CHAPTER-WISE PREVIOUS YEARS' QUESTIONS
MATHEMATICS
HINTS & SOLUTIONS
Class X (CBSE)
,
, MATHEMATICS
Chapter - 1 : Real Numbers
1. 2 This contradicts the fact that 2 is irrational.
a 3 So, our assumption is incorrect. [½]
b
7
Hence, 5 3 2 is an irrational number.
7. Since 7344 > 1260
Let assume the missing entries be a, b.
7344 = 1260 × 5 + 1044 [½]
b = 3 × 7 = 21 [½]
Since remainder 0
a = 2 × b = 2 × 21 = 42 [½]
1260 = 1044 × 1 + 216
2. Given two numbers 100 and 190.
1044 = 216 × 4 + 180 [½]
HCF × LCM = 100 × 190 [½]
216 = 180 × 1 + 36
= 19000 [½]
180 = 36 × 5 + 0 [½]
441
3. Given a rational number 5 7 2 . The remainder has now become zero.
2 5 7
HCF of 1260 and 7344 is 36. [½]
441 9
5 7 2
5 7 [½] 8. Let a be positive odd integer.
2 5 7 2 5
Using division algorithm on a and b = 4 [½]
Since, the denominator is in the form of 2m 5n.
So, the rational number has terminating decimal a = 4q + r
expansion. [½] Since 0 r < 4, the possible remainders are
4. Smallest prime number is 2. 0, 1, 2 and 3. [½]
Smallest composite number is 4. a can be 4q or 4q + 1 or 4q + 2 or 4q + 3,
where q is the quotient.
Therefore, HCF is 2. [1]
Since a is odd, a cannot be 4q and 4q + 2.
5. Rational number lying between 2 and 3 is
[½]
15 3
1.5 [½] Any odd integer is of the form 4q + 1 or
10 2
4q + 3, where q is some integer. [½]
[∵ 2 1.414 and 3 1.732] [½]
9. Let 'a' be any positive integer and b = 3.
6.
Let us assume that 5 3 2 is rational. Then
We know a = bq + r, 0 r < b.
there exist co-prime positive integers a and b
Now, a = 3q + r, 0 r < 3.
such that
The possible remainder = 0, 1 or 2
a
53 2 [½] Case (i) a = 3q
b
a2 = 9q2
a
3 2 5 = 3 × (3q2)
b
= 3m (where m = 3q2) [1]
a 5b
2 [½] Case (ii) a = 3q + 1
3b
a2 = (3q + 1)2
2 is irrational.
= 9q2 + 6q + 1
a 5b
[∵ a, b are integers, is rational]. = 3(3q2 + 2q) + 1
3b
[½] = 3m + 1 (where m = 3q2 + 2q) [1]
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
, 2 Mathematics Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10)
Case (iii) a = 3q + 2 Common prime factor = 2, Least exponent = 2
a2 = (3q + 2)2 HCF = 22 = 4 [1]
= 9q2 + 12q + 4 To find the LCM, we list all prime factors of 404
= 3(3q2 + 4q + 1) + 1 and 96 and their greatest exponent as follows :
= 3m + 1 (where m = 3q2 + 4q + 1) Prime factors of Greatest Exponent
404 and 96
From all the above cases it is clear that square
of any positive integer (as in this a2) is either of 2 5
the form 3m or 3m + 1. [1] 3 1
10. Let assume 3 2 is a rational number. 101 1
p LCM = 25 × 31 × 1011
3 2
q = 25 × 31 × 1011
{p, q are co-prime integers and q 0} [1]
= 9696 [1]
p
2 3 Now,
q
HCF × LCM = 9696 × 4 = 38784
p 3q
2 [1] Product of two numbers = 404 × 96 = 38784
q
Therefore, HCF × LCM = Product of two
p 3q numbers. [1]
Since, is a rational number but we know
q
2 is an irrational. 13. Let 2 be rational. Then, there exist positive
Irrational rational a
integers a and b such that 2 . [Where a
b
3 2 is not a rational number. [1] and b are co-prime, b 0]. [½]
2
11. Let assume 2 3 5 is a rational number. 2 a
( 2) [½]
p b
23 5 ,
q a2
2
(where p, q are co-prime integers and q 0) b2
p 2b2 = a2
2 3 5 [1]
q 2 divides a2
2q p 2 divides a
5 ...(i)
3q
Let a = 2c for some integer c. [½]
2q p a2 = 4c2
Since, is a rational number but we also
3q
2b2 = 4c2
know 5 is an irrational [1]
b2 = 2c2
Rational irrational.
2 divides b2
Our assumption is wrong.
2 divides b ...(ii) [½]
2 3 5 is an irrational number. [1]
From (i) and (ii), we get
12. Using the factor tree for the prime factorization of
2 is common factor of both a and b.
404 and 96, we have
But this contradicts the fact that a and b have
404 = 22 × 101 and 96 = 25 × 3
no common factor other than 1. [½]
To find the HCF, we list common prime factors
Our supposition is wrong.
and their smallest exponent in 404 and 96 as
under : Hence, 2 is an irrational number. [½]
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
MATHEMATICS
HINTS & SOLUTIONS
Class X (CBSE)
,
, MATHEMATICS
Chapter - 1 : Real Numbers
1. 2 This contradicts the fact that 2 is irrational.
a 3 So, our assumption is incorrect. [½]
b
7
Hence, 5 3 2 is an irrational number.
7. Since 7344 > 1260
Let assume the missing entries be a, b.
7344 = 1260 × 5 + 1044 [½]
b = 3 × 7 = 21 [½]
Since remainder 0
a = 2 × b = 2 × 21 = 42 [½]
1260 = 1044 × 1 + 216
2. Given two numbers 100 and 190.
1044 = 216 × 4 + 180 [½]
HCF × LCM = 100 × 190 [½]
216 = 180 × 1 + 36
= 19000 [½]
180 = 36 × 5 + 0 [½]
441
3. Given a rational number 5 7 2 . The remainder has now become zero.
2 5 7
HCF of 1260 and 7344 is 36. [½]
441 9
5 7 2
5 7 [½] 8. Let a be positive odd integer.
2 5 7 2 5
Using division algorithm on a and b = 4 [½]
Since, the denominator is in the form of 2m 5n.
So, the rational number has terminating decimal a = 4q + r
expansion. [½] Since 0 r < 4, the possible remainders are
4. Smallest prime number is 2. 0, 1, 2 and 3. [½]
Smallest composite number is 4. a can be 4q or 4q + 1 or 4q + 2 or 4q + 3,
where q is the quotient.
Therefore, HCF is 2. [1]
Since a is odd, a cannot be 4q and 4q + 2.
5. Rational number lying between 2 and 3 is
[½]
15 3
1.5 [½] Any odd integer is of the form 4q + 1 or
10 2
4q + 3, where q is some integer. [½]
[∵ 2 1.414 and 3 1.732] [½]
9. Let 'a' be any positive integer and b = 3.
6.
Let us assume that 5 3 2 is rational. Then
We know a = bq + r, 0 r < b.
there exist co-prime positive integers a and b
Now, a = 3q + r, 0 r < 3.
such that
The possible remainder = 0, 1 or 2
a
53 2 [½] Case (i) a = 3q
b
a2 = 9q2
a
3 2 5 = 3 × (3q2)
b
= 3m (where m = 3q2) [1]
a 5b
2 [½] Case (ii) a = 3q + 1
3b
a2 = (3q + 1)2
2 is irrational.
= 9q2 + 6q + 1
a 5b
[∵ a, b are integers, is rational]. = 3(3q2 + 2q) + 1
3b
[½] = 3m + 1 (where m = 3q2 + 2q) [1]
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
, 2 Mathematics Hints & Solutions - Chapter-wise Previous Years' Questions (Class-10)
Case (iii) a = 3q + 2 Common prime factor = 2, Least exponent = 2
a2 = (3q + 2)2 HCF = 22 = 4 [1]
= 9q2 + 12q + 4 To find the LCM, we list all prime factors of 404
= 3(3q2 + 4q + 1) + 1 and 96 and their greatest exponent as follows :
= 3m + 1 (where m = 3q2 + 4q + 1) Prime factors of Greatest Exponent
404 and 96
From all the above cases it is clear that square
of any positive integer (as in this a2) is either of 2 5
the form 3m or 3m + 1. [1] 3 1
10. Let assume 3 2 is a rational number. 101 1
p LCM = 25 × 31 × 1011
3 2
q = 25 × 31 × 1011
{p, q are co-prime integers and q 0} [1]
= 9696 [1]
p
2 3 Now,
q
HCF × LCM = 9696 × 4 = 38784
p 3q
2 [1] Product of two numbers = 404 × 96 = 38784
q
Therefore, HCF × LCM = Product of two
p 3q numbers. [1]
Since, is a rational number but we know
q
2 is an irrational. 13. Let 2 be rational. Then, there exist positive
Irrational rational a
integers a and b such that 2 . [Where a
b
3 2 is not a rational number. [1] and b are co-prime, b 0]. [½]
2
11. Let assume 2 3 5 is a rational number. 2 a
( 2) [½]
p b
23 5 ,
q a2
2
(where p, q are co-prime integers and q 0) b2
p 2b2 = a2
2 3 5 [1]
q 2 divides a2
2q p 2 divides a
5 ...(i)
3q
Let a = 2c for some integer c. [½]
2q p a2 = 4c2
Since, is a rational number but we also
3q
2b2 = 4c2
know 5 is an irrational [1]
b2 = 2c2
Rational irrational.
2 divides b2
Our assumption is wrong.
2 divides b ...(ii) [½]
2 3 5 is an irrational number. [1]
From (i) and (ii), we get
12. Using the factor tree for the prime factorization of
2 is common factor of both a and b.
404 and 96, we have
But this contradicts the fact that a and b have
404 = 22 × 101 and 96 = 25 × 3
no common factor other than 1. [½]
To find the HCF, we list common prime factors
Our supposition is wrong.
and their smallest exponent in 404 and 96 as
under : Hence, 2 is an irrational number. [½]
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
