Class 10 Maths Chapter 1 Previous Year Questions - Real Numbers

Ans: (d)

Ans: (c)
The numbers p and q are prime numbers, 
∴ HCF (p, q) = 1 
Here, LCM(p, q) = 221 
∴ As, p > q
p = 17, q = 13 
(As p × q = 221) 
Now, 3p – q = 3 × 17 – 13 
= 51 – 13
= 38

Ans:(c)
Here √3 and √27 both are irrational numbers.
The product of √3 x √27 = √3 x 27

= √81
= 9​ ,= pq;q ≠ 0
∴ 9 is a rational number.

Ans:(a)
HCF(2520, 6600) = 40
LCM(2520, 6600) = 252 × k 
∴ HCF × LCM = Ist No. × IInd No. 
∴ 40 × 252 × k = 2520 × 6600
⇒ k = 2520 x 660040 x 252
⇒ k = 1650

Ans:
(A)

So least prime no. used by students = 3(because 2 is announced by the teacher, so the least number used by the students is 3)
(B)As the last student got  173250 = 2 × 3 × 3 × 5 × 5 × 5 × 7 × 11
there are 7 factors other than 2, which is announced by teacher. So, Number of student = 7
OR
Highest prime number used by student = 11
(C)Prime number 5 is used maximum times i.e., 3 times.

Ans: (b) 17 x 500 


850 = 17 × 5² × 2
500 = 5³ × 2²
LCM (850, 500) = 17 × 5³ × 2²
= 17 × 500
=8500

Ans: Let us assume that 6 - 4√5 be a rational number
Let  ...[b ≠ 0; a and b are integers]

We know that,
is a rational number.
But this contradicts the fact that √5 is an irrational number.
So, our assumption is wrong.
Therefore,  6 - 4√5 is an irrational number.

Ans: We have
11 × 19 × 23 + 3 × 11
⇒ 11(19 × 23 + 3)
⇒ 11(437 + 3)
⇒ 11(440)
⇒ 11(2 × 2 × 2 × 5 × 11)
⇒ 2 × 2 × 2 × 5 × 11 × 11
As it can be represented as a product of more than two primes (1 and number itself). So, it is not a prime number.

Ans: Given,
p = 18 a²b¹ and
q = 20 a³b²
LCM (p, q) = LCM (18 a²b¹, 20 a³b²)
= 180 a³b²

Ans: Assuming 5 – 2√3 to be a rational number. 
 
Here RHS is rational but LHS is irrational.  
Therefore our assumption is wrong.  
Hence, 5 – 2√3 is an irrational number. 

Ans: The numbers are prime numbers and composite numbers.
Prime numbers can be divided by 1 and itself. A composite number has factors other than 1 and itself.
(5 × 11 × 17) + (3 × 11)
= (85 × 11) + (3 × 11)
= 11 × (85 + 3)
= 11 × 88
= 11 × 11 × 8
= 2 × 2 × 2 × 11 × 11
The given expression has 2 and 11 as its factors. Therefore, it is a composite number.

Ans: The number of rooms will be minimum if each room accommodates a maximum number of teachers. Since in each room the same number of teachers are to be seated and all of them must be of the same subject. Therefore, the number of teachers in each room must be HCF of 48, 80, and 144.
The prime factorisations of 48, 80 and 144 are as under 
48 = 2⁴ × 3¹
80 = 2⁴ × 5¹
144 = 2⁴ × 3²
∴ HCF of 48, 80 and 144 is 24 = 16
Therefore, in each room 16 teachers can be seated.

Ans: Assertion (A): "If the graph of a polynomial touches the x-axis at only one point, then the polynomial cannot be a quadratic polynomial."
A quadratic polynomial is of the form ax² + bx + c. It can have two real roots, one real root (if the discriminant is zero), or no real roots (if the discriminant is negative).
If the graph of a quadratic polynomial touches the x-axis at exactly one point, this means it has a repeated real root (a double root), which is possible for quadratic polynomials. Hence, the assertion is false.
Reason (R): "A polynomial of degree n (n > 1) can have at most n zeroes."
A polynomial of degree n can have at most n real or complex zeroes. This is a true statement, so the reason is true.
Given the above analysis, the correct answer is: (d) Assertion (A) is false but Reason (R) is true.

Sol: Least composite number = 4
Least prime number = 2
∴ HCF = 2, LCM = 4
∴ Required ratio = HCF / LCM = 2/4
i.e. 1 : 2

Ans: 55

Given, least number which when divided by 12, 16 and 24 leaves remainder 7 in each case
∴  Least number = LCM(12, 16, 24) + 7
= 48 + 7
= 55

Ans:30

Let the two numbers be 2x and 3x
LCM of 2x and 3x = 6x, HCF(2x, 3x) = x
Now, 6x = 180
⇒ x = 180/6
x = 30

Ans: Let us assume that √3 is a rational number.

Then √3 = a/b; where a and b ( ≠ 0) are co-prime positive integers.
Squaring on both sides, we get
3 = a²/b²
⇒ a² = 3b² 
⇒ 3 divides a²
⇒ 3 divides a   _________(i)
= a = 3c, where c is an integer
Again, squaring on both sides, we get
a² = 9c²
⇒3b² = 9c²
⇒b² = 3c²
⇒ 3 divides b²
⇒ 3 divides b   _________(ii)
From (i) and (ii), we get 3 divides both a and b.
⇒ a and b are not co- prime integers.
This contradicts the fact that a and b are co-primes.
Hence,  √3 is an irrational number.

Ans:(a)
Sol: Given, HCF = 12
Let two numbers be 12a and 12b
So. 12a x 12b = 6336 
⇒ ab = 44
We can write 44 as product of two numbers in these ways:
ab = 1 x 44 = 2 x 22 = 4x  11
Here, we will take a = 1 and b = 44 ; a = 4 and b = 11.
We do not take ab = 2  x 22 because 2 and 22 are not co-prime to each other.

For a = 1 and b = 44, 1^st no. = 12a = 12, 2^nd no. = 12b = 528
For a = 4 and b = 11, 1^st no. = 12a = 48, 2^nd no. = 12b = 132
Hence, we get two pairs of numbers, (12, 528) and (48, 132).

Ans: (d)
Sol: 

• For any natural number n, the expression (12)ⁿ cannot end with the digit 0.
• This is because the number 12 does not contain the prime factor 5, which is necessary for a number to end in 0.
• Thus, regardless of the value of n, (12)ⁿ will never end with 0.

Ans: (b)
Sol: We have,

∴ Prime factorisation of 385 = 5 x 7 x 11

Ans: We have, 2 x 3 x 5 + 5 and 5 x 7 x 11 + 7 x 5.
We can write these numbers as:
2 x 3 x 5 + 5 = 5(2 x 3 + 1)
= 1 x 5 x 7
and 5 x 7 x 11 + 7 x 5 = 5 x 7(11 + 1)
= 5 x 7 x 12 = 1 x 5 x 7 x 12
Since, on simplifying. we find that both the numbers have more than two factors. 
So. these are composite numbers.

Ans: (c)
Sol:We have, 
12 = 2 x 2 x 3 = 2² x 3
21= 3 x 7
15 = 3 x 5
∴ HCF (12, 21, 15) = 3
and LCM (12, 21 ,15 ) = 2² x 3 x 5 x 7
= 420

Ans: Let the other number be x
As, HCF (a, b) x  LCM (a, b) = a  x  b
⇒ 13  x  182= 26x
⇒ x = 13 x 182 / 26
= 91
Hence, other number is 91.

Ans: We know that
LCM × HCF = Product of two numbers
∴ LCM (135, 225) = Product of 135 and 225 / HCF(135, 225)
= 135 x 225 / 45
= 675
So, LCM (135, 225) = 675

Ans: Using the formula:  HCF (a, b) x LCM (a, b) = a x b
∴ HCF (336, 54) x  LCM (336, 54) = 336 x 54
⇒ 6 x LCM(336, 54) = 18144
⇒ LCM (336, 54) = 18144 / 6
= 3024

Ans: We know that HCF (a, b) x  LCM (a, b)=a x b
⇒ 5 x 200 = ab
⇒  ab = 1000

Ans: Since, HCF (65 ,117) = 13
Given HCF ( 65, 117 ) = 65n - 117
13 = 65n - 117
⇒  65n = 13 +117
⇒  n = 2

Ans: Prime factorisation of 612 and 1314 are
612 = 2 x 2 x 3 x 3 x 17
1314 = 2 x 3 x 3 x 73
∴ HCF (612, 1314) = 2 x 3 x 3
= 18

Ans: Let us assume that √5 is a rational number.
Then √5 = a/b where a and b (≠ 0} are co-prime integers,
if Squaring on both sides, we get
5 = a²b² ⇒ a² = 5b²
⇒ 5 divides a²
⇒ 5 divides a  ----------(i)
⇒ a = 5c, where c is an integer
Again, squaring on both sides, we get
a² = 25c²
⇒ 5b² = 25c²  
⇒ b² = 5c²
⇒ 5 divides b² ----------(ii)
⇒ 5 divides b
From (i) and {ii), we get 5 divides both a and b.
⇒  a and b are not co-prime integers.
Hence, our supposition is wrong.
Thus, √5 is an irrational number.

Ans: Let us assume √2 be a rational number.
Then, √2 = p/q where p, q (q ≠ 0) are integers and co-prime. ;
On squaring both sides. we get
2 = p²q² ⇒ p² = 2q² ------------(i)
⇒ 2 divides p²
⇒ 2 divides p -----------(ii)
So, p = 2a, where a is some integer.
Again squaring on both sides, we get
p² = 4a²
⇒ 2q² = 4a² (using (i))
⇒ q² = 2a²
⇒ 2 divides q²
⇒ 2 divides q       -----------(iii)
From (ii) and (iii), we get
2 divides both p and q.
∴ p and q are not co-prime integers.
Hence, our assumption is wrong.
Thus √2 is an irrational number.

Ans: Suppose 2 + 5√3 is a rational number.
We can find two integers a, b (b ≠ 0) such that
2 + 5√3 = a/b, where a and b are co -prime integers.
5√3 = ab - 2 ⇒ √3 = 15 [ a b - 2]
⇒ √3 is a rational number.

[ ∵ a, b are integers, so  15 [ a b - 2] is a rational number]
But this contradicts the fact that √3 is an irrational number.
Hence, our assumption is wrong.
Thus, 2 + 5√3 is an irrational number.

Ans: Given numbers are 306 and 657. 
The smallest number divisible by 306 and 657 = LCM(306, 657) 
Prime factors of 306 = 2 × 3 × 3 × 17 
Prime factors of 657 = 3 × 3 × 73 
LCM of (306, 657) = 2 × 3 × 3 × 17 × 73 
= 22338 
Hence, the smallest number divisible by 306 and 657 is 22,338.