × y × yand xy = x × y\ LCM (x3y2, xy) = x × x × x × y × y = x3y2

and HCF (x3y2, xy) = x × y = xy7. We have, 5 × 7 × 11 + 7 = 7 × (5 × 11 + 1) = 7 × 56, which is a composite number, because the given number contains two factors 7 and 56 other than 1 and the number itself.8. Product of two numbers = 980Their HCF = 2Q HCF × LCM = Product of the numbers⇒ 2 × LCM = 980⇒ LCM = 4909. The prime factorisation of 40, 42 and 45 is40 = 2 × 2 × 2 × 5 = 23 × 542 = 2 × 3 × 745 = 3 × 3 × 5 = 32 × 5\ LCM (40, 42, 45) = 23 × 32 × 5 × 7 = 2520\ Required distance = 2520 cm or 0.0252 km10. There are 4 such pairs i.e., (1134, 81), (1053, 162), (891, 324), (648, 567).11. Since, 126 > 90.\ Applying Euclid's division lemma, we get126 = 90 × 1 + 3690 = 36 × 2 + 1836 = 18 × 2 + 0Here, remainder is zero when divisor is 18.Thus, HCF of 126 and 90 is 18.Also, prime factorisation of 90 = 2 × 32 × 5and prime factorisation of 126 = 2 × 32 × 7\ LCM of 90 and 126 = 2 × 32 × 5 × 7 = 630Again, LCM × HCF = 630 × 18 = 11340Product of two numbers = 90 × 126 = 11340Hence, LCM × HCF = Product of two numbers

12. In order to arrange the books as required, we have tofindthelargestnumberthatdivides96,240and336exactly. Clearly, such a number is their HCF.We have, 96 = 25 × 3, 240 = 24 × 3 × 5 and 336 = 24 × 3 × 7\ HCF of 96, 240 and 336 = 24 × 3 = 48So, there must be 48 books in each stack.

\ Number of stacks of English books = 9648 = 2

Number of stacks of Hindi books = 24048 = 5

Number of stacks of Mathematics books = 33648

= 7

13. The required number of seconds after 8 a.m., when the lights change simultaneously is the LCM of 48, 72 and 108.The prime factorisation of 48, 72 and 108 is48 = 24 × 3, 72 = 23 × 32, 108 = 22 × 33

Thus, LCM of 48, 72 and 108 = 24 × 33

= 432 seconds = 7 minutes 12 seconds\ Required time = 08 : 07 : 12 a.m.14. Circumference of circular track = 1080 mSpeed of two cyclists is 6 m/s and 9 m/s\ Time taken to go around the track once by Rohan

= 10806

= 180 sec.

Time taken to go around the track once by Sumeet

= 1080

9 = 120 sec.

The required number of minutes, when they meet again at the starting point is LCM of 180 and 120.The prime factorisation of 180 and 120 is,180 = 22 × 32 × 5, 120 = 23 × 3 × 5\ LCM of 180 and 120 = 23 ×32 × 5 = 360 seconds = 6 minutesHence, they will meet again at the starting point after 6 minutes.15. Time taken by A to complete 1 round

= 1980330 = 6 mins

Time taken by B to complete 1 round

= 1980198

= 10 mins

Time taken by C to complete 1 round

= 1980220

= 9 mins

\ Required number of minutes, when the three cyclists will meet at the starting point again is LCM (6, 10, 9) minutes.Q 6 = 2 × 3, 10 = 2 × 5 and 9 = 3 × 3\ LCM (6, 10, 9) = 2 × 5 × 32 = 90 minutesSo, they will meet after 90 minutes or 1 hour 30 mins.

11Real Numbers


TOPIC 3 : REVISITING IRRATIONAL NUMBERSEarlier, we have learnt about irrational numbers and their properties. In this topic, we shall prove that p , where p is a prime number, are irrational numbers, by using the following theorem.

TheoremIf a prime number p divides a2, then p divides a, where a is a positive integer.Proof : Let prime factorisation of a is given by a = p1 × p2 × ..... × pn, where p1, p2, ...., pn are primes, not necessarily distinct.⇒ a2 = {p1 × p2 × ..... × pn}2 = p1

2 × p22 × ..... × pn

2

Now if p divides a2, then p is a factor of p12 × p2

2 × .... × pn2.

Now, as primes in the factorisation p12 × p2

2 × ..... × pn2 are unique [By Fundamental theorem of Arithmetic]

⇒ p is a prime, which is one out of primes p1, p2, ...., pn.Suppose p = pk for some value of k varying from 1 to n.Now, pk divides {p1 × p2 × ..... × pn}⇒ pk divides a ⇒ p divides a [Q pk = p]Hence, the theorem is proved.

Method of Proving Irrationality of NumbersIn order to prove irrationality of the numbers, we use the method of contradiction as discussed in the following illustrations.

ILLUSTRATIONS

Show that 2 is an irrational number.Sol. Let us assume that 2 is a rational number.So,wecanfindintegersa and b (b ≠ 0 and a, b are co-prime) such that 2 = a/b ⇒ 2b a=Squaring both sides, we get 2b2 = a2 ....(i)\ 2 divides a2 ⇒ 2 divides a ....(ii)\ a can be written as 2m, where m is an integer.Putting a = 2m in (i), we get2b2 = (2m)2 ⇒ 2b2 = 4m2 ⇒ b2 = 2m2

\ 2 divides b2 ⇒ 2 divides b ....(iii)From (ii) and (iii), 2 is a common factor of a and b which contradicts the fact that a and b are co-prime.\ Our supposition that 2 is rational is wrongHence, 2 is irrational.

Prove that 5 2 is an irrational number.Sol. Assume that 5 2 is rational. Then, we can findintegersa and b (b ≠ 0 and a, b are co-prime) such that

5 2 = ab

⇒ 25

= ab

, which is a rational number.

⇒ 2 is rational.But this contradicts the fact that 2 is irrational.So, 5 2 is irrational.

Show that 5 – 3 is an irrational number. (NCERT)

Sol. Assume that 5 – 3 is rational. Then, we can findintegersa and b (b≠0anda, b are co-prime) such that5 3− = a b/ ⇒ 5 – a b/ = 3Since, 5 and a/b are rational, so their difference will be rational.

⇒ 3 is rational.But this contradicts the fact that 3 is irrational.Hence, 5 – 3 is irrational.

Prove that 2 3+ is irrational. (NCERT Exemplar)

Sol. Let us assume that 2 3+ is rational.Then,wecanfindco-primepositiveintegersa and

b (b ≠ 0 and a, b are co-prime) such that 2 3+ = ab

⇒ ab−

=2 3

⇒ ab

ab

2

2 2 2 2 3+ − = [Squaring both sides]

⇒ ab

ab

2

2 1 2 2− = ⇒ a bab

2 2

22− =

Here, a bab

2 2

2− is rational [Q a, b are integers]

⇒ 2 is rational.which contradicts the fact that 2 is irrational number.\ Our assumption is wrong.Hence, 2 3+ is irrational.

12

YOURSELF15. Show that 3 is irrational.

16. Show that 3 2+ is irrational.

EXERCISE - 1.3

1. Prove that 5 is irrational.

2. Prove that 3 2 5+ is irrational.

3. Prove that the following are irrationals.

(i) 12

(ii) 7 5 (iii) 6 2+

Objective Type QuestionsMCQs (1 Mark)

1. 2 5 is(a) an integer (b) an irrational number(c) a rational number (d) none of these

2. ( )2 2+ is(a) an integer(b) a rational number(c) an irrational number(d) none of these

Fill in the Blanks (1 Mark)3. If y2 = 23/25, then y is a/an ______ number.

VSA Type Questions (1 Mark)4. Is (p – 3.14) a rational number, an irrational number or zero?

Short Answer Type QuestionsSA Type I Questions (2 Marks)

5. Show that 3 2 is an irrational number.

SA Type II Questions (3 Marks)

6. Show that ( )3 5 2+ is an irrational number.

7. Prove that p q+ is irrational, where p and q are primes. (NCERT Exemplar)

8. If ab is an irrational number, then prove that ( )a b+ is an irrational number.


9. Show that 7 is irrational.

10. Prove that 7 – 2 3 is an irrational number.

1. (b) : 2 is rational number and 5 is an irrational number. And the product of a rational number and an irrational number is irrational number.\ 2 5 is an irrational number.

2. (c) : Here, 2 is rational number and 2 is irrational number. And the sum of a rational number and an irrational number is an irrational number.\ ( )2 2+ is an irrational number.

For YOURSELF SolutionsVisit http://bit.ly/mtg-100-mt-tryyourself


13Real Numbers


3. Given, y2 = 23/25

⇒ y = ± 235

, which is an irrational number.

4. As p is an irrational number and 3.14 is a rational number. Hence, (p – 3.14) is an irrational number.

5. Let 3 2 be a rational number. Then, it will be of the form p/q, where p, q are co-prime integers and q ≠ 0.

Now, 3 2 23

= ⇒ =pq

pq

...(i)

So, p/3q is a rational number.⇒ 2 is a rational number.But, this contradicts the fact that 2 is an irrational number. Hence, 3 2 is an irrational number.

6. If possible, let ( )3 5 2+ be a rational number.

⇒ ( )3 5 2+ = pq

, where q ≠ 0; p, q are co-prime integers.

⇒ 3 + 5 + 2 3 5 = pq

⇒ = −2 15 8pq

⇒ = −2 15 8p qq

⇒ = −15 82p qq

, which is rational.

But this contradicts the fact that 15 is an irrational number.\ Our supposition is wrong.Hence, ( )3 5 2+ is an irrational number.

7. Let us suppose that p q+ be a rational number.Let p q r+ = , where r is rational.∴ = −p r qSquaring both sides, we have

p = r2 + q – 2r q ⇒ = + −q r q pr

2

2Since, r is rational ⇒ r2 is rational.Also, q and p are both rationals. [ p, q are primes.]

\ r q pr

2

2+ − is rational.

⇒ q is rational, which contradicts the fact that q , where q is prime, is an irrational number\ Our supposition is wrong.Hence, p q+ is an irrational number.

8. If possible let, ( )a b+ be a rational number, say r.Then, a b r+ = where, a and b are rational. ...(i)⇒ + =( )a b r2 2

⇒ a + b + 2 ab = r2 [Squaring both sides]

⇒ = − − ⇒ = − −22

22

ab r a b ab r a b ...(ii)

Since, r, a and b are rational numbers.

⇒ − +r a b2

2( ) is a rational number.

⇒ ab is a rational number.\ Our supposition is wrong, because ab is an irrational number.Hence, ( )a b+ is an irrational number.

9. Let us assume 7 be a rational number.

So,wecanfindintegersa and b (b ≠ 0) such that 7 = ab

,where a and b (≠ 0) are co-prime positive integers.So, 7b a= …(i)Squaring both sides, we get7b2 = a2. Therefore, 7 divides a2.⇒ 7 divides a. …(ii)So, we can write a = 7c, for some integer c.Putting a = 7c in (i), we get 7 7b c=⇒ 7b2 = 49c2 ⇒ b2 = 7c2 [Squaring both sides]\ 7 divides b2 ⇒ 7 divides b. …(iii)From (ii) and (iii), a and b have 7 as a common factor. But this contradicts the fact that a and b are co-primes.\ Our assumption that 7 is rational, is wrong.Hence, 7 is irrational.

10. Suppose 7–2 3 is a rational number\ Wecanfindtwoco-primeintegersa, b (b ≠ 0) such

that 7–2 3 = ab

⇒ − =7 2 3ab

⇒ −

=1

27 3a

b

Q 12

7 −

ab

is a rational number

⇒ 3 is a rational number.But this contradict the fact that 3 is an irrational number.Hence, our supposition is wrong.

Thus, 7 – 2 3 is an irrational number.

TOPIC 4 : REVISITING RATIONAL NUMBERS AND THEIR DECIMAL EXPANSIONS

In this topic, we will consider a rational number, say pqq( )↑ 0 and explore exactly when the decimal expansion of p

q

is terminating and when it is non-terminating repeating, without actually performing long division.Note : (i) In terminating decimal expansion, the number terminates after a finite number of steps in the process of division. For example, 2.375.(ii) In non-terminating repeating decimal expansion, the number does not terminate but repeats the particular number or period of numbers again and again in the process of division. For example, 3.232323.....

14

Some Important Theorems

Theorem 1

Let x be a rational number whose decimal expansion terminates, then x can be expressed in the form pq

, where

p and q are co-prime and the prime factorisation of q is of the form 2m × 5n, where m and n are non-negative integers.For example,

(i) 0.25 can be written as 25100

14

12 52 0= =

×

(ii) 0.773 can be written as 7731000

7732 2 2 5 5 5

7732 53 3=

× × × × ×=

×

Theorem 2 (Converse of Theorem 1)

If a rational number x is of the form pqq( )↑ 0 such that prime factorisation of q is of the form 2m × 5n, where m, n are

non-negative integers, then x has a terminating decimal expansion.

For example, 78

= 72 53 0 , which is of the form p

m n2 5, where m = 3, n = 0 and p = 7. So, 7

8 has a terminating decimal

expansion.

Theorem 3Let x = p/q be a rational number, such that the prime factorisation of q is not of the form 2m × 5n, where m, n are non-negative integers. Then x has a decimal expansion which is non-terminating repeating (recurring).

For example, consider 1935

195 7

=×

, since the denominator given rational number is not of the form 2m × 5n, where

m and n are non-negative integers, therefore 19/35 has a non-terminating repeating decimal expansion.

Method to Write Decimal Expansion of those Rational Numbers which have Terminating Decimal ExpansionLet p/q be a rational number in the lowest form such that the prime factorisation of q is of the form 2m × 5n, where m, n are non-negative integers.To write decimal expansion of p/q, convert p/q to an equivalent rational number of the form c/d, where d is a power of 10. We have the following cases :Case I : When m = n

In this case, we have pq

p p pm n m m m=×

=×

=2 5 2 5 10( )

Case II : When m > n, say m = n + a, where a is a positive integer.


p pm n

a

m n a=×

= ×× +2 5

52 5

[Multiplying numerator and denominator by 5a]

=××

=××

=p p ca

m m

a

m m5

2 55

2 5 10( ), where c = p × 5a

Case III : When m < n, say n = m + a, where a is a positive integer.


p pm n

a

m a n=×

= ××+2 52

2 5 [Multiplying numerator and denominator by 2a]

= ××

= ××

=p p ca

n n

a

n n2

2 52

2 5 10( ), where c = p × 2a

For example,

(i) 78

72

7 52 5

7 1252 5

87510

0 8753

3

3 3 3 3= = ××

= ××

= =( )

. (ii) 21391250

21392 5

2139 22 5

2139 82 5

1711210

1 711 4

3

4 4 4 4=×

= ××

= ××

= =( )

. 112

15Real Numbers


ILLUSTRATIONS

Express 33.340 in the form of pq

, where

p and q are co-prime and q is of the form 2m × 5n, where m, n are non-negative integers.Sol. We have a rational number 33.340 whose decimal expansion terminates.

Now, 33.340 = 333401000

3334102= = ×

×=

×2 16672 5

16672 52 2 2

Thus, 33.340 can be expressed as 16672 52×

, whose

denominator is of the form 2m × 5n, where m and n are non-negative integers.

Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion.

(i) 31625 (ii)

192 5 72 3 5. .

Sol. (i) Since 625 = 20 × 54, i.e., denominator 625 is of the form 2m × 5n (where m and n are non-

negative integers) therefore 31625

will have a

terminating decimal expansion.(ii) Since the denominator of given rational number is 22 ⋅ 53 ⋅ 75, which is not of the form

2m × 5n. So 192 5 72 3 5⋅ ⋅

will have non-terminating

repeating decimal.

Check whether 13125 has a terminating

decimalexpansionornot.Ifithas,findthedecimal expansion of it.Sol. Given, rational number is 13

125.

Prime factorisation of 125 = 53

Thus, the denominator of 13125

is of the form

2m × 5n, (where m and n are non-negative integers)

therefore 13125

will have a terminating decimal

expansion.

Now, 13125

135

13 25 23

3

3 3= = ××

= =10410

0 1043 . ,

which is the required decimal expansion.

Write the decimal expansion of 17172 52 3×

without actual division. (2017)

Sol. We have, 17172 52 3×

= ××

= =1717 22 5

343410

3 4343 3 3( ).

What can you say about the prime factors of the denominators of the given rationals:

(i) 44.123 (ii) 4.5678

Sol. (i) Since, 44.123 has terminating decimal expansion. So, its denominator will be of the form 2m × 5n, where m, n are non-negative integers and therefore it will have prime factors 2 or 5.

(ii) Since, 4.5678 has non-terminating repeating decimal expansion. So, its denominator will not be of the form 2m × 5n, where m, n are non-negative integers and therefore it will have prime factors other than 2 and 5.

YOURSELF17. Express 23.3408 in the form p/q, where p and q are co-prime and q is of the form 2m × 5n,

and m, n are non-negative integers. Ans. 145882 50 4×

18. Without actually performing long division, state whether 1292 152 3×

has a terminating or non-terminating repeating decimal expansion.

(Ans. Non-terminating repeating decimal expansion)

19. Find the smallest rational number by which 1/3 should be multiplied so that its decimal expansion terminates after one place of decimal. (Ans. 3/10)


16

1. Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:

(i) 133125

(ii) 178

(iii) 64455

(iv) 151600

(v) 29343

(vi) 232 53 2 (vii) 129

2 5 72 7 5 (viii) 6

15

(ix) 3550

(x) 77210

2. Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.3. The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form p/q, what can you say about the prime factors of q ?(i) 43.123456789(ii) 0.120120012000120000 ....(iii) 43.123456789

EXERCISE - 1.4

Objective Type QuestionsMCQs (1 Mark)1. The decimal expansion of the rational

number 372 52 ×

will terminate after

(a) one decimal place (b) two decimal places(c) three decimal places (d) four decimal places2. 2.13113111311113... is(a) an integer (b) a rational number(c) an irrational number(d) None of these3. 1.2348 is(a) an integer(b) a rational number(c) an irrational number(d) None of these

Fill in the Blanks (1 Mark)4. The decimal expansion of 9/125 is ______.5. The decimal expansion of the rational number 14587/1250 will terminate after ______ places of decimal. (NCERT Exemplar)

VSA Type Questions (1 Mark)6. Comment on the prime factorisation of the denominators of the 62.521521521....

Short Answer Type QuestionsSA Type I Questions (2 Marks)7. After how many places of decimals will the

decimal expansion of the rational number 272 53 4⋅terminate?

8. The decimal representation of 72 510 15×

will

terminate after how many places of decimals?

9. Write whether 2 45 3 203 5

+ is a rational or an irrational number.

10. State whether 111 3235 34

. +

is a rational

number or not.

SA Type II Questions (3 Marks)11. (i) Write the decimal expansion of 76/6250 without actual division.


17Real Numbers


1. (b) : We have,

372 5

37 52 5

1852 5

18510

185100

1 852 2 2 2 2×= ×

×=

×= = =

( ) ( ).

2. (c) : 2.13113111311113...... is a non-terminating, non-repeating decimal. So, it is an irrational number.

3. (b) : 1 2348. = 1.234848....... which is non-terminating repeating decimal expansion.

\ 1 2348. is a rational number.

4. We have, 9125

95

9 22 5

7210

0 0723

3

3 3 3= = ××

= = .

5. Q 1250 = 53 × 10

\ 145871250

14587 25 2 10

14587 210

3

3 3

3

4= ×× ×

= ×( )

So, the decimal expansion of the given rational number terminates after four decimal places.6. As 62.521521521... has non-terminating repeating decimal expansion, so its denominator has factor other than 2 or 5.

7. We have, 272 53 4⋅

Multiplying and dividing by 2, we get

27 22 2 5

542 5

5410

5410000

0 00543 4 4 4 4×

× ×=

×= = = .

Thus, the given expression will terminate after 4 decimal places.

8. We have, 72 5

7 22 2 510 15

5

10 5 15×= ×

× ×

= ××

= ××

= ×7 22 5

7 22 5

7 210

5

15 15

5

15

5

15( )

\ 7 210

5

15× will terminate after 15 decimal places.

9. We have, 2 45 3 203 5

2 3 5 3 2 53 5

+ = × + ×

= + =6 5 6 53 5

12 53 5

= 4, which is a rational number.10. Here, 111.3235 = 111.323535.... is a non-terminating repeating decimal.

So, 111.3235 is a rational number and 3/4 is in the form

of pq

, where q ≠ 0. Thus, 34

is also a rational number.

We know that, the sum of two rational numbers is a rational number.

Therefore, 111 3235 34

. +

is a rational number.

11. (i) We have, 766250

762 5

76 22 2 55

4

4 5=×

= ×× ×

=×

12162 55 5

=( )

= =121610

1216100000

0 012165 .

(ii) 271250

272 5

27 22 2 54

3

3 4=×

= ×× ×

= ××

27 82 54 4 = = =216

10216

100000 02164( )

.

12. Let us assume that 2 5+ is a rational number. Then,wecanfindtwoco-primepositiveintegersa and b(b ≠ 0) such that

2 5+ = ab

⇒ − =ab

2 5

⇒ −

=ab

2 52

2( ) [Squaring both sides]

⇒ − + =ab

ab

2

22 2 2 5 ⇒ − =a

bab

2

2 3 2 2

⇒ − =a bab

2 232

2

(ii) Write the decimal expansion of 271250

without

actual division.

12. Prove that 2 5+ is an irrational number.


13. Express 2.4178 in the form p/q and hence comment on the prime factorisation of q.

14. Prove that, if a, b, c and d are positive rationals such that a b c d+ = + , then either a = c and b = d or b and d are squares of rational number.15. The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational and of the form p/q, what can you say about the prime factors of q ?(i) 0.737373 … (ii) 2.317317 …(iii) 324.281456

18

⇒ 2 is a rational number

Q a b a b

ab, are integers is rational∴ −

2 232

This contradicts the fact that 2 is irrational. So, our assumption is wrong.Hence, 2 5+ is an irrational number.

13. Let x = 2.4178 = 2.4178178...Multiplying both sides by 10, we get10x = 24.178178.. ...(i)Multiplying both sides of (i) by 1000, we get10000x = 24178.178178... ...(ii)Subtracting (i) from (ii), we get9990x = 24154

⇒ x = 241549990

= 120774995

Hence, 2 4178 120774995

120775 9 111

. = =× ×

Hence, 4995 has prime factors other than 2 and 5.

14. Given, a b c d+ = +If a = c, then b d=⇒ b = d [Squaring both sides]If a ≠ c, then there exists a positive rational number x such that a = c + x.Now, a b c d+ = + ⇒ c + x + b c d= +⇒ x + b d= ...(i)⇒ (x + b )2 = ( )d 2 [Squaring both sides]

⇒ x2 + b + 2x b = d ⇒ x2 + 2x b + b – d = 0

⇒ 2x b = d – b – x2 ⇒ b d b xx

= − − 2

2Since, d, x and b are rational numbers and x > 0.

So, d b xx

− − 2

2is a rational number.

⇒ b is a rational number.⇒ b is the square of a rational number.From (i), d x b= +

⇒ d is a rational number

⇒ d is the square of a rational number.Hence, either a = c and b = d or b and d are the squares of rational number.15. (i) Q The given decimal expansion is non-terminating repeating.\ It is a rational number.Let x = 0.737373.... = 0 73. ...(i)⇒ 100x = 73.737373 = 73 73. ...(ii)On subtracting (i) from (ii), we get

⇒ 99x = 73 ⇒ x = 7399

=pq

Hence, p = 73 and q = 99, which are co-prime.Here, q = 32 × 11, i.e., prime factors of q are not of the form 2m × 5n, where m and n are non-negative integers.(ii) Q The given decimal expansion is non-terminating repeating.\ It is a rational number.Let x = 2.317317… …(i) 1000x = 2317.317317… …(ii)Subtracting (i) and (ii), we get 999x = 2315⇒ x = 2315/999 = p/qHence, p = 2315, q = 999, which are co-prime.Here, q = 32 × 111, i.e., prime factors of q are not of the form 2m × 5n, where m, n are non-negative integers.(iii) Q The given decimal expansion is a non-terminating repeating.\ It is a rational number.

Let pq = x = 324.281456 …(i)

Multiplying both sides by 1000000, we have 1000000x = 324281456. 281456 …(ii)Subtracting (i) from (ii), we get 1000000x – x = 324281456.281456 – 324.281456

⇒ x = 324281132

999999= pq

Hence, p = 324281132, q = 999999Here, q = 32 × 111111, i.e., prime factors of q are not of the form 2m × 5n, where m, n are non-negative integers.

19Real Numbers


Method of Proving Irrationality of NumbersTheorem

20

Real numbers are the collection of all rational numbers and irrational numbers.

Lemma : A lemma is a proven statement used for proving another statement.Algorithm : An algorithm is a series of well defined steps which gives a procedure for solving a problem.

Here, we have discussed about how to prove the irrationality

of numbers.

Euclid’s Division Lemma

• “Given positive integers a and b, there exist unique integers q and r satisfying a = bq + r, 0 ≤ r < b.”

• When a and b are positive integers, then q and r can take values only from whole numbers.

• If a < b, then in this case quotient, q = 0 and remainder, r = a.

If a prime number p divides a2, then p divides a, where a is a positive integer.

In order to prove irrationality of the numbers, we use the method of contradiction. i.e., first we assume that given number is rational and after reaching at contradiction, we say (prove) that given number is irrational.

Euclid’s Division Algorithm

• Euclid’s division algorithm is a technique to compute the Highest Common Factor (HCF) of two given positive integers. According to this, the HCF of any two positive integers a and b, with a > b, is obtained as follows :

Step 1 : Apply Euclid’s division lemma to a and b i.e., find whole numbers, q and r such that a = bq + r, 0 ≤ r < b.

Step 2 : If r = 0, b is the HCF of a and b. If r ≠ 0, apply the division lemma to b and r. Step 3 : Continue the process till the remainder is zero. The divisor at this stage will be the required HCF.• To find HCF of three numbers say a > b > c by Euclid’s Division lemma, first find HCF (a, b) = d(say) then find

HCF (d, c), which will give the required HCF of (a, b, c).

Euclid’s Division Lemma and Algorithm

Revisiting Irrational Numbers

The Fundamental Theorem of Arithmetic

Revisiting Rational Numbers and their

Decimal Expansions

Real numbers

Rational numbers

Whole numbers

Integers

Natural numbers

Irrational numbers

–10 –9 –8 –7 –6 –5 –4

–4 –1.5 8

1/2 = 0.5 p = 3.14159…

0

–3 –2 –1 1 2 3 4 5 6 7 8 9 100

Topic Topic

Topic Topic

Relationship between Numbers and their HCF and LCM

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. It is also known as unique factorisation theorem.

Here, we have discussed about when the decimal expansion of a rational number is terminating and when it is non-terminating repeating, without actually performing division.

Factor Tree

A chain of factors, which is represented in the form of a tree is called factor tree.

Theorem 1

Let x be a rational number whose decimal expansion terminates, then x can be expressed in the form p/q, where p and q are co-prime and the prime factorisation of q is of the form 2m × 5n, where m, n are non-negative integers.

HCF and LCM Using Prime Factorisation

• HCF = Product of the smallest power of each common prime factor in the numbers.

• LCM = Product of the greatest power of each prime factor involved in the numbers.

• If there is no common prime factor, then HCF of given numbers is 1.

z HCF (a, b) × LCM (a, b) = a × b

z HCF (a, b, c) = a b c a b ca b b c c a

× × ×× ×

LCMLCM LCM LCM

( , , )( , ) ( , ) ( , ) and LCM (a, b, c) = a b c a b c

a b b c c a× × ×

× ×HCF

HCF HCF HCF( , , )

( , ) ( , ) ( , )

Theorem 2

If a rational number x is of the form pqq( )↑ 0

such that prime factorisation of q is of the form 2m × 5n, where m, n are non-negative integers, then x has a terminating decimal.

Theorem 3

Let x = p/q be a rational number, such that the prime factorisation of q is not of the form 2m × 5n, where m, n are non-negative integers. Then, x has decimal expansion which is non-terminating repeating (recurring).

Method to write decimal expansion of those rational numbers which have terminating decimal expansion

Let p/q be a rational number in the lowest form such that the prime factorisation of q is of the form 2m × 5n, where m, n are non-negative integers.To write decimal expansion of p/q, convert p/q to an equivalent rational number of the form c/d, where d is a power of 10.


1. Show that square of any positive odd integer is of the form 8m + 1, for some integer m.Sol. Let a be any positive odd integer and b = 4.By division algorithm, there exists integers q and r such thata = 4q + r, where 0 ≤ r < 4Put r = 0, 1, 2, 3, we havea = 4q, a = 4q + 1, a = 4q + 2 and a = 4q + 3Since, a is an odd integer.\ a = 4q + 1 or a = 4q + 3Case I : When a = 4q + 1\ a2 = (4q + 1)2 = 16q2 + 8q + 1 = 8(2q2 + q) + 1 = 8m + 1, where m is an integer.Case II : When a = 4q + 3\ a2 = (4q + 3)2 = 16q2 + 24q + 9 = 16q2 + 24q + 8 + 1= 8(2q2 + 3q + 1) + 1 = 8m + 1, where m is an integer.Hence, square of any positive odd integer is of the form 8m + 1, where m is an integer.

2. Show that the cube of a positive integer is of the form 6q + r, where q is an integer and r = 0, 1, 2, 3, 4, 5.Sol. Let a be an arbitrary positive integer, then by Euclid’s division lemma, corresponding to the positive integers a and 6, there exist non-negative integers q and r such thata = 6q + r, where 0 ≤ r < 6Now, a3 = (6q + r)3 = 216q3 + 108q2r + 18qr2 + r3, where 0 ≤ r < 6 ...(i)Case I : If r = 0, then from (i), we geta3 = 216q3 = 6(36q3) = 6m + 0, where m = 36q3 is an integer.Case II : If r = 1, then from (i), we geta3 = 216q3 + 108q2 +18q + 1 = 6(36q3 + 18q2 + 3q) + 1 = 6m + 1, where m = 36q3 + 18q2 + 3q is an integerCase III : If r = 2, then from (i), we get

a3 = 216q3 + 216q2 + 72q + 8 = 6(36q3 + 36q2 + 12q + 1) + 2 = 6m + 2, where m = 36q3 + 36q2 + 12q + 1 is an integerCase IV : If r = 3, then from (i), we get a3 = 216q3 + 324q2 + 162q + 27 = 6(36q3 + 54q2 + 27q + 4) + 3 = 6m + 3, where m = 36q3 + 54q2 + 27q + 4 is an integerCase V : If r = 4, then from (i), we geta3 = 216q3 + 432q2 + 288q + 64 = 6(36q3 + 72q2 + 48q + 10) + 4 = 6m + 4,where m = 36q3 + 72q2 + 48q + 10 is an integerCase VI : If r = 5, then from (i), we geta3 = 216q3 + 540q2 + 450q + 125

= 6(36q3 + 90q2 + 75q + 20) + 5= 6m + 5, where m = 36q3 + 90q2 + 75q + 20 is an integer

Hence, cube of a positive integer is of the form 6q + r, where q is an integer and r = 0, 1, 2, 3, 4, 5.

3. Find the HCF of 81 and 237 and express it as a linear combination of 81 and 237.Sol. Given integers are 81 and 237 such that 81 < 237.Applying Euclid's division lemma to 81 and 237, we get237 = 81 × 2 + 75 ...(i)81 = 75 × 1 + 6 ...(ii)75 = 6 × 12 + 3 ...(iii)6 = 3 × 2 + 0Since, remainder is zero. So, 3 is the HCF of 81 and 237.Now, from (iii), we have 3 = 75 – (6 × 12)= 75 – (81 – 75 × 1) × 12 [Using (ii)]= 75 – 12 × 81 + 12 × 75 = [1 + 12] 75 – 12 × 81= 13 × 75 – 12 × 81= 13 × (237 – 81 × 2) – 12 × 81 [Using (i)]= 13 × 237 – 26 × 81 – 12 × 81= 13 × 237 – [26 + 12] × 81 = 13 × 237 – 38 × 81\ 3 = 237 a + 81 b, where a = 13 and b = –38

4. Find the HCF of 55 and 210. Represent it as a linear combination of 55 and 210 as 210a + 55b, for some a and b. Does this representation unique, if not, justify your answer.Sol. Here, 55 < 210. So, applying Euclid's division lemma to 210 and 55, we get210 = (55 × 3) + 45 ...(i)55 = (45 × 1) + 10 ...(ii)45 = (10 × 4) + 5 ...(iii) 10 = 5 × 2 + 0Here, remainder = 0. Thus, HCF of 210 and 55 = 5From (iii), we have 5 = 45 – (10 × 4) = 45 – [(55 – 45 × 1) × 4] [Using (ii)]= 45 – 55 × 4 + 45 × 4 = 5 × 45 – 55 × 4= 5 × (210 – 55 × 3) – 55 × 4 [Using (i)]= 210 × 5 – 55 × 19 ...(iv)⇒ 5 = 210a + 55b, where, a = 5 and b = –19From (iv), we have 5 = 210 × 5 – 55 × 19 = 210 × 5 – 55 × 19 + 210 × 55 – 210 × 55 [Adding and subtracting 210 × 55]= (210 × 5 + 210 × 55) + (–55 × 19 – 210 × 55)= 210(5 + 55) + 55(–19 – 210) ⇒ 5 = 210 × 60 + 55(–229)\ a = 60, b = –229Hence, a and b are not unique.

5. If the HCF of 657 and 963 is expressible in the form of 657x + 963 (–15), then find x.Sol. Given, HCF of 657 and 963 = 657 x + 963 (–15) ...(i)By using Euclid's division lemma to 657 and 963, we get963 = (657 × 1) + 306657 = (306 × 2) + 45

22

306 = (45 × 6) + 3645 = (36 × 1) + 936 = (9 × 4) + 0Here, remainder is zero when divisor is 9.\ HCF of 657 and 963 = 9 ...(ii)From (i) and (ii), we have, 9 = 657x + 963(–15)⇒ 657x = 9 + 963 × 15 = 9 + 14445

⇒ 657x = 14454 ⇒ x = 14454657

= 22

6. Can the number 16n, n being a natural number, end with the digit 0? Give reason.Sol. Given, n is a natural number.Let 16n ends with 0, then 16n is divisible by 2 and 5. But prime factors of 16 are 2 × 2 × 2 × 2\ 16n = (2 × 2 × 2 × 2)n = 24n

Thus, prime factorisation of 16n does not contain 5. So, by the uniqueness of fundamental theorem of arithmetic, which guarantees that there is no other prime in the factorisation of 16n. So, 5 does not occur in the prime factorisation of 16n. Hence, 16n can never end with digit 0, for any natural number.

7. If x = 28 + (1 × 2 × 3 × 4 × ... × 16 × 28) and y = 17 + (1 × 2 × 3 × ... × 17), then show that x + y is a composite number.Sol. Given, x = 28 + (1 × 2 × 3 × 4 × ...... × 16 × 28)= 28 [1 + (1 × 2 × 3 × ...... × 16)]\ x is a composite number, as it has 28 as a factorAlso, y = 17 + (1 × 2 × 3 × ...... × 17) = 17 [1 + (1 × 2 × ..... × 16)]\ y is a composite number, as it has 17, as a factorNow, x + y = (28 + 17) [1 + (1 × 2 × 3 × ...... × 16)] = 45 [1 + (1 × 2 × 3 × ..... 16)]\ x + y is a composite number as it has 45 as a factor.

8. Prove that 2 3 5+ is an irrational number. Also, check whether ( )( )2 3 5 2 3 5+ − is rational or irrational.Sol. Let 2 3 5+ is rational.Then, 2 3 5+ = a, where a is rational⇒ = −2 3 5aSquaring both sides, we get ( ) ( )2 3 52 2= −a⇒ 12 = a2 + 5 – 2a 5 ⇒ 2a 5 = a2 + 5 – 12

⇒ 5 72

2= −a

a

But, ( )aa

2 72− is a rational number, as a is a rational

number.⇒ 5 is a rational number.This contradicts the fact that 5 is an irrational numberSo, our assumption is wrong.

\ 2 3 5+ is an irrational number

Now, ( )( ) ( ) ( )2 3 5 2 3 5 2 3 52 2+ − = − = 12 – 5 = 7,

which is a rational number. So, ( )( )2 3 5 2 3 5+ − is a rational number.

9. If a and b are two rational numbers such that

a > b. Then, show that a a b

a a b

a a b

a a b

+ −

− −+ − −

+ −

2 2

2 2

2 2

2 2

is also a rational number.

Sol. We have, a a b

a a b

a a b

a a b

+ −

− −+ − −

+ −

2 2

2 2

2 2

2 2

=+ −( ) + − −( )

− −( )a a b a a b

a a b

2 22

2 22

2 2 22

= + − + − + + − − −− +

a a b a a b a a b a a ba a b

2 2 2 2 2 2 2 2 2 2

2 2 22 2

= − = −4 2 2 22 2

2

2 2

2a bb

a bb

( )

Q a, b are rational numbers and a > bSo, the given expression represents a rational number.

10. Let a, b and c be rational numbers such that p is not a perfect cube. If a + bp1/3 + cp2/3 = 0, then prove that a = b = c.Sol. We have, a + bp1/3 + cp2/3 = 0 ...(i)Multiplying both sides by p1/3, we getap1/3 + bp2/3 + cp = 0 ...(ii)Multiplying (i) by b and (ii) by c and subtracting, we get

(ab + b2 p1/3 + bcp2/3) – (acp1/3 + bcp2/3 + c2p) = 0⇒ (b2 – ac)p1/3 + ab – c2p = 0⇒ b2 – ac = 0 and ab – c2p = 0 [ p1/3 is irrational]⇒ b2 = ac and ab = c2p⇒ b2 = ac and a2b2 = c4p2

⇒ a2(ac) = c4p2 ⇒ a3c – p2c4 = 0⇒ (a3 – p2c3)c = 0⇒ a3 – p2c3 = 0 or c = 0Now, a3 – p2c3 = 0

⇒ = ⇒ =

p a

cp a

c2

3

32 1 3

3

3

1 3

( ) //

⇒ =

⇒ =( ) ( )/

//p a

cp a

c1 3 2

3 1 31 3 2

This is not possible as p1/3 is irrational and ac

is rational.

\ a3 – p2c3 ≠ 0Hence, c = 0Putting c = 0 in b2 – ac = 0, we get b = 0Putting b = 0 and c = 0 in (i), we get a = 0Hence, a = b = c = 0.

23Real Numbers


Multiple Choice Questions1. P = 2(4)(6)...(20) and Q = 1(3)(5) ... (19). What is the HCF of P and Q?(a) 33 . 5 . 7 (b) 34 . 5(c) 34 . 52 . 7 (d) 33 . 52

2. The least number which is a perfect square and is divisible by each of 16, 20 and 24 is(a) 240 (b) 2400 (c) 1600 (d) 36003. The following are the steps involved in finding the LCM of 72 and 48 by prime factorisation method. Arrange them in sequential order from first to last.(A) 72 = 23 × 32 and 48 = 24 × 31

(B) LCM = 24 × 32

(C) All the distinct factors with highest exponents are 24 and 32

(a) ABC (b) ACB (c) CAB (d) BCA4. If (– 1)n + (– 1)4n = 0, then n is(a) any positive integer(b) any negative integer(c) any odd natural number(d) any even natural number.

5. If a b c= − = + = +6 5 6 5 6 58 8 8 8 6 6, , ,

d = +6 54 4 and e = +6 5 , then which of the following is a rational number?(a) abcde (b) abde(c) abc (d) cd6. Find the value of x and state which of the given statements is/are required for it.I. The LCM of x and 18 is 36.II. The HCF of x and 18 is 2.(a) Only statement-I is required(b) Only statement-II is required(c) Both statement-I and II are required(d) None of these7. Which of the following rational numbers are terminating decimals?

(a) 17

2 53 2× (b)

253 22 3×

(c) 68

2 5 72 2 2× ×

(d) 1253 73 2×

8. Which of the following rational numbers have non-terminating repeating decimal expansion?

(a) 313125

(b) 71512

(c) 23200 (d) None of these

9. P is the LCM of 2, 4, 6, 8, 10, Q is the LCM of 1, 3, 5, 7, 9 and L is the LCM of P and Q. Then, which of the following is true?(a) L = 21P (b) L = 4Q(c) L = 63P (d) L = 16Q10. If a natural number ‘a’ is divided by 7, the remainder is 5. If a natural number ‘b’ is divided by 7, the remainder is 3. The remainder is ‘r’, if

a + b is divided by 7. Find the value of 3 54

r + .

(a) 7 (b) 2 (c) 8 (d) 11

Match the ColumnsMatch Column I with Column II and select the correct answer by choosing an appropriate option.11. Match the HCF given in Column II for the given pair of numbers given in Column I. Column I Column IIP. 105 and 120 1. 9Q. 196 and 38220 2. 24R. 240 and 6552 3. 15S. 963 and 657 4. 196(a) P-3, Q-4, R-2, S-1 (b) P-4, Q-2, R-3, S-1(c) P-2, Q-1, R-3, S-4 (d) P-3, Q-4, R-1, S-212. Column I Column IIP. 63 1. is a non-terminating but repeating decimal.Q. 0.3465 2. terminates after 3 places of decimal.R.

1234625 3. is an irrational

number.S.

3416125 4. terminates after 4

places of decimal.(a) P-2, Q-3, R-4, S-1 (b) P-1, Q-3, R-2, S-4(c) P-3, Q-1, R-4, S-2 (d) P-4, Q-2, R-3, S-1

24

Comprehension TypeDirection (Q. No. 13 to 15) : Read the given passage and answer the following questions.Every composite number can be expressed as a product of primes and this factorisation is unique except the order in which prime factors occurs.13. Express 13915 as a product of prime factors.(a) 3 × 53 × 7 × 11 (b) 32 × 52 × 7 × 11(c) 5 × 112 × 23 (d) 5 × 32 × 2114. Determine the prime factorisation of 20570.(a) 2 × 5 × 112 × 17 (b) 10 × 112 × 17(c) 5 × 34 × 121 (d) 102 × 17 × 1115. Write the prime factorisation of 58500.(a) 49 × 13 × 192 × 3 (b) 22 × 32 × 53 × 13(c) 7 × 13 × 17 × 91 (d) 7 × 132 × 17 × 19 × 21Direction (Q. No. 16 to 18) : Read the given passage and answer the following questions.For two numbers, a and b, relationship between the HCF, LCM and product of the numbers is given by the following relationship :HCF (a, b) × LCM (a, b) = a × b16. The HCF of two numbers, 1261 and 1067 is 97. Find their LCM.(a) 18222 (b) 1212(c) 13871 (d) 3647017. The LCM and HCF of two numbers is 5665 and 103 respectively. If one of the numbers is 515, then find the other number.(a) 1123 (b) 1133(c) 1130 (d) 114318. The sum of two numbers is 135 and their HCF is 27. If their LCM is 162, then the numbers are(a) 108, 27 (b) 72, 54(c) 81, 54 (d) 99, 36

Assertion & ReasonEach of the following questions contains two statements : Statement-I (Assertion) and Statement-II (Reason) followed by four options, out of which only one is correct. Select the correct option.(a) Both Statement-I and Statement-II are true. (b) Both Statement-I and Statement-II are false. (c) Statement-I is true, Statement-II is false.(d) Statement-I is false, Statement-II is true.

19. Statement-I : HCF (11, 21) = 3

Statement-II : A real number 2 3 72 5 3 7

2 2 3

2 3 5 4× ×

× × ×

is a terminating decimal.20. Statement-I : If product of two integers is 15 × 72 and their LCM is 360, then their HCF = 3.Statement-II : LCM × Product of integers = HCF21. Statement-I : HCF of (12, 21) is 3.Statement-II : If p and q are primes, then HCF (p, q) = 1

22. Statement-I : 1236250 is a terminating

decimal.Statement-II : The rational number is a terminating decimal, if denominator is of the form (2m × 5n) for some whole numbers m and n.

23. Statement-I : 7 is a rational number.Statement-II : If p be a prime, then p is an irrational number.

Integer / Numerical Value

24. Find the minimum number by which 980110500

must be multiplied to make it a terminating decimal.25. On dividing 13401 by a certain number, we get 82 as quotient, 35 as remainder and (14)2 – k as divisor. Find k.26. The greatest number that will divide 103, 127 and 175, so as to leave remainder 55 in each case is k × 22. Find the value of k.27. Three numbers are in the ratio 5 : 6 : 7 and their HCF is 7. Then, find the square root of largest number.28. The length, breadth and height of a room are 8 m 25 cm, 6 m 75 cm and 4 m 50 cm respectively. Find the length (in cm) of the longest rod that can measure the three dimensions of the room exactly.29. If x3 + y = 2249, where x and y are natural numbers and HCF of x and y is not 1, then find the value of (x + y).30. The lowest common multiple of two numbers is 14 times their highest common factor. The sum of LCM and HCF is 600. If one number is 80, then other number is

25Real Numbers


1. (c) : We have, P = 2 ⋅ 4 ⋅ 6 ⋅ 8 ⋅ 10 ⋅ 12 ⋅ 14 ⋅ 16 ⋅ 18 ⋅ 20= (2) ⋅ (22) (2 × 3) (23) (2 × 5) (22 × 3) (2 × 7) (24) (2 × 32) ⋅ (22 × 5) = 218 × 34 × 52 × 7Similarly, Q = 1 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 9 ⋅ 11 ⋅ 13 ⋅ 15 ⋅ 17 ⋅ 19= 34 × 52 × 7 × 11 × 13 × 17 × 19\ HCF of P and Q = 34 × 52 × 72. (d) : LCM of 16, 20 and 24 is 240.The least multiple is 240 but it is not a perfect square. Similarly, 2400 is also ruled out because it is not a perfect square. 1600 is the divided by 16 and 20 but not by 24. Therefore, 3600 is the least number, which is a perfect square and divisible by 16, 20 and 24.3. (b) : ACB is the correct sequential order.4. (c) : (– 1)n + (– 1)4n = 0 will be possible only, when n is any odd natural number.5. (b) : Clearly, cd is an irrational number.Now, ab = −( ) +( )6 5 6 58 8 8 8

= −( )6 54 4 , which is irrational.Also, ( )( )ab d = −( ) +( )6 5 6 54 4 4 4

= −6 5 , which is irrational.Also, abde = 6 5 6 5−( ) +( )= 6 – 5 = 1, which is a rational number.6. (c) : We know that, HCF (a, b) × LCM (a, b) = a × bHCF(x, 18) × LCM (x, 18) = x × 18⇒ 2 × 36 = x × 18 ⇒ x = 4\ Both the statements are required.

7. (a) : Clearly, 172 53 2×

is the only terminating decimal

as denominator is of the form 2m × 5n and the remaining are non-terminating decimals.8. (d) : 3125, 512 and 200 has factorisation of the form 2m × 5n, m and n are non-negative integers, so fractions have terminating decimal expansion.9. (a) : P is the LCM of 2, 4, 6, 8, 10.\ P = 3(8)(5)Q is the LCM of 1, 3, 5, 7, 9.\ Q = 5(7)(9)Now, L is the LCM of P and Q\ L = 3(8)(5)21 or 5(7)(9)8, i.e., 21P or 8Q.10. (b) : We have, a = 7l1 + 5 …(i)and b = 7l2 + 3 …(ii)where l1 and l2 are whole numbers.Adding (i) and (ii), we get a + b = 7(l1 + l2) + 8⇒ a + b = 7(l1 + l2) + 7 + 1 = 7(l1 + l2 + 1} + 1 = 7l + 1, where l = l1 + l2 + 1 is whole number.⇒ (a + b) when divided by 7 will give remainder 1.⇒ r = 1

Thus, 3 54

3 1 54

2r + = × + =

11. (a) : (P) Prime factors of 105 = 3 × 5 × 7 Prime factors of 120 = 2 × 2 × 2 × 3 × 5\ HCF (105, 120) = 3 × 5 = 15(Q) Prime factors of 196 = 2 × 2 × 7 × 7 Prime factors of 38220 = 2 × 2 × 3 × 5 × 7 × 7 × 13\ HCF (196, 38220) = 2 × 2 × 7 × 7 = 196(R) Prime factors of 240 = 2 × 2 × 2 × 2 × 3 × 5 Prime factors of 6552 = 2 × 2 × 2 × 3 × 3 × 7 × 13\ HCF (240, 6552) = 2 × 2 × 2 × 3 = 24(S) Prime factors of 963 = 3 × 3 × 107 Prime factors of 657 = 3 × 3 × 73\ HCF (963, 657) = 3 × 3 = 9

12. (a) : (P) 63 is an irrational number.(Q) 0.3465 is non-terminating but repeating decimal.

(R) We have, 1234625

2345 2

2 1234 16104 4

44= 1

×× = ×

( )So, it terminates after 4 places of decimal.

(S) 3416125

34165 2

2 3416 8103 3

33=

×× = ×

( )So, it terminates after 3 places of decimal.13. (c) : 5 13915

11 278311 25323 23

1

\ 13915 = 5 × 112 × 2314. (a) : 2 20570

5 1028511 205711 18717 17

1\ 20570 = 2 × 5 × 112 × 17

15. (b) : 2 585002 292503 146253 48755 16255 3255 65

13 131

\ 58500 = 22 × 32 × 53 × 1316. (c) : Given, HCF = 97

Now, LCM = Product of given two numbersHCF

= × =1261 106797

13871

17. (b) : Given, LCM = 5665, HCF = 103Also, one number = 515

26

Let other number be x.

⇒ x = 5665 103515

1133× =

18. (c) : HCF = 27So, let the numbers be 27a and 27bNow, 27a + 27b = 135⇒ a + b = 5 ... (i)Also, 27a × 27b = 27 × 162⇒ ab = 6 ... (ii)Now, (a – b)2 = (a + b)2 – 4ab⇒ (a – b)2 = (5)2 – 4(6) [Using (i) and (ii)]⇒ a – b = ±1 ... (iii)Solving (i) and (iii), we geta = 3, b = 2 or b = 3, a = 2So, numbers are 27 × 3, 27 × 2 i.e., 81, 54.19. (b) : HCF (11, 21) = 1So, Statement-I is false.Since the denominator of the given real number is not of the form 2m × 5n.So, it is a non-terminating repeating decimal.\ Statement-II is also false.20. (c) : Q LCM × HCF = Product of two numbers

⇒ 360 × HCF = 15 × 72 ⇒ HCF = × =15 72360

321. (a) : Clearly, statement-II is trueNow, prime factorisation of 12 = 22 × 3Prime factorisation of 21 = 3 × 7\ HCF (12, 21) = 3, which is true

22. (a) : 1236250

1232 3125

1232 51 5=

×=

×Q Denominator is of the form 2m × 5n,

∴ 1236250

is terminating decimal. So, Statement-I is true.

23. (d)

24. (7) : We have, 980110500

= 3 112 5 7 3

3 112 5 7

4 2

2 3

3 2

2 3×

× × ×= ×

× ×.

The denominator is not the form of 2m × 5n, where m, n are whole numbers, so it is not a terminating decimal. To make it a terminating decimal least number to be multiplied is 7.

25. (33) : By Euclid division lemma, we have a = bq + rHere a = 13401, q = 82 and r = 35\ 13401 = b × 82 + 35 ⇒ b = 163Now, 163 = (14)2 – k⇒ 163 = 196 – k ⇒ k = 196 – 163 = 3326. (6) : Required number= HCF of (103 – 55), (127 – 55) and (175 – 55)= HCF of 48, 72 and 120Now, prime factorisation of 48, 72 and 120 is,48 = 2 × 2 × 2 × 2 × 372 = 2 × 2 × 2 × 3 × 3120 = 2 × 2 × 2 × 3 × 5\ HCF of 48, 72 and 120 = 2 × 2 × 2 × 3 = 6 × 22

⇒ k × 22 = 6 × 22 ⇒ k = 627. (7) : Let the required numbers be 5x, 6x and 7x.Then their HCF = x.So, x = 7 (Given)\ The numbers are 35, 42 and 49.\ Square root of largest number = 49 7= .28. (75) : The required length of the rod is the HCF of 825 cm, 675 cm, 450 cm.LetusfirstfindHCFoffirsttwonumbers825and675.825 = 675 × 1 + 150675 = 150 × 4 + 75150 = 75 × 2 + 0\ HCF (825, 675) = 75Now,letusfindHCFof75and450.Now, 450 = 75 × 6 + 0\ HCF (450, 75) = 75Hence, HCF (825, 675, 450) = 75Hence, the length of the longest required rod is 75 cm.29. (65) : We have, x3 + y = 2249 …(i)Put x = 13; 133 = 2197\ From (i), y = 2249 – 2197 = 52HCF (13, 52) = 13 ≠ 1 \ x + y = 13 + 52 = 6530. (280) : Let HCF be x, then LCM = 14xWe have, 14x + x = 600 ⇒ x = 40Let other number be yAs we know that, product of two numbers = product of their LCM and HCF⇒ y × 80 = 40 × 14 × 40 ⇒ y = 280

27Real Numbers


SECTION - A

Multiple Choice Type Questions1. Find the HCF of the numbers 125 and 240 using Euclid’s division algorithm.(a) 5 (b) 10(c) 15 (d) 252. A positive integer n when divided by 9, gives 7 as remainder. What will be the remainder, when (3n – 1) is divided by 9?(a) 1 (b) 2(c) 3 (d) 43. The sum of three non-zero prime numbers is 100. If one of them exceeds the other by 36, then the largest number is(a) 73 (b) 91(c) 67 (d) 574. The values of x and y in the given figure are :

y 3

7

4

x

(a) x = 10 ; y = 14 (b) x = 21 ; y = 84(c) x = 25 ; y = 21 (d) x = 14 ; y = 405. Two numbers are in the ratio of 15:11. If their HCF is 13, then numbers will be(a) 195 and 143 (b) 190 and 140(c) 185 and 163 (d) 185 and 1436. Find the least number, which when divided by 12 leaves 7 as a remainder, when divided by 15 leaves 10 as a remainder and when divided by 16 leaves the remainder as 11.(a) 115 (b) 230(c) 247 (d) 235

Time: 1 hr 30 mins. Max. Marks: 44

(i) All the questions are compulsory.(ii) The question paper consists of 22 questions.(iii) Section A comprises of 10 questions of 1 mark each. Section B comprises of 5 questions

of 2 marks each. Section C comprises of 4 questions of 3 marks each. Section D comprises of 3 questions of 4 marks each.

(iv) Use of calculators is not permitted.

Fill in the Blanks7. If two positive integers a and b are written as a = x5y3 and b = x3y4, where x, y are prime numbers, then HCF (a, b) = ______.8. If least prime factor of p is 3 and least prime factor of q is 5, then the least prime factor of (p + q) is ______.

VSA Type Questions9. If p is prime, then find the HCF and LCM of p and (p + 1).

10. Is 27 a terminating or non-terminating decimal expansion?

SECTION - B

11. Find the greatest possible speed in (km/h) at which a bird should fly to cover a distance of 45 km and 336 km in exact number of hours.12. Express each number as a product of its prime factors.(i) 429(ii) 732513. Explain whether the following number is prime or composite number (3 × 5 × 13 × 46) + 23.14. Without actually performing the long

division, state whether 543225

has a terminating

decimal expansion or non-terminating recurring decimal expansion.15. Which of the integers (99, 101, 176, 182) has most number of divisors ?

28

SECTION - C16. Use Euclid's division algorithm to find the HCF of 595 and 721.17. Use Euclid's division algorithm to find the HCF of 4182 and 15540.18. Express 1.37 in the form of p/q, q ≠ 0 and hence discuss about prime factors of q.19. If p is prime, n is a positive integer, n + p = 2000 and LCM of n and p is 21879, then find the value of n and p.

SECTION - D

20. Show that one and only one out of n, n + 2 or n + 4 is divisible by 3, where n is any positive integer.

21. If d is the HCF of 56 and 72, find a, b satisfying d = 56a + 72b. Also, show that a and b are not unique.

22. Show that there is no positive integer n, for which n n+ + −1 1 is rational.

For EXAM DRILL Solutions visit http://bit.ly/mtg-100-mt-examdrill

29Real Numbers

ACTIVITYAimTo find the HCF of two numbers experimentally based on Euclid Division Lemma.

Pre-requisite Knowledge(i) Euclid's Division Lemma : Given two positive integers a and b, there exist unique integers q and r satisfying

a = bq + r, 0 ≤ r < b.(ii) Euclid's Division Algorithm : To obtain the HCF of two positive integers, say a and b, a > b, follow the steps

given below. Step 1 : Apply Euclid's division lemma to a and b, to find whole numbers q and r, such that a = bq + r, 0 ≤ r < b. Step 2 : If r = 0, b is the HCF of a and b. If r ≠ 0, apply the division lemma to b and r. Step 3 : Continue the process till the remainder is zero. The divisor at this stage will be the required HCF.

Objects Required(i) Cardboard sheets (ii) Glazed papers of different colours (iii) Scissors (iv) Ruler (v) Sketch pen (vi) Glue.

Steps of Construction(i) Cut out one strip of length x units, one strip of length y units (y < x), two strips each of length z units (z < y), one

strip of length u units (u < z) and two strips each of length v units (v < u) from the cardboard such that x = y + z, y = 2z + u, z = u + v and u = 2v.

(ii) Cover these strips using glazed papers of different colours as shown in Fig. 1 to Fig. 5:

x unitsFig. 1

Fig. 2y units

LAB-ACTIVITIES

Fig. 3 Fig. 4z units z units u units

Fig. 5v units v units

(iii) Stick these strips on the other cardboard sheet as shown in Fig. 6 to Fig. 9.

y unitsFig. 6

x units

z units

y units

Fig. 7

z units z units

u units

Fig. 9Fig. 8

z units

u units

v units

u units

v units v units

DemonstrationAs per Euclid Division Lemma,Fig. 6 depicts x = y × 1 + z (Here, q = 1, r = z) ...(i)Fig. 7 depicts y = z × 2 + u (Here, q = 2, r = u) ...(ii)Fig. 8 depicts z = u × 1 + v (Here, q = 1, r = v) ...(iii)and Fig. 9 depicts u = v × 2 + 0 (Here, q = 2, r = 0) ...(iv)As per the Euclid Division Algorithm,HCF of x and y = HCF of y and z = HCF of z and u = HCF of u and vBut HCF of u and v is equal to v, from (iv) above.So, HCF of x and y = v.

ObservationOn actual measurement (in mm) x = ..........., y = ........., z = ........., u = .........., v = ........So, HCF of ................ and ................ = ................

UsesThe Euclid‘s Division Lemma can be used for finding the HCF of two or more positive integers, which is known as finding HCF of positive integers by Division Method.

31Lab-Activities

1. What do you mean by HCF?Ans. HCF of two positive integers a and b is the largest positive integer d that divides both a and b exactly.

2. What is the HCF of smallest composite number and the smallest prime number?Ans. The smallest composite number is 4 and the smallest prime number is 2. Therefore, their HCF is 2.

3. Can the value of q in Euclid's division lemma be zero? Give recision.

Ans. Yes, when a < b then a is written as a = b· 0 + a, where q = 0 and r = a.

4. Can Euclid's division algorithm be extended for integers?Ans. Yes, but except 0.

5. If a = 5q + r, then how many values r can take?Ans. 5, because r is a whole number and 0 ≤ r < 5.

VOCEVIVAThis section is specially designed to prepare you for viva voce.

32 100PERCENTMathematics Class-10

real numbers - snd public school

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