Important Questions for Class 10
Mathematics
Chapter 1 -Real Numbers
Very Short Answer Questions.
1 Mark
1.Use Euclid’s division lemma to show that the square of any positive
integer is either of 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 .]
Ans: Let a be any positive integer, then we can write a = 3q + r for some
integer q  0 …..(1)
Clearly, in the expression (1) we are dividing a by 3 with quotient q and
remainder r , r = 0,1,2 because 0  r  3 . ..…(2)
Therefore from (1) and (2) we can get, a = 3q or 3q + 1 or 3q + 2 . ……(3)
Squaring both sides of equation (3) we get,
a 2 = (3q) 2 or (3q + 1) 2 or (3q + 2) 2
 a 2 = 9q 2 or 9q 2 + 6q + 1 or 9q 2 + 12q + 4 …..(4)
Taking 3 common from LHS of equation (4) we get,
a 2 = 3  (3q 2 ) or 3(3q 2 + 2q) + 1 or 3(3q 2 + 4q + 1) + 1. …..(5)
Equation (5) could be written as
a 2 = 3k1 or 3k 2 + 1 or (3k 3 + 1) for some positive integers k1 , k 2 and k 3 .
Hence, it can be said that the square of any positive integer is either of the form
3m or 3m + 1 .
2.Express each number as product of its prime factors.
(i)
140
Ans: The product of prime factors of number 140 is 140 = 2  2  5  7 = 22  5  7
(ii) 156
Ans: The product of prime factors of number 156 is 156 = 2  2  3 13 = 22  3 13
(iii) 3825
Ans: The product of prime factors of number 3825 is
3825 = 3  3  5  5 17 = 32  52 17 .
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(iv) 5005
Ans: The product of prime factors of number 5005 is 5005 = 5  7 1113 .
(v)
7429
Ans: The product of prime factors of number 7429 is 7429 = 17 19  23 .
3.Given that HCF (306,657) = 9 , find LCM (306,657) .
Ans: It is given that the HCF of the two numbers 306, 657 is 9 . We have to
find its LCM.
We know that, LCM  HCF = Product of two numbers. Therefore,
LCM  HCF = 306  657
 LCM =
306  657
HCF
 LCM =
306  657
9
LCM = 22338
4.Check whether 6n can end with the digit 0 for any natural number n .
Ans: If any number ends with the digit 0 , it should be divisible by 10 .
Let us check this for natural number n = 2 .
Clearly, 6 2 = 36 , which is not divisible by 10 . Therefore, 6n cannot end with
the digit 0 for any natural number n .
(
5.Prove that 3 + 2 5
) is irrational.
Ans: We will prove this by contradiction. To solve this problem, suppose that
3 + 2 5 is rational i.e.,  two co-prime integers a and b (b  0) such that
(
)
a
= 3 + 2 5 …..(1)
b
Subtracting 3 from both sides of the equation (1) and simplifying it further we
get,
a
−3= 2 5
b
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a − 3b
=2 5
b
a − 3b

= 5
2b

…..(2)
Since we know that
5 is irrational. Therefore, from (2)
irrational but it could be written in
a − 3b
should be
2b
p
form which means it is rational. It is not
q
possible and hence our assumption is wrong and (3 + 2 5) cannot be rational.
Hence, (3 + 2 5) is irrational.
6.The number 7  11  13  15 + 15 is a
a.
Composite Number
b.
Whole Number
c.
Prime Number
d.
None of these
Ans: (a) and (b) both
7.For what least value of 'n' a natural number, (24)n is divisible by 8 ?
0
a)
−1
b)
1
c)
d)
No value of 'n' is possible
Ans: (c) 1
8.The sum of a rational and an irrational number is
a)
Rational
b)
Irrational
c)
Both (a) & (b)
d)
Either (a) & (b)
Ans: (b) Irrational
9.HCF of two numbers is 113, their LCM is 56952 . If one number is 904 ,
then other number is:
7719
a)
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7119
b)
7791
c)
7911
d)
Ans: (b) 7119
10.A lemma is an axiom used for providing
a)
other statement
b)
no statement
c)
contradictory statement
d)
none of these
Ans: (a) other statement
11.If HCF of two numbers is 1 , the two numbers are called relatively
______ or _______.
a)
prime, co-prime
b)
composite, prime
c)
Both (a) and (b)
d)
None of these
Ans: (a) prime, co-prime
12. The number 2.35 is
a)
a terminating decimal number
b)
a rational number
c)
an irrational number
d)
Both (a) and (b)
Ans: (b) a rational number
13.The number 2.13113111311113...... is
a)
a rational number
b)
a non-terminating decimal number
c)
an irrational number
d)
Both (a) & (c)
Ans: (b) a non-terminating decimal number
14.The smallest composite number is
1
a)
2
b)
3
c)
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4
d)
Ans: (d) 4
15.The number 1.2348 is
a)
an integer
b)
an irrational number
c)
a rational number
d)
None if these
Ans: (c) a rational number
16.The number  is
a)
a rational number
b)
an irrational number
c)
Both (a) and (b)
d)
neither rational nor irrational
Ans: (b) an irrational number
17. The number (2 + 5) is
a)
a rational number
b)
an irrational number
c)
an integer
d)
not real number
Ans: (b) an irrational number
Short Answer Questions.
2 Marks
1.Show that any positive odd integer is of the form 6q + 1, or 6q + 3 or
6q + 5 , where q is some integer.
Ans: Let a be any odd positive integer, then by Euclid’s algorithm we can write
a = 6q + r for some integer q  0 .
…..(1)
Clearly, in the expression (1) we are dividing a by 6 with quotient q and
remainder r ,
r = 1,3,5 because 0  r  6 and a is odd. …..(2)
Therefore from (1) and (2) we can get,
a = 6q + 1 or 6q + 3 or 6q + 5
And therefore, any odd integer can be expressed in the form 6q + 1,6q + 3, or
6q + 5 .
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2. An army contingent of 616 members are 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?
Ans: To solve this problem, we have to find the HCF of two numbers (616,32)
To find the HCF, we can use Euclid’s algorithm which states that if there are
any two integers a and b , then there exists q and r such that it satisfies the
condition a = bq + r where 0  r  b .
For a = 616 , 616 = 32 19 + 8
…..(1)
For a = 32 , 32 = 8  4 + 0
…..(2)
Therefore, from (1) and (2), the HCF (616,32) is 8 and hence they can march
in 8 columns each.
3.Use Euclid’s division lemma to show that the cube of any positive integer
is of the form 9m,9m + 1 or 9m + 8 .
Ans: Let a be any positive integer, then by Euclid’s algorithm we can write
a = 3q + r, for some integer q  0 . …..(1)
Clearly, in the expression (1) we are dividing a by 3 with quotient q and
remainder r , r = 0,1,2 because 0  r  3 . …..(2)
Therefore from (1) and (2) we can get,
a = 3q or 3q + 1 or 3q + 2 …..(3)
Cubing both sides of equation (3) we get,
Case 1: When a = 3q ,
a 3 = 27q 3
 a 3 = 9(3q 3 )
 a 3 = 9m
Where m is an integer such that m = 3q 3
Case 2: When a = 3q + 1
a 3 = (3q + 1)3
 a 3 = 9q 3 + 9q 2 + 9q + 1
 a 3 = 9 ( q3 + q 2 + q ) + 1
 a 3 = 9m + 1
Where m is an integer such that m = (3q + 3q + q)
Case 3: When a = 3q + 2
3
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a 3 = (3q + 2)3
 a 3 = 27q 3 + 54q 2 + 36q + 8
 a 3 = 9(3q 3 + 6q 2 + 4q) + 8
 a 3 = 9m + 8
Where m is an integer such that m = (3q 3 + 6q 2 + 4q)
Therefore, from case 1,2 and 3 we conclude that the cube of any positive integer
is of the form 9m,9m + 1 or 9m + 8 .
4.Find the LCM and HCF of the following pairs of integers and verify that
LCM  HCF = product of two numbers.
(i)
26 and 91
Ans: Let us first find the HCF of the two given numbers.
Writing the prime factorization of both of the numbers we get,
…..(1)
26 = 2 13 and 91 = 7 13 .
Since 13 is common in the prime factorization of both of the numbers we get
HCF = 13 . …..(2)
From (1), LCM = 2  7 13=182 …..(3)
Product of two numbers, 26  91 = 2366 …..(4)
From (2) and (3), HCF  LCM =13 182=2366 . …..(5)
Hence from (4) and (5), product of two numbers = HCF  LCM .
510 and 92
Ans: Let us first find the HCF of the two given numbers.
Writing the prime factorization of both of the numbers we get,
510 = 2  3  5 17 and 92 = 2  2  23 . …..(1)
Since 2 is common in the prime factorization of both of the numbers we get
…..(2)
HCF = 2 .
From (1), LCM = 2  2  3  5 17  23 = 23460 …..(3)
Product of two numbers, 510  92 = 46920 …..(4)
From (2) and (3), HCF  LCM = 2  23460 = 46920 . …..(5)
Hence from (4) and (5), product of two numbers = HCF  LCM .
(ii)
336 and 54
Ans: Let us first find the HCF of the two given numbers.
Writing the prime factorization of both of the numbers we get,
336 = 2  2  2  2  3  7 = 24  3  7 and 54 = 2  3  3  3 = 2  33 . …..(1)
Since 2,3 are common in the prime factorization of the numbers,
HCF = 2  3 = 6 . .....(2)
(iii)
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From (1), LCM = 24  33  7 = 3024
…..(3)
Product of two numbers = 336  54 = 18144
…..(4)
From (2) and (3), HCF  LCM = 6  3024 = 18144 . …..(5)
Hence from (4) and (5), product of two numbers = HCF  LCM .
5.Find the LCM and HCF of the following integers by applying the prime
factorisation method.
(i)
12,15 and 221
Ans: Let us first find the HCF of the two given numbers.
Writing the prime factorization of all the numbers we get,
12 = 22  3 , 15 = 3  5 and 21 = 3  7 . …..(1)
Since 3 is common in the prime factorization of all the numbers we get
HCF = 3 .
From (1), LCM = 22  3  5  7 = 420 .
17,23 and 29
Ans: Let us first find the HCF of the two given numbers.
Writing the prime factorization of all the numbers we get,
17 = 117 , 23 = 1 23 and 29 = 1 29 . …..(1)
Since 1 is common in the prime factorization of all the numbers we get HCF = 1
From (1), LCM = 17  23  29 = 11339 .
(ii)
8,9 and 25
Ans: Let us first find the HCF of the two given numbers.
Writing the prime factorization of all the numbers we get,
8 = 2  2  2 = 23 , 9 = 3  3 = 32 and 25 = 5  5 = 52 . …..(1)
Since 1 is common in the prime factorization of all the numbers we get HCF = 1
From (1), LCM = 23  32  52 = 1800 .
(iii)
6.Explain why 7  11  13 + 13 and 7  6  5  4  3  2  1 + 5 are composite
numbers.
Ans: Numbers are of two types – prime and composite.
Prime numbers can be divided by 1 and only itself, whereas composite
numbers have factors other than 1 and itself.
It can be observed that 7 1113 + 13 = 13  (7 11 + 1)
The given expression has 13 as one of the factors. Therefore, it is a composite
number.
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Also, 7  6  5  4  3  2 1 + 5 = 5  (7  6  4  3  2 1 + 1) .
The given expression has 5 as one of the factors. Therefore, it is a composite
number.
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?
Ans: It can be observed that Ravi takes lesser time than Sonia for completing
one round of the circular path. As they are going in the same direction, they
will meet again at the time that will be the LCM of time taken by Sonia and
Ravi for completing 1 round of circular path respectively i.e., LCM of 18
minutes and 12 minutes.
Let us now calculate LCM of 18 minutes and 12 minutes.
18 = 2  3  3 and 12 = 2  2  3 …..(1)
LCM (12,18 ) = 2  2  3  3 = 36
Therefore, Ravi and Sonia will meet together at the starting point after 36
minutes.
8.Prove that 5 is irrational.
Ans: We will prove this by contradiction.
Let us suppose that 5 is rational. It means that we have co-prime integers a
a
and b (b  0) such that 5 =
b
 b 5 = a …..(1)
Squaring both sides of (1), we get 5b = a 2 …..(2)
From (2) we can conclude that 5 is factor of a 2 , this means that 5 is also factor
of a . …..(3)
From (3) we can write a = 5c for some integer c . …..(4)
Substituting value of (4) in (2) we get,
5b2 = 25c2
 b2 = 5c2
It means that 5 is factor of b 2 , this means that 5 is also factor of b . …..(5)
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From (3),(5) , we can say that 5 is factor of both a,b , which is a contradiction
to our assumption. Therefore, our assumption was wrong and
5 is irrational.
9.Write down the decimal expansion of those rational numbers in Question
1 which have terminated decimal expansion.
Ans:(i).The decimal expansion of
13
13 13  25 13  25 416
is =
=
= 5 = 0.00416
3125 55 55  25
105
10
(ii) The decimal expansion of
17 17 17  53 17  53 2125
is 3 = 3 3 =
=
= 2.215
8
2
2 5
103
103
(iii)The decimal expansion of
15
15
15  54
15  54 9375
is 6 2 =
=
=
= 0.009375
1600
2 5
2  52  5 4
106
106
(iv)The decimal expansion of
23
23  51
23  51 115
is
=
= 3 = 0.115
23  52
23  52  51
103
10
(v)The decimal expansion of
6
2 2 2 4
is =
= = 0.4
15
5 5  2 10
(vi)The decimal expansion of
35
7
= 0.7
is
50 10
10. The following real numbers have decimal expansions as given below.
In each case, decide whether they are rational or not. If, they are rational,
p
and of form
, what can you say about the prime factors of q ?
q
(i)
Number - 43.123456789
Ans: Since the decimal expansion is terminating so it is a rational number and
p
therefore it can be expressed in form.
q
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43123456789
109
Hence, the factors of q are of the form 2 n  5m where n and m are nonnegative integers.
43.123456789 =
(ii) Number - 0.1201120012000120000...
Ans: Since the decimal expansion is neither terminating nor non-terminating
repeating so it is an irrational number.
(iii) Number - 43.123456789
Ans: Since the decimal expansion is non-terminating repeating so it is a rational
p
number and therefore it can be expressed in form.
q
 123456789   123456789   123456789 
43.123456789 = 43  


  ...
9
9
9
10
10
10

 
 

Hence, the factors of q are of the form 2 n  5m where n and m are nonnegative integers.
11.Show that every positive integer is of the form 2q and that every positive
odd integer is of the form 2q + 1 for some integer q .
Ans: Let a be any positive integer, then by Euclid’s algorithm we can write
a = 2q + r, for some integer q  0 . …..(1)
Clearly, in the expression (1) we are dividing a by 2 with quotient q and
remainder r , r = 0,1 because 0  r  2 . …..(2)
Therefore from (1) and (2) we can get,
a = 2q or 2q + 1 …..(3)
If a = 2q (which is even)
If a = 2q + 1 (which is odd)
So, every positive even integer is of the form 2q and odd integer is of the form
2q + 1 .
12. Show that any number of the form 4n , n  N can never end with the
digit 0 .
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Ans: The prime factorization of 10 is 2  5 . Therefore, for a number to end
with 0 it should have both 2 and 5 as a factor.
For the given number of the form 4n = [22 ]n = 22n , we can observe that it has
only 2 as a factor and not 5 . So 4n ,n  N can never end with the digit 0.
13.Use Euclid’s Division Algorithm to find the HCF of 4052 and 12576
Ans: Step 1: Since12576  4052 , we will apply Euclid’s division lemma such
that 12576 = 4052  3 + 420 …..(1)
Step 2: Since the remainder in (1) is not zero so we have to apply the Euclid’s
division lemma to 4052 and 420 , such that 4052 = 420  9 + 272 …..(2)
Step 3: Again, the remainder in (2) is not zero so we have to apply the Euclid’s
division lemma to 420 and 272 , such that 420 = 272 1 + 148 . …..(3)
Step 4: Again, the remainder in (3) is not zero so we have to apply the Euclid’s
division lemma to 148 and 272 , such that 272 = 148 1 + 124 . …..(4)
Step 5: Again, the remainder in (4) is not zero so we have to apply the Euclid’s
division lemma to 148 and 124 , such that 148 = 124 1 + 24 . …..(5)
Step 6: Again, the remainder in (5) is not zero so we have to apply the Euclid’s
division lemma to 24 and 124 , such that 124 = 24  5 + 4 . …..(6)
Step 7: Again, the remainder in (6) is not zero so we have to apply the Euclid’s
division lemma to 24 and 4 , such that 24 = 4  6 + 0 . …..(7)
Now the remainder in equation (7) is zero. Therefore, HCF of 12576 and 4052
is '4'.
14.Given that HCF of two numbers is 23 and there LCM is 1449 . If one
of the numbers is 161 , find the other.
Ans: It is given that the HCF of the two numbers is 23 and their LCM is 1449
We have to find the other number if one of the numbers is 161 . Let the other
number be a . We know that, LCM  HCF = Product of two numbers. Therefore,
1449  23 = a 161
1449  23
a =
161
a = 207
15.Show that every positive odd integer is of the form (4q + 1) or (4q + 3)
for some integer q .
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Ans: Let a be any positive integer, then by Euclid’s algorithm we can write
a = 4q + r, for some integer q  0 . …..(1)
Clearly, in the expression (1) we are dividing a by 4 with quotient q and
remainder r , r = 0,1,2,3 because 0  r  4 . …..(2)
Therefore from (1) and (2) we can get,
a = 4q or 4q + 1 or 4q + 2 or 4q + 3 …..(3)
If a = 4q = 2 ( 2q ) (which is even)
If a = 4q + 1 = 2 ( 2q ) + 1 (which is odd)
If a = 4q + 2 = 2(2q + 1) (which is even)
If a = 4q + 3 = 2(2q + 1) + 1 (which is odd)
So, every positive odd integer is of the form (4q + 1) or (4q + 3) .
16.Show that any number of the form 6x ,x  N can never end with the digit
0.
Ans: The prime factorization of 10 is 2  5 . Therefore, for a number to end
with 0 it should have both 2 and 5 as a factor.
For the given number of the form 6n = (2  3)n = 2n  3 n , we can observe that
.
it has only 2 as a factor and not 5 . So 6x , x  N can never end with the digit
0.
17.Find HCF and LCM of 18 and 24 by the prime factorization method.
Ans: The prime factorization of 18 is 18 = 2  3  3 = 2  32 …..(1)
The prime factorization of 24 is 24 = 2  2  2  3 = 23  3 …..(2)
From (1) and (2) we can see that 18,24 has prime factors 2,3 in common.
HCF = 2  3 = 6
From (1) and (2) we can see that 3 appears twice in prime factorization of 18
and 2 appears thrice in prime factorization of 24 .
 LCM = ( 32 )  ( 23 ) = 72
18.The HCF of two numbers is 23 and their LCM is 1449 . If one of the
numbers is 161, find the other.
Ans: It is given that the HCF of the two numbers is 23 and their LCM is 1449
. We have to find the other number if one of the numbers is 161 . Let the other
number be a .
We know that, LCM  HCF = Product of two numbers. Therefore,
1449  23 = a 161
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1449  23
161
a = 207
a =
19.Prove that the square of any positive integer of the form 5g + 1 is of the
same form.
Ans: Let a be any positive integer, of the form a = 5g + 1
…..(1)
Squaring both sides of equation (1) we get,
a 2 = (5g + 1)2
 a 2 = 25g 2 + 10g + 1
 a 2 = 5(5g 2 + 2g) + 1
 a 2 = 5m + 1
Hence, the square of any positive integer of the form 5g + 1 is of the same form.
20.Use Euclid’s Division Algorithm to find the HCF of 4052 and 12576 .
Ans: Step 1: Since12576  4052 , we will apply Euclid’s division lemma such
that 12576 = 4052  3 + 420 …..(1)
Step 2: Since the remainder in (1) is not zero so we have to apply the Euclid’s
division lemma to 4052 and 420 , such that 4052 = 420  9 + 272 …..(2)
Step 3: Again, the remainder in (2) is not zero so we have to apply the Euclid’s
division lemma to 420 and 272 , such that 420 = 272 1 + 148 . …..(3)
Step 4: Again, the remainder in (3) is not zero so we have to apply the Euclid’s
division lemma to 148 and 272 , such that 272 = 148 1 + 124 . …..(4)
Step 5: Again, the remainder in (4) is not zero so we have to apply the Euclid’s
division lemma to 148 and 124 , such that 148 = 124 1 + 24 . …..(5)
Step 6: Again, the remainder in (5) is not zero so we have to apply the Euclid’s
division lemma to 24 and 124 , such that 124 = 24  5 + 4 . …..(6)
Step 7: Again, the remainder in (6) is not zero so we have to apply the Euclid’s
division lemma to 24 and 4 , such that 24 = 4  6 + 0 . …..(7)
Now the remainder equation (7) is zero. Therefore, HCF of 12576 and 4052
is '4'.
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21.Find the largest number which divides 245 and 1029 leaving remainder
5 in each case.
Ans. Since we want the remainder to be 5 , therefore the required number is the
HCF of (245 − 5) and (1029 − 5) i.e., 240 and 1024 .
Step 1: Since1024  240 , we will apply Euclid’s division lemma such that
…..(1)
1024 = 240  4 + 64
Step 2: Since the remainder in (1) is not zero so we have to apply the Euclid’s
division lemma to 240 and 64 , such that 240 = 64  3 + 48 …..(2)
Step 3: Again, the remainder in (2) is not zero so we have to apply the Euclid’s
division lemma to 64 and 48 , such that 64 = 48 1 + 16 . …..(3)
Step 4: Again, the remainder in (3) is not zero so we have to apply the Euclid’s
division lemma to 48 and 16 , such that 48 = 16  3 + 0 . …..(4)
Now the remainder in equation (4) is zero. Therefore, HCF of 1024 and 240
is '16' .
The largest number which divides 245 and 1029 leaving remainder 5 in each
case is 16 .
22.A shopkeeper has 120 litres of petrol, 180 litres of diesel and 240
litres of kerosene. He wants to sell oil by filling the three kinds of oils in
tins of equal capacity. What should be the greatest capacity of such a tin?
Ans: The required greatest capacity is the HCF of 120,180 and 240 . First let
us find the HCF of 240,180 by Euclid’s division algorithm.
Step 1: Since 240  180 , we will apply Euclid’s division lemma such that
…..(1)
240 = 180 1 + 60
Step 2: Since the remainder in (1) is not zero so we have to apply the Euclid’s
division lemma to 180 and 60 , such that 180 = 60  3 + 0 …..(2)
Now the remainder in equation (2) is zero. Therefore, HCF of 180 and 240 is
60 .
Let us now find the HCF of 60,120 . Clearly, 120 = 60  2 + 0 .
 HCF of 120,180 and 240 is 60 and hence the required capacity is 60 litres.
Short Answer Questions
3 Marks
1.
Use Euclid’s division algorithm to find the HCF of:
(i)
135 and 225
Ans: Step 1: Since 225  135 , we will apply Euclid’s division lemma such that
225 = 135 1 + 90 …..(1)
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Step 2: Since the remainder in (1) is not zero so we have to apply the Euclid’s
division lemma to 135 and 90 , such that 135 = 90 1 + 45 …..(2)
Step 3: Again, the remainder in (2) is not zero so we have to apply the Euclid’s
division lemma to 90 and 45 , such that 90 = 2  45 + 0 . …..(3)
Now the remainder in equation (3) is zero. Therefore, the HCF of 135 and 225
is 45 .
(ii) 196 and 38220
Ans: Since 38220  196 , we will apply Euclid’s division lemma such that
38220 = 196 195 + 0 …..(1)
Since, the remainder in (1) is zero. Therefore, the HCF of 38220 and 196 is
196 .
(iii) 867 and 255
Ans: Step 1: Since 867  255, , we will apply Euclid’s division lemma such
that 867 = 255  3 + 102
…..(1)
Step 2: Since the remainder in (1) is not zero so we have to apply the Euclid’s
division lemma to 255 and 102 , such that 255 = 102  2 + 51 …..(2)
Step 3: Again, the remainder in (2) is not zero so we have to apply the Euclid’s
division lemma to 102 and 51 , such that 102 = 51 2 + 0 . …..(3)
Now the remainder is zero. Therefore, the HCF of 867 and 255 is 51 .
2.Find the greatest number of 6 digits exactly divisible by 24,15 and 36.
Ans: To find the greatest number of 6 digits exactly divisible by 24,15 and 36
first we have to find the LCM of 24,15 and 36 .
The prime factorization of 24,15 and 36 are:
24 = 2  2  2  3 …..(1)
15 = 5  3 …..(2)
36 = 2  2  3  3 …..(3)
Hence, from (1), (2) and (3), LCM (15,24,36 ) = 2  2  2  5  3  3 = 360 .….(4)
Now, we know that the greatest number of 6 digits is 999999. …..(5)
From (4) and (5), using Euclid’s division lemma we get,
999999 = 360  2777 + 279 .
 999999 − 279 = 360  2777
 999720 = 360  2777
Hence, the greatest number of 6 digits exactly divisible by 24,15 and 36 is
999720 .
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3.Prove that the square of any positive integer is of the form 4m or 4m + 1
for some integer m .
Ans: Let a be any positive integer, then by Euclid’s algorithm we can write
a = 4q + r, for some integer q  0 .
…..(1)
Clearly, in the expression (1) we are dividing a by 4 with quotient q and
remainder r , r = 0,1,2,3 because 0  r  4 . …..(2)
Therefore from (1) and (2) we can get,
a = 4q or 4q + 1 or 4q + 2 or 4q + 3 …..(3)
Case 1: If a = 4q then, squaring both sides we get,
a 2 = 16q 2
 a 2 = 4 ( 4q 2 ) …..(4)
Equation (4) could be written as a 2 = 4m for some integer m ….. (5)
Case 2: If a = 4q + 1 then, squaring both sides we get,
a 2 = 16q 2 + 8q + 1
 a 2 = 4 ( 4q 2 + 2q ) + 1 …..(6)
Equation (6) could be written as a 2 = 4m + 1 for some integer m ….. (7)
Case 3: If a = 4q + 2 then, squaring both sides we get,
a 2 = 16q 2 + 16q + 4
 a 2 = 4 ( 4q 2 + 4q + 1) …..(8)
Equation (8) could be written as a 2 = 4m for some integer m ….. (9)
Case 4: If a = 4q + 3 then, squaring both sides we get,
a 2 = 16q 2 + 24q + 9
a 2 = 4 ( 4q 2 + 6q + 2 ) + 1 …..(10)
Equation (10) could be written as a 2 = 4m + 1 for some integer m ….. (11)
Therefore, from equations (5), (7), (9) and (11) it can be said that the square of
any positive integer is either of the form 4m or 4m + 1 for some integer m .
4.There are 144 cartons of coke can and 90 cartons of Pepsi can to be
stacked in a canteen. If each stack is of the same height and is to contain
cartoons of the same drink. What would be the greater number of cartons
each stack would have?
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Ans: Given 144 cartons of coke can and 90 cartons of Pepsi can are to be
stacked in a canteen. For stack to be of the same height and is to contain
cartoons of the same drink, we have to find the HCF of 144 and 90 .
Using Euclid’s division algorithm,
Step 1: Since 144  90, , we will apply Euclid’s division lemma such that
…..(1)
144 = 90 1 + 54
Step 2: Since the remainder in (1) is not zero so we have to apply the Euclid’s
division lemma to 90 and 54 , such that 90 = 54 1 + 36 …..(2)
Step 3: Again, the remainder in (2) is not zero so we have to apply the Euclid’s
division lemma to 54 and 36 , such that 54 = 36 1 + 18 …..(3)
Step 4: Again, the remainder in (3) is not zero so we have to apply the Euclid’s
division lemma to 18 and 36 , such that 36 = 18  2 + 0 . …..(4)
Now the remainder in (4) is zero. Therefore, the HCF of 144 and 90 is 18 .
So, greatest number of cartoons is 18.
5.Prove that product of three consecutive positive integers is divisible by 6
Ans: Let us consider three consecutive positive integers x,(x + 1) and (x + 2) .
Using Euclid’s lemma, let us write x = 6q + r …..(1)
Clearly, in the expression (1) we are dividing a by 6 with quotient q and
remainder r , r = 0,1,2,3,4,5 because 0  r  6 . …..(2)
Therefore from (1) and (2) we can get,
x = 6q,6q + 1,6q + 2,6q + 3,6q + 4,6q + 5
Case 1: If x = 6q then product
x(x + 1)(x + 2) = 6q(6q + 1)(6q + 2) , which is divisible by 6 . …..(3)
Case 2: If x = 6q + 1 then product
x(x + 1)(x + 2) = (6q + 1)(6q + 2)(6q + 3)
 x(x + 1)(x + 2) = 2(3q + 1).3(2q + 1)(6q + 1)
 x(x + 1)(x + 2) = 6(3q + 1).(2q + 1)(6q + 1)
which is divisible by 6 . …..(4)
Case 3: If x = 6q + 2 then product
x(x + 1)(x + 2) = (6q + 2)(6q + 3)(6q + 4)
 x(x + 1)(x + 2) = 3(2q + 1).2(3q + 1)(6q + 4)
 x(x + 1)(x + 2) = 6(2q + 1)(3q + 1)(6q + 4)
which is divisible by 6 . …..(5)
Case 4: If x = 6q + 3 then product
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x(x + 1)(x + 2) = (6q + 3)(6q + 4)(6q + 5)
 x(x + 1)(x + 2) = 6(2q + 1)(3q + 2)(6q + 5)
which is divisible by 6 . …..(6)
Case 5: If x = 6q + 4 then product
x(x + 1)(x + 2) = (6q + 4)(6q + 5)(6q + 6)
 x(x + 1)(x + 2) = 6(6q + 4)(6q + 5)(q + 1)
which is divisible by 6 . …..(7)
Case 6: If x = 6q + 5 then product
x(x + 1)(x + 2) = (6q + 5)(6q + 6)(6q + 7)
 x(x + 1)(x + 2) = 6(6q + 5)(q + 1)(6q + 7)
which is divisible by 6 . …..(8)
From equations (3) to (8) we can conclude that the product of three consecutive
positive integers is divisible by 6 .
(
6.Prove that 3 − 5
) is an irrational number.
Ans: We will prove this by contradiction. To solve this problem, suppose that
3 − 5 is rational i.e.,  two co-prime integers a and b (b  0) such that
(
)
a
=3− 5
…..(1)
b
Subtracting 3 from both sides of the equation (1) and simplifying it further we
get,
a
−3= − 5
b
3b − a

= 5
…..(2)
b
3b − a
Since we know that 5 is irrational. Therefore, from (2)
should be
b
p
irrational but it could be written in form which means it is rational. It is not
q
(
)
possible and hence our assumption is wrong and 3 − 5 cannot be rational.
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7.Prove that if x and y are odd positive integers, then x 2 + y 2 is even but
not divisible by 4 .
Ans: Let x = 2p + 1 and y = 2q + 1 be two odd positive integers p,q  .
…..(1)
Squaring x, y from (1) and adding them we get,
x 2 + y 2 = (2p + 1) 2 + (2q + 1) 2
 x 2 + y 2 = ( 4p 2 + 4p + 1) + ( 4q 2 + 4q + 1)
 x 2 + y 2 = 2(2p 2 + 2q 2 + 2p + 2q + 1)
 x 2 + y 2 = 2m
…..(2)
Where m = (2p 2 + 2q 2 + 2p + 2q + 1)
From (2), it is clear that x 2 + y 2 is even but not divisible by 4 .
7.Show that one and only one out of n,(n + 2) or (n + 4) is divisible by 3 ,
where n  N.
Ans: Let n be a number such that n = 3q + r . …..(1)
Clearly, in the expression (1) we are dividing n by 3 with quotient q and
remainder r , r = 0,1,2 because 0  r  3 . …..(2)
Therefore from (1) and (2) we can get, n = 3q,3q + 1,3q + 2
Case 1: If n = 3q then, numbers are 3q,(3q + 2)(3q + 4) out of these 3q is
divisible by 3 . …..(1)
Case 2: If n = 3q + 1 then, numbers are (3q + 1),(3q + 3),(3q + 5) out of these
(3q + 3) is divisible by 3 . …..(2)
Case 3: If n = 3q + 2 then, numbers are (3q + 2),(3q + 4),(3q + 6) out of these
(3q + 6) is divisible by 3 . …..(3)
 from (1), (2) and (3) we can conclude that out of n,(n + 2) and (n + 4) only
one is divisible by 3 .
9.Use Euclid’s Division Lemma to show that the square of any positive
integer of the form 3m or (3m + 1) for some integer q .
Ans: Let a be any positive integer, then we can write a = 3q + r for some
integer q  0 …..(1)
Clearly, in the expression (1) we are dividing a by 3 with quotient q and
remainder r , r = 0,1,2 because 0  r  3 . ..…(2)
Therefore from (1) and (2) we can get, a = 3q or 3q + 1 or 3q + 2 . ……(3)
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Squaring both sides of equation (3) we get,
a 2 = (3q) 2 or (3q + 1) 2 or (3q + 2) 2
 a 2 = 9q 2 or 9q 2 + 6q + 1 or 9q 2 + 12q + 4 …..(4)
Taking 3 common from LHS of equation (4) we get,
a 2 = 3  (3q 2 ) or 3(3q 2 + 2q) + 1 or 3(3q 2 + 4q + 1) + 1. …..(5)
Equation (5) could be written as
a 2 = 3k1 or 3k 2 + 1 or (3k 3 + 1) for some positive integers k1 , k 2 and k 3 .
Hence, it can be said that the square of any positive integer is either of the form
3m or 3m + 1 .
10.Prove that if n is not a rational number, if n is not a perfect square.
Ans: We will prove this by contradiction.
Let us suppose that n is rational. It means that we have co-prime integers a
a
and b (b  0) such that n =
b
 b n = a …..(1)
Squaring both sides of (1), we get nb 2 = a 2 …..(2)
From (2) we can conclude that n is factor of a 2 , this means that n is also factor
of a . …..(3)
From (3) we can write a = nc for some integer c . …..(4)
Substituting value of (4) in (2) we get,
nb 2 = n 2c2
 b 2 = nc2
It means that n is factor of b 2 , this means that n is also factor of b . …..(5)
From (3),(5) , we can say that n is factor of both a,b , which is a contradiction
to our assumption. Therefore, our assumption was wrong and
n is irrational.
11.Prove that the difference and quotient of (3 + 2 3) and (3 − 2 3)
are irrational.
Ans: Difference of (3 + 2 3) and (3 − 2 3) is 3 + 2 3 − 3 − 2 3 = 4 3 .
(
) (
)
We will prove this by contradiction.
Let us suppose that 4 3 is rational. It means that we have co-prime integers a
a
and b (b  0) such that 4 3 =
b
 4b 3 = a …..(1)
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Squaring both sides of (1), we get 48b2 = a 2 …..(2)
From (2) we can conclude that 48 is factor of a 2 , this means that 48 is also
factor of a . …..(3)
From (3) we can write a = 48c for some integer c . …..(4)
Substituting value of (4) in (2) we get,
4 3b 2 = 48c 2
 b 2 = 4 3c 2
It means that 48 is factor of b 2 , this means that 48 is also factor of b . …..(5)
From (3),(5) , we can say that 48 is factor of both a,b , which is a contradiction
to our assumption. Therefore, our assumption was wrong and 4 3 is irrational.
(
)(
)
Now, the quotient of 3 + 2 3 , 3 − 2 3 is
3+ 2 3 3+ 2 3 3+ 2 3
=

3−2 3 3−2 3 3+ 2 3
3+ 2 3

= − 5+ 4 3
3− 2 3
(
)
(
)
Similarly, using the approach used above, we can prove that − 5 + 4 3 is also
irrational.
12.Show that (n 2 − 1) is divisible by 8 , if n is an odd positive integer.
Ans: Let n = 4q + 1 be an odd positive integer. …..(1)
Then from (1),
n 2 − 1 = (4q + 1) 2 − 1
 n 2 − 1 = 16q 2 + 8q …..(2)
Taking 8 common from RHS of equation (2) we get,
n 2 − 1 = 8 ( 2q 2 + q )
 n 2 − 1 = 8m where m = 2q 2 + q . …..(3)
Hence, from (3) we can conclude that (n 2 − 1) is divisible by 8 , if n is an odd
positive integer.
13.Use Euclid Division Lemma to show that cube of any positive integer is
either of the form 9m,(9m + 1) or (9m + 8)
Ans: Let a be any positive integer, then we can write a = 3q + r for some
integer q  0 …..(1)
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Clearly, in the expression (1) we are dividing a by 3 with quotient q and
remainder r , r = 0,1,2 because 0  r  3 . ..…(2)
Therefore from (1) and (2) we can get, a = 3q or 3q + 1 or 3q + 2 . ……(3)
Cubing both sides of equation (3) we get,
a 3 = (3q)3 or (3q + 1)3 or (3q + 2)3
 a 3 = 27q 3 or 27q 3 + 27q 2 + 9q + 1 or 27q 3 + 54q 2 + 36q + 8 …..(4)
Taking 9 common from RHS of equation (4) we get,
a 3 = 9 ( 3q 3 ) or 9 ( 3q 3 + 3q 2 + q ) + 1 or 9 ( 3q 3 + 6q 2 + 4q ) + 8 …..(5)
Equation (5) could be written as
a 3 = 9k1 or 9k 2 + 1 or (9k 3 + 1) for some positive integers k1 , k 2 and k 3 .
Hence, it can be said that the square of any positive integer is either of the form
9m,9m + 1 or 3m + 8 .
Long Answer Questions
1.Prove that the following are irrationals.
1
(i)
2
4 Marks
1
is rational.
2
1
a
=
It means that we have co-prime integers a and b (b  0) such that
2 b
b
…..(1)
 = 2
a
b
Since we know that 2 is irrational. Therefore, from (1) should be irrational
a
p
but it could be written in
form which means it is rational. It is not possible
q
1
and hence our assumption is wrong and
cannot be rational.
2
(ii) 7 5
Ans: (i) We will prove this by contradiction. Let us suppose that
Ans: We will prove this by contradiction. Let us suppose that 7 5 is rational.
a
It means that we have co-prime integers a and b (b  0) such that 7 5 =
b
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 5=
a
7b
…..(1)
Since we know that
5 is irrational. Therefore, from (1)
irrational but it could be written in
a
should be
7b
p
form which means it is rational. It is not
q
possible and hence our assumption is wrong and 7 5 cannot be rational.
(iii)
6+ 2
Ans: We will prove this by contradiction. To solve this problem, suppose that
6 + 2 is rational i.e.,  two co-prime integers a and b (b  0) such that
a
…..(1)
=6+ 2
b
Subtracting 6 from both sides of the equation (1) and simplifying it further we
get,
a
−6= 2
b
a − 6b
…..(2)

= 2
b
a − 6b
Since we know that 2 is irrational. Therefore, from (2)
should be
b
p
irrational but it could be written in form which means it is rational. It is not
q
possible and hence our assumption is wrong and 6 + 2 cannot be rational.
2.Without actually performing the long division, state whether the
following rational numbers will have a terminating decimal expansion or
a non-terminating decimal expansion.
13
(i)
3125
p
Ans: Theorem: Let x be a rational number expressed in the form , where p
q
and q are coprime, and the prime factorisation of q is of the form 2 n  5m ,
where m and n are non-negative integers. Then x has a decimal expansion
which terminates.
…..(1)
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For rational number
p
13
, the prime factorization of
=
q 3125
q = 3125 = 5  5  5  5  5 = 55 .
q
is
…..(2)
From (2) we can see that the denominator q = 3125 is of the form 2 n  5m ,
where m = 5 and n = 0 .
13
Therefore, it follows from theorem (1), the rational number
has a
3125
terminating decimal expansion.
(ii)
17
8
p
where p
q
and q are coprime, and the prime factorisation of q is of the form 2 n  5m ,
where m and n are non-negative integers. Then x has a decimal expansion
which terminates.
…..(1)
p 17
For rational number = , the prime factorization of q is
q 8
Ans: Theorem: Let x be a rational number expressed in the form ,
q = 8 = 2  2  2 = 23 .
…..(2)
From (2) we can see that the denominator q = 8 is of the form 2 n  5m , where
m = 0 and n = 3 .
17
Therefore, it follows from theorem (1), the rational number
has a
8
terminating decimal expansion.
(iii)
64
455
Ans: Theorem: Let x be a rational number expressed in the form ,
p
where p
q
and q are coprime, and the prime factorisation of q is of the form 2 n  5m ,
where m and n are non-negative integers. Then x has a decimal expansion
which terminates.
…..(1)
p 64
For rational number =
, the prime factorization of q is
q 455
q = 455 = 5  91. …..(2)
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From (2) we can see that the denominator q = 455 is not of the form 2 n  5m .
64
Therefore, it follows from theorem (1), the rational number
has a non455
terminating repeating decimal expansion.
(iv)
15
1600
Ans: Theorem: Let x be a rational number expressed in the form ,
p
where p
q
and q are coprime, and the prime factorisation of q is of the form 2 n  5m ,
where m and n are non-negative integers. Then x has a decimal expansion
which terminates.
…..(1)
p
15
For rational number =
, the prime factorization of q is
q 1600
q = 320 = 2  2  2  2  2  2  5 = 26  5 .
…..(2)
From (2) we can see that the denominator q = 8 is of the form 2 n  5m , where
m = 1 and n = 6 .
15
Therefore, it follows from theorem (1), the rational number
has a
1600
terminating decimal expansion.
(v)
29
343
p
where p
q
and q are coprime, and the prime factorisation of q is of the form 2 n  5m ,
where m and n are non-negative integers. Then x has a decimal expansion
which terminates.
…..(1)
p 29
For rational number =
, the prime factorization of q is
q 343
q = 343 = 7  7  7 . …..(2)
Ans: Theorem: Let x be a rational number expressed in the form ,
From (2) we can see that the denominator q = 343 is not of the form 2 n  5m .
29
Therefore, it follows from theorem (1), the rational number
has a non343
terminating repeating decimal expansion.
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(vi)
23
2  52
3
Ans: Theorem: Let x be a rational number expressed in the form ,
p
where p
q
and q are coprime, and the prime factorisation of q is of the form 2 n  5m ,
where m and n are non-negative integers. Then x has a decimal expansion
which terminates.
…..(1)
23
For rational number 3 2 , the prime factorization of q is
2 5
3
2
q = 2  5 . …..(2)
From (2) we can see that the denominator q = 23  52 is of the form 2 n  5m ,
where m = 2 and n = 3 .
23
Therefore, it follows from theorem (1), the rational number 3 2 has a
2 5
terminating decimal expansion.
(vii)
129
2  57  7 5
2
Ans: Theorem: Let x be a rational number expressed in the form ,
p
where p
q
and q are coprime, and the prime factorisation of q is of the form 2 n  5m ,
where m and n are non-negative integers. Then x has a decimal expansion
which terminates.
…..(1)
129
For rational number 2 7 5 , the prime factorization of q is
2 5 7
2
7
5
q = 2  5  7 . …..(2)
From (2) we can see that the denominator q = 23  52 is not of the form 2 n  5m
129
Therefore, it follows from theorem (1), the rational number 2 7 5 has a
2 5 7
non-terminating repeating decimal expansion.
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(viii)
6 2
=
15 5
Ans: Theorem: Let x be a rational number expressed in the form ,
p
where p
q
and q are coprime, and the prime factorisation of q is of the form 2 n  5m ,
where m and n are non-negative integers. Then x has a decimal expansion
which terminates.
…..(1)
p 6 2
For rational number = = , the prime factorization of q is q = 5 = 51 .
q 15 5
…..(2)
From (2) we can see that the denominator q = 5 is of the form 2 n  5m , where
m = 1 and n = 0 . Therefore, it follows from theorem (1), the rational number
6 2
= has a terminating decimal expansion.
15 5
(ix)
35 7
=
50 10
Ans: Theorem: Let x be a rational number expressed in the form ,
p
where p
q
and q are coprime, and the prime factorisation of q is of the form 2 n  5m ,
where m and n are non-negative integers. Then x has a decimal expansion
which terminates. …..(1)
35 7
= , the prime factorization of q is
For rational number
50 10
q = 10 = 2  5 = 21  51 . …..(2)
From (2) we can see that the denominator q = 5 is of the form 2 n  5m , where
m = 1 and n = 1. Therefore, it follows from theorem (1), the rational number
35
has a terminating decimal expansion.
50
(x)
77 11
=
210 30
Ans: Theorem: Let x be a rational number expressed in the form ,
p
where p
q
and q are coprime, and the prime factorisation of q is of the form 2 n  5m ,
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where m and n are non-negative integers. Then x has a decimal expansion
which terminates. …..(1)
77 11
For rational number
= , the prime factorization of q is
210 30
q = 30 = 2  3  5 . …..(2)
From (2) we can see that the denominator q = 23  52 is not of the form 2 n  5m
77 11
Therefore, it follows from theorem (1), the rational number
has a
=
210 30
non-terminating repeating decimal expansion.
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