1 REAL NUMBERS CHAPTER

CHAPTER
1
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REAL NUMBERS
Points to Remember :
1. Euclid’s division lemma : Given positive integers a and b, there exists whole numbers q and r satisfying
a = bq + r, 0  r < b.
2. Euclid’s division algorithm : This is based on Euclid’s division lemma. According to this, the HCF of any
two positive integers a and b, with a > b, is obtained as follows :
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Apply Euclid’s division lemma to find q and r where a = bq + r, 0  r < b.
If r = 0, the HCF is b. If r  0, apply the Euclid’s lemma to b and r.
Continue the process till the remainder is zero. The divisor at this stage will be HCF (a, b).
3. The fundamental theorem of arithmetic : Every composite number can be expressed (factorised) as a
product of primes, and this factorisation is unique, except for the order in which the prime factors occur.
4. For any two positive integers a and b, HCF (a, b) × LCM (a, b) = a × b.

ab
HCF (a, b) 
and
LCM(a, b)
LCM (a, b) 
a b
HCF(a, b)
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5. For any three positive integes a, b and c, we have
and
B
HCF (a, b, c) 
a  b  c  LCM (a, b, c)
LCM (a, b)  LCM (b, c)  LCM (a, c )
LCM (a, b, c) 
a  b  c  HCF (a, b, c )
HCF (a, b)  HCF (b, c )  HCF (a, c)
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6. Let a be a positive integer and p be a prime number such that p/a2, then p/a.
7. If p is a positive prime, then
p is an irrational number..
p
8. Let x be a rational number, whose decimal expansion terminates. Then we can express x in the form q ,
where p and q are co-prime and the prime factorisation of q is of the form 2n5m, where n, m are nonnegative integers.
p
be a rational number, such that the prime factorisation of q is of the form 2n5m, where n, m are
q
non-negative integers. Then x has a decimal expansion which terminates.
9. Let x 
10. Let x 
p
be a rational number, such that the prime factorisation of q is not of the form 2n5m, where n,
q
m are non-negative integers. Then x has a decimal expansion which is non terminating repeating (recurring).
MATHEMATICS–X
REAL NUMBERS
1
ILLUSTRATIVE EXAMPLES
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Example 1. Use Eculid’s division algorithm to find the HCF of 135 and 225.
Solution. Since 225 > 135. Let a = 225, b = 135.
135
1
135
90
Applying Eculid’s division algorithm, we get
225 = 135 × 1 + 90
Since the remainder 90  0. So, we apply the euclid’s division
algorithm to 135 and 90, to get
90 135 1
90
45
135 = 90 × 1 + 45
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Again, since the remainder is non-zero, so we apply the euclid’s division algorithm to 90 and 45, to
get
45 90 2
90
0
90 = 45 × 2 + 0
We observe that remainder at this stage is zero.  the divisor at this stage i.e. 45 is the HCF of 135
and 225.
B
Symbolically we write HCF (135, 225) = 45 Ans.
Example 2. 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?
[NCERT]
Solution.
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Clearly, the maximum number of columns in which members can march is the HCF of 616 and 32. So,
let us find the HCF of 616 and 32 by Euclid’s division algorithm.
32 616 19
32
Since, 616 > 32. Let a = 616, b = 32
296
288
Applying Euclid’s division algorithm, we get
8
616 = 32 × 19 + 8
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Since the remainder is non-zero, so we apply the Euclid’s
division algorithm to 32 and 8, to get
32 = 8 × 4 + 0
8 32 4
32
0
We observe that the remainder is zero at this stage.
 The divisor at this stage i.e. 8 is the HCF of 616 and 32.
Hence, the required maximum number of columns is 8.
Example 3. Use Eculid’s division demma, to show that the cube of any positive integer is of the form
9m, 9m + 1 or 9m + 8.
[NCERT]
Solution.
Let x be any positive integer. Then, it is of the form 3q or, 3q + 1 or, 3q + 2.
So, we have the following cases :
Case I : When x = 3q.
then, x3 = (3q)3 = 27q3 = 9 (3q3) = 9m, where m = 3q3.
2
REAL NUMBERS
MATHEMATICS–X
Case II : When x = 3q + 1
then, x3 = (3q + 1)3
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= 27q3 + 27q2 + 9q + 1
= 9 q (3q2 + 3q + 1) + 1
= 9m + 1, where m = q (3q2 + 3q + 1)
Case III. When x = 3q + 2
then, x3 = (3q + 2)3
= 27 q3 + 54q2 + 36q + 8
= 9q (3q2 + 6q + 4) + 8
= 9 m + 8, where m = q (3q2 + 6q + 4)
Hence, x is either of the form 9 m or 9 m + 1 or, 9 m + 8.
3
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Example 4. Express each of the following number as a product of its prime factors :
(i) 140
(ii) 3825
(iii) 5005
Solution.
(i)
Using the factor tree for prime factorisation, we have
140
70
2
B
35
2
5
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7
 140 = 2 × 2 × 5 × 7 = 22 × 5 × 7
(ii) Using the factor tree for prime factorisation, we have
3825
1275
3
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
3
[NCERT]
425
85
5
17
5
3825 = 3 × 3 × 5 × 5 × 17
= 32 × 52 × 17
(iii) Using the factor tree for prime factorisation, we have

5005
1001
5
143
7
11
13
5005 = 5 × 7 × 11 × 13
MATHEMATICS–X
REAL NUMBERS
3
Example 5. Find the HCF of 96 and 404 by prime factorisation method. Hence, find their LCM.
Solution.
We have, 96 = 25 × 3 and 404 = 22 × 101
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 HCF = 22 = 4.
Now, HCF × LCM = 96 × 404

LCM 
96  404 96  404

 96  101 = 9696 Ans.
HCF
4
Example 6. Check whether 6n can end with the digit 0 for any natural number n.
Solution.
n
n
n
[NCERT]
n
We have, 6 = (2 × 3) = 2 × 3 . Therefore, prime factorisation of 6 does not contain 5 as a factor.
Hence, 6n can never end with the digit 0 for any natural number n.
Example 7. Prove that
Solution.
n
2 is an irrational number..
Let us assume, to the contrary, that
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2 is rational. So, we can find integers r and s ( 0) such that
r
2  . Suppose r and s have a common factor other than 1. Then, we divide by the common
s
factor to get
2
a where a and b are co-prime.
,
b
B
So, b 2  a.
Squaring on both sides, we get 2b2 = a2. Therefore, 2 divides a2  2 divides a.
So, we can write a = 2c for some integer c.
Substituting for a, we get 2b2 = 4c2, or b2 = 2c2. This means that 2 divides b2, and so 2 divides b.
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Therefore, a and b have atleast 2 as a common factor. But this contradicts the fact that a and b
have no common factors other than 1.
This contradiction has arisen because of our incorrect assumption that
So, we conclude that
2 is irrational. Hence proved.
Example 8. Prove that 3+ 2 5 is irrational.
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Solution.
2 is rational.
[NCERT]
Let us assume on the contrary that 3  2 5 is rational. Then there exists co-prime integers a and
b such that,
32 5 
a
b
a
3
b

2 5

5

5 is rational.
a  3b
2b
[ a, b are integers, 
This contradicts the fact that
a  3b
is a rational]
2b
5 is irrational. So, our supposition is incorrect.
Hence, 3  2 5 is an irrational number..
4
REAL NUMBERS
MATHEMATICS–X
Example 9. Without actually performing the long division, state whether the following rational number will
have a terminating decimal expansion or a non-terminating repeating decimal expansion :
(i)
Solution.
13
3125
(ii)
17
8
(iii)
64
455
(iv)
15
1600
(v)
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29
343
[NCERT]
13
13
13


3125 55 20  55
So, the denominator is of the form 2m × 5n. Hence, it has terminating decimal expansion.
(i)
17 17
17
 3  3 0
8 2
2 5
So, the denominator is of the form 2m × 5n. Hence, it has terminating decimal expansion.
(ii)
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64
64

455 5  7  13
Clearly, the denominator is not of the form 2m × 5n. Hence, it has non-terminating repeating decimal
expansion.
(iii)
(iv)
15
3
3

 6 1
1600 320 2  5
B
So, the denominator is of the form 2m × 5n. Hence, it has terminating decimal expansion.
(v)
29 29

343 35
Clearly, the denominator is not of the form 2m × 5n. Hence, it has non-terminating repeating decimal
expansion.
Example 10. The following real numbers have decimal expression as given below. In each case decide whether
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they are rational or not. If they are rational, and of the form
prime factors of q?
Solution.
(i)
43.123456789
(i)
p
43.123456789 is a rational number of the form q and q is of the form 2m × 5n.
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(ii) 0.120120012000120000....
p
, what can you say about the
q
(iii) 43.123456789
[NCERT]
(ii) 0.120120012000120000 .... is not a rational number since it has non-recurring and non-terminating decimal and q is not of the form 2m × 5n.
(iii) 43.123456789 is a rational number having recurring and non-terminating decimal expansion
since q is not of the form 2m × 5n.
PRACTICE EXERCISE
1. Using Euclid’s division algorithm, find the HCF of :
2.
(i) 210 and 55
(ii) 117 and 65
(iii) 240 and 1024
(iv) 391 and 425
(v) 1288 and 575
(vi) 155 and 1385
Find the largest number which exactly divides 280 and 1245 leaving remainders 4 and 3, respectively.
3. Find the HCF of 300, 540, 890 by using Euclid’s division algorithm.
MATHEMATICS–X
REAL NUMBERS
5
4. A merchant has 105 litres of oil of one kind, 140 litres of another kind and 175 litres of third kind. He wants
to sell the oil by filling the three kinds of oil in tins of equal capacity. What should be the greatest
capacity of such a tin?
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5. The length, breadth and height of a room are 8 m 25 cm, 6 m 75 cm and 4 m 50 cm respectively. Determine
the length of largest rod which can measure the three dimensions of the room exactly.
6. In a seminar, the number of participants in Hindi, English and Mathematics are 60, 84 and 108, respectively. Find the minimum number of participants are to be seated and all of them being in the same
subject.
7. Show that every positive even integer is of the form 2q, and that every positive odd integer is of the form
2q + 1, where q is some integer.
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8. Show that any positive integer is of the form 3q or, 3q + 1, or 3q + 2 for some integer q.
9. Show that any positive odd integer is of the form 4q + 1, or 4q + 3, where q is some integer.
10.
11.
12.
13.
Show that the square of any positive integer is of the form 3 m or, 3 m + 1 for some integer m.
Show that one and only one out of n, n + 2 or, n + 4 is divisible by 3, where n is any positive integer.
Show that one and only one out of n, n + 3, n + 6, n + 9 is divisible by 4.
Express each of the following positive integers as the product of its prime factors :
(i) 60
(ii) 1386
(iii) 6435
(iv) 2184
(v) 8085
(vi) 14850
14. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 ×3 × 2 × 1 + 5 are composite numbers?
B
15. Check whether 8n can end with the digit zero for any natural number n.
16. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = Product of the
integers.
17.
18.
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(i) 63 and 168
(ii) 144 and 160
(iii) 510 and 92
(iv) 252, 488
Find the LCM and HCF of the following integers by applying the prime factorisation method :
(i) 12, 15 and 21
(ii) 15, 24 and 36
(iii) 225, 336 and 360
(iv) 240, 1024, 1536
The HCF and LCM of two numbers are 12 and 240 respectively. If one of these numbers is 48, find the
other number.
The product of two numbers is 20736 and their HCF is 54. Find their LCM.
The LCM of two numbers is 192 and their product is 3072. Find their HCF.
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19.
20.
21. Prove that
3 is an irrational number..
22. Show that the following numbers are irrational :
(i) 5
(ii) 2  3
(iii) 4  5
(iv) 3 5
(v)
(vi) 2  3 5
(vii)
1
2
(viii)
2 3
3
(viii) 3  5 2
7
23. (i) If p is a prime number, prove that
(ii) Show that
6
p is irrational.
n  1  n  1 is irrational for every natural number n.
REAL NUMBERS
MATHEMATICS–X
24. Without actually performing the long division, state whether the following rational numbers will have
terminating decimal expansion or a non-terminating repeating decimal expansion :
(i)
13
121
(iv)
108
250
(ii)
57
128
(iii)
19
45
(v)
113
175
(vi)
517
2000
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25. Write down the decimal expansions of the following rational numbers by writing their denominators in
the form 2m × 5n, where m, n are non-negative integers :
(i)
5
8
(iv)
123
625
(ii)
17
125
(iii)
13
80
(v)
7014
400
(vi)
17
2000
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HINTS TO SELECTED QUESTIONS
2.
4.
5.
6.
Required number = HCF (280 – 4, 1245 – 3) = HCF (276, 1242).
Greatest capacity = HCF (105 l, 140 l, 175 l).
Length of largest rod = HCF (825 cm, 675 cm, 450 cm)
Number of participants in each room = HCF (60, 84, 108) = 12.
Now, total number of participants = 60 + 84 + 108 = 252.
B
252
 21.
12
10. Let a be any positive integer. Then, it is of the form 3q or, 3q + 1 or, 3q + 2. So, we have the following
cases :
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 number of rooms required 
Case I : When a = 3q, here a 2  (3q ) 2  9q 2  3(3q )  3m, where m  3q.
Case II : When a = 3q + 1, a2 = (3q + 1)2 = 9q2 + 6q + 1 = 3q (3q + 2) + 1, where m = q (3q + 2).
Case III : When a = 3q + 2, here, a 2  (3q  2) 2  9 q 2  12q  4
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 3(3q 2  4q  1)  1
 3m  1, where m  3q 2  4 q  1.
Hence, a is of the form 3m or 3m + 1.
14. 7 × 11 × 13 + 13 = 13 × (7 × 11 + 1) = 13 × 78, which is a composite no.
also, 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5 × (7 × 6 × 4 × 3 × 2 × 1 + 1)
= 5 × 1009, which is a composite no.
23. (i) Let p be a prime number and if possible, let
p is rational. Let its simplest form be
p
m
,
n
where m and n are integers having no common factor other than 1, and n  0.
m
m2
 p  2  pn2  m2
n
n
 p divides m2  p divides m.
Now,
p
MATHEMATICS–X
...(i)
REAL NUMBERS
7
Let m = pq for some integer q.
putting m = pq in (i), we get
pn2 = p2q2  n2 = pq2  p divides n2  p divides n.
...(ii)
from (i) and (ii), we observe that p is a common factor of m and n, which contradicts our
assumption.
Hence,
(ii) Let
n  1  n  1 is rational. Let
Now,
a
 n 1  n 1
b

b

a

2b
 n 1  n 1
a
1
n 1  n 1
a
 n  1  n  1, where a, b are positive integers.
b
n 1 
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...(i)

b
n 1  n 1


a ( n  1)2  ( n  1) 2
n  1  n 1
2
...(ii)
adding (i) and (ii) and subtracting (ii) from (i), we get

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p is irrational.
a 2  2b2
a 2  2b 2
and n  1 
2ab
2ab
B
a 2  2b 2
a 2  2b2
and
are rationals.]
n  1 and n  1 are rationals. [ a, b are integers, 
2ab
2ab
 (n + 1) and (n – 1) are perfect squares of positive integers, which is not possible as any two
perfect squares differ atleast by 3.
Hence,
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n  1  n  1 is irrational.
MULTIPLE CHOICE QUESTIONS
Mark the correct alternative in each of the following :
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1. The product of three consecutive positive integers is always divisible by :
(a) 4
(b) 5
(c) 6
(d) 12
2. The HCF of 2048 and 960 is :
(a) 32
(b) 64
(c) 128
(d) none of these
3. The HCF of 657and 963 is :
(a) 3
(b) 6
(c) 9
(d) none of these
4. The largest number which divides 615 and 963 leaving remainder 6 in each case is :
(a) 29
(b) 87
(c) 116
(d) none of these
5. The HCF of 405, 840 and 960 is
(a) 9
(b) 15
(c) 45
(d) none of these
6. The GCD of 144, 418 and 112 is ;
(a) 2
(b) 4
(c) 8
(d) 12
7. The prime factorisation of 468 is :
(a) 2 × 33 × 13
(b) 22 × 3 × 13
(c) 22 × 32 × 13
(d) none of these
8
REAL NUMBERS
MATHEMATICS–X
8. The least number that is divisible by all the numbers between 1 and 10 (both inclusive) is:
(a) 252
(b) 2520
(c) 1260
(d) none of these
9. The HCF of two numbers is 145 and their LCM is 2175. If one number is 725, the other number is :
(a) 2075
(b) 870
(c) 87
(d) 435
10. If the sum of two numbers is 1215 and their HCF is 81, the total number of such pairs is :
(a) 2
(b) 3
(c) 4
(d) 5
11. Which of the following number is rational :
(a)
12.
2 3
(b) (7  5) (7  5)
(c)
3
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1
, when expressed as a non-terminating and recurring decimal is given by :
7
(a) 0.142857
(b) 0.428571
(c) 0.857142
13. Which of the following number is given by 0.1236 :
(a)
17
275
(b)
34
550
(c)
34
275
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(d) ( 3  7) 2
5
B
(d) none of these
(d)
13
825
(d)
5
224
14. Which of the following number have a terminating decimal representation :
(a)
7
80
(b)
3
345
(c)
8
17
15. Which of the following number do not have a terminating decimal representation :
(a)
7
50
(b)
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19
1000
(c)
12
105
(d)
113
2500
VERY SHORT ANSWER TYPE QUESTIONS (1 MARK QUESTIONS)
1. State Euclid’s division lemma.
2. State fundamental theorem of arithmetic.
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3. Write the condition to be satisfied by q so that a rational number
p
has a terminating decimal expansion.
q
4. Write the condition to be satisfied by q so that a rational number p has a non-terminating recurring
q
decimal expansion.
5. Find the HCF of 26 and 91.
6. If HCF (72, 120) = 24, then what is the LCM (72, 120)?
7. LCM of two numbers is 2079 and their HCF is 27. If one of the numbers is 297, find the other number.
8. If LCM (24, 80) = 240, then what is the HCF (24, 80)?
9. Why the number 4n, where n is a natural number, cannot end with 0?
10. Why 5 × 7 × 11 × 13 × 17 + 13 is a composite number?
11. Express 156 as a product of its prime factors.
MATHEMATICS–X
REAL NUMBERS
9
12. Find the missing number in the following factor tree:
5005
143
7
13
11
13. Give an example of two irrational numbers whose sum is a rational number.
14. Give an example of two irrational numbers whose product is a rational number.
15. Without actually performing long division, state why
expansion.
16. Write down the decimal expansion of
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?
5
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32
has a non-terminating repeating decimal
455
7.
80
3
is a rational number or not.
4
18. How many prime factors are there in the prime factorisation of 420.
19. By which smallest irrational number 28 be multiplied so as to get a rational number?
17. State whether x  2.23 
B
20. Which number should be multiplied to ( 7  5) to get a rational number?
PRACTICE TEST
M.M : 30
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General Instructions :
Time : 1 hour
Q. 1-4 carry 2 marks, Q. 5-8 carry 3 marks and Q. 9-10 carry 5 marks each.
1. Write 504 as a product of its prime factors.
2. Find the LCM and HCF of 26 and 91.
3. Is (5  3)2  (5  3)2 is a rational number? Justify..
5
in decimal form.
6
Use Euclid’s division algorithm to find HCF of 867 and 255.
Prove that one of every three consecutive positive integers is divisible by 3.
Find the HCF and LCM of 144, 180 and 192 by prime factorisation method.
Two tankers contains 850 litres and 680 litres of petrol respectively. Find the maximum capacity of a
container which can measure the petrol of either tanker in exact number of times.
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4. Write the rational number
5.
6.
7.
8.
9. Prove that 5 is an irrational number..
10. 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)
7
13
(iv)
10
131
2  52  7
3
(ii)
131
420
(v)
134
2500
REAL NUMBERS
(iii)
23
500
MATHEMATICS–X
ANSWERS OF PRACTICE EXERCISE
1. (i) 5
(ii) 13
2. 138
3. 10
2
(iii) 16
(iv) 17
(v) 23
4. 35
5. 75 cm
6. 21
2
13. (i) 2 × 3 × 5
(v) 3 × 5 × 72 × 11
2
(ii) 2 × 3 × 7 × 11
(iii) 3 × 5 × 11 × 13
(vi) 2 × 33 × 52 × 11
15.
16. (i) LCM = 504, HCF = 21
(vi) 5
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(iv) 23 × 3 × 7 × 13
No
(ii) LCM = 1440, HCF = 16
(iii) LCM = 23460, HCF = 2
(ii) LCM = 360, HCF = 3
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(iv) LCM = 30744, HCF = 4
17. (i) LCM = 420, HCF = 3
(iii) LCM = 25200, HCF = 3
(iv) LCM = 15360, HCF = 16
18. 60
19. 384
20. 16
24. (i) non-terminating repeating
(ii) terminating
(iii) non-terminating repeating
(iv) terminating
(v) non-terminating repeating
25. (i) 0.625
(ii) 0.136
(iv) 0.1968
(v) 17.535
(vi) terminating
(iii) 0.1625
(vi) 0.0085
B
ANSWERS OF MULTIPLE CHOICE QUESTIONS
1. (c)
6. (a)
11. (b)
2. (b)
7. (c)
12. (a)
3. (c)
8. (b)
13. (c)
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4. (b)
9. (d)
14. (a)
5. (b)
10. (c)
15. (c)
ANSWERS OF VERY SHORT ANSWER TYPE QUESTIONS
3. The prime factorisation of q should be of the form 2n 5m, where n, m are non-negative integers.
5. 13
6. 360
7. 189
M
A
12. 1001
13. 2  3 and 2  3
17. rational
18. 4
19.
7
8. 8
14.
2 and 8
20.
7 5
11. 22 × 3 × 13
16. 0.0875
ANSWERS OF PRACTICE TEST
1. 23 × 32 × 7
4.
0.83
3. LCM = 182, HCF = 13
3. yes
5. 51
7. LCM = 2880, HCF =12
8. 170 litres
10. (i) non-terminating repeating
(iii) Terminating
(ii) non-terminating repeating
(iv) non-terminating repeating
(v) Terminating
MATHEMATICS–X
REAL NUMBERS
11